My brother asked me this (for some reason).
My solution is:
(34n−24n)mod
(81^{n}-16^{n})\bmod{65}=
((81\bmod{65})^{n}-16^{n})\bmod{65}=
(16^{n}-16^{n})\bmod{65}=
0\bmod{65}
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"?
Answer
You can use the formula a^n-b^n = (a-b)\left(a^{n-1}+a^{n-2}b+a^{n-3}b^2+\ldots+ab^{n-2}+b^{n-1}\right)
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