# 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.

___________________________________________________________________________________________________________________

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.

___________________________________________________________________________________________________________________

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.

___________________________________________________________________________________________________________________

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

## 3 thoughts on “Introducing Carmichael numbers”

1. I am currently taking computational number theory and have been working with Carmichael numbers recently! Very fun!