orl wrote:

Find all the pairs of positive integers $(x,p)$ such that p is a prime, $x \leq 2p$ and $x^{p - 1}$ is a divisor of $(p - 1)^{x} + 1$ .

First the case where $p=2$ : $x \mid 2$ . So there are solutions $(p,x) = (2,1)$ and $(p,x) = (2,2)$ .
Then the case where $x=1$ . (I will use that $x$ has prime divisors later) $1^{p-1} \mid (p-1)^1 + 1 \iff 1 \mid p$ always holds. So $(1,p)$ are also solutions.
$x^{p-1} \mid (p-1)^x + 1 \mid (p-1)^{2x} - 1$ . Let $q$ be the smallest primedivisor of $x$ . Then $(p-1)^{2x} \equiv 1 \bmod q \Rightarrow (p-1)^{(2x,q-1)} \equiv 1 \bmod q$ . But $(2x,q-1) \le 2(x,q-1) = 2$ . So $q \mid (p-1)^2-1 = p(p-2)$ .
If $q \mid p$ then $q=p$ . Since $x$ is odd and $x \leq 2p$ we see that $x=p > 2$ . Then $p^{p-1} \mid (p-1)^p + 1 = (p-1)^p - (-1)^p$
It is well known that $v_p(a^n-b^n) = v_p(n) + v_p(a-b)$ when $a \equiv b \bmod p$ and $(ab,p)=1$ .
Hence $v_p( (p-1)^p - (-1)^p ) = v_p(p) + v_p(p-1-(-1))=2$ . Hence $p-1 \le 2$ . And then $p=3$ gives the solution $(x,p)=(3,3)$ .
Now assume that $q \mid p-2$ . Then $q^{p-1} \mid (p-1)^{2x} - 1^{2x}$ .
So $v_q( (p-q)^{2x} - 1^{2x} ) = v_q(p-2) + v_q(2x) \ge p-1$ .
But since $q$ is odd: $v_q(2x) = v_q(x)$ . And $q \ge 3$ . Hence $v_q(n) \le log_3(n)$ when $n \in \mathbb{N}$ . So:
$log_3(p-2) + log_3(x) \ge p-1$ .
$x \le 2p$ : $log_3(p-2) + log_3(2p) \ge p-1 \iff 2p^2-4p \ge 3^{p-1}$
Let $f(p) = 3^{p} - 2(p+1)^2+4(p+1) = 3^p - 2p^2+2$ . Then $f(p-1) \le 0$ .
$f'(p) = \ln (3) \cdot 3^p -4p$ and $f''(p) = \ln^2(3) \cdot 3^p > 0$
But $f'(2) = \ln (3) \cdot 3^2 - 4 \cdot 2 > 3^2 - 4 \cdot 2 = 1 > 0$ . Since $f''(p) > 0$ we see that $f(p)$ is increasing when $p \ge 2$ . But $f(2) = 3^2-2\cdot2^2+2 = 3 > 0$
Hence: When $p \ge 3$ we see that $f(p) > 0$ . So $q \mid p-2$ doesn't give any solutions. QED

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