# (a^m^2 + 1) | (a^n^2 - 1) ?

1. ### JdotAckdot

4
(a^m^2 + 1) | (a^n^2 - 1) ???

I'm sure there is a quick trick I'm missing somewhere... but anyone have any ideas on how to prove:

(a^m^2 + 1) | (a^n^2 - 1) , for n > m.

[Show [a^(n^2) -1] is divisible by [a^(m^2) +1]

Thanks a lot. . .

(I've tried letting k=n-m, and other stuff like that... kept going in circles. I'm guessing Fermat's Little Thm comes in somewhere?)

2. ### Muzza

695
It's not true. A counterexample exists among the very small integers...