# Proving Chinese Remainder Theorem

In this post, we consider systems of linear congruence equations with respect to pairwise relatively prime moduli. The theorem we prove here is known as the Chinese remainder theorem (see Theorem 1 below). It is useful in many contexts. For an example, see the blog post Introducing Carmichael numbers (used in proving the Korselt’s Criterion). Theorem 2 below is a useful corollary of the Chinese remainder theorem. For examples of applications, see Primitive roots of twice the powers of odd primes and The primitive root theorem (used in proving the primitive root theorem).

Theorem 1 (Chinese Remainder Theorem)

Let $n_1,n_2,\cdots,n_k$ be positive integers that are pairwise relatively prime. Let $a_1,a_2,\cdots,a_k$ be any integers. Then the following system of linear congruence equations

$x \equiv a_1 \ (\text{mod} \ n_1)$

$x \equiv a_2 \ (\text{mod} \ n_2)$

$x \equiv a_3 \ (\text{mod} \ n_3)$

$\cdots$
$\cdots$
$\cdots$

$x \equiv a_k \ (\text{mod} \ n_k)$

has a solution $x=b$.

Furthermore, $x=b_0$ is another solution to the above system of equations if and only if $b_0 \equiv b \ (\text{mod} \ n_1 n_2 \cdots n_k)$.

Proof of Theorem 1
Let $n=n_1 n_2 \cdots n_k$. For each $i=1,\cdots,k$, let $\displaystyle t_i=\frac{n}{n_i}$. Note that $t_i$ is the product of all $n_j$ where $j \ne i$. So $t_i$ and $n_i$ are relatively prime. Thus for each $i$, $t_i$ has a multiplicative inverse modulo $n_i$. Let $(t_i)^{-1}$ be the inverse of $t_i$. Now define $g_i=t_i \cdot (t_i)^{-1}$ for each $i$.

It is clear that for each $i$, $g_i \equiv 1 \ (\text{mod} \ n_i)$ for each $i$ and $g_i \equiv 0 \ (\text{mod} \ n_j)$ for all $j \ne i$. Now define $b$ as the following:

$b=g_1 \cdot a_1+g_2 \cdot a_2+\cdots+g_k \cdot a_k$

It follows that $b \equiv a_i \ (\text{mod} \ n_i)$ for each $i=1,\cdots,k$. Thus $x=b$ is a solution that we are looking for.

It remains to show that solutions to the above system of equations are unique modulo $n$. Suppose that $b_0 \equiv b \ (\text{mod} \ n)$. This means that $b_0=b+n_1 n_2 \cdots n_k \cdot y$ for some integer $y$. It follows from this equation that $b_0 \equiv b \ (\text{mod} \ n_i)$ for each $i$. Thus $b_0 \equiv a_i \ (\text{mod} \ n_i)$ for each $i$.

On the other hand, suppose that $x=b_0$ is another solution to the above system of equations. Then $b_0 \equiv b \ (\text{mod} \ n_i)$ for each $i$. So $n_i \ \lvert \ (b_0-b)$ for each $i$. We have the following equations

$b_0-b=n_1 \cdot c_1$

$b_0-b=n_2 \cdot c_2$

$b_0-b=n_3 \cdot c_3$

$\cdots$
$\cdots$
$\cdots$

$b_0-b=n_k \cdot c_k$

for some integers $c_1, c_2, \cdots, c_k$. From these equations, it follows that $b_0-b=n_1 n_2 \cdots n_k \cdot c$ for some integer $c$. This is due to the fact that the integers $n_1,n_2,\cdots,n_k$ are relatively prime. For example, from the first two equations, $n_2 \ \lvert \ n_1 \cdot c_1$. Since $n_1$ and $n_2$ are relatively prime, $n_2 \ \lvert \ c_1$, we have $b_0-b=n_1 \cdot n_2 \cdot c_{1,2}$ for some integer $c_{1,2}$. Then consider this new equation along with the third equation above, we have $n_3 \ \lvert \ n_1 \cdot n_2 \cdot c_{1,2}$. Because the integers $n_j$ are pairwise relatively prime, $n_3 \ \lvert \ c_{1,2}$. So we can further factor $c_{1,2}$ into $n_3 \cdot c_{1,3}$ for some integer $c_{1,3}$. Continuing this process, we can obtain $b_0-b=n_1 n_2 \cdots n_k \cdot c$. It follows that $b_0 \equiv b \ (\text{mod} \ n_1 n_2 \cdots n_k)$. $\blacksquare$

Theorem 2 (Corollary of Chinese Remainder Theorem)

Let $n_1,n_2,\cdots,n_k$ be positive integers that are pairwise relatively prime. Then for any integers $c$ and $d$, the following linear congruences hold

$c \equiv d \ (\text{mod} \ n_1)$

$c \equiv d \ (\text{mod} \ n_2)$

$c \equiv d \ (\text{mod} \ n_3)$

$\cdots$
$\cdots$
$\cdots$

$c \equiv d \ (\text{mod} \ n_k)$

if and only of $c \equiv d \ (\text{mod} \ n_1 n_2 \cdots n_k)$.

Proof of Theorem 2
This is an easy consequence of Theorem 1. Let $d=a_1=a_2=\cdots=a_k$. Then $x=d$ would be a solution to the resulting system of linear congruence equations in Theorem 1. Theorem 2 then follows from the fact that any other solution to the system is congruent to $d$ modulo $n_1 n_2 \cdots n_k$. $\blacksquare$

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$\copyright \ 2013 \text{ by Dan Ma}$

# Introducing Carmichael numbers

This is an introduction to Carmichael numbers. We first discuss Carmichael numbers in the context of Fermat primality test and then discuss several basic properties. We also prove Korselt’s criterion, which gives a useful characterization of Carmichael numbers.

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Fermat Primality Test

Fermat’s little theorem states that if $p$ is a prime number, then $a^p \equiv a \ (\text{mod} \ p)$ for any integer $a$. Fermat primality test refers to the process of using Fermat little theorem to check the “prime vs. composite” status of an integer.

Suppose that we have a positive integer $n$ such that the “prime vs. composite” status is not known. If we can find an integer $a$ such that $a^n \not \equiv a \ (\text{mod} \ n)$, then we know for certain that the modulus $n$ is composite (or not prime). For example, let $n = \text{8,134,619}$. Note that $2^{8134619} \equiv 3024172 \ (\text{mod} \ 8134619)$. So we know right away that $n = \text{8,134,619}$ is not prime, even though we do not know what its prime factors are just from applying this test.

Given a positive integer $n$, whenever $a^n \not \equiv a \ (\text{mod} \ n)$, we say that $a$ is a Fermat witness for (the compositeness of) the integer $n$. Thus $2$ is a Fermat witness for $n = \text{8,134,619}$.

What if we try one value of $a$ and find that $a$ is not a witness for (the compositeness of) $n$? Then the test is inconclusive. The best we can say is that $n$ is probably prime. It makes sense to try more values of $a$. If all the values of $a$ we try are not witnesses for $n$ (i.e. $a^n \equiv a \ (\text{mod} \ n)$ for all the values of $a$ we try), then it “seems likely” that $n$ is prime. But if we actually declare that $n$ is prime, the decision could be wrong!

Take $n=\text{10,024,561}$. For several randomly chosen values of $a$, we have the following calculations:

$\displaystyle 5055996^{10024561} \equiv 5055996 \ (\text{mod} \ 10024561)$

$\displaystyle 4388786^{10024561} \equiv 4388786 \ (\text{mod} \ 10024561)$

$\displaystyle 4589768^{10024561} \equiv 4589768 \ (\text{mod} \ 10024561)$

$\displaystyle 146255^{10024561} \equiv 146255 \ (\text{mod} \ 10024561)$

$\displaystyle 6047524^{10024561} \equiv 6047524 \ (\text{mod} \ 10024561)$

The above calculations could certainly be taken as encouraging signs that $n=\text{10,024,561}$ is prime. With more values of $a$, we also find that $a^{10024561} \equiv a \ (\text{mod} \ 10024561)$. However, if we declare that $n=\text{10,024,561}$ is prime, it turns out to be a wrong conclusion.

In reality, $n=\text{10,024,561}$ is composite with $\text{10,024,561}=71 \cdot 271 \cdot 521$. Furthermore $a^{10024561} \equiv a \ (\text{mod} \ 10024561)$ for any integer $a$. So there are no witnesses for $n=\text{10,024,561}$. Any composite positive integer that has no Fermat witnesses is called a Carmichael number, in honor of Robert Carmichael who in 1910 found the smallest such number, which is 561.

Fermat primality test is always correct if the conclusion is that the integer being tested is a composite number (assuming there is no computational error). If the test says the number is composite, then it must be a composite number. In other words, there are no false negatives in using Fermat primality test as described above.

On the other hand, there can be false positives as a result of using Fermat primality test. If the conclusion is that the integer being tested is a prime number, it is possible that the conclusion is wrong. For a wrong conclusion, it could be that there exists a witness for the number being tested and that we have missed it. Or it could be that the number being tested is a Carmichael number. Though Carmichael numbers are rare but there are infinitely many of them. So we cannot ignore them entirely. For these reasons, Fermat primality test as described above is often not used. Instead, other extensions of the Fermat primality test are used.

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Carmichael Numbers

As indicated above, a Carmichael number is a positive composite integer that has no Fermat witnesses. Specifically, it is a positive composite integer that satisfies the conclusion of Fermat’s little theorem. In other words, a Carmichael number is a positive composite integer $n$ such that $a^n \equiv a \ (\text{mod} \ n)$ for any integer $a$.

Carmichael numbers are rare. A recent search found that there are $\text{20,138,200}$ Carmichael numbers between $1$ and $10^{21}$, about one in 50 trillion numbers (documented in this Wikipedia entry on Carmichael numbers). However it was proven by Alford, Granville and Pomerance in 1994 that there are infinitely many Carmichael numbers (paper).

The smallest Carmichael number is $561=3 \cdot 11 \cdot 17$. A small listing of Carmichael numbers can be found in this link, where the example of $n=\text{10,024,561}$ is found.

Carmichael numbers must be odd integers. To see this, suppose $n$ is a Carmichael number and is even. Let $a=-1$. By condition (1) of Theorem 1, we have $(-1)^n=1 \equiv -1 \ (\text{mod} \ n)$. On the other hand, $-1 \equiv n-1 \ (\text{mod} \ n)$. Thus $n-1 \equiv 1 \ (\text{mod} \ n)$. Thus we have $n \equiv 2 \ (\text{mod} \ n)$. It must be the case that $n=2$, contradicting the fact that $n$ is a composite number. So any Carmichael must be odd.

The following theorem provides more insight about Carmichael numbers. A positive integer $n$ is squarefree if its prime decomposition contains no repeated prime factors. In other words, the integer $n$ is squarefree means that if $\displaystyle n=p_1^{e_1} p_2^{e_2} \cdots p_t^{e_t}$ is the prime factorization of $n$, then all exponents $e_j=1$.

Theorem 1 (Korselt’s Criterion)

Let $n$ be a positive composite integer. Then the following conditions are equivalent.

1. The condition $a^n \equiv a \ (\text{mod} \ n)$ holds for any integer $a$.
2. The condition $a^{n-1} \equiv 1 \ (\text{mod} \ n)$ holds for any integer $a$ that is relatively prime to $n$.
3. The integer $n$ is squarefree and $p-1 \ \lvert \ (n-1)$ for any prime divisor $p$ of $n$.

Proof of Theorem 1

$1 \Longrightarrow 2$
Suppose that $a$ is relatively prime to the modulus $n$. Then let $b$ be the multiplicative inverse of $a$ modulo $n$, i.e., $ab \equiv 1 \ (\text{mod} \ n)$. By (1), we have $a^n \equiv a \ (\text{mod} \ n)$. Multiply both sides by the multiplicative inverse $b$, we have $a^{n-1} \equiv 1 \ (\text{mod} \ n)$.

$2 \Longrightarrow 3$
Let $\displaystyle n=p_1^{e_1} p_2^{e_2} \cdots p_t^{e_t}$ be the prime factorization of $n$ where $p_i \ne p_j$ for $i \ne j$ and each exponent $e_j \ge 1$. Since $n$ must be odd, each $p_j$ must be an odd prime.

We first show that each $e_j=1$, thus showing that $n$ is squarefree. To this end, for each $j$, let $a_j$ be a primitive root modulo $p_j^{e_j}$ (see Theorem 4 in the post Primitive roots of powers of odd primes). Consider the following system of linear congruence equations:

$x \equiv a_1 \ (\text{mod} \ p_1^{e_1})$

$x \equiv a_2 \ (\text{mod} \ p_2^{e_2})$

$\cdots$
$\cdots$
$\cdots$

$x \equiv a_t \ (\text{mod} \ p_t^{e_t})$

Since the moduli $p_j^{e_j}$ are pairwise relatively prime, this system must have a solution according to the Chinese Remainder Theorem (a proof is found here). Let $a$ one such solution. For each $j$, since $a_j$ is a primitive root modulo $p_j^{e_j}$, $a_j$ is relatively prime to $p_j^{e_j}$. Since $a \equiv a_j \ (\text{mod} \ p_j^{e_j})$, $a$ is relatively prime to $p_j^{e_j}$ for each $j$. Consequently, $a$ is relatively prime to $n$. By assumption (2), we have $a^{n-1} \equiv 1 \ (\text{mod} \ n)$.

Now fix a $j$ with $1 \le j \le t$. We show that $e_j=1$. Since $a^{n-1} \equiv 1 \ (\text{mod} \ n)$, $a^{n-1} \equiv 1 \ (\text{mod} \ p_j^{e_j})$. Since $a \equiv a_j \ (\text{mod} \ p_j^{e_j})$, we have $a_j^{n-1} \equiv 1 \ (\text{mod} \ p_j^{e_j})$. Note that the order of $a_j$ modulo $p_j^{e_j}$ is $\phi(p_j^{e_j})=p_j^{e_j-1}(p_j-1)$. Thus we have $p_j^{e_j-1}(p_j-1) \ \lvert \ (n-1)$. If $e_j>1$, then $p_j \ \lvert \ (n-1)$, which would mean that $p_j \ \lvert \ 1$. So it must be the case that $e_j=1$. It then follows that $(p_j-1) \ \lvert \ (n-1)$.

$3 \Longrightarrow 1$
Suppose that $n=p_1 p_2 \cdots p_t$ is a product of distinct prime numbers such that for each $j$, $(p_j-1) \ \lvert \ (n-1)$.

Let $a$ be any integer. First we show that $a^n \equiv a \ (\text{mod} \ p_j)$ for all $j$. It then follows that $a^n \equiv a \ (\text{mod} \ n)$.

Now fix a $j$ with $1 \le j \le t$. First consider the case that $a$ and $p_j$ are relatively prime. According to Fermat’s little theorem, $a^{p_j-1} \equiv 1 \ (\text{mod} \ p_j)$. Since $(p_j-1) \ \lvert \ (n-1)$, $a^{n-1} \equiv 1 \ (\text{mod} \ p_j)$. By the Chinese Remainder Theorem, it follows that $a^{n-1} \equiv 1 \ (\text{mod} \ n)$ and $a^n \equiv a \ (\text{mod} \ n)$. $\blacksquare$

Examples
With Korselt’s criterion, it is easy to verify Carmichael numbers as long as the numbers are factored. For example, the smallest Carmichael number is $561=3 \cdot 11 \cdot 17$. The number is obviously squarefree. furthermore $560$ is divisible by $2$, $10$ and $16$.

The number $\text{10,024,561}= 71 \cdot 271 \cdot 521$ is discussed above. We can also verify that this is a Carmichael number: $70 \ \lvert \ \text{10,024,560}$, $270 \ \lvert \ \text{10,024,560}$ and $520 \ \lvert \ \text{10,024,560}$.

Here’s three more Carmichael numbers (found here):

$\text{23,382,529} = 97 \cdot 193 \cdot 1249$

$\text{403,043,257} = 19 \cdot 37 \cdot 43 \cdot 67 \cdot 199$

$\text{154,037,320,009} = 23 \cdot 173 \cdot 1327 \cdot 29173$

We end the post by pointing out one more property of Carmichael numbers, that Carmichael numbers must have at least three distinct prime factors. To see this, suppose that $n=p \cdot q$ is a Carmichael number with two distinct prime factors $p$ and $q$. We can express $n-1$ as follows:

$n-1=pq-1=(p-1)q+q-1$

Since $n$ is Carmichael, $p-1 \ \lvert \ (n-1)$. So $n-1=(p-1)w$ for some integer $w$. Plugging this into the above equation, we see that $p-1 \ \lvert \ (q-1)$. By symmetry, we can also show that $q-1 \ \lvert \ (p-1)$. Thus $p=q$, a contradiction. So any Carmichael must have at least three prime factors.

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$\copyright \ 2013 \text{ by Dan Ma}$

# The primitive root theorem

The primitive root theorem identifies all the positive integers for which primitive roots exist. The list of such integers is a restrictive list. This post along with two previous posts give a complete proof of this theorem using only elementary number theory. We prove the following theorem.

Main Theorem (The Primitive Root Theorem)

There exists a primitive root modulo $m$ if and only if $m=2$, $m=4$, $m=p^t$ or $m=2p^t$ where $p$ is an odd prime number and $t$ is a positive integer.

The theorem essentially gives a list of the moduli that have primitive roots. Any modulus outside this restrictive list does not have primitive roots. For example, any integer that is a product of two odd prime factors is not on this list and hence has no primitive roots. In the post Primitive roots of powers of odd primes, we show that the powers of an odd prime have primitive roots. In the post Primitive roots of twice the powers of odd primes, we show that the moduli that are twice the power of an odd prime have primitive roots. It is easy to verify that the moduli $2$ and $4$ have primitive roots. Thus the direction $\Longleftarrow$ of the primitive root theorem has been established. In this post we prove the direction $\Longrightarrow$, showing that if there exists a primitive root modulo $p$, then $p$ must be one of the moduli in the list stated in the theorem.

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LCM

The proof below makes use of the notion of the least common multiple. Let $a$ and $b$ be positive integers. The least common multiple of $a$ and $b$ is denoted by $\text{LCM}(a,b)$ and is defined as the least positive integer that is divisible by both $a$ and $b$. For example, $\text{LCM}(16,18)=144$. We can also express $\text{LCM}(a,b)$ as follows:

$\displaystyle \text{LCM}(a,b)=\frac{a \cdot b}{\text{GCD}(a,b)} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$

where $\text{GCD}(a,b)$ is the greatest common divisor of $a$ and $b$. The above formula reduces the calculation of LCM to that of calculating the GCD. To compute the LCM of two numbers, we can simply remove the common prime factors between the two numbers. When the number $a$ and $b$ are relatively prime, i.e., $\text{GCD}(a,b)=1$, we have $\displaystyle \text{LCM}(a,b)=a \cdot b$.

Another way to look at LCM is that it is the product of multiplying together the highest power of each prime number. For example, $48=2^4 \cdot 3$ and $18=2 \cdot 3^2$. Then $\text{LCM}(16,18)=2^4 \cdot 3^2=144$.

The least common divisor of the numbers $a_1,a_2,\cdots,a_n$ is denoted by $\text{LCM}(a_1,a_2,\cdots,a_n)$ and is defined similarly. It is the least positive integer that is divisible by all $a_j$. Since the product of all the numbers $a_j$ is one integer that is divisible by each $a_k$, we have:

$\displaystyle \text{LCM}(a_1,a_2,\cdots,a_n) \le a_1 \cdot a_2 \cdots a_n \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)$

As in the case of two numbers, the LCM of more than two numbers can be thought as the product of multiplying together the highest power of each prime number. For example, the LCM of $48=2^4 \cdot 3$, $18=2 \cdot 3^2$ and $45=3^2 \cdot 5$ is $2^4 \cdot 3^2 \cdot 5=720$.

For a special case, there is a simple expression of LCM.

Lemma 1

Let $a_1,a_2,\cdots,a_n$ be positive integers.

Then $\displaystyle \text{LCM}(a_1,a_2,\cdots,a_n)=a_1 \cdot a_2 \cdots a_n$ if and only if the numbers $a_1,a_2,\cdots,a_n$ are pairwise relatively prime, i.e., $a_i$ and $a_j$ are relatively prime whenever $i \ne j$.

Proof of Lemma 1
$\Longleftarrow$
Suppose the numbers are pairwise relatively prime. Then there are no common prime factors in common between any two numbers on the list. Then multiplying together the highest power of each prime factor is the same as multiplying the individual numbers $a_1,a_2 \cdots,a_n$.

$\Longrightarrow$
Suppose $a_i$ and $a_j$ are not relatively prime for some $i \ne j$. As a result, $d=\text{GCD}(a_i,a_j)>1$. It follows that

$\displaystyle \text{LCM}(a_1,a_2,\cdots,a_n) \le \frac{a_1 \cdot a_2 \cdots a_n}{d} < a_1 \cdot a_2,\cdots a_n \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3)$

To give a sense of why the above is true, let’s look at a simple case of $d=\text{GCD}(a_i,a_j)=p^u$ where $p$ is a prime number and $u \ge 1$. Assume that $a_i=a_1$, $a_j=a_2$ and $p^u$ is part of the prime factorization of $a_1$. Furthermore, note that $\displaystyle \text{LCM}(\frac{a_1}{p^u},a_2,\cdots,a_n)$ is identical to $\displaystyle \text{LCM}(a_1,a_2,\cdots,a_n)$. The following derivation confirms (3):

$\displaystyle \text{LCM}(a_1,a_2,\cdots,a_n)=\text{LCM}(\frac{a_1}{p^u},a_2,\cdots,a_n) \le \frac{a_1}{p^u} \cdot a_2 \cdots a_n < a_1 \cdot a_2,\cdots a_n$

With the above clarification, the lemma is established. $\blacksquare$

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Other Tools

We need to two more lemmas to help us prove the main theorem.

Lemma 2

Let $m$ and $n$ be positive integers that are relatively prime. Then $a \equiv b \ (\text{mod} \ m)$ and $a \equiv b \ (\text{mod} \ n)$ if and only if $a \equiv b \ (\text{mod} \ mn)$.
Lemma 3

The number $a$ is a primitive root modulo $m$ if and only if $\displaystyle a^{\frac{\phi(m)}{q}} \not \equiv 1 \ (\text{mod} \ m)$ for all prime divisors $q$ of $\phi(m)$.

Lemma 2 a version of the Chinese Remainder Theorem and is proved a previous post (see Theorem 2 in Primitive roots of twice the powers of odd primes or see Theorem 2 in Proving Chinese Remainder Theorem). Lemma 3 is also proved in a previous post (see Lemma 2 in More about checking for primitive roots).

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Breaking It Up Into Smaller Pieces

The proof of the direction $\Longrightarrow$ of the primitive root theorem is done in the following lemmas and theorems.

Lemma 4

Let $\displaystyle m=p_1^{e_1} p_2^{e_2} \cdots p_t^{e_t}$ be the prime factorization of the positive integer $m$. Let $a$ be a primitive root modulo $m$. Then the numbers $\phi(p_1^{e_1}), \phi(p_2^{e_2}),\cdots,\phi(p_t^{e_t})$ are pairwise relatively prime.

Proof of Lemma 4
Note that $a$ is relatively prime to $m$. So $a$ is relatively prime to each $p_j^{e_j}$. By Euler’s theorem, we have $\displaystyle a^{\phi(p_j^{e_j})} \equiv 1 \ (\text{mod} \ p_j^{e_j})$ for each $j$. Let $\displaystyle W=\text{LCM}(\phi(p_1^{e_1}),\phi(p_2^{e_2}),\cdots,\phi(p_t^{e_t}))$.

By definition of LCM, $\phi(p_j^{e_j}) \ \lvert \ W$ for each $j$. So $\displaystyle a^{W} \equiv 1 \ (\text{mod} \ p_j^{e_j})$ for each $j$. By the Chinese remainder theorem (Lemma 2 above), $\displaystyle a^{W} \equiv 1 \ (\text{mod} \ m)$. Since $a$ is a primitive root modulo $m$, it must be that $\phi(m) \le W$. Interestingly, we have:

$\displaystyle \phi(p_1^{e_1}) \phi(p_2^{e_2}) \cdots \phi(p_t^{e_t})=\phi(m) \le W \le \phi(p_1^{e_1}) \phi(p_2^{e_2}) \cdots \phi(p_t^{e_t})$

Thus $\displaystyle \text{LCM}(\phi(p_1^{e_1}),\phi(p_2^{e_2}),\cdots,\phi(p_t^{e_t}))=\phi(p_1^{e_1}) \phi(p_2^{e_2}) \cdots \phi(p_t^{e_t})$. By Lemma 1, the numbers $\phi(p_j^{e_j})$ are relatively prime. $\blacksquare$

The following theorems follow from Lemma 4. The main theorem is a corollary of these theorems.

Theorem 5

If there exists a primitive root modulo $m$, then $m$ cannot have two distinct prime divisors.

Proof of Theorem 5
Let $\displaystyle m=p_1^{e_1} p_2^{e_2} \cdots p_t^{e_t}$ be the prime factorization of $m$ where $t \ge 2$.

If $p_i$ and $p_j$ are odd prime with $i \ne j$, then $\phi(p_i^{e_i})=p_i^{e_i-1}(p_i-1)$ and $\phi(p_j^{e_j})=p_j^{e_j-1}(p_j-1)$ are both even and thus not relatively prime. If there exists a primitive root modulo $m$, $\phi(p_i^{e_i})$ and $\phi(p_j^{e_j})$ must be relatively prime (see Lemma 4). Since we assume that there exists a primitive root modulo $m$, $m$ cannot have two distinct odd prime divisors. $\blacksquare$

Theorem 6

Suppose that there exists a primitive root modulo $m$ and that $m$ has exactly one odd prime factor $p$. Then $m$ must be of the form $p^e$ or $2p^e$ where $e \ge 1$.

Proof of Theorem 6
By Theorem 5, the prime factorization of $m$ must be $m=2^{e_1} p^{e_2}$ where $e_1 \ge 0$ and $e_2 \ge 1$.

We claim that $e_1=0$ or $e_1=1$. Suppose $e_1 \ge 2$. Then $\phi(2^{e_1})=2^{e_1-1}$ and $\phi(p^{e_2})=p^{e_2-1}(p-1)$ are both even and thus not relatively prime. Lemma 4 tells us that there does not exist a primitive root modulo $m$. So if there exists a primitive root modulo $m$, then it must be the case that $e_1=0$ or $e_2=1$.

If $e_1=0$, then $m=p^{e_2}$. If $e_1=1$, then $m=2p^{e_2}$. $\blacksquare$

Lemma 7

Let $n=2^k$ where $k \ge 3$. Then $\displaystyle a^{\frac{\phi(n)}{2}} \equiv 1 \ (\text{mod} \ n)$ for any $a$ that is relatively prime to $n$.

Proof of Lemma 7
We prove this by induction on $k$. Let $k=3$. Then $n=8$ and $\displaystyle \frac{\phi(8)}{2}=2$. For any odd $a$ with $1 \le a <8$, it can be shown that $a^2 \equiv 1 \ (\text{mod} \ 8)$.

Suppose that the lemma holds for $k$ where $k \ge 3$. We show that it holds for $k+1$. Note that $\phi(2^k)=2^{k-1}$ and $\displaystyle \frac{\phi(2^k)}{2}=2^{k-2}$. Since the lemma holds for $k$, we have $\displaystyle v^{2^{k-2}} \equiv 1 \ (\text{mod} \ 2^k)$ for any $v$ that is relatively prime to $2^k$. We can translate this congruence into the equation $v^{2^{k-2}}=1+2^k y$ for some integer $y$.

Note that $\phi(2^{k+1})=2^{k}$ and $\displaystyle \frac{\phi(2^{k+1})}{2}=2^{k-1}$. It is also the case that $(v^{2^{k-2}})^2=v^{2^{k-1}}$. Thus we have:

$\displaystyle v^{2^{k-1}}=(1+2^k y)^2=1+2^{k+1} y+2^{2k} y^2=1+2^{k+1}(y+2^{k-1} y^2)$

The above derivation shows that $\displaystyle v^{\frac{\phi(2^{k+1})}{2}} \equiv 1 \ (\text{mod} \ 2^{k+1})$ for any $v$ that is relatively prime to $2^k$.

On the other hand, $a$ is relatively prime to $2^{k+1}$ if and only if $a$ is relatively prime to $2^k$. So $\displaystyle a^{\frac{\phi(2^{k+1})}{2}} \equiv 1 \ (\text{mod} \ 2^{k+1})$ for any $a$ that is relatively prime to $2^{k+1}$. Thus the lemma is established. $\blacksquare$

Theorem 8

Suppose that there exists a primitive root modulo $m$ and that $m=2^e$ where $e \ge 1$. Then $m=2^e$ where $e=1$ or $e=2$.

Proof of Theorem 8
Suppose $m=2^e$ where $e \ge 3$. By Lemma 7, $\displaystyle a^{\frac{\phi(m)}{2}} \equiv 1 \ (\text{mod} \ n)$ for any $a$ relatively prime to $m$. Since $2$ is the only prime divisor of $m$, by Lemma 3, there cannot be primitive root modulo $m$. Thus if there exists a primitive root modulo $m$ and that $m=2^e$ where $e \ge 1$, then the exponent $e$ can be at most $2$. $\blacksquare$

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Putting It All Together

We now put all the pieces together to prove the $\Longrightarrow$ of the Main Theorem. It is a matter of invoking the above theorems.

Proof of Main Theorem
Suppose that there exists a primitive root modulo $m$. Consider the following three cases about the modulus $m$.

1. $m$ has no odd prime divisor.
2. $m$ has exactly one odd prime divisor.
3. $m$ has at least two odd prime divisors.

$\text{ }$
Suppose Case 1 is true. Then $m=2^e$ where $e \ge 1$. By Theorem 8, $m=2$ or $m=4$.

Suppose Case 2 is true. Then Theorem 6 indicates that $m$ must be the power of an odd prime or twice the power of an odd prime.

Theorem 5 indicates that Case 3 is never true. Thus the direction $\Longrightarrow$ of the Main Theorem is proved. $\blacksquare$

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$\copyright \ 2013 \text{ by Dan Ma}$

# Primitive roots of twice the powers of odd primes

In a previous post, we show that there exist primitive roots modulo the power of an odd prime number (see Primitive roots of powers of odd primes). In this post we show that there exist primitive roots modulo two times the power of an odd prime number. Specifically we prove the following theorem.

Theorem 1

Let $p$ be an odd prime number. Let $j$ be any positive integer. Then there exist primitive roots modulo $p^j$.

We make use of the Chinese Remainder Theorem (CRT) in proving Theorem 1. We use the following version of CRT (also found in this post)

Theorem 2 (CRT)

Let $m$ and $n$ be positive integers that are relatively prime. Then $a \equiv b \ (\text{mod} \ m)$ and $a \equiv b \ (\text{mod} \ n)$ if and only if $a \equiv b \ (\text{mod} \ mn)$.

Proof of Theorem 2
$\Longrightarrow$
Suppose $a \equiv b \ (\text{mod} \ m)$ and $a \equiv b \ (\text{mod} \ n)$. Converting these into equations, we have $a=b+mx$ and $a=b+ny$ for some integers $x$ and $y$. It follows that $mx=ny$. This implies that $m \ \lvert \ ny$. Since $m$ and $n$ and relatively prime, $m \ \lvert \ y$ and $y=mt$ for some integer $t$. Now the equation $a=b+ny$ can be written as $a=b+mnt$, which implies that $a \equiv b \ (\text{mod} \ mn)$.

$\Longleftarrow$
Suppose $a \equiv b \ (\text{mod} \ mn)$. Then $a=b+mns$ for some integer $s$, which implies both congruences $a \equiv b \ (\text{mod} \ m)$ and $a \equiv b \ (\text{mod} \ n)$. $\blacksquare$

Proof of Theorem 1
Let $a$ be a primitive root modulo $p^j$ (shown to exist in the post Primitive roots of powers of odd primes). When $a$ is odd, we show that $a$ is a primitive root modulo $2p^j$. When $a$ is even, we show that $a+p^j$ is a primitive root modulo $2p^j$.

First the odd case. Since $a$ is a primitive root modulo $p^j$, $a^k \not \equiv 1 \ (\text{mod} \ p^j)$ for all positive $k<\phi(p^j)$. Since $a$ is odd, $a^k$ is odd for all integers $k \ge 1$. So $a^k \equiv 1 \ (\text{mod} \ 2)$ for all integers $k \ge 1$. By CRT (Theorem 2), $a^k \not \equiv 1 \ (\text{mod} \ 2p^j)$ for all positive $k<\phi(p^j)=\phi(2 p^j)$. This implies that $a$ is a primitive root modulo $2p^j$.

Now the even case. Note that $a+p^j$ is odd (even + odd is odd). It is also the case that $(a+p^j)^k$ is odd for all $k \ge 1$. Thus $(a+p^j)^k \equiv 1 \ (\text{mod} \ 2)$ for all $k \ge 1$.

In expanding $(a+p^j)^k$ using the binomial theorem, all terms except the first term $a^k$ is divisible by $p^j$. So $(a+p^j)^k \equiv a^k \ (\text{mod} \ p^j)$. Furthermore, $a^k \not \equiv 1 \ (\text{mod} \ p^j)$ for all positive $k<\phi(p^j)$ since $a$ is a primitive root modulo $p^j$. So $(a+p^j)^k \not \equiv 1 \ (\text{mod} \ p^j)$ for all positive $k<\phi(p^j)$.

By CRT (Theorem 2), we have $(a+p^j)^k \not \equiv 1 \ (\text{mod} \ 2p^j)$ for all positive $k<\phi(p^j)=\phi(2p^j)$. This implies that $a+p^j$ is a primitive root modulo $2p^j$. $\blacksquare$

The remainder of the proof of the primitive root theorem is found in The primitive root theorem.

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$\copyright \ 2013 \text{ by Dan Ma}$

# Primitive roots of powers of odd primes

Let $p$ be an odd prime number. There exist primitive roots modulo $p$ (in fact there are $\phi(p-1)$ many). There are various strategies for finding primitive roots of a prime modulus, other than simply trying all candidates (discussed here and here). In this post we discuss how to obtain primitive roots of moduli that are power of odd primes. Starting from a given primitive root modulo $p$, we show how to get a primitive root modulo $p^2$. Next, starting with a primitive root modulo $p^2$, we show how to get a primitive root modulo $p^n$ for any integer $n \ge 3$. It follows from these results that there exist primitive roots modulo any power of any odd prime number. The results discussed in this post build up to the primitive root theorem. See the following posts for the rest of the proof of the primitive root theorem.

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Example

We start the discussion with an example. There are two primitive roots for the odd prime modulus $p=7$. They are $3$ and $5$. Are $3$ and $5$ also primitive roots modulo $7^2=49$?

Note that $\phi(49)=\phi(7^2)=7(7-1)=42$. One way to find out is to check

$3^x \ (\text{mod} \ 49)$ and $5^x \ (\text{mod} \ 49)$

by letting $x$ be the proper divisors of $42$. The proper divisors of $42$ are $1$, $2$, $3$, $6$, $7$, $14$, and $21$. However, it is not necessary to check all $7$ proper divisors of $42$.

There is a better check that requires only checking $3$ divisors of $42$. According to a previous post, we only need to check the following:

$\displaystyle 3^{\frac{\phi(49)}{2}} \ (\text{mod} \ 49)$ and $\displaystyle 3^{\frac{\phi(49)}{3}} \ (\text{mod} \ 49)$ and $\displaystyle 3^{\frac{\phi(49)}{7}} \ (\text{mod} \ 49)$

$\displaystyle 5^{\frac{\phi(49)}{2}} \ (\text{mod} \ 49)$ and $\displaystyle 5^{\frac{\phi(49)}{3}} \ (\text{mod} \ 49)$ and $\displaystyle 5^{\frac{\phi(49)}{7}} \ (\text{mod} \ 49)$

To see why this works, see Check #3 in the post More about checking for primitive roots. Even this better check can be improved.

According to Lemma 1 discussed below, all we need to do is to check the following:

$3^6 \ (\text{mod} \ 49)$ and $5^6 \ (\text{mod} \ 49)$

If the congruence $\not \equiv 1 \ (\text{mod} \ 49)$, then the number in question is a primitive root modulo $49$. Otherwise, it is not a primitive root. Note that the exponent is $\phi(7)=6$.

We have $3^6 \equiv 43 \ (\text{mod} \ 49)$ and $5^6 \equiv 43 \ (\text{mod} \ 49)$. So the two numbers $3$ and $5$ are primitive roots modulo $49$.

Remarks
In general, given that $a$ is a primitive root modulo $p$, to check whether $a$ is a primitive root modulo $p^2$, all we need to do is to check $a^{p-1} \ (\text{mod} \ p^2)$. If it is $\not \equiv 1 \ (\text{mod} \ p^2)$, then $a$ is a primitive root modulo $p^2$. Otherwise, it is not a primitive root modulo $p^2$.

What makes this works is that if $a$ is a primitive root modulo $p$, the order of $a$ modulo $p^2$ can only be $p-1$ or $p(p-1)=\phi(p^2)$ (see Lemma 1 below). When $a^{p-1} \not \equiv 1 \ (\text{mod} \ p^2)$, the order of $a$ modulo $p^2$ is not $p-1$ and is $p(p-1)=\phi(p^2)$, which means that $a$ is a primitive root modulo $p^2$. When $a^{p-1} \equiv 1 \ (\text{mod} \ p^2)$, the order of $a$ modulo $p^2$ is $p-1$, which means that $a$ is not a primitive root modulo $p^2$.

Furthermore, in the case that the order of $a$ modulo $p^2$ is $p-1$, even though the number $a$ is not a primitive root modulo $p^2$, the number $a+p$ is a primitive root modulo $p^2$ (see Theorem 2 below).

As an example, $31 \equiv 3 \ (\text{mod} \ 7)$. So $31$ is also a primitive root modulo $7$. But $31^6 \equiv 1 \ (\text{mod} \ 49)$. So $31$ is not a primitive root modulo $49$. However $38$ is a primitive root modulo $49$. Note that $38^6 \equiv 15 \ (\text{mod} \ 49)$. For more details, see Theorem 2 below.

Theorem 4 below states that any primitive root modulo $p^2$ is also a primitive root modulo any higher power of $p$. For example, $38$ is a primitive root modulo $7^n$ for any $n \ge 3$.

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Square of an Odd Prime

We now show that there exist primitive roots modulo the square of an odd prime (Theorem 2 below). The starting point is that there is a given primitive root modulo a prime number (the existence is proved in Primitive roots of prime moduli).

Lemma 1

Let $p$ be a prime number. Let $g$ be a primitive root modulo $p$. Let $t$ be the order of $g$ modulo $p^2$. Then either $t=p$ or $t=p(p-1)$.

Proof of Lemma 1
Immediately we know that $t \ \lvert \ \phi(p^2)=p(p-1)$. Furthermore, it follows that $g^t \equiv 1 \ (\text{mod} \ p^2)$, which implies $g^t=1+p^2 y$ for some integer $y$. In turn this equation implies that $g^t \equiv 1 \ (\text{mod} \ p)$. We also have we have $p-1 \ \lvert \ t$ since the order of $g$ modulo $p$ is $p-1$.

So we have $p-1 \ \lvert \ t$ and $t \ \lvert \ p(p-1)$. In words, the integer $t$ is at least $p-1$ and at the same time $t$ is a divisor of $p(p-1)$. The only possibilities of $t$ are $t=p-1$ or $t=p(p-1)$. $\blacksquare$

Theorem 2

Let $p$ be an odd prime number. Let $a$ be a primitive root modulo $p$. Then either $a$ or $a+p$ is a primitive root modulo $p^2$. Thus there exist primitive roots modulo $p^2$.

Proof of Theorem 2
Let $k$ be the order of $a$ modulo $p^2$. By Lemma 1, we have $k=p-1$ or $k=\ p(p-1)$. Note that $\phi(p^2)=p(p-1)$. Thus if it is the case that $k=\ p(p-1)$, then $a$ is a primitive root modulo $p^2$. So in the remainder of the proof, we assume that $k=p-1$.

Since $a+p \equiv a \ (\text{mod} \ p)$, the number $a+p$ is a primitive root modulo $p$. Let $h$ be the order of $a+p$ modulo $p^2$. Using Lemma 1 again, there are only two possibilities for $h$, namely $h=p-1$ or $h=\ p(p-1)=\phi(p^2)$. If we can show that the case $h=p-1$ is not possible, then $a+p$ must be a primitive root modulo $p^2$. To this end, we show $(a+p)^{p-1} \not \equiv 1 \ (\text{mod} \ p)$.

Using the binomial theorem, we expand $(a+p)^{p-1}$ as follows:

$\displaystyle (a+p)^{p-1}=a^{p-1}+\binom{p-1}{1}a^{p-2}p+\binom{p-1}{2}a^{p-3}p^2+\binom{p-1}{3}a^{p-4}p^3 +\cdots+p^{p-1}$

All terms except the first two terms are divisible by $p^2$. Recall that we assume above that $k=p-1$ (the order of $a$ modulo $p^2$). So in the following derivation, we use $a^{p-1} \equiv 1 \ (\text{mod} \ p^2)$.

\displaystyle \begin{aligned} (a+p)^{p-1}&\equiv a^{p-1}+\binom{p-1}{1}a^{p-2}p \ (\text{mod} \ p^2) \\&\equiv a^{p-1}+(p-1)a^{p-2}p \ (\text{mod} \ p^2) \\&\equiv a^{p-1}+p^2 a^{p-2}-p a^{p-2} \ (\text{mod} \ p^2) \\&\equiv 1+0-p a^{p-2} \ (\text{mod} \ p^2) \\&\equiv 1-p a^{p-2} \ (\text{mod} \ p^2) \end{aligned}

We now need to show that $1-p a^{p-2} \not \equiv 1 \ (\text{mod} \ p^2)$. Suppose that $1-p a^{p-2} \equiv 1 \ (\text{mod} \ p^2)$. Then we have $-p a^{p-2} \equiv 0 \ (\text{mod} \ p^2)$.

Because $a^{p-1} \equiv 1 \ (\text{mod} \ p^2)$, $a^{p-2}$ is the multiplicative inverse of $a$ modulo $p^2$. Now multiply both sides of $-p a^{p-2} \equiv 0 \ (\text{mod} \ p^2)$ by $a$, we get $-p \equiv 0 \ (\text{mod} \ p^2)$, which implies $p$ is divisible by $p^2$, a contradiction. So $(a+p)^{p-1} \equiv 1-p a^{p-2} \not \equiv 1 \ (\text{mod} \ p^2)$. Thus $h=p-1$ is not possible. Then $h=p(p-1)$, which means that $a+p$ is a primitive root modulo $p^2$. $\blacksquare$

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Higher Powers of an Odd Prime

In this section, we show that there exist primitive roots modulo any higher power of an odd prime.

Lemma 3

Let $p$ be an odd prime number. If $g$ be a primitive root modulo $p^2$, then $g$ is also a primitive root modulo $p$.

Proof of Lemma 3
Let $g$ be a primitive root modulo $p^2$ where $p$ is an odd prime number. To show that $g$ be a primitive root modulo $p$, we use the following result:

(*) A number $h$ is a primitive root modulo $m$ if and only if $\displaystyle h^{\frac{\phi(m)}{t}} \not \equiv 1 \ (\text{mod} \ m)$ for all prime divisor $t$ of $\phi(m)$.

Let $t$ be a prime divisor of $\phi(p)=p-1$. Then $t$ is a prime divisor of $\phi(p^2)=p(p-1)$. By the result indicated by (*), we have $\displaystyle g^{\frac{p(p-1)}{t}} \not \equiv 1 \ (\text{mod} \ p^2)$. It follows that $\displaystyle g^{\frac{p-1}{t}} \not \equiv 1 \ (\text{mod} \ p)$. Thus by the result indicated by (*), $g$ is a primitive root modulo $p$. Note that if $\displaystyle g^{\frac{p-1}{t}} \equiv 1 \ (\text{mod} \ p)$, then $\displaystyle (g^{\frac{p-1}{t}})^p \equiv 1 \ (\text{mod} \ p)$. $\blacksquare$

Theorem 4

Let $p$ is an odd prime number. Let $a$ be a primitive root modulo $p^2$. Then $a$ is a primitive root modulo $p^{j}$ for all integers $j \ge 3$.

Proof of Theorem 4
Note that $\phi(p^2)=p(p-1)$. Thus $a^{p-1} \not \equiv 1 \ (\text{mod} \ p^2)$. By Lemma 3, the number $a$ is also a primitive root modulo $p$. Thus $a^{p-1} \equiv 1 \ (\text{mod} \ p)$. Thus $a^{p-1}=1+wp$ for some integer $w$. It is the case that $w$ cannot be a multiple of $p$. Otherwise, we would have $a^{p-1} \equiv 1 \ (\text{mod} \ p^2)$. We will use that fact that $w$ cannot be a multiple of $p$ at a later step in the proof.

Let $j$ be any integer with $j \ge 3$. Note that $\phi(p^j)=p^{j-1}(p-1)$. We establish the following three claims.

Claim 1
Let $k$ be the order of $a$ modulo $p^j$. The possibilities for $k$ are $p^n(p-1)$ where $n=1,2,3,\cdots,j-1$.

Claim 2
It is the case that $k \ne p^{j-2}(p-1)$. To establish this, we show $\displaystyle a^{p^{j-2}(p-1)} \not \equiv 1 \ (\text{mod} \ p^j)$.

Claim 3
It is the case that $k \ne p^{n}(p-1)$ for $n=1,2,3,\cdots,j-3$.

Once the three claims are established, the order of $a$ modulo $p^j$ must be $\phi(p^j)=p^{j-1}(p-1)$. Hence $a$ is a primitive root modulo $p^j$.

Claim 3 follows from Claim 2. Note that if $\displaystyle a^{p^{n}(p-1)} \equiv 1 \ (\text{mod} \ p^j)$ where $n=0,1,2,3,\cdots,j-3$, then we can raise both sides of the equation by an appropriate power of $p$ to get $\displaystyle a^{p^{j-2}(p-1)} \equiv 1 \ (\text{mod} \ p^j)$.

Proof of Claim 1. Since $k$ is the order, we have $\displaystyle a^{k} \equiv 1 \ (\text{mod} \ p^j)$, which leads to $\displaystyle a^{k} \equiv 1 \ (\text{mod} \ p^2)$. These two congruences imply $k \ \lvert \ \phi(p^j)=p^{j-1}(p-1)$ and $p(p-1) \ \lvert \ k$.

Thus $k$ is at least $p(p-1)$ and divides $p^{j-1}(p-1)$. With $p$ being a prime, $k$ can only be $p(p-1)$, $p^{2}(p-1)$, $\cdots$, $p^{j-1}(p-1)$.

Proof of Claim 2.
We show that $\displaystyle a^{p^{j-2}(p-1)} \not \equiv 1 \ (\text{mod} \ p^j)$. With $a^{p-1}=1+wp$ derived at the beginning of the proof, we have $\displaystyle a^{p^{j-2}(p-1)}=(1+wp)^{p^{j-2}}$. Upon using the binomial theorem to expand $\displaystyle (1+wp)^{p^{j-2}}$, we see that all terms except the first two are divisible by $p^j$. We can discard them since we take congruence modulo $p^j$. The following shows the first two terms:

\displaystyle \begin{aligned} a^{p^{j-2}(p-1)}&= (1+wp)^{p^{j-2}} \\&\equiv 1+p^{j-2} wp \ (\text{mod} \ p^j) \\&\equiv 1+wp^{j-1} \ (\text{mod} \ p^j) \end{aligned}

We claim that $1+wp^{j-1} \not \equiv 1 \ (\text{mod} \ p^j)$. Suppose $1+wp^{j-1} \equiv 1 \ (\text{mod} \ p^j)$. Then $wp^{j-1} \equiv 0 \ (\text{mod} \ p^j)$, which implies $wp^{j-1}=p^j c$ for some integer $c$. Cancelling out $p^{j-1}$, we have $w=p c$, which contradicts the fact that $w$ cannot be a multiple of $p$. So $1+wp^{j-1} \not \equiv 1 \ (\text{mod} \ p^j)$. This establish Claim 3 and the theorem is also established. $\blacksquare$

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$\copyright \ 2013 \text{ by Dan Ma}$