If $m>n$, prove that $a^{2^n}+1$ is a divisor of $a^{2^m}-1$. Find $(a^{2^n}+1, a^{2^m}+1)$, when $a,m,n$ are positive and $m\neq n$.
I TRY LIKE THIS:
Since $m>n$, $m=n+k$, $$a^{2^m}-1=a^{2^n\cdot 2^k}-1$$ $$=(a^{2^n})^{2^k}-1$$ here I stops and can't able to proceed further. Please give me a hint and not answer. THANKS
We can write $k=m-n$, then note that $$a^{2^m}-1 = (a^{2^{n}2^{k-1}}+1)(a^{2^{n}2^{k-1}}-1).$$ If $k=1$, we have the result. If not, write $$(a^{2^{n}2^{k-1}}-1) = (a^{2^{n}2^{k-2}}-1)(a^{2^{n}2^{k-2}}+1)$$ if $k=2$, we have the result, if else repeat the process. Since $k$ is finite, in a finite number of steps we will have $$(a^{2^m}-1) = (a^{2^{n}}-1)(a^{2^{n}}+1)(a^{2^{n}2^1}+1)(a^{2^{n}2^{2}}+1)\cdots(a^{2^{n}2^{k-1}}+1)$$ and hence the result.