Let $a>1,b>1,p$ be integers with $p$ an odd prime.
I am trying to find a concise way to show that $2^a \ne p^b+1$.
Is this a valid way to prove it? Is there a better way or more standard way?
Here is my argument:
(1) Assume that $2^a = p^b + 1$
(2) $b$ is odd
Since $a \ge 2$, $p^b \equiv -1 \pmod 4$ which is only possible if $p \equiv -1 \pmod 4$ and $b$ is odd.
(3) Since $b \ge 3$, it follows that:
$$2^a = (p+1)(p^{b-1} - p^{b-2} + \dots + 1)$$
(4) $p^{b-1} - p^{b-2} + \dots + 1 = \frac{p^b+1}{p+1} > 1$
$(p^b + 1) - (p + 1) = p(p^{b-1} - 1) > 1$ so that $p^b+1 > p+1$
(5) But now we have a contradiction because:
- $p^{b-1} - p^{b-2} + \dots + 1 \equiv 1 \pmod 2$ since $b-1$ is even
- $p^{b-1} - p^{b-2} + \dots + 1 > 1$
Edit: Made change based on comment received.
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Edit 2: In my first edit, I asked an additional question which is clearly wrong. I have removed the additional question.
I think it is ok until $\frac{p^b + 1}{p+1} > 1$ which is almost obviously, but I would put a little more efort to prove this. Say $$\frac{p^b + 1}{p+1}\geq \frac{p^2 + 1}{p+1} > 1$$ since $p^2>p$.