Does (a^m^2 + 1) Divide (a^n^2 - 1) for n > m?

[JdotAckdot]

(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?)

 

[Muzza]

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

 

[tacman]

One approach to proving this statement is to use mathematical induction. Let's start by assuming that the statement holds for some value of n, i.e. (a^m^2 + 1) | (a^n^2 - 1) for n > m. Now, we need to show that it also holds for n + 1, i.e. (a^m^2 + 1) | (a^(n+1)^2 - 1).

First, we can rewrite (a^(n+1)^2 - 1) as (a^n^2 + 2a^n + 1 - 1), and then factor out (a^n^2 - 1) to get (a^n^2 + 2a^n). By our assumption, we know that (a^m^2 + 1) | (a^n^2 - 1), so we can rewrite (a^n^2 + 2a^n) as (a^m^2 + 1) * (a^(n-m)^2 + 2a^(n-m)).

Now, we need to show that (a^(n-m)^2 + 2a^(n-m)) is also divisible by (a^m^2 + 1). We can do this by using the fact that (a^m^2 + 1) | (a^n^2 - 1) for n > m. We can rewrite (a^(n-m)^2 + 2a^(n-m)) as (a^m^2 * a^(n-m)^2 + 2a^(n-m)), and then factor out a^m^2 to get (a^m^2 * (a^(n-m)^2 + 2)).

Since (a^m^2 + 1) | (a^n^2 - 1) for n > m, we know that (a^m^2 + 1) | (a^(n-m)^2 - 1). Therefore, we can rewrite (a^m^2 * (a^(n-m)^2 + 2)) as (a^m^2 * (a^(n-m)^2 + 2) + (a^(n-m)^2 - 1)), which is equal to (a^m^2 * a^(n-m)^2 + 2a^(n-m)