Let $m,n \in \mathbb{Z}^+$ and $p$ be an odd prime number. Show that $n=1$ if $2^m = p^n +1$.

Question

I could show the case that $m$ is even as follows, but not show the other case. Please give me hints!!!

When $m$ is even, put $m=2k$. $2^m-1=(2^k-1)(2^k+1)$ By Euclidean algorithm, $2^k-1$ and $2^k+1$ are coprime. So, these are integers to n-th power. Put $2^k-1=a^n, 2^k+1=b^n$ ($1≦a<b,a$ and $b$ are odd). $2=(b-a)(b^{n-1}+...+a^{n-1})≧2n$ ∴$n=1$

Answer 1

From $p^n + 1 \ge p+1 \ge 4$ we have $m \ge 2$.

Hence $p^n = 2^m - 1 \equiv -1\pmod 4$, and thus $p^n$ cannot be a square.

This forces $n$ to be odd.

From $p^n + 1 = (p+1)(p^{n-1} - p^{n-2} + \dots + 1) = 2^m$, we see that

$$(p^{n-1} - p^{n-2} + \dots + 1) \mid 2^m$$

Since $p$ is odd, this factor, which is a sum of $n$ odd numbers, is odd as well.

This forces $p^{n-1} - p^{n-2} + \dots + 1 = 1$, and thus $p^n+1 = (p+1)(1)$.

Answer 2

If $n$ is odd, so $$p^n+1=(p+1)(p^{n-1}-p^{n-2}+...+1)=2^m,$$ which says $$p\equiv-1(\operatorname{mod}2^k)$$ and from here $$p^{n-1}-p^{n-2}+...-p+1\equiv n(\operatorname{mod}2^k).$$ Can you end it now?