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From what I know, the question can be solved from congruence modulo or without. Since the question is before the concept of congruence modulo in the book which I am using, I guess a solution without using modulo should be the answer.

Here's what I have tried till now -

Let $a^{2^n}$ = k

$a^{2^m} = a^{2^{n+r}}$ (where $r ≠ 0$)

$a^{2^m} = a^{2^n\cdot 2^r} + 1 = k^{2^r} + 1$

The expression then simplifies to $\gcd(k+1, k^{2^r} + 1)$

Ideally, there should be two cases - one where $k$ is odd and one where $k$ is even. But I am unable to find a solution on that basis. Thanks in advance!

Source : Challenge and Thrill of Pre-College Mathematics

Gary
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  • Hint before scrolling down to the answer: let $p$ be a prime dividing the gcd. Then $ a^{2^{m}} = -1 \pmod{p}, \quad a^{2^{n}} = -1 \pmod{p}.$ Try to deduce that $p=2$. – koifish Feb 26 '24 at 03:37
  • Welcome to Math SE. FYI, I found the duplicate using an Approach0 search. Note that duplicate question was closed as a duplicate of $3$ other questions, and there are quite a few additional duplicates on this site. In addition, the Approach0 search results include quite a few AoPS threads, e.g., Pre College Question, How solve this NT problem, etc. – John Omielan Feb 26 '24 at 04:25

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