Show that, for all integers m and n, mn(m^{420} − n^{420}) is divisible by 446617991732222310.

Firstly, we obtain the prime factorization of 446617991732222310; for example, at Dario Alpern's Factorization Engine. We get:

446617991732222310 = 2 × 3 × 5 × 7 × 11 × 13 × 29 × 31 × 43 × 61 × 71 × 211 × 421.

We must show that mn(m^{420} − n^{420}) is divisible by each of these prime factors.

Let p be one of the above prime factors. We consider two mutually exclusive cases:

- If mn is divisible by p, then clearly mn(m
^{420}− n^{420}) is also divisible by p. - If mn is
*not*divisible by p, then both m and n are not divisible by p.

Hence, since p is prime, m and p and n and p are relatively prime, and, by Fermat's Little Theorem, m^{p−1}n^{p−1}1 (modulo p).

Now note that, for each p, p − 1 divides 420.^{ }

Hence m^{420}n^{420}1 (mod p), or m^{420}− n^{420}0 (mod p).

That is, m^{420}− n^{420}is divisible by p.

Thus, in both cases, mn(m^{420} − n^{420}) is divisible by each prime factor p.

Therefore, for all integers m and n, mn(m^{420} − n^{420}) is divisible by 446617991732222310.

Source: Traditional