Show divisibility conditions

Question

Let $ a, m $ and $ n $ be integers such that $ m> n \geq 0 $.

i) Show that $a ^ {2n} +1$ divides $ a ^ {2^m} -1$.

ii) Show that $ (a ^ {2^m} + 1, a ^ {2^n} + 1) = $ 1, if a even; 2, if a odd (Use elementary techniques)

Attemp: $i) $ Let $m=n+k $ where $k\ge 1$. See that the problem is show that $a^{2^m}=(a^{2^n})^{2^k}\equiv 1\mod ((a^2)^n+1) $. I think you can work this by induction and show that $a^{2^n}\equiv \pm 1\mod ((a^2)^n+1) $, but I only can show it for $a=2$.

$ii) $ Suppose that $p $ is a prime that divides $a^{2^m}+1$ and $a^{2^n}+1$ then $p\mid a^{2^m}-a^{2^n}=a^{2^n}(a^{2^k}-1)\implies p\mid a^{2^k}-1 $ this yields $a^{2^k}\equiv 1\mod p $ wich since $p\mid a^{2^m} +1$ we obtain: $(a^{2^k})^{2^n}+1\equiv 2\equiv 0\mod p $

Then $p\mid 2$ and $\gcd (a^{2^m}+1,a^{2^n}+1)=\boxed {1,2} $

i) $a^{2n}\equiv-1\bmod a^{2n}+1$; if $m>n\ge0$ then $2^m-2n$ is even so $(a^{2n})^{2^m-2n}\equiv1\bmod a^{2n}+1$ — J. W. Tanner, Dec 10 '19 at 09:27
See my post in https://math.stackexchange.com/questions/123524/fermat-numbers-are-coprime — lab bhattacharjee, Dec 10 '19 at 09:45

score 0 · Accepted Answer · answered Dec 11 '19 at 09:04

Your answer to part (ii) is correct. In fact both parts can be solved in the same way using the fact that $a^{2^n}$ is $-1$ modulo $a^{2^n}+1$.

(i) $a^{2^m}-1=(a^{2^n})^{2^k}-1\equiv 1-1\equiv 0\mod {a^{2^n}}+1$

(ii) $a^{2^m}+1=(a^{2^n})^{2^k}-1\equiv 1+1\equiv 2\mod {a^{2^n}}+1$

So $ (a ^ {2^m} + 1, a ^ {2^n} + 1)$ is either $1$ or $2$ and it is $2$ if and only if $a$ is odd.

Show divisibility conditions

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