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Let $m, n$ be positive integers, with $m > n$ and $2^{2^m}+1$ and $2^{2^n}+1$, Fermat numbers.

To prove that both Fermat numbers are coprime, it's sufficient to state $$\begin{align}(2^{2^m}+1, 2^{2^n}+1) =&\\ (2^{2^m}+1-(2^{2^n}+1), 2^{2^n}+1) =& \\(2^{2^m}-2^{2^n}, 2^{2^n}+1) =&\\ (2k,2q+1) \overset{(1)}{=}& \quad 1\end{align}$$.

Where $2k = 2^{2^m}-2^{2^n}$ and $2q+1=2^{2^n}+1$.

$(a,b)$ is a shorthand notation for $gcd(a,b)$.

Is this proof incorrect? Why so?

In particular, is the setp $(1)$ wrong?

I'm unable to see what I'm missing here.

Thanks in advance.

agus
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    How do you arrive at the last statement, that (2k, 2q+ 1)= 1? I presume, though you did not say, that $k= 2^{2^{m-1}}- 2^{2^{n-1}}$ and that $q= 2^{2^{n-1}}$. But how does it follow that (2k, 2q+1)= 1? – George Ivey Jun 12 '22 at 12:22
  • If it's not wrong, it's surely incomplete. It's almost an induction step, but even that needs some fleshing out. – B. Goddard Jun 12 '22 at 12:35
  • @GeorgeIvey thanks! I see the error now! Certainly it doesn't follow. I'm posting the answer to my own question now, but I don't know how to give you creditin the "answer your question" section. Thanks. – agus Jun 12 '22 at 13:19

1 Answers1

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Thanks to @GeorgeIvey. Certainly assuming that for every $q,k$ we have $(2q,2k+1)=1$ is false, since $(42,3) = 3$, since $42 = 2 \cdot 7 \cdot 3$.

agus
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