Proving that $p ^ {p + 2} + (p + 2)^{p} \equiv 0\pmod{2p + 2}$ for a prime $p>2$

Question

So, the task is to prove the following:

$$p ^ {p + 2} + (p + 2)^{p} \equiv 0 \pmod{2p + 2}$$

For all prime numbers $p > 2$.

I haven't got any clue on how to do this, but, for some reason, I feel like it is somehow related to the Euler's theorem or its 'friend' Fermat's little theorem.

One observation I have made is that this expression can be refactored as $a ^ b + b ^ a \equiv 0 \pmod{a + b}$ which is maybe another way to think about the problem.

Any help would be appreciated!

Answer 1

We can assume $p$ odd.

Since $p^{p+2}+(p+2)^p$ is a sum of two odd numbers, it is even. Thus, it is enough to prove that the sum is divisible by $p+1$. Modulo $p+1$, the expression is $(-1)^{p+2}+1^p=0$ since $p$ is odd. Hence the original expression can be divided by $2p+2$.

Answer 2

As $p,p+2$ are odd, so it's sufficient to prove $p+1$ divides $p^{2m+1}+(p+2)^n$

Now $p\equiv-1\pmod{p+1}\implies p^{2m+1}\equiv?$

and $p+2\equiv1\pmod{p+1}\implies(p+2)^n\equiv?$

Answer 3

Let $p$ be an odd prime. Note that by Fermat's Little Theorem, $2p+2$ divides $p^{p} + (p+2)^{p}$, thus it suffices to prove that $2p+2$ divides $p^{p+2}-p^p$ since $$p^{p+2}+(p+2)^p=(p^{p+2}-p^p)+p^p+(p+2)^p.$$ Notice that the task is now easy. Hint: $p^{p+2}-p^p=p^{p}(p+1)(p-1)$.

Answer 4

$p^{p+2} +(p+2)^p \equiv p^{p+2} + (-p)^p \equiv p^{p+2}-p^p\equiv p^p(p^2 -1)$

$p^p(p+1)(p-1)\equiv p^p(2p + 2)\frac {p-1}2 \equiv 0\pmod {2p+2}$

Unless I did something horribly wrong this will be true for any odd number $p$ whether prime or not.