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This question already has an answer here:

Prove that $\gcd(g_a,g_b) = 1$ given that for $n \in Z^{\geq 0}$, define $g_n = 2^{2^n} + 1$. I have already proved that $g_0\cdot g_1\cdots g_{n-1} = g_n -2$ if this hint is useful in this proof.

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marked as duplicate by user26857, Joel Reyes Noche, Jyrki Lahtonen Oct 2 '15 at 3:31

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  • $\begingroup$ If your relation is true (which I haven't checked), then let's say a > b, and we know g(a) = k g(b) + 2. Also, neither g(n) is divisible by 2 (except for g(0)). Can you finish the argumentation? $\endgroup$ – SometimesBlind Oct 1 '15 at 11:03
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    $\begingroup$ math.stackexchange.com/questions/1458929/… $\endgroup$ – Zircht Oct 1 '15 at 11:42
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If $g_n=2^{2^n}+1$ and $g_n=g_0\cdot g_1 \cdots g_{n-1}+2$.

The gcd must divide 2, but it cannot be 2 as $g_n$ cannot be divided by 2. Hence $\gcd(g_a,g_b)=1$

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All the $g_i$ are odd.
If $(g_a,g_b)=d>1$ then there must be an odd prime $p$ which divides both $g_a$ and $g_b$.
Assume $a>b$. Then $g_a=g_1\cdots g_b\cdots g_{a-1}-2$ which means that if $p\mid g_a ,p\mid g_b$ then $ p\mid g_a- g_1\cdots g_b\cdots g_{a-1}$ which means $p\mid 2$ a contradiction.

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  • $\begingroup$ I do not understand where the $p|2$ comes from in this, could you please explain? $\endgroup$ – user275825 Oct 2 '15 at 1:13

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