1. Case: $p=2$, then we want $n|1^n+1=2$, thus $n=2$. 2. Case: $p$ odd. Let $q$ be smallest prime divisor of $n$ and let $o$ be the order of $p-1 \mod q$. Since $(p-1)^{2n} \equiv 1 \mod q$, we get $o|2n$, and by Fermat's little theorem, we get $o|q-1$. But if $o$ would have a factor in common with $n$, this factor would be $\leq q-1 <q$, impossible. But then $o|2n$ reduces to $o|2$, yielding $(p-1)^2 \equiv 1 \mod q$ or equivalent $q|p(p-2)$. Case $q|p-2$; We get $(p-1)^n+1 \equiv 1^n+1=2 \mod q$. But we need $q$ to divide this number, thus $q=2$. But then $p$, is even, too, so $p=2$, which we excluded. Case $q|p$; We write $n=p^k s, p \nmid s , k>0$. It's well known [I can give a proof if desired] that for odd primes $p$ if $v_p(a-b)>0, p \nmid a,b$, then $v_p(a^t-b^t)=v_p(a-b)+v_p(t)$ (here $v_p(m)$ is as always the $p$-adic valuation of $m$, thus the number of times $m$ is divisible by $p$). With $a=p-1, b=-1, t=n$ this gives us $v_p((p-1)^{n}+1)=k+1$ and has to be $\geq k(p-1)$ to enable $p^{k(p-1)}|n^{p-1}|(p-1)^{p^ks}+1$. Now $k+1 \geq k(p-1) \iff 2k+1 \geq kp \iff 2+\frac 1k \geq p$. Since we assumed $p\geq 3$, we have $k=1$ and $p=3$. If $s=1$, we have $n=3$ giving a solution. So let's assume $s>1$, let $q$ be a smallest prime divisor of $s$, and by $n=3$ we want that $s|9s^2=n^2|2^n+1=8^s+1$. Following the same idea than before, we take $o$ as the order of $8 \mod q$ and get again $o|2$ by the same argument. In both cases $q|8^2-1=63$, thus $q=7$. But $8^s +1 \equiv 1^s+1 =2 \mod 7$, contradiction. Conclusion: all solutions are given by $(n,p)=(2,2),(3,3)$.

ZetaX wrote: [I can give a proof if desired] that for odd primes $ p$ if $ v_p(a - b) > 0, p \nmid a,b$, then $ v_p(a^t - b^t) = v_p(a - b) + v_p(t)$ Proof would be welcome Note to self: checked up to there.

What about $n=1$ ?? $p$ could be any prime.

shinichiman wrote: What about $n=1$ ?? $p$ could be any prime. N>1 ..