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.

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- Nov 5th 2009, 02:19 PMsirellwoodgcd odd and even proof
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. - Nov 6th 2009, 07:46 AMMedia_ManFormulation?
Not sure if this question is formulated correctly. I am reading:

For all

This would mean that for a even, any two elements in the sequence should be relatively prime without exception: . As you can see, the conjecture is simply wrong. - Nov 7th 2009, 04:57 AMMedia_ManExponents not Associative
Theorem: WLOG, for

Proof: Start by noticing -- see proof at http://www.mathhelpforum.com/math-he...s-2-n-1-a.html. So we can rewrite as , for 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 . If a is odd, k is even, and . QED