Show that if m does not = n then
gcd((a^2)^n) + 1, ((a^2)^m + 1) = 1, if a is even
and 2, if a is odd.
Not sure if this question is formulated correctly. I am reading:
For all $\displaystyle m\neq n, gcd(a^{2m}+1,a^{2n}+1)=\begin{cases}1 & \text{a even} \\2 & \text{a odd} \\ \end{cases}$
This would mean that for a even, any two elements in the sequence $\displaystyle c(a)_k=a^{2k}+1$ should be relatively prime without exception: $\displaystyle c(2)=\{5,17,65,257,1025,4097,16385,...\}$. As you can see, the conjecture is simply wrong.
Theorem: WLOG, for $\displaystyle
m> n, \gcd(a^{2^m}+1,a^{2^n}+1)=\begin{cases}1 & \text{a even} \\2 & \text{a odd} \\ \end{cases}
$
Proof: Start by noticing $\displaystyle a^{2^n}+1|a^{2^m}-1$ -- see proof at http://www.mathhelpforum.com/math-he...s-2-n-1-a.html. So we can rewrite as $\displaystyle \gcd(kx+2,k)$, for $\displaystyle k=a^{2^n}+1$ and some x. It can be easily seen that k has the opposite parity of a. So if a is even, k is odd, and $\displaystyle \gcd(kx+2,k)=1$. If a is odd, k is even, and $\displaystyle \gcd(kx+2,k)=2$. QED