Competition problem — Number theory

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Number theory Difficulty 5.1 AIME, harder Prove it

4. Prove: When $n>1$ , there do not exist an odd prime $p$ and a positive integer $m$ such that $p^{n}+1=2^{m}$ holds.

Solution

If $n$ is odd and $n>1$ , then $p^{n}+1=2^{m}$ has no prime solution $p$ , thus, $n$ is even.
Let $n=2 s(s \geqslant 1)$ . Therefore, $m>2$ .
Hence, $p^{2 s}+1 \equiv 0(\bmod 4)$ .
Since the odd prime $p=4 k \pm 1$ , we have,
$p^{2 s}+1 \equiv( \pm 1)^{2 s}+1 \equiv 2(\bmod 4) \text {, }$

contradiction.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.