My brother asked me this (for some reason).

My solution is:

$(3^{4n}-2^{4n})\bmod{65}=$

$(81^{n}-16^{n})\bmod{65}=$

$((81\bmod{65})^{n}-16^{n})\bmod{65}=$

$(16^{n}-16^{n})\bmod{65}=$

$0\bmod{65}$

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I think that this solution is mathematically flawless (please let me know if you think otherwise).

But I'm wondering if there's another way, perhaps with the binomial expansion of $(81-16)^{n}$.

In other words, something like:

$3^{4n}-2^{4n}=$

$81^{n}-16^{n}=$

$(81-16)^{n}+65k=$

$65^{n}+65k=$

$65(65^{n-1}+k)$

How would I go from "$81^{n}-16^{n}$" to "$(81-16)^{n}+65k$"?