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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.

user26857
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user275825
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  • 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? – SometimesBlind Oct 01 '15 at 11:03
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    http://math.stackexchange.com/questions/1458929/prove-that-gcdg-a-g-b-1-for-g-n-22n-1 –  Oct 01 '15 at 11:42

2 Answers2

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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 $\gcd(g_a,g_b)=1$

MrYouMath
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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.

Konstantinos Gaitanas
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