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

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The discussion centers on the divisibility of the expression (a^m^2 + 1) by (a^n^2 - 1) for integers n and m where n > m. Participants explored various approaches, including the application of Fermat's Little Theorem, but ultimately concluded that the statement is not universally true, as a counterexample exists among small integers. The exploration highlights the complexity of number theory and the necessity of rigorous proof in mathematical claims.

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  • Basic algebraic manipulation of polynomial expressions.
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(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?)
 
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It's not true. A counterexample exists among the very small integers...
 

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