gcd odd and even proof

sirellwood · #1 $Old$ November 5th, 2009, 03:19 PM

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.

Media_Man · #2 $Old$ November 6th, 2009, 08:46 AM

Not sure if this question is formulated correctly. I am reading:

For all $m\neq n, gcd(a^{2m}+1,a^{2n}+1)=\begin{cases}1 & \text{a even} \\2 & \text{a odd} \\ \end{cases}$

This would mean that for a even, any two elements in the sequence $c(a)_k=a^{2k}+1$ should be relatively prime without exception: $c(2)=\{5,17,65,257,1025,4097,16385,...\}$ . As you can see, the conjecture is simply wrong.

Media_Man · #3 $Old$ November 7th, 2009, 05:57 AM

Theorem: WLOG, for $m> n, \gcd(a^{2^m}+1,a^{2^n}+1)=\begin{cases}1 & \text{a even} \\2 & \text{a odd} \\ \end{cases}$

Proof: Start by noticing $a^{2^n}+1|a^{2^m}-1$ -- see proof at Show that if m > n then ((a^2)^n) + 1 divides ((a^2)^n)− 1. So we can rewrite as $\gcd(kx+2,k)$ , for $k=a^{2^n}+1$ 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 $\gcd(kx+2,k)=1$ . If a is odd, k is even, and $\gcd(kx+2,k)=2$ . QED