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}

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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 <m 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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Comments

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}

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:

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}