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I have been working a bit with the following question:

"Prove that all numbers given by $a_n=2^{2^n}+1$ are coprime to each other"

I think I have a proof but I can't quite finish the last step.

We can immediately tell that if $a_n$ has a divisor $q$, then q has to be odd (since $2^{2^n}$ is always even). Furthermore, we can tell that the remainder when dividing $2^{2^n}$ by this number $q=2k+1, k\in\mathbb{Z}$ has to be exactly $2k$. Because of this, we can say that $2^{2^n} \equiv_{2k+1}2k$.

Now, let's add some arbitrary integer $p\ge1$ to the index which gives us $2^{{2^{n+p}}}\equiv_{2k+1} (2k)^{2^p}$. Then, we ask if this can ever be congruent with $2k$ (since if it can't for some $k$ which is not $0$ or $-1$, they have to be coprime). This gives us the following congruence: $(2k)^{2^p}\equiv _{2k+1}2k\equiv _{2k+1}-1$. Intuatively, it seems like one should be able to use this to show that the only k's satisfying the congruence are $0$ and $-1$. But I can't get it to work. Help is appreciated!

Bill Dubuque
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naytte2
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1 Answers1

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Great work! You can write $2k \equiv -1 \pmod{2k+1}$. Thus, you have $(-1)^{2^p} \equiv -1 \pmod{2k+1}$. Do you see how to finish the proof? You may want to see how this is pretty much equivalent to the proof linked to in the comments.

Haran
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