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For $a, m, n \in \mathbb{N}$, prove that $$\gcd(a^{2^m} + 1, a^{2^n} + 1) = \begin{cases} 1, & \text{if $a$ is even}\\ 2, & \text{if $a$ is odd} \end{cases}$$ given $m \ne n$.

My attempt:

Let $\gcd(a^{2^m} + 1, a^{2^n} + 1) = d$. Let $p$ be a prime factor of $d$. Clearly, $p \nmid a \implies \gcd(p,a) = 1$. $\tag*{}$ But, $ \ \ \ a^{2^m} \equiv -1 \pmod{p} \\ \implies a^{2^{m+1}} \equiv 1 \pmod{p} \\ \implies \text{ord}_p(a) = 2^{m+1}$ $\tag*{}$ But by similar logic, $\text{ord}_p(a) = 2^{n+1}$ $\implies m = n$ $\tag*{}$ Therefore, no such prime exists. $\implies d = 1$.

But obviously, there's something wrong with this, as, for an odd $a \text{, } 2 \mid d$.

After some head-scratching, I realized that $-1 \equiv 1 \pmod{2}$. And for all odd positive integers, $\text{ord}_2(a) = 1$.

So my question is, while working with orders, do I need to treat $\text{ord}_2(x)$ as an exception? I just can't figure a general working rule for it.

Edit: Just to calrify, $\text{ord}_p(a) = 2^{m+1} \text{ because, }\text{ord}_p(a) \mid 2^{m+1} \text{ but, } \text{ord}_p(a) \nmid 2^{m}$.

Bill Dubuque
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    Knowing that $a^k\equiv 1 \pmod p$ does not tell you that the order of $a\pmod p$ is $k$. $k$ could be a multiple of the order. – lulu Sep 13 '22 at 14:03
  • I didn't think it was necessary to mention as it is easy to see here that $\text{ord}_p(a) \mid 2^{m+1} \text{ but, } \text{ord}_p(a) \nmid 2^{m} \implies \text{ord}_p(a) = 2^{m+1}$, and similarly for $n$. I'll edit it in. – Nick Larry Sep 13 '22 at 14:07
  • Yes, that's good. And it does indeed show that no odd prime can divide the gcd. But, as you remark, the argument breaks down when $p=2$, so you still have to rule out the case where the gcd is divisible by $4$. – lulu Sep 13 '22 at 14:11
  • Yes, but I can't figure at what step in the solution, does it imply that either $p = 2$ or $m = n$. – Nick Larry Sep 13 '22 at 14:14
  • Maybe we can do something like: $\text{ord}_p(a) \mid 2^{m+1} \text{ and, } \text{ord}_p(a) \mid 2^{n+1} \text{ but, } \text{ord}_p(a) \nmid 2^{m} \text{ and, } \text{ord}_p(a) \nmid 2^{n} \implies \text{ either } m = n \text{ or } -1 \equiv 1 \pmod{p} \implies p = 2$. – Nick Larry Sep 13 '22 at 14:16
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    If $p$ is odd then I agree you have shown that the order is simultaneously $2^{n+1}$ and $2^{m+1}$ which forces $n=m$ as you say. However, your argument breaks down for $2$. There we only know that, say, $a^{2^{n}}+1$ is even which only tells us that $a$ is odd. – lulu Sep 13 '22 at 14:18
  • @DavidC.Ullrich It is given that $m \ne n$. – Nick Larry Sep 13 '22 at 14:28
  • To prove that $d$ does not contain more than one factor of $p = 2$, we can again argue that order of $a$ modulo $d$ would make $m = n$ if $d \ne 2$. – Nick Larry Sep 13 '22 at 15:39
  • The argument implicitly uses the Order Test as here in the first linked dupe. Of course you do need to pay attention to cases where the general argument degenerates (here when $-1\equiv 1)\ \ $. – Bill Dubuque Sep 13 '22 at 18:05

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