Fermat Might Like This One...

We have that $p^2$ ( $p$ is a prime), don't divide $n$ because $p^2 \nmid p^{13}-p$

Then $n$ can be factoring in primes. As $n|a^{13}-a$ for all $a>1$ , $n$ has to divide at $2^{13}-2=2\times3^2\times5\times7\times13$ .

Apply Fermat's Little Theorem such that $a^{13}\equiv a \mod p$ for $(p=2,3,5,7,13) \therefore a^{13} \equiv a \mod (2\times3\times5\times7\times13)$ for all $a$ .

Then, $n$ that we search are $\times3\times5\times7\times13$ divisors except $1$

Apply Tao and have $2^5-1=\boxed{31}$ that satisfy this.

Alternative solution(?)

Let us consider a positive integer $A = p^x q^y$ , where $p$ and $q$ are primes. Then the number of divisors of A is $(x+1)(y+1)$ .
For example, we have 500. Note that $500 = 2^2 5^3$ . So 100 has a total of $(2+1)*(3+1) = 12$ factors. These are ${1, 2, 4, 5, 10, 20, 25, 50, 100, 125, 250, 500}$ .
This set of number, however, includes 1 as a factor.

Now, $a^{13} - a$ can be factored as: $a^{13} - a = a ( a^{12} - 1)$

Further factoring of $(a^{12} - 1)$ leads to

$(a^{12} - 1) = ((a^{4})^{3} -1 ) = (a^4 - 1 ) (a^8 + a^4 + 1) = (a-1)(a+1)(a^2+1) (a^8 + a^4 + 1)$

Thus, the factors of $a^{13} - a$ are: $a, (a-1), (a+1), (a^2+1), (a^8 + a^4 + 1)$

This leads to a total of $(1+1)(1+1)(1+1)(1+1)(1+1) = 2^5 = 32$ factors, which includes 1. since we need only to consider factors $n > 1$ , then we only have $32-1 =$ 31 factors.

In particular n divides 2¹³-2 and 3¹³-3. Decomposition of these 2 numbers in prime factors yields 2¹³-2 = 2 3² 5 7 13 and 3¹³-3 = 2^4 3 5 7 13 73

So n is a divisor of 2 3 5 7 13 (the smaller of the 2 exponents for each prime number is kept for the common divisor).

Now the problem is to determine if all of these 5 prime numbers always divide a¹³-a for all a > 1. a¹³-a = a(a¹²-1) = a[a^6-1][a^6+1] = a[(a³)²-1][(a²)³+1] = a[a³-1][a³+1][a²+1][(a²)²-(a²)+1] = a[a-1][a²+a+1][a+1][a²-a+1][a²+1][(a²)²-(a²)+1] = [a-1]a[a+1][a²-a+1][a²+1][a²+a+1][(a²)²-(a²)+1] (rearranged)

Among a[a+1], there is always a multiple of 2 Among [a-1]a[a+1], there is always a multiple of 3

Modulo 5, if a ≡ -2, -1, 0, 1, 2 respectively, then [a²+1], [a+1], a, [a-1], [a²+1] ≡ 0 respectively. So 5 is a common prime factor

Modulo 7, if a ≡ -3, -2, -1, 0, 1, 2, 3 respectively, then [a²+a+1], [a²-a+1], [a+1], a, [a-1], [a²+a+1], [a²-a+1] ≡ 0 respectively. So 7 is a common prime factor

Modulo 13, if a ≡ -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6 respectively, then [(a²)²-(a²)+1], [a²+1], [a²+a+1], [a²-a+1], [(a²)²-(a²)+1], [a+1], a, [a-1], [(a²)²-(a²)+1], [a²+a+1], [a²-a+1], [a²+1], [(a²)²-(a²)+1] ≡ 0 respectively. So 13 is a common prime factor.

Therefore all divisors of 2 3 5 7 13 (except 1) are the required divisors. Nb of such divisors = 2^5 - 1 = 31.

First, to ease things out, we should factor that polynomial $a^{13} - a = a(a^{12} - 1) = a(a^6 + 1)(a^6 - 1) = a(a^6 + 1)(a^3 + 1)(a^3 - 1)$

Then, notice that $p^2 \not| p^{13} - p$ , so no $n$ is a prime power greater than $1$ .

We start looking at primes up to $13$ for values of $n$ , using the factorization and Fermat's Little Theorem (and the corollary that $p | a^{p - 1} - 1$ ) :

$2 | a(a^{12} - 1)$

$3 | (a^3 - 1)(a)(a^3 + 1)$ by Fermat and " $n$ divides one amongst $n$ consecutive numbers"

$5 | a(a^{12} - 1)$ notice that $a^{12} = (a^3)^{4}$

$7 | a(a^6 - 1)$

$11$ does not

$13 | a^{13} - a$

And as we have $5$ prime factors (to the first power), using combinatorics we get $2^5 - 1 = 31$ divisors

a^13-a = a(a^12-1). By Fermat’s little theorem it is therefore divisible by 13. Substitute b = a^2. You get a^12-1 = b^6-1. This shows (again using Fermat’s little theorem) that this factor is divisible by 7. In similar fashion (using c=a^3, d=a^6) it can be shown that a^12-1 is divisible by 3 and 5.

The full expression a^13-a is also divisible by 2 for all a. This is because a^13 and a either are both even or both odd. In either case the difference is even.

It remains to be shown that 2,3,5,7 and 13 are the only prime factors for all values of a and that there is no higher power of any that divides the original expression for all a. To do so it is sufficient to find a countrrexample.

For the first part, let a=2. 2^13-2=8190. The prime factorization is 2 *3^2 * 5 * 7 * 13. This shows that ONLY these five primes are prime factors of a^13-a for all a.

This also shows that for all a, the only factor that could possibly appear as a power greater than the first for all a is 3. Next consider 3^13-3. Dividing by 3 gives 3^12-1, which is not divisible by 3. Therefore 3^13-3 is divisible by 3 but not 3^2.

This shows that a^13-a is divisible only by 2 3 5 7 13 and any of its factors. Any factor of this value may be expressed as the product of these five primes raised to an exponent of either 1 or 0. Five factors with 2 choices of exponents gives 2^5 = 32 possible values. The case where all exponents are zero yields 1, though, so there are 31 integers >1 that divide a^13-a for all integer values of a.