Show that $n^{n^{n^4}} - n^{n^{n^2}} \equiv 0 (\mod 547)$

Question

While doing problems around Fermat's Little Theorem I encountered this problem: For all $n \in \mathbb{N}$ show that: $$ n^{n^{n^4}} - n^{n^{n^2}} \equiv 0 \quad(\mod\quad 547) $$ I tried to approach it with changing exponents, but I weren't able to see where to use FLT and the given hint was to use it.

Answer 1

$547$ is a prime number.

Fermat's Little Theorem implies:

$$n^m \equiv n^{m \text{ mod }(p-1)} \text{ (mod } p)$$

So it is sufficient if we prove:

$$n^{n^4} \equiv n^{n^2} \text{ (mod 546)}$$

$546=2\times3\times7\times13$. Let's prove for $p=2,3,7,13$:

$$n^{n^4} \equiv n^{n^2} \text{ (mod }p)$$ $$\Longleftarrow n^4\equiv n^2\text{ (mod }p-1)$$ $$\Longleftrightarrow n^2(n+1)(n-1)\equiv 0\text{ (mod }p-1)$$

We know $n^2(n+1)(n-1)\equiv 0 \text{ (mod 3, 4)}$, so $n^2(n+1)(n-1)\equiv 0 \text{ (mod 12)}$.

$p-1=1,2,6,12$ are all divisors of $12$. The proof is done.