# How to Chinese remainder, part 1

This is the third post in a series of posts on the Chinese remainder theorem (first post and second post). In this and the next post, we highlight two algorithms for obtaining the solution of a system of linear congruence equations. The Chinese remainder theorem guarantees that, under certain condition, any system of linear congruence equations has a unique solution (unique modulo the product of all the moduli of the congruence equations). The theorem is usually abbreviated as CRT. Even though the solution is unique, there are more than one way to solve systems of simultaneous linear congruence equations. In this post, we extract an algorithm from a constructive proof of CRT that is found in many textbooks (used here in this previous post). In the next post, we present another algorithm for solving systems of linear congruence equations.

The statement of CRT only guarantees the existence of a solution and does not show how to derive the unique solution. In many situations, it is sufficient to know that solutions exist (e.g. using CRT to prove theorems). In other situations where actual solutions are sought, we need more than the statement of CRT. In these situations, an algorithm is needed to actually find a solution. It is critical to have an algorithm for solving CRT problems, especially in computer implementation.

The algorithm discussed here is to solve the following is the version of the Chinese remainder theorem. This version has been shown to be equivalent to several other versions in this previous post.

Theorem F (Chinese Remainder Theorem)
Suppose $m_1,m_2,\cdots,m_t$ are positive integers that are pairwise relatively prime. Let $m=m_1 \cdot m_2 \cdots m_t$. Suppose we have the following system of linear congruence equations

$a_1 \ x \equiv b_1 \ (\text{mod} \ m_1)$

$a_2 \ x \equiv b_2 \ (\text{mod} \ m_2)$

$\cdots$

$a_{t-1} \ x \equiv b_{t-1} \ (\text{mod} \ m_{t-1})$

$a_t \ x \equiv b_t \ (\text{mod} \ m_t)$

such that each of the linear congruence equations has a unique solution, i.e. for each $i$, $a_i$ and $m_i$ are relatively prime. Then the system of linear congruence equations has a unique solution modulo $m=m_1 \cdot m_2 \cdots m_t$.

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Examples

Example 1
Solve the following system of linear congruence equations.

$x \equiv 3 \ (\text{mod} \ 17)$

$x \equiv 10 \ (\text{mod} \ 21)$

$x \equiv 15 \ (\text{mod} \ 29)$

Both 17 and 29 are prime. The factors of 21 are 3 and 7. The three moduli are pairwise relatively prime (i.e. no two of which have a common factor other than 1). Thus CRT implies that the three equations have a solution $x_0$. The solution is unique modulo 10353 = 17 x 21 x 29. This means that if $y$ is another solution, then $y\equiv x_0 \ (\text{mod} \ 10353)$.

The solution is $x_0=3 \cdot 609 \cdot 11+10 \cdot 493 \cdot 19+15 \cdot 357 \cdot 13 \ (\text{mod} \ 10353)$ where

$609=21 \cdot 29$ and $609 \cdot 11 \equiv 1 \ (\text{mod} \ 17)$

$493=17 \cdot 29$ and $493 \cdot 19 \equiv 1 \ (\text{mod} \ 21)$

$357=17 \cdot 21$ and $357 \cdot 13 \equiv 1 \ (\text{mod} \ 29)$

The number 609 is the product of the remaining moduli (all the modulo not 17). Because 609 and the modulus 17 are relatively prime, the number 609 has a multiplicative inverse modulo 17. The multiplicative inverse is 11, found using the extended Euclidean algorithm. The other two pairs “493 and 19” and “357 and 13” are found similarly. Thus the solution $x_0$ is the sum of three numbers. The first is 3 x 609 x 11, where 3 is the solution to the first equation, and 609 and 11 multiplicative inverses to each other modulo 17. The other two numbers 10 x 493 x 19 and 15 x 357 x 13 are obtained similarly.

After reducing modulo 10353, the solution is

\displaystyle \begin{aligned} x_0&\equiv 3 \cdot 609 \cdot 11+10 \cdot 493 \cdot 19+15 \cdot 357 \cdot 13 \ (\text{mod} \ 10353) \\&\equiv 20097+93670+69615 \ (\text{mod} \ 10353) \\&\equiv 9744+493+7497 \ (\text{mod} \ 10353) \\&\equiv 17734 \ (\text{mod} \ 10353) \\&\equiv 7381 \ (\text{mod} \ 10353) \end{aligned}

The algorithm demonstrated here requires the computation of three multiplicative inverses. One try and true method for finding multiplicative inverse in the extended Euclidean algorithm (i.e. applying the Euclidean algorithm and then work backward). We demonstrate how to solve $609y \equiv 1 \ (\text{mod} \ 17)$. To this end, solve the linear diophantine equation $609y+17z=1$. Then work backward.

\displaystyle \begin{aligned} &609=17 \cdot 35+1 \\&17=14 \cdot 1+3 \\&14=3 \cdot 4+2 \\&3=2 \cdot 1+1 \\&2=1 \cdot 2+0 \\&\text{ } \\&\text{ } \end{aligned} \ \ \ \ \ \ \ \ \ \displaystyle \begin{aligned} 1&=3-2 \\&=3-(14-3 \cdot 4) \\&=14(-1)+3(5) \\&=14(-1)+(17-14)5 \\&=17(5)+14(-6) \\&=17(5)+(609-17 \cdot 35) (-6) \\&=609(-6)+17(215) \end{aligned}

The calculation on the left shows the repeated applications of divisions to show that $(609,17)=1$. The calculation on the right works backward to solve $609y+17z=1$. The solution is $y=-6$, the least residue of which is $y=11$. $\square$

Example 2
Solve the following system of linear congruence equations.

$3 \ x \equiv 1 \ (\text{mod} \ 5)$

$4 \ x \equiv 6 \ (\text{mod} \ 14)$

$5 \ x \equiv 11 \ (\text{mod} \ 3)$

The algorithm used in Example 1 works here. The only difference is that each equation needs to be solved individually. The numbers here are small. The equations can be solved by inspection. Otherwise, each equation can also be solved by solving an appropriate linear diophantine equation (as discussed here). The following are the solutions to the above equations (individually).

$x \equiv 2 \ (\text{mod} \ 5)$

$x \equiv 5 \ (\text{mod} \ 14)$

$x \equiv 1 \ (\text{mod} \ 3)$

The problem is then to solve the system of the above 3 linear congruence equations. The solution is $x_0=2 \cdot 42 \cdot 8+5 \cdot 15 \cdot 1+1 \cdot 70 \cdot 1$ where:

$42=14 \cdot 3$ and $42 \cdot 8 \equiv 1 \ (\text{mod} \ 5)$

$15=5 \cdot 3$ and $15 \cdot 1 \equiv 1 \ (\text{mod} \ 14)$

$70=5 \cdot 14$ and $70 \cdot 1 \equiv 1 \ (\text{mod} \ 3)$

The above calculation requires solving three linear congruence equations individually and finding three multiplicative inverses. Both tasks can be done using the extended Euclidean algorithm. After further reduction modulo 210, the unique solution is:

\displaystyle \begin{aligned} x_0&\equiv 2 \cdot 42 \cdot 8+5 \cdot 15 \cdot 1+1 \cdot 70 \cdot 1 \ (\text{mod} \ 210) \\&\equiv 672+75+70 \ (\text{mod} \ 210) \\&\equiv 42+75+70 \ (\text{mod} \ 210) \\&\equiv 187 \ (\text{mod} \ 210) \end{aligned}

The above gives the unique solution to the given system of linear equations. $\square$

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An algorithm for CRT

Now we describe the algorithm that is described by the above two examples. Suppose that the following system of equations such that the moduli $m1,m_2,\cdots,m_t$ are pairwise relatively prime and such that each equation by itself has a unique solution, i.e. $a_i$ and $m_i$ are relatively prime for each $i$.

$a_1 \ x \equiv b_1 \ (\text{mod} \ m_1)$

$a_2 \ x \equiv b_2 \ (\text{mod} \ m_2)$

$\cdots$

$a_{t-1} \ x \equiv b_{t-1} \ (\text{mod} \ m_{t-1})$

$a_t \ x \equiv b_t \ (\text{mod} \ m_t)$

The following three steps describe how to obtain the solution to this system of equations.

Step 1
Solve each equation $a_i \ x \equiv b_i \ (\text{mod} \ m_i)$ individually to obtain the solution $x \equiv c_i \ (\text{mod} \ m_i)$. The original system is equivalent to the following equivalent system of equations, i.e. any solution to one system is the solution to the other.

$x \equiv c_1 \ (\text{mod} \ m_1)$

$x \equiv c_2 \ (\text{mod} \ m_2)$

$\cdots$

$x \equiv c_{t-1} \ (\text{mod} \ m_{t-1})$

$x \equiv c_t \ (\text{mod} \ m_t)$

Step 2
For each $i$, let $n_i$ be the product of all moduli $m_j$ where $j \ne i$. Each $n_i$ is relatively prime to $m_i$. Thus $n_i$ has a multiplicative inverse modulo $m_i$. Equivalently solve the following equations individually.

$n_1 \ y \equiv 1 \ (\text{mod} \ m_1)$

$n_2 \ y \equiv 1 \ (\text{mod} \ m_2)$

$\cdots$

$n_{t-1} \ y \equiv 1 \ (\text{mod} \ m_{t-1})$

$n_t \ y \equiv 1 \ (\text{mod} \ m_t)$

For each $i$, let $w_i$ be the inverse of $n_i$ modulo $m_i$, i.e. $n_i^{-1} \equiv w_i \ (\text{mod} \ m_i)$.

Step 3
The solution is given by

$x_0 \equiv c_1 \cdot n_1 \cdot w_1+c_2 \cdot n_2 \cdot w_2+\cdots+c_t \cdot n_t \cdot w_t \ (\text{mod} \ m)$

where $m=m_1 \cdot m_2 \cdots m_t$.

Let’s look at what the algorithm entails computationally. Step 1 requires solving a set of linear congruence equations individually (equivalently, solving linear diophantine equations individually), unless the equations are given in the form $x \equiv c_i \ (\text{mod} \ m_i)$. The extended Euclidean algorithm is an excellent approach, especially in computer implementation.

Step 2 involves finding multiplicative inverses of the numbers $n_i$. The extended Euclidean algorithm is also an excellent approach, especially in computer implementation.

Step 3 is to gather up the results from Step 1 and Step 2. The computation here is to reduce modulo $m=m_1 \cdot m_2 \cdots m_t$.

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Why does the algorithm work?

The number $x_0 \equiv c_1 \cdot n_1 \cdot w_1+c_2 \cdot n_2 \cdot w_2+\cdots+c_t \cdot n_t \cdot w_t \ (\text{mod} \ m)$ is a solution to each of the equations in the given system of linear congruence equations. To see that $x_0$ is a solution to the equation $a_1 \ x \equiv b_1 \ (\text{mod} \ m_1)$, note that:

\displaystyle \begin{aligned} a_1 \cdot x_0&\equiv a_1 \cdot (c_1 \cdot n_1 \cdot w_1+c_2 \cdot n_2 \cdot w_2+\cdots+c_t \cdot n_t \cdot w_t) \\&\equiv a_1 \cdot c_1 \cdot n_1 \cdot w_1+a_1 \cdot c_2 \cdot n_2 \cdot w_2+\cdots+a_1 \cdot c_t \cdot n_t \cdot w_t \\&\equiv a_1 \cdot c_1 \cdot n_1 \cdot w_1 \\&\equiv a_1 \cdot c_1 \\&\equiv b_1 \ (\text{mod} \ m_1) \end{aligned}

In the above derivation, all the terms containing $n_j$ with $j \ne 1$ drop out. This is because $n_j$ has $m_1$ as a factor. The product $n_1 \cdot w_1$ drops out since $w_1$ is the multiplicative inverse if $n_1$ modulo $m_1$. Finally $a_1 \cdot c_1 \equiv b_1 \ (\text{mod} \ m_1)$ since $c_1$ is a solution of the equation $a_1 \ x \equiv b_1 \ (\text{mod} \ m_1)$. By the same reasoning, $x_0$ is the solution to all the other equations in the system.

To see that the solution $x_0$ is unique, see the proof in this previous post.

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$\copyright \ 2015 \text{ by Dan Ma}$

# Versions of the Chinese remainder theorem

This post is the second post in a series of posts on the Chinese remainder theorem (CRT). The previous post sets up the scene by discussing the fundamental theorem of arithmetic. In this post, we derive the various versions of CRT from a lemma (Lemma A below) that is equivalent to the fundamental theorem of arithmetic.

Links to the other posts in the series: first post, third post, fourth post.

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The starting point

Lemma A and Theorem B are discussed in the previous post. Lemma A is shown to be equivalent to the fundamental theorem of arithmetic. Lemma A makes CRT possible.

Lemma A
Let $a$, $b$ and $d>0$ be integers. Suppose that $\text{GCD}(a,d)=1$, i.e. the greatest common divisor of $a$ and $d$ is 1. If $d \lvert (a \cdot b)$, then $d \lvert b$.

Theorem B
The following conditions are equivalent:

1. The statement of Lemma A.
2. (Euclid’s lemma) If $p$ is a prime number and $p \lvert (a \cdot b)$, then either $p \lvert a$ or $p \lvert b$.
3. Every positive integer $n>1$ can be written as a product of prime numbers and that this product (called a factorization) is unique.

In this post, we start from Lemma A and derive several versions of CRT. First, let’s look at a consequence of Lemma A.

Lemma C
Let $m_1,m_2,\cdots,m_t$ be positive integers that are pairwise relatively prime. Let $M=m_1 \cdot m_2 \cdots m_t$. Then for any integer $n$,

$M \lvert n$ if and only if $m_i \ \lvert \ n$ for each $i$.

Proof
The direction $\rightarrow$ is clear.

We show the direction $\leftarrow$. Suppose that for each $i$, $m_i \lvert n$. Then $n=m_1 \cdot r_1$ for some integer $r_1$. Note that $m_2 \lvert n=m_1 \cdot r_1$. By Lemma 1, $m_2 \lvert r_1$. We can write $n=m_1 \cdot m_2 \cdot r_2$ for some integer $r_2$. Continue the same argument, it follows that $n=m_1 \cdot m_2 \cdots m_t \cdot r$ for some integer $r$. This means that $M \lvert n$. $\square$

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The Chinese remainder theorem

The following are several statements of the Chinese remainder theorem. The first version is a re-statement of Lemma C using congruence notation.

Theorem D (Chinese Remainder Theorem)
Let $m_1,m_2,\cdots,m_t$ be positive integers that are pairwise relatively prime. Let $M=m_1 \cdot m_2 \cdots m_t$. Then for any integers $a$ and $b$,

$a \equiv b \ (\text{mod} \ M)$ if and only if $a \equiv b \ (\text{mod} \ m_i)$ for each $i$.

The proof of Lemma C would take care of Theorem D. Note that $a \equiv b \ (\text{mod} \ y)$ means that $y \lvert (a-b)$.

Theorem E (Chinese Remainder Theorem)
Let $m_1,m_2,\cdots,m_t$ be positive integers that are pairwise relatively prime. Let $M=m_1 \cdot m_2 \cdots m_t$. Then for any integers $a$ and $b$,

the integer $x_0$ is a solution of the linear congruence equation $a \cdot x \equiv b \ (\text{mod} \ M)$ if and only if the integer $x_0$ is satisfies simultaneously the following linear congruence equations:

$a \cdot x \equiv b \ (\text{mod} \ m_1)$

$a \cdot x \equiv b \ (\text{mod} \ m_2)$

$\cdots$

$a \cdot x \equiv b \ (\text{mod} \ m_t)$

Proof
By Theorem D, $a \cdot x_0 \equiv b \ (\text{mod} \ M)$ if and only if $a \cdot x_0 \equiv b \ (\text{mod} \ m_i)$ for each $i$. $\square$

Theorem F (Chinese Remainder Theorem)
Suppose $m_1,m_2,\cdots,m_t$ are positive integers that are pairwise relatively prime. Let $m=m_1 \cdot m_2 \cdots m_t$. Suppose we have the following system of linear congruence equations

$a_1 \ x \equiv b_1 \ (\text{mod} \ m_1)$

$a_2 \ x \equiv b_2 \ (\text{mod} \ m_2)$

$\cdots$

$a_{t-1} \ x \equiv b_{t-1} \ (\text{mod} \ m_{t-1})$

$a_t \ x \equiv b_t \ (\text{mod} \ m_t)$

such that each of the linear congruence equations has a unique solution, i.e. for each $i$, $a_i$ and $m_i$ are relatively prime. Then the system of linear congruence equations has a unique solution modulo $m=m_1 \cdot m_2 \cdots m_t$.

Proof
First, a solution is produced. Then it is shown that the solution is unique. To produce the solution, the first step is solve each equation individually. Because $a_i$ and $m_i$ are relatively prime, the equation $a_i \ x \equiv b_i \ (\text{mod} \ m_i)$ has a unique solution $c_i$. Thus $a_i \ c_i \equiv b_i \ (\text{mod} \ m_i)$ for each $i$.

Next, let $N_i$ be the product of all the moduli $m_j$ where $j \ne i$. Note that $N_i$ and $m_i$ are still relatively prime. There exists a unique $n_i$ such that $N_i \ n_i \equiv 1 \ (\text{mod} \ m_i)$. In other words, $n_i$ is the multiplicative inverse of $N_i$ modulo $m_i$.

The proposed solution is $x_0=c_1 \cdot N_1 \cdot n_1+c_2 \cdot N_2 \cdot n_2+\cdots+c_t \cdot N_t \cdot n_t$. The following shows that $x_0$ is the solution to each equation.

\displaystyle \begin{aligned} a_i x_0 &\equiv a_i (c_1 \cdot N_1 \cdot n_1+c_2 \cdot N_2 \cdot n_2+\cdots+c_i \cdot N_i \cdot n_i+ \cdots +c_t \cdot N_t \cdot n_t) \\&\equiv a_i \cdot c_1 \cdot N_1 \cdot n_1+a_i \cdot c_2 \cdot N_2 \cdot n_2+\cdots+a_i \cdot c_i \cdot N_i \cdot n_i +\cdots + a_i \cdot c_t \cdot N_t \cdot n_t \\&\equiv a_i \cdot c_i \cdot N_i \cdot n_i \\&\equiv a_i \cdot c_i \\&\equiv b_i \ (\text{mod} \ m_i) \end{aligned}

All the terms containing $N_j$ with $j \ne i$ drop out since $N_j$ contains $m_i$ as a factor. The product $N_i \cdot n_i$ drops out since it is congruent to 1 modulo $m_i$. Then $a_i \cdot c_i$ is congruent to $b_i$ modulo $m_i$.

To show the uniqueness, suppose $x$ is also a solution to the system of linear congruence equations. Then $a_i \ x \equiv a_i \ x_0 \ (\text{mod} \ m_i)$ for each $i$. Multiplying the inverse of $a_i$ on both sides, we have and $x \equiv x_0 \ (\text{mod} \ m_i)$ for each $i$. By Theorem E, $x \equiv x_0 \ (\text{mod} \ m)$. $\square$

Theorem G (Chinese Remainder Theorem)
Suppose $m_1,m_2,\cdots,m_t$ are positive integers that are pairwise relatively prime. Let $m=m_1 \cdot m_2 \cdots m_t$. For any sequence of integers $c_1,c_2,\cdots,c_t$, the following system of linear congruence equations

$x \equiv c_1 \ (\text{mod} \ m_1)$

$x \equiv c_2 \ (\text{mod} \ m_2)$

$\cdots$

$x \equiv c_{t-1} \ (\text{mod} \ m_{t-1})$

$x \equiv c_t \ (\text{mod} \ m_t)$

has a unique solution modulo $m=m_1 \cdot m_2 \cdots m_t$.

Proof
This follows directly from Theorem F. $\square$

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To complete the loop

To complete the loop, we show that Theorem G implies Theorem D. Suppose that $a \equiv b \ (\text{mod} \ m_i)$ for each $i$. Consider the following system of congruence equations:

$x \equiv b \ (\text{mod} \ m_1)$

$x \equiv b \ (\text{mod} \ m_2)$

$\cdots$

$x \equiv b \ (\text{mod} \ m_t)$

The number $b$ is clearly a solution to this system of equations. By Theorem G, this solution is unique. So any other solution to this system must be congruent to $b$ modulo $m=m_1 \cdot m_2 \cdots m_t$. By assumption, $a$ is a solution to the system. So $a \equiv b \ (\text{mod} \ m)$. $\square$

The loop $D \rightarrow E \rightarrow F \rightarrow G \rightarrow D$ is now complete. Theorem G is the usual statement of CRT. Based on the loop, any one of the theorems can be called the Chinese remainder theorem. Note that the loop by itself does not establish the Chinese remainder theorem. For that to happen, some condition outside the loop must imply one condition in the loop. The above discussion shows that the fundamental theorem of arithmetic (in the form of Lemma A) implies Theorem D.

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CRT is a versatile tool and is found to be useful in many areas of mathematics. One approach in applying CRT is that of a divide and conquer idea. The original problem is divided into smaller problems, which can be solved independently of one another. At the end, the results of the smaller problems are combined to form the solution of the original problem. For example, in solving a linear congruence equation $a \cdot x \equiv b \ (\text{mod} \ M)$ for a large modulus $M$, CRT in the form of Theorem E suggests that the problem can be broken up into a set of linear congruence equations with smaller moduli. For this reason, CRT is important in computing (both theory and applications), especially in computing intensive areas such as coding theory and cryptography.

In the next posts, we discuss two algorithms for solving CRT simultaneous systems of equations and look at a few applications.

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$\copyright \ 2015 \text{ by Dan Ma}$

# Another look at the fundamental theorem of arithmetic

This is the first post of a series of blog posts on the Chinese remainder theorem, often abbreviated CRT. In this post, we take another look at the fundamental theorem of arithmetic. This post is background for the next post, which will show that CRT can be built step by step from the fundamental theorem of arithmetic.

Links to the other posts in the series:
second post, third post, fourth post.

Every integer can be factored into a product of primes in essentially one way. This fact is called the fundamental theorem of arithmetic. For example, the number 84 is 2 x 2 x 3 x 7. The ordering of the primes is not important here since, for example, 2 x 2 x 3 x 7 is the same as 3 x 2 x 7 x 2. The example of 84 seems to suggest that the fundamental theorem of arithmetic is simply an exercise at the elementary school. In general, factoring a number is a very hard problem. For integers with hundreds or thousands of digits, finding the factors may take more seconds than the number of atoms in the universe! The fundamental theorem of arithmetic guarantees that such factorization exists for any integer, large or small. Because of this, the prime numbers are the building blocks of the integers (the atoms of arithmetic).

What is the importance of knowing that every integer can be factored into prime numbers in a unique way? A short answer is that the fundamental theorem of arithmetic is a foundation result. A vast array of mathematical structures are built on this foundation. The fundamental theorem of arithmetic is the beginning point of many mathematical stories. In this and subsequent posts, we highlight one such example. We show that the Chinese remainder theorem follows from the fundamental theorem of arithmetic. CRT combines beauty and utility. Even though CRT has been known for at least 1,000 years, it continues to present new applications in many areas, e.g. in coding theory, cryptography and the theory of computing. Surprisingly, the proof of CRT is quite simple. CRT follows from a lemma that is equivalent to the fundamental theorem of arithmetic. As a result, the discussion in this and the subsequent posts is a clear demonstration of the power of the fundamental theorem of arithmetic.

In this post, we discuss the fundamental theorem of arithmetic. In the next posts, we discuss CRT.

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The lemma

The starting point is the following lemma.

Lemma A
Let $a$, $b$ and $d>0$ be integers. Suppose that $\text{GCD}(a,d)=1$, i.e. the greatest common divisor of $a$ and $d$ is 1. If $d \lvert (a \cdot b)$, then $d \lvert b$.

Proof
Suppose that $\text{GCD}(a,d)=1$. By the extended Euclidean algorithm, the linear diophantine equation $ax+dy=1$ is solvable in integers. Let $x$ and $y$ be integers that satisfy this equation. Multiply this equation by $b$ to obtain $abx+dby=b$. Since $d \lvert (a \cdot b)$, $d$ divides both terms on the left hand side of the last equation. Thus $d \lvert b$. $\square$

Lemma A is a simple statement. The proof is also simple, simply using the extended Euclidean algorithm, which is the Euclidean algorithm working backward. the Euclidean alogorithm consists, at heart, of a series of divisions to derive the greatest common divisor. As discussed below, the simple Lemma A leads to the fundamental theorem of arithmetic and Lemma A can also be derived from the fundamental theorem of arithmetic.

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The fundamental theorem of arithmetic

The fundamental theorem of arithmetic states that any positive integer greater than 1 can be expressed as a product of primes and that the factorization is unique. The theorem is proved in this previous post. Here, we show that it is equivalent to Lemma A.

Theorem B
The following conditions are equivalent:

1. The statement of Lemma A.
2. (Euclid’s lemma) If $p$ is a prime number and $p \lvert (a \cdot b)$, then either $p \lvert a$ or $p \lvert b$.
3. (Fundamental Theorem of Arithmetic) Every positive integer $n>1$ can be written as a product of prime numbers and that this product (called a factorization) is unique.

Condition 2 is usually referred to as the Euclid’s lemma. Condition 3 is, of course, a statement of the fundamental theorem of arithmetic. The story we would like to tell is that Lemma A is equivalent to the fundamental theorem of arithmetic. Thus any consequence of Lemma A is also a consequence of the fundamental theorem of arithmetic.

The proof of $1 \rightarrow 2 \rightarrow 3$ is done in this previous post. We also like to point out one additional argument is needed to establish the equivalence of the three conditions. The proof to establish the fundamental theorem of arithmetic in the previous post uses an induction argument to show that every integer is a product of prime numbers. So condition 2 plus the induction argument establish condition 3. The role of condition 2 is to establish the uniqueness of the product of primes. We only need to show $3 \rightarrow 1$.

Proof
$3 \rightarrow 1$
Suppose that $a$ and $d$ have no prime factors in common and that $d \lvert (a \cdot b)$. So we have $a \cdot b=d \cdot m$ for some integer $m$. By condition 3, we can express $a$ and $d$ as a product of primes as follows:

$\displaystyle p_1^{\alpha_1} \cdot p_2^{\alpha_2} \cdots p_t^{\alpha_t} \times b=q_1^{\delta_1} \cdot q_2^{\delta_2} \cdots q_r^{\delta_r} \times m$

The numbers $p_i$ are the prime factors of $a$ and the numbers $q_i$ are the prime factors of $d$. The exponents $\alpha_i$ and $\delta_j$ are positive integers. Note that $p_i \ne q_j$ for any $i$ and $j$. Each $q_i^{\delta_i}$ must appear in the prime factorization of the left-hand side. Since $q_i^{\delta_i}$ cannot appear in the factorization of $a$, it must be in the factorization of $b$. $\square$

The loop $1 \rightarrow 2 \rightarrow 3 \rightarrow 1$ shows the equivalence of the three statements. From a logical standpoint, any one of these statements is the fundamental theorem of arithmetic. As a demonstration of the power of the fundamental theorem of arithmetic, the next post shows that the Chinese remainder theorem is derived using Lemma A.

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$\copyright \ 2015 \text{ by Dan Ma}$

# Euler’s phi function is multiplicative

This post gives a proof that Euler’s phi function is multiplicative. The proof is a simpler and more elegant proof, as compared to the one presented here. The combinatorial argument is greatly simplified using the Chinese remainder theorem (abbreviated CRT).

The letter $m$ is used below to denote the modulus in modular arithmetic. The phi function $\phi(m)$ is defined to be the count of all the integers $0 \le a \le m-1$ such that $a$ and $m$ are relatively prime. If $m$ is prime, then $\phi(m)=m-1$ since all integers from $1$ to $m-1$ have no factors in common with $m$ (other than 1). If $m=10$, then $\phi(10)=4$ since 1, 3, 7, and 9 are the only numbers that are relatively prime to $m=10$.

To properly work with the phi function, conmsider the following setting. Given a modulus $m$, a set of interest is $\mathbb{Z}_m=\left\{0,1,2,\cdots,m-1 \right\}$. This is the set of all least resdues modulo $m$, i.e., every integer is congruent modulo $m$ to exactly one element of $\mathbb{Z}_m$. Another set of interest is $\mathbb{Z}_m^*$, which is the set of all elements $a$ in $\mathbb{Z}_m$ such that $a$ and $m$ are relatively prime. In other words, $\phi(m)$ is defined to be the cardinality of the set $\mathbb{Z}_m^*$.

The set $\mathbb{Z}_m$ is called the ring of integers modulo $m$ since it satisfies the definition of a ring with regard to addition and multiplication modulo $m$. The set $\mathbb{Z}_m^*$ with the multiplication modulo $m$ is a group, i.e. every element of $\mathbb{Z}_m^*$ has a multiplicative inverse with respect to the binary operation of multiplication modulo $m$. The set $\mathbb{Z}_m^*$ is called the multiplicative group of integers modulo $m$. The fact that $\mathbb{Z}_m^*$ is a group is established by the following theorem, which is proved here.

Theorem 1
Let $a$ be an integer in $\mathbb{Z}_m$. The following conditions are equivalent.

1. The numbers $a$ and $m$ are relatively prime, i.e. $\text{GCD}(a,m)=1$.
2. There is a $b \in \mathbb{Z}_m$ such that $a \cdot b \equiv 1 \ (\text{mod} \ m)$.
3. Some positive power of $a$ modulo $m$ is $1$, i.e., $a^n \equiv 1 \ (\text{mod} \ m)$ for some positive integer $n$.

Another interesting fact to point out is that when $m$ is a prime number, $\mathbb{Z}_m$ is a field, i.e. it is a commutative ring in which every non-zero element has a multiplicative inverse. Another terminology that is used by some authors is that the elements in $\mathbb{Z}_m$ that have multiplicative inverses are called units. Thus $\mathbb{Z}_m^*$ is also called the group of units of the ring of integers modulo $m$. Our notation here is not standard. The usual notation for the ring $\mathbb{Z}_m$ is $\mathbb{Z}/m\mathbb{Z}$. The usual notation for $\mathbb{Z}_m^*$ is $(\mathbb{Z}/m\mathbb{Z})^*$.

The phi function is multiplicative in the following sense.

Theorem 2
Let $m$ and $n$ be positive integers such that they are relatively prime. Then $\phi(m \times n)=\phi(m) \times \phi(n)$.

Theorem 2 is established by the following results.

Lemma 3
Let $m$ and $n$ be positive integers such that they are relatively prime. Then the set $\mathbb{Z}_{mn}$ and the set $\mathbb{Z}_m \times \mathbb{Z}_n$ have the same cardinality.

Proof
To prove that the two sets have the same cardinality, we define a bijection $f: \mathbb{Z}_{mn} \rightarrow \mathbb{Z}_m \times \mathbb{Z}_n$, i.e. $f$ is a one-to-one function that maps $\mathbb{Z}_{mn}$ onto the set $\mathbb{Z}_m \times \mathbb{Z}_n$. For each $a \in \mathbb{Z}_{mn}=\left\{0,1,2,\cdots,m \cdot n-1 \right\}$, define $f(a)=(c, d)$ where $c \in \mathbb{Z}_m$ with $c \equiv a \ (\text{mod} \ m)$ and $d \in \mathbb{Z}_n$ with $d \equiv a \ (\text{mod} \ n)$.

To see that $f$ is one-to-one, let $f(a)=f(b)$. Then we have $a \equiv b \ (\text{mod} \ m)$ and $a \equiv b \ (\text{mod} \ n)$. By the Chinese remainder theorem, $a \equiv b \ (\text{mod} \ mn)$. Since $a,b \in \mathbb{Z}_{mn}$, $a=b$.

To show that $f$ maps $\mathbb{Z}_{mn}$ onto the set $\mathbb{Z}_m \times \mathbb{Z}_n$, let $(c,d) \in \mathbb{Z}_m \times \mathbb{Z}_n$. Consider the equations $x \equiv c \ (\text{mod} \ m)$ and $x \equiv d \ (\text{mod} \ n)$. By CRT, these two equations have a simultaneous solution $a$ that is unique modulo $m \times n$. This means that $f(a)=(c, d)$. $\square$

Lemma 4
Let $m$ and $n$ be positive integers such that they are relatively prime. Then the set $\mathbb{Z}_{mn}^*$ and the set $\mathbb{Z}_m^* \times \mathbb{Z}_n^*$ have the same cardinality.

Proof
The same mapping $f$ defined in the proof of Lemma 3 is used. We show that when $f$ is restricted to the set $\mathbb{Z}_{mn}^*$, it is a bijection from $\mathbb{Z}_{mn}^*$ onto the set $\mathbb{Z}_m^* \times \mathbb{Z}_n^*$. First, we show that for any $a \in \mathbb{Z}_{mn}^*$, $f(a)$ is indeed in $\mathbb{Z}_m^* \times \mathbb{Z}_n^*$. To see this, note that $a$ and $m \times n$ are relatively prime. So it must be that $a$ and $m$ are relatively prime too and that $a$ and $n$ are relatively prime.

The function $f$ is one-to-one as shown above. The remaining piece is that for each $(c,d) \in \mathbb{Z}_m^* \times \mathbb{Z}_n^*$, there is some $a \in \mathbb{Z}_{mn}^*$ such that $f(a)=(c,d)$. As in the proof of Lemma 3, there is some $a \in \mathbb{Z}_{mn}$ such that $f(a)=(c,d)$. Note that $a$ and $m$ are relatively prime and $a$ and $n$ are relatively prime. This means that $a$ and $m \times n$ are relatively prime (see Lemma 4 here). Thus the function $f$ is a one-to-one from $\mathbb{Z}_{mn}^*$ onto $\mathbb{Z}_m^* \times \mathbb{Z}_n^*$. $\square$

Proof of Theorem 2
Lemma 4 shows that the set $\mathbb{Z}_{mn}^*$ and the set $\mathbb{Z}_m^* \times \mathbb{Z}_n^*$ have the same cardinality. First, $\phi(m \times n)$ is the cardinality of the set $\mathbb{Z}_{mn}^*$. Theorem 2 is established by noting that $\phi(m) \times \phi(n)$ is the cardinality of $\mathbb{Z}_m^* \times \mathbb{Z}_n^*$. $\square$

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Evaluating the phi function

The combinatorial argument for $\phi(m \times n)=\phi(m) \times \phi(n)$ is greatly simplified by using the Chinese remainder theorem. Compare the above proof with the lengthier proof in an earlier post (see Theorem 3 here). With the multiplicative property of the phi function, we can evaluate the phi function for any positive integer.

For any positive integer $m$, consider its unique prime factorization $m=p_1^{e_1} \cdot p_2^{e_2} \cdots p_t^{e_t}$. Then we have:

\displaystyle \begin{aligned} \phi(m)&=\phi(p_1^{e_1} \cdot p_2^{e_2} \cdots p_t^{e_t}) \\&=\phi(p_1^{e_1}) \times \phi(p_2^{e_2}) \times \cdots \times \phi(p_t^{e_t}) \\&=p_1^{e_1-1} \cdot (p_1-1) \times p_2^{e_2-1} \cdot (p_2-1) \times \cdots \times p_t^{e_t-1} \cdot (p_t-1) \\&=p_1^{e_1} \cdot \biggl(1-\frac{1}{p_1}\biggr) \times p_2^{e_2} \cdot \biggl(1-\frac{1}{p_2}\biggr) \times \cdots \times p_t^{e_t} \cdot \biggl(1-\frac{1}{p_t}\biggr) \\&=p_1^{e_1} \cdot p_2^{e_2} \cdots p_t^{e_t} \cdot \biggl(1-\frac{1}{p_1}\biggr) \times \cdot \biggl(1-\frac{1}{p_2}\biggr) \times \cdots \times \cdot \biggl(1-\frac{1}{p_t}\biggr) \\&=m \cdot \biggl(1-\frac{1}{p_1}\biggr) \times \cdot \biggl(1-\frac{1}{p_2}\biggr) \times \cdots \times \cdot \biggl(1-\frac{1}{p_t}\biggr)\end{aligned}

In the above evaluation, this fact is used. For any prime $p$, $\phi(p^n)=p^{n-1} \times (p-1)$ (see see Theorem 2 here). We also use the extended version of Theorem 2: if $m_1,m_2,\cdots,m_t$ are pairwise relatively prime, then

$\phi(m_1 \times m_2 \times \cdots \times m_t)=\phi(m_1) \times \phi(m_2) \times \cdots \times \phi(m_t)$.

The extended result is to extend the product to more than two numbers. It is derived by a simple inductive argument.

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A formula for the phi function

The above evaluation leads us to the following way to express the phi function. For $m=p_1^{e_1} \cdot p_2^{e_2} \cdots p_t^{e_t}$, we have the following:

$\displaystyle \phi(m)=m \ \prod \limits_{p \lvert m} \biggl( 1-\frac{1}{p} \biggr)$

In the above product, the $p$ ranges over all prime divisors of $m$. A quick example: for $33075=3^3 \cdot 5^2 \cdot 7^2$,

\displaystyle \begin{aligned}\phi(33075)&=33075 \times \biggl( 1-\frac{1}{3} \biggr) \times \biggl( 1-\frac{1}{5} \biggr) \times \biggl( 1-\frac{1}{7} \biggr) \\&=33075 \times \frac{2}{3} \times \frac{4}{5} \times \frac{6}{7} \\&=315 \times 2 \times 4 \times 6 \\&=15120 \end{aligned}

One comment. For the above formula to work, we only need to know the prime divisors of the number, not necessarily the powers of the primes. For example, for 33075, we only need to know that the prime divisors are 3, 5 and 7.

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$\copyright \ 2015 \text{ by Dan Ma}$

# The primitive root theorem revisited

The primitive root theorem lists out all possible values of $m$ for which primitive roots exist in the modulo $m$ arithmetic. The theorem is proved in this previous post and several blog posts leading to it. In this post, we reorganize the proof and add some additional information. The proof of the primitive root theorem detailed below shows that the list of values of the moduli $m$ is very restrictive and, in addition, that such moduli $m$ are such that there are no extra square root beside 1 and -1. The crux of the argument is a lemma that indicates that most moduli have a third square root of 1 in addition to 1 and -1. The proof is also an opportunity for applying the Chinese remainder theorem (usually abbreviated CRT).

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The primitive root theorem

In this post, $m$ is a positive integer that serves as the modulus. Given $m$, it is of interest to know all the integers $a$ where $1 \le a and that $a$ and the modulus $m$ are relatively prime. The number of such values of $a$ is denoted by $\phi(m)$ (this is called the phi function). For example, for $m=11$, $\phi(11)=10$. For $m=10$, $\phi(10)=4$ since there are only 4 numbers that are relatively prime to 10, namely 1, 3, 7, and 9.

A positive integer $g$ is said to be a primitive root modulo $m$ if $\phi(m)$ is the least positive integer $x$ such that $g^x \equiv 1 \ (\text{mod} \ m)$. If $g$ is said to be a primitive root modulo $m$, it is necessary that $g$ and the modulus $m$ are relatively prime. This is because of this basic fact: $a$ and the modulus $m$ are relatively prime if and only $a \cdot b \equiv 1 \ (\text{mod} \ m)$ if and only of $a^t \equiv 1 \ (\text{mod} \ m)$ for some integer $t$. The middle condition is saying that $a$ has a multiplicative modulo $m$.

The following lemma is alluded to at the beginning. The idea will used through the proof of the primitive root theorem. So it is extracted as a lemma to make the argument easier to follow.

Lemma 1
Let $c$ and $d$ be integers such that $c>2$ and $d>2$ and such that $c$ and $d$ are relatively prime. Let $m=c \cdot d$. Then there exist $x_0$ such that $x_0 \not \equiv \pm 1 \ (\text{mod} \ m)$ and such that $x_0^2 \equiv 1 \ (\text{mod} \ m)$, i.e. $x_0$ is a third square root of 1 modulo $m$ that is neither 1 nor -1.

Proof
Consider the two equations $x \equiv 1 \ (\text{mod} \ c)$ and $x \equiv -1 \ (\text{mod} \ d)$. By the Chinese remainder theorem, there is a solution $x_0$ to this system of equations. We have $x_0^2 \equiv 1 \ (\text{mod} \ c)$ and $x_0^2 \equiv 1 \ (\text{mod} \ d)$. By the Chinese remainder theorem again, $x_0^2 \equiv 1 \ (\text{mod} \ m)$. There are three possibilities for $x_0$, 1, -1 or another value. If $x_0 \equiv 1 \ (\text{mod} \ m)$, then $1 \equiv -1 \ (\text{mod} \ d)$, which means $d \lvert 2$, contradicting that $d>2$. Similarly, $x_0 \equiv -1 \ (\text{mod} \ m)$ would lead to $c \lvert 2$, which contradicts $c>2$. The only possibility is that $x_0 \not \equiv \pm 1 \ (\text{mod} \ m)$. $\square$

The proof of Lemma 1 makes use of CRT twice. Lemma 1 will be used several times in the proof of $2 \longrightarrow 3$ in the primitive root theorem.

The Primitive Root Theorem
Let $m$ be a positive integer. Then the following conditions are equivalent:

1. There exists a primitive root modulo $m$.
2. The equation $x^2 \equiv 1 \ (\text{mod} \ m)$ has no solution outside of $\pm 1$ modulo $m$. In other words, the only square roots of 1 modulo $m$ are $\pm 1$.
3. The only possibilities for $m$ are:
• $m=2$,
• $m=4$,
• $m$ is the power of an odd prime,
• $m$ is twice the power of an odd prime.

Proof
$1 \longrightarrow 2$
Suppose that $g$ is a primitive root modulo $m$. Suppose that condition 2 does not hold. As a result, there exists some positive integer $a \not \equiv \pm 1 \ (\text{mod} \ m)$ such that $a^2 \equiv 1 \ (\text{mod} \ m)$. Since $g$ is a primitive root modulo $m$, $a \equiv g^h \ (\text{mod} \ m)$ for some integer integer $h$ with $1 \le h < \phi(m)$ (see Theorem 5 here). If $h = \phi(m)$, then $a \equiv 1 \ (\text{mod} \ m)$. Thus $1 \le h < \phi(m)$.

Furthermore, $a^2 \equiv g^{2h} \equiv 1 \ (\text{mod} \ m)$. We have $\phi(m) \le 2h$, since $\phi(m)$ is the least power $x$ such that $g^x$ is congruent to 1 modulo $m$. Since $g^{2h}=g^{\phi(m)} g^{2h-\phi(m)}$, $g^{2h} \equiv g^{2h-\phi(m)} \equiv 1 \ (\text{mod} \ m)$. We have $\phi(m) \le 2h-\phi(m)$ since $\phi(m)$ is the least power $x$ such that $g^x$ is congruent to 1 modulo $m$. The last inequality leads to $\phi(m) \le h$, contradicting the earlier observation of $h < \phi(m)$. Thus condition 2 must holds if condition 1 is true.

$2 \longrightarrow 3$
We show that for any $m$ outside of the four possibilities listed in condition 3, condition 2 does not hold. This means that if condition 2 holds for $m$, $m$ must be one of the four possibilities. We consider the cases of $m$ even and $m$ odd separately.

Case 1. The modulus $m$ is odd. Since $m$ is not the power of one single odd prime, it must be the product of two or more powers of odd primes. Let $m=c \cdot d$ where $c$ a power of an odd prime and $d$ is the product of one ore more powers of odd primes. It is clear that $c$ and $d$ are relatively prime. It is also the case that $c>2$ and $d>2$ since each of them has at least one odd prime factor. By Lemma 1, there exists a square root $x_0$ of 1 such that $x_0 \not \equiv \pm 1 \ (\text{mod} \ m)$.

For the second case, $m$ is even. We break this up into two cases. One is that $m$ is a power of 2. The other is that $m$ is not a power of 2. In these two cases, the goal is still to show that if $m$ is outside of the four possibilities in condition 3, then condition 2 does not hold.

Case 2a. The modulus $m$ is a power of 2.
Then $m=2^h$ where $h \ge 3$. In this case, we can actually exhibit a square root that is neither 1 nor -1. Define $x_0=2^{h-1}-1$. Since $h \ge 3$, $x_0>1$. Consider the following derivation:

$\displaystyle x_0^2=(2^{h-1}-1)^2=2^{2(h-1)}-2^h+1=2^h (2^{h-2}-1)+1$

The above derivation shows that $x_0^2 \equiv 1 \ (\text{mod} \ m=2^h)$. It is clear that $x_0 \not \equiv \pm 1 \ (\text{mod} \ 2^h)$.

Case 2b. The modulus $m$ is not a power of 2.
Since $m$ is even and $m$ is not a power of 2, it must be that $m=2^k \times q$ where $q$ is odd. Either $q$ is the power of an odd prime or it is not. If $q$ is the power of an odd prime, then $k \ge 2$. If $q$ is not the power of an odd prime (i.e. $q$ is of case 1 above), then $k \ge 1$. To make the argument clear, we consider the two sub cases separately.

Case 2b-1. The modulus $m=2^k \times p^j$ where $k \ge 2$, $j \ge 1$ and $p$ is an odd prime.
Let $c=2^k$ and $d=p^j$. Note that $m=c \cdot d$. By Lemma 1, there exists a square root $x_0$ of 1 modulo $m$ such that $x_0 \not \equiv \pm 1 \ (\text{mod} \ m)$.

Case 2b-2. The modulus $m=2^k \times b$ where $k \ge 1$ and $b$ is an odd integer that is not the power of an odd prime.
Consider the prime factorization of $b$, say $b=p_1^{e_1} \cdot p_2^{e_2} \cdots p_t^{e_t}$ where $t \ge 2$. Let $c=2^k \cdot p_1^{e_1}$ and $d=p_2^{e_2} \cdots p_t^{e_t}$. We can also apply Lemma 1 to obtain a square root of 1 modulo $m=c \cdot d$ that is neither 1 nor -1.

$3 \longrightarrow 1$
It is clear that there are primitive roots modulo both $m=2$ and $m=4$. For the case of power of an odd prime, see this previous post. For the case of twice the power of an odd prime, see this previous post. $\square$

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Lemma 1 plays a prominent role in the proof of the direction $2 \longrightarrow 3$. Lemma 1 shows that it is easy to find moduli that have a third square root of 1 (other than 1 and -1). As the primitive root theorem shows, having a third square root of 1 is the condition that kills the possibility of having a primitive root. Lemma 1 and the primitive root theorem speak to different sides of the same coin. One tells us that moduli with no primitive roots are easy to find. The other says that only in rare cases do you find moduli that admit primitive roots, namely the moduli that are not product of two factors, each of which is greater than 2, that are relatively prime. The proof of $2 \longrightarrow 3$ may seem tedious (by listing out all cases that satisfy Lemma 1). The key to understanding the proof is through Lemma 1.

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$\copyright \ 2015 \text{ by Dan Ma}$

# Counting Fermat witnesses

For the Fermat primality test, looking for a Fermat witness is the name of the game. Given an integer $n$ for which the “prime or composite” status is not known, if you can find a Fermat witness for $n$, you can conclude decisively that $n$ is not a prime number. If $n$ is a composite number in reality (but the status is not known to you), how likely is it to find a Fermat witness? In other words, when you use the Fermat primality test on a composite integer, what is the probability of the test giving the correct answer? Or what is the probability of making a mistake? In this post we answer these questions by having a detailed look at Fermat witnesses. This is done through a theorem (see Theorem 1 below) and by counting the numbers of Fermat witnesses of a series of composite integers.

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Fermat witness

What is a Fermat witness? A Fermat witness is a number that violates the conclusion of Fermat’s little theorem. Here’s the theorem:

Fermat’s little theorem
If $n$ is a prime number and if $a$ is an integer that is relatively prime to $n$, then $a^{n-1} \equiv 1 \ (\text{mod} \ n)$.

If we can find a number $a$ that is relatively prime to $n$ such that $a^{n-1} \not \equiv 1 \ (\text{mod} \ n)$, then we know for sure that $n$ is composite. Such a number $a$ is said to be a Fermat witness for the compositeness of $n$. For convenience, we say that a potential Fermat witness for the compositeness of $n$ is any integer $a$ in the interval $1 that is relatively prime to $n$.

To use the Fermat primality test on the integer $n$, we examine a random sample of potential Fermat witnesses for $n$. If one of the potential Fermat witnesses in the sample turns out to be a Fermat witness for $n$, we know with certainty that $n$ is composite. If none of the potential Fermat witnesses in the random sample is a Fermat witness, then $n$ is a likely a prime number.

As with most diagnostic tests, a test can make two types of mistakes – false positives or false negatives. For primality testing, we define a positive result as the outcome that says the number being tested is a prime number and a negative result as the outcome that says the number being tested is a composite number. Thus a false positive is identifying a composite number as a prime number and a false negative is identifying a prime number as a composite number. For the Fermat test, there is no false negative (see Case 1 in the next section). If the Fermat test gives a negative result, it would be a true negative.

On the other hand, the Fermat test can give a false positive. There are two cases for false positives. In one case (Case 2a), the probability of a false positive can be made as low as possible. For the other case (Case 2b), the probability of a false positive is virtually 100%.

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Looking at different cases

If $n$ is a prime number in reality (but the status is not known before the testing), the Fermat test will always give the correct result. The Fermat test will never make the mistake of declaring a prime number as composite. What if is $n$ is a composite number in reality? How would the test behave? It turns out that if $n$ is a composite number that has a Fermat witness, the test is an effective probabilistic primality test. On the other hand, if $n$ is a composite number that has no Fermat witness, the test will identify $n$ as a prime number (so the test fails completely). Here’s the cases we just describe:

• Case 1. $n$ is a prime number.
• Case 2. $n$ is a composite number.
• Case 2a. $n$ is a composite number that has a Fermat witness.
• Case 2b. $n$ is a composite number that has no Fermat witness.

Let’s look at the cases in more details. If $n$ is a prime number in reality, then it satisfies Fermat’s little theorem. The congruence $a^{n-1} \equiv 1 \ (\text{mod} \ n)$ is always true. So the Fermat test will always give the correct result when the number being tested is a prime number in reality. In Case 1, it will never identify a prime number as a composite number. As mentioned above, the probability of a false negative is zero.

If $n$ is a composite number that has at least one Fermat witness, there is a chance that the Fermat test can identify $n$ as a prime (i.e. a false positive). However, the probability of making a mistake can be reducdd by increasing the number of potential witnesses to be calculated. Indeed, if we sample $k$ potential witnesses, there is at most a $2^{-k}$ chance of getting a wrong result. In Case 2a, the probability of error can be made so small that it is practically zero for all practical purposes. In this case, the Fermat test can work truly as a probabilistic primality test. In this case, the probability of a false positive can be made as small as possible.

The last case (Case 2b) is the weakest link in the Fermat test. Any composite number $n$ such that $a^{n-1} \equiv 1 \ (\text{mod} \ n)$ for all $a$ relatively prime to $n$ is said to be a Carmichael number. When the Fermat test is applied on such a number, it will never give the correct conclusion (i.e. it will always give a false positive). It does not matter how many potential Fermat witnesses that are calculated. In fact, calculating $a^{n-1} \ (\text{mod} \ n)$ for a large number of values of $a$ can lead one to believe that this number $n$ is a prime number. In this case, the probability of a false positive is virtually 100%. So the case of Carmichael numbers cannot be totally ignored. There are infinitely many Carmichael numbers (proved in 1994). Fortunately, it is harder and harder to find Carmichael numbers as you move up in the number line. For an illustration that Carmichael numbers are rare, see a discussion in this previous post.

We will see below that for the composite numbers in Case 2a, the Fermat test has a very small probability of a false positive (identifying a composite number as a prime number). On the other hand, for the composite numbers in Case 2b, the Fermat test has a 100% probability of a false positive. Of course, when you test a number for primality, you do not know in advance what case it is. Thus the presence of Carmichael numbers is a concern.

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A floor for the count of Fermat witnesses

Theorem 1 below sets a floor for Fermat witnesses. Once again, in this theorem we only care about the composite numbers that have at least one Fermat witness. Recall that a potential Fermat witness for the compositeness of $n$ is an integer $a$ in the interval $1 such that $a$ and $n$ are relatively prime, i.e., $\text{GCD}(a,n) = 1$.

Theorem 1
Let $n$ be a composite integer that has at least one Fermat witness. Then at least half of the potential witnesses for the compositeness of $n$ are Fermat witnesses.

Proof of Theorem 1
Let $a$ be a Fermat witness for $n$. We claim that if $b$ is a potential Fermat witness that is not a Fermat witness, then $a \cdot b$ is a Fermat witness for $n$. First of all, $a \cdot b$ is relatively prime to $n$. Note that the product if two numbers, each of which is relatively prime to $n$, is once again a number that is relatively prime to $n$ (see Lemma 4 in this previous post). Thus $a \cdot b$ is a potential Fermat witness for $n$. The following shows that $a \cdot b$ is a Fermat witness for $n$.

$(a \cdot b)^{n-1}=a^{n-1} \cdot b^{n-1} \equiv a^{n-1} \not \equiv 1 \ (\text{mod} \ n)$

In the above derivation, we use the fact that $a$ is a Fermat witness for $n$ and that $b$ is not a Fermat witness for $n$, which means that $b^{n-1} \equiv 1 \ (\text{mod} \ n)$.

If all the potential Fermat witnesses are Fermat witnesses, then we are done since the conclusion of the theorem is true. Assume that at least one potential Fermat witness is not a Fermat witness. In fact, the following enumerates all potential Fermat witnesses that are not Fermat witnesses:

$b_1,\ b_2,\ \cdots, \ b_k$

where $k \ge 1$ and $b_i \not \equiv b_j \ (\text{mod} \ n)$ for all $i \ne j$. By the claim established earlier, the following numbers are all Fermat witnesses for $n$.

$a \cdot b_1, \ a \cdot b_2, \ \cdots, \ a \cdot b_k$

The above Fermat witnesses are all distinct. If $ab_i \equiv ab_j \ (\text{mod} \ n)$ where $i \ne j$, then we can multiply both sides by $a^{-1}$ and obtain $b_i \equiv b_j \ (\text{mod} \ n)$, which is not possible. What we have proved is that if there exists one Fermat witness for $n$ and if there are $k$ many distinct potential Fermat witnesses that are not Fermat witnesses, then there are at least $k$ many Fermat witnesses for $n$. This means that at most half of the potential witnesses are non-Fermat witnesses (if there are more than half, we get a contradiction). Thus at least half of the potential Fermat witnesses are Fermat witnesses. $\blacksquare$

There is another way to state Theorem 1. Recall that Euler’s phi function $\phi(n)$ is defined to be the number of integers $a$ in the interval $1 that are relatively prime to $n$. In other words, $\phi(n)$ is the count of the potential Fermat witnesses for $n$. With this in mind, Theorem 1 can be restated as the following:

Corollary 2
Let $n$ be a composite integer that has at least one Fermat witness. Then the number of Fermat witnesses for the compositeness of $n$ is at least $\displaystyle \frac{\phi(n)}{2}$.

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The significance of Theorem 1

Theorem 1 gives a floor on Fermat witnesses for one type of composite numbers, namely the composite numbers that have at least one Fermat witness (put another way, the composite numbers that are not Carmichael numbers). More importantly, Theorem 1 allows us to estimate the probability of error when using the Fermat test on such composite numbers.

When applying the Fermat test on a composite integer $n$ that has at least one Fermat witness, the only scenario in which the Fermat test can make an error (aside from calculation errors of course) is that all the random selections of $a$ are not Fermat witnesses. So you are so unlucky that you happen to pick all values of $a$ that are not Fermat witnesses. What is the probability of that?

Randomly select a potential Fermat witness for $n$, there is at least 50% chance that it is a Fermat witness and at most 50% chance that it is not a Fermat witness. We have the following statement:

If $n$ is a composite number that has at least one Fermat witness, and if you sample one potential Fermat witness for $n$, there is at most a 50% chance that it is not a Fermat witness.

If you pick two values of $a$ at random, there is at most $0.5^2=0.25=25 \%$ chance that both are not Fermat witnesses. If you pick three values of $a$ at random, there is at most $0.5^3=0.125=12.5 \%$ chance that you pick non-Fermat witnesses three times in a row. If you pick 10 potential witnesses at random, there is at most

$0.5^{10}=0.000977=0.0977 \%$

chance that all ten values of $a$ are not Fermat witnesses. In general, we can make the following statement:

If $n$ is a composite number that has at least one Fermat witness, then when sampling $k$ many potential Fermat witnesses for $n$, the probability that none of the $k$ potential witnesses is Fermat witness is $0.5^k$.

Recall that the only scenario in which the Fermat test can make a mistake when testing a composite number belonging to Case 2a is that the random sample of values of $a$ contains only non-Fermat witnesses. Thus when the sample size is large, the probability of error can be made very small.

The bottom line is that the more potential witnesses you sample, the less likely that you won’t pick a Fermat witness (i.e., it is more likely that you will pick one). Picking a Fermat witness is essentially a coin toss. If you toss a fair coin many times, it is not likely to get all tails (it is likely to get at least one head).

The calculation for each potential witness $a$ is $a^{n-1} \ (\text{mod} \ n)$. Exponentiation in modular arithmetic can be done using the fast powering algorithm, which is an efficient algorithm that involves repeated squarings and multiplications. Thus if the composite number $n$ happens to be a composite number with a Fermat witness, the Fermat test is very efficient (when used with the fast powering algorithm) and accurate.

Of course, when you test the primality of a number, you do not know which case it belongs to. If you want to avoid the case of Carmichael numbers entirely, you might want to try a primality test that can detect Carmichael numbers (e.g. the Miller-Rabin test).

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Counting examples

To reinforce the discussion in the previous sections, we count the Fermat witnesses for 10 integers. They are all small numbers (the largest one is a little over 200,000). These integers are composite integers that are not Carmichael numbers. So each has at least one Fermat witness. To count the witnesses, we create a computer program to determine the witness status for all $a$ in $1 that are relatively prime to $n$. The counts are shown in the following matrix.

Composite integers that are not Carmichael numbers

$\left[\begin{array}{rrrrrrrrr} \text{ } & \text{ } & \text{ } & \text{ } & \text{Fermat Witness} & \text{ } & \text{Fermat Witness} & \text{ } & \text{Fermat Witness} \\ n & \text{ } & \phi(n) & \text{ } & \text{Yes} & \text{ } & \text{No} & \text{ } & \text{Yes} \ \% \\ \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } \\ 91 & \text{ } & 72 & \text{ } & 36 & \text{ } & 36 & \text{ } & 50 \% \\ 221 & \text{ } & 192 & \text{ } & 176 & \text{ } & 16 & \text{ } & 91.67 \% \\ 341 & \text{ } & 300 & \text{ } & 200 & \text{ } & 100 & \text{ } & 66.67 \% \\ 5,777 & \text{ } & 5,616 & \text{ } & 5,600 & \text{ } & 16 & \text{ } & 99.72 \% \\ 10,873 & \text{ } & 10,660 & \text{ } & 10,656 & \text{ } & 4 & \text{ } & 99.96 \% \\ 21,809 & \text{ } & 21,504 & \text{ } & 21,248 & \text{ } & 256 & \text{ } & 98.81 \% \\ 50,113 & \text{ } & 42,948 & \text{ } & 42,912 & \text{ } & 36& \text{ } & 99.92 \% \\ 73,861 & \text{ } & 73,312 & \text{ } & 73,296 & \text{ } & 16 & \text{ } & 99.98 \% \\ 100,097 & \text{ } & 99,396 & \text{ } & 99,392 & \text{ } & 4 & \text{ } & 100 \% \\ 201,217 & \text{ } & 200,260 & \text{ } & 200,256 & \text{ } & 4 & \text{ } & 100 \% \end{array}\right]$

The first column is the 10 integers from 91 to 201,217. The second column is Euler’s phi function, which is the count of all integers $a$ that are relatively prime to $n$, which is the count of the potential Fermat witnesses for $n$. For $n=91$, there are 72 potential Fermat witnesses where exactly half are Fermat witnesses. For the other numbers on the list, the percentages of Fermat witnesses are a lot more than 50%. In fact, for most of them, the Fermat witness percentages are over 99%. For the last two numbers on the list the percentage is virtually 100%.

Consider 201,217, the last number on the list. Applying the Fermat test on the number, it will be hard not to pick up a Fermat witness. With only 4 potential witnesses that are not Fermat witness, it is certain to find one Fermat witness when more than 4 numbers are sampled. In fact, even if just 4 numbers are sampled, there is a virtually 100% chance that a Fermat witness will be found.

Take the number in the above table that has the largest number of non-Fermat witnesses ($n=21,809$), which has 256 non-Fermat witnesses. It has 21,504 potential witnesses. Out of a random sample of 20 such potential Fermat witnesses, the probability that none of them is a Fermat witness is $3.24 \cdot 10^{-39}$, which is practically zero.

The above calculations show that for composite numbers that have at least one Fermat witness, the Fermat test is very accurate with a very high probability. The key is to sample a large enough number of potential witnesses. However for Carmichael numbers, it is totally different story.

As discussed earlier, a Carmichael number is a composite number $n$ that has no Fermat witnesses. Specifically a composite number $n$ is a Carmichael number if $a^{n-1} \equiv 1 \ (\text{mod} \ n)$ for all integers $a$ such that $1 and $a$ and $n$ are relatively prime. The following table is a demonstration of this property.

The first ten Carmichael numbers

$\left[\begin{array}{rrrrrrrrr} \text{ } & \text{ } & \text{ } & \text{ } & \text{Fermat Witness} & \text{ } & \text{Fermat Witness} & \text{ } & \text{Fermat Witness} \\ n & \text{ } & \phi(n) & \text{ } & \text{Yes} & \text{ } & \text{No} & \text{ } & \text{Yes} \ \% \\ \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } \\ 561 & \text{ } & 320 & \text{ } & 0 & \text{ } & 320 & \text{ } & 0 \% \\ 1,105 & \text{ } & 768 & \text{ } & 0 & \text{ } & 768 & \text{ } & 0 \% \\ 1,729 & \text{ } & 1,296 & \text{ } & 0 & \text{ } & 1,296 & \text{ } & 0 \% \\ 2,465 & \text{ } & 1,792 & \text{ } & 0 & \text{ } & 1,792 & \text{ } & 0 \% \\ 2,821 & \text{ } & 2,160 & \text{ } & 0 & \text{ } & 2,160 & \text{ } & 0 \% \\ 6,601 & \text{ } & 5,280 & \text{ } & 0 & \text{ } & 5,280 & \text{ } & 0 \% \\ 8,911 & \text{ } & 7,128 & \text{ } & 0 & \text{ } & 7,128& \text{ } & 0 \% \\ 10,585 & \text{ } & 8,064 & \text{ } & 0 & \text{ } & 8,064 & \text{ } & 0 \% \\ 15,841 & \text{ } & 12,960 & \text{ } & 0 & \text{ } & 12,960 & \text{ } & 0 \% \\ 29,341 & \text{ } & 25,920 & \text{ } & 0 & \text{ } & 25,920 & \text{ } & 0 \% \end{array}\right]$

The above table lists out the first ten Carmichael numbers. Though we can show that each one of them is a Carmichael number by using the Korselt’s Criterion (see here but you have to first factor the numbers). We calculate the Fermat witness status for each potential witness for each $n$ in the table. The above table is a visual demonstration of the fact that the column for the Fermat witnesses is entirely zero! So if you happen to test the primality on such numbers using the Fermat test, you will conclude that they are prime numbers unless you happen to sample a value of $a$ whose GCD with $n$ is greater than one (i.e., the only way you can determine the compositeness of a Carmichael number is to stumble into a GCD witness).

Fermat test does what it does well with the exception of Carmichael numbers. As mentioned earlier, if you want to avoid the trap of working with Carmichael numbers (as rare as they are), you can always switch to a test that will always detect Carmichael numbers (e.g. Miller-Rabin test).

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$\copyright \ 2014 \text{ by Dan Ma}$

# Factorization versus primality testing

Let $n$ be a large positive integer whose “prime versus composite” status is not known. One way to know whether $n$ is prime or composite is to factor $n$ into its prime factors. If there is a non-trivial factor (one that is neither 1 nor $n$), it is composite. Otherwise $n$ is prime. This may sound like a reasonable approach in performing primality testing – checking whether a number is prime or composite. In reality, factoring and primality testing, though related, are very different problems. For a very large number (e.g. with at least 300 decimal digits), it is possible that, even with the state of the art in computing, factoring it may take more than a few million years. On the other hand, it will take a modern computer less than a second to determine whether a 300-digit number is prime or composite. Interestingly this disparity is one reason that makes the RSA work as a practical and secure cryptosystem. In this post, we use the RSA cryptosystem as an example to give a sense that factoring is a “hard” problem while primality testing is an “easy” problem. The primality test used in the examples is the Fermat primality test.

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The brute force approach

There is a natural and simple approach in factoring, which is to do trial divisions. To factor the number $n$, we divide $n$ by every integer $a$ in the range $1. Once a factor $a$ is found, we repeat the process with the complementary factor $\frac{n}{a}$ until all the prime factors of $n$ are found. This is simple in concept and is sure to produce the correct answer. For applications in cryptography, this brute force approach is essentially useless since the amount of time to try every candidate factor is prohibitively huge. The amount of time required may be more than the age of the universe if the brute force approach is used.

The brute force approach can be improved upon slightly by doing the trial divisions using candidate factors up to $\sqrt{n}$. It is well known that if a composite integer $n$ is greater than one, then it has a prime divisor $d$ such that $1. So instead of dividing $n$ by every number $a$ with $1, we can divide $n$ by every prime number $a$ with $1. But even this improved brute force approach will still take too long to be practical.

Let’s look at a quick example for brute force factoring. Let $n=96638243$. Note that $\sqrt{n}=\sqrt{96638243}=9676$. There are 1192 odd primes less than 9676. In dividing $n$ by these primes, we stop at 127 and we have $n=96638243=127 \cdot 737309$. We now focus the attention on $737309$. Note that $\sqrt{737309}=858.67$ and there are 147 odd primes less than 858. Dividing $737309$ by these 147 candidate factors, we find that none of them is a factor. We can conclude $737309$ is prime. Then we have the factorization $n=96638243=127 \cdot 737309$.

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Example of RSA

RSA is a public-key cryptosystem and is widely used for secure data transmission. The RSA public key consists of two parts. One is the modulus that is the product of two distinct prime factors. Suppose the modulus is called $N$ and we have $N=pq$ where $p$ and $q$ are distinct prime numbers. How large does $N$ have to be? The larger the $N$ is, the more secure RSA is. The current practice is that for corporate use the modulus is at least a 1024-bit number (the bit size is called the key length). If data is extra sensitive or if the data needs to be retained for a long time, then a larger key length should be used (e.g. 2048-bit). With a 1024-bit modulus $N=pq$, each prime factor is a 512-bit number. The other part of the RSA public key is the encryption key, which is an integer $e$ that is relatively prime to the integer $(p-1) \cdot (q-1)$.

Let’s say we want to generate a 1024-bit modulus. There are two challenges with a key of this size. One is that a reliable way is needed to obtain two prime numbers that are 512-bit long. Given a large integer that is at least 512-bit long, how do we determine reliably that it is prime? Is it possible to factor a 512-bit integer using the brute force approach? The other challenge is from the perspective of an attacker – successful factoring the 1024-bit modulus would break RSA and allow the attacker to read the secret message. Let’s look at the problem of the attacker trying to factor a 1024-bit number. A 1024-bit number is approximately $2^{1024}$. The following calculation converts it to a decimal basis:

$\displaystyle 2^{1024}=(10^{\text{log(2)}})^{1024} \approx 10^{308.25}$

We use $\text{log}(x)$ to denote the logarithm of base 10. Note that $1024 \cdot \text{log}(2)=308.25$. So a 1024-bit number has over 300 digits.

Let’s see what the challenge is if you want to factor a 1024-bit number. Suppose your chosen large number $n$ is such that $n \approx 10^{308}$. Note that $\sqrt{10^{308}}=10^{154}$. According to the improved brute force approach described above, in effect you will need to divide $n$ by every prime number less than $10^{154}$.

Now let’s get an estimate on the number of prime numbers less than $10^{154}$. According to the prime number theorem, the number of prime numbers at most $x$ is approximately

$\displaystyle \pi(x) \approx \frac{x}{\text{ln}x}$

where $\pi(x)$ is the number of primes at most $x$. Then $\pi(10^{154}) \approx 2.82 \cdot 10^{151}$. This is certainly a lot of prime numbers to check.

It is hard to comprehend such large numbers. Let’s put this into perspective. Currently the world population is about 7 billion. Let’s say each person in the world possesses a supercomputer that can check $10^{40}$ prime numbers per second (i.e. to check whether they are factors of the number $n$). This scenario clearly far exceeds the computing resources that are currently available. Suppose that the 7 billion supercomputers are available and that each one can check $10^{40}$ many primes per second. Then in each second, the following is the number of prime numbers that can be checked by the 7 billion supercomputers.

$\displaystyle 7 \cdot 10^9 \cdot 10^{40}=7 \cdot 10^{49} \text{ prime numbers per second}$

The following is the number of seconds it will take to check $2.82 \cdot 10^{151}$ many prime numbers:

$\displaystyle \frac{2.82 \cdot 10^{151}}{7 \cdot 10^{49}} \approx 4 \cdot 10^{101} \text{ seconds}$

The universe is estimated to be about 13 billion years old. The following calculation converts it to seconds.

$13 \text{ billion years}=13 \cdot 10^9 \cdot 365 \cdot 24 \cdot 3600 \approx 4 \cdot 10^{17} \text{ seconds}$

With 7 billion fast suppercomputers (one for each person in the world) running in the entire life of the universe, you can only finish checking

$\displaystyle \frac{4 \cdot 10^{17}}{4 \cdot 10^{101}}=\frac{1}{10^{84}}$

of the $2.82 \cdot 10^{151}$ many prime numbers. Note that $\frac{1}{10^{84}}$ is a tiny portion of 1%. So by taking the entire life of the universe to run the 7 billion supercomputers, each checking $10^{40}$ many candidate prime factors per second, you would not even make a dent in the problem!

The security of RSA rests on the apparent difficulty of factoring large numbers. If the modulus $N=pq$ can be factored, then an eavesdropper can obtain the private key from the public key and be able to read the message. The difficulty in factoring means there is a good chance that RSA is secure. In order to break RSA, an attacker would probably have to explore other possible vulnerabilities instead of factoring the modulus.

By carrying out a similar calculation, we can also see that factoring a 512-bit number by brute force factoring is also not feasible. Thus in the RSA key generation process, it is not feasible to use factoring as a way to test primality. The alternative is to use efficient primality tests such as Fermat test or Miller-Rabin test. The computation for these tests is based on the fast powering algorithm, which is a very efficient algorithm.

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The story told by RSA numbers

The required time of more than the life of the universe as discussed above is based on the naïve brute force approach of factoring. There are many other factoring approaches that are much more efficient and much faster, e.g., the quadratic sieve algorithm, the number field sieve algorithm, and the general number field sieve algorithm. For these methods, with ample computing resources at the ready, factoring a 1024-bit or 2048-bit number may not take the entire life of the universe but make take decades or more. Even with these better methods, the disparity between slow factoring and fast primality testing is still very pronounced and dramatic.

The best evidence of slow factoring even with using modern methods is from the RSA numbers. The RSA numbers are part of the the RSA Factoring Challenge, which was created in 1991 to foster research in computational number theory and the practical difficulty of factoring large integers. The challenge was declared inactive in 2007. The effort behind the successful factorization of some of these numbers gives us an idea of the monumental challenges in factoring large numbers.

According to the link given in the above paragraph, there are 54 RSA numbers, ranging from 330 bits long to 2048 bits long (100 decimal digits to 617 decimal digits). Each of these numbers is a product of two prime numbers. Of these 54 numbers, 18 were successfully factored (as of the writing of this post). They were all massive efforts involving large groups of volunteers (in some cases using hundreds or thousands of computers), spanning over months or years. Some of methods used are the quadratic sieve algorithm, the number field sieve algorithm, and the general number field sieve algorithm.

The largest RSA number that was successfully factored is the RSA-768, which is 768 bits long and has 232 decimal digits (completed in December 2009). The method used was the Number Field Sieve method. There were 4 main steps in this effort. The first step is the polynomial selection, which took half a year using 80 processors. The second step is the sieving step, which took almost two years on many hundreds of machines. If only using a single core 2.2 GHz AMD Opteron processor with 2 GB RAM, the second step would take about 1500 years! The third step is the matrix step, which took a couple of weeks on a few processors. The final step took a few days, which involved a great deal of debugging.

The number field sieve method is the fastest known method for factoring large numbers that are a product of two primes (i.e. RSA moduli). The effort that went into factoring RSA-768 was massive and involved many years of complicated calculations and processing. This was only a medium size number on the list!

Another interesting observation that can be made is on the RSA numbers that have not been factored yet. There are 36 unfactored numbers in the list. One indication that RSA is secure in the current environment is that the larger numbers in the list are not yet factored (e.g. RSA-1024 which is 1024-bit long). Successful factorization of these numbers has important security implication for RSA. The largest number on the list is RSA-2048, which is 2048-bit long and has 617 digits. It is widely believed that RSA-2048 will stay unfactored in the decades to come, barring any dramatic and significant advance in computing technology.

The factoring challenge for the RSA numbers certainly provides empirical evidence that factoring is hard. Of course, no one should be complacent. We should not think that factoring will always be hard. Technology will continue to improve. A 768-bit RSA modulus was once considered secure. With the successful factorization of RSA-768, key size of 768 bits is no longer considered secure. Currently 1024 bit key size is considered secure. The RSA number RSA-1024 could very well be broken in within the next decade.

There could be new advances in factoring algorithm too. A problem that is thought to be hard may eventually turn out to be easy. Just because everyone thinks that there is no fast way of factoring, it does not mean that no such method exists. It is possible that someone has discovered such a method but decides to keep it secret in order to maintain the advantage. Beyond the issue of factoring, there could be some other vulnerabilities in RSA that can be explored and exploited by attackers.

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Fermat primality test

We now give some examples showing primality testing is a much better approach (over factoring) if the goal is to check the “prime or composite” status only. We use Fermat primality test as an example.

Example 1
Let $n=15144781$. This is a small number. So factoring would be practical as a primality test. We use it to illustrate the point that the “prime or composite” status can be determined without factoring. One option is to use Fermat’s little theorem (hence the name of Fermat primality test):

Fermat’s little theorem
If $n$ is a prime number and if $a$ is an integer that is relatively prime to $n$, then $a^{n-1} \equiv 1 \ (\text{mod} \ n)$.

Consider the contrapositive of the theorem. If we can find an $a$, relatively prime to $n$ such that $a^{n-1} \not \equiv 1 \ (\text{mod} \ n)$, then we know for sure $n$ is not prime. Such a value of $a$ is said to be a Fermat witness for the compositeness of $n$.

If a Fermat witness is found, then we can say conclusively that $n$ is composite. On the other hand, if $a$ is relatively prime to $n$ and $a^{n-1} \equiv 1 \ (\text{mod} \ n)$, then $n$ is probably a prime. We can then declare $n$ is prime or choose to run the test for a few more random values of $a$.

The exponentiation $a^{n-1} \ (\text{mod} \ n)$ is done using the fast powering algorithm, which involves a series of squarings and multiplications. Even for large moduli, the computer implementation of this algorithm is fast and efficient.

Let’s try some value of $a$, say $a=2$. Using an online calculator, we have

$2^{15144780} \equiv 1789293 \not \equiv 1 \ (\text{mod} \ 15144781)$

In this case, one congruence calculation tells us that $n=15144781$ is not prime (if it were, the congruence calculation would lead to a value of one). It turns out that $n=15144781$ is a product of two primes where $n=15144781=3733 \cdot 4057$. Of course, this is not a secure modulus for RSA. The current consensus is to use a modulus that is at least 1024-bit long.

Example 2
Let $n=15231691$. This is also a small number (in relation what is required for RSA). Once again this is an illustrative example. We calculate $a^{15231690} \ (\text{mod} \ 15231691)$ for $a=2,3,4,5,6,7$, the first few values of $a$. All such congruence values are one. We suspect that $n=15231691$ may be prime. So we randomly choose 20 values of $a$ and compute $a^{15231690} \ (\text{mod} \ 15231691)$. The following shows the results.

$\left[\begin{array}{rrr} a & \text{ } & a^{n-1} \ \text{mod} \ n \\ \text{ } & \text{ } & n=15,231,691 \\ \text{ } & \text{ } & \text{ } \\ 3,747,236 & \text{ } & 1 \\ 370,478 & \text{ } & 1 \\ 12,094,560 & \text{ } & 1 \\ 705,835 & \text{ } & 1 \\ 10,571,714 & \text{ } & 1 \\ 15,004,366 & \text{ } & 1 \\ 12,216,046 & \text{ } & 1 \\ 10,708,300 & \text{ } & 1 \\ 6,243,738 & \text{ } & 1 \\ 1,523,626 & \text{ } & 1 \\ 10,496,554 & \text{ } & 1 \\ 10,332,033 & \text{ } & 1 \\ 10,233,123 & \text{ } & 1 \\ 3,996,691 & \text{ } & 1 \\ 4,221,958 & \text{ } & 1 \\ 3,139,943 & \text{ } & 1 \\ 1,736,767 & \text{ } & 1 \\ 12,672,150 & \text{ } & 1 \\ 12,028,143 & \text{ } & 1 \\ 8,528,642 & \text{ } & 1 \end{array}\right]$

For all 20 random values of $a$, $a^{15231690} \equiv 1 \ (\text{mod} \ 15231691)$. This represents strong evidence (though not absolute proof) that $n=15231691$ is a prime. In fact, we can attach the following probability statement to the above table of 20 random values of $a$.

If $n=15231691$ were a composite number that has at least one Fermat witness, there is at most a 0.0000953674% chance that 20 randomly selected values of $a$ are not Fermat witnesses.

In other words, if $n=15231691$ were a composite number that has at least one Fermat witness, there is at most a 0.0000953674% chance of getting 20 1’s in the above computation.

In general, if $n$ has at least one Fermat witness, the probability that all $k$ randomly selected values of $a$ with $1 are not Fermat witnesses is at most $0.5^k$. For $k=20$, $0.5^{20}=0.000000953674$, which is 0.0000953674%. The probability statement should give us enough confidence to consider $n=15231691$ a prime number.

There is a caveat that has to be mentioned. For the above probability statement to be valid, the number $n$ must have at least one Fermat witness. If a number $n$ is composite, we would like the test to produce a Fermat witness. It turns out that there are composite numbers that have no Fermat witnesses. These numbers are called Carmichael numbers. If $n$ is such a number, $a^{n-1} \equiv 1 \ (\text{mod} \ n)$ for any $a$ that is relatively prime to the number $n$. In other words, the Fermat test will always indicate “probably prime” for Carmichael numbers. Unless you are lucky and randomly pick a value of $a$ that shares a common prime factor with $n$, the Fermat test will always incorrectly identify a Carmichael number $n$ as prime. Fortunately Carmichael numbers are rare, even though there are infinitely many of them. In this previous post, we estimate that a randomly selected 1024-bit odd integer has a less than one in $10^{88}$ chance of being a Carmichael number!

The Fermat test is a powerful test when the number being tested is a prime number or a composite number that has a Fermat witness. For Carmichael numbers, the test is likely to produce a false positive (identifying a composite number as prime). Thus the existence of Carmichael numbers is the biggest weakness of the Fermat test. Fortunately Carmichael numbers are rare. Though they are rare, their existence may still make the Fermat test unsuitable in some situation, e.g., when you test a number provided by your adversary. If you really want to avoid situations like these, you can always switch to the Miller-Rabin test.

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$\copyright \ 2014 \text{ by Dan Ma}$