My brother asked me this (for some reason).
My solution is:
(34n−24n)mod65=
(81n−16n)mod65=
((81mod65)n−16n)mod65=
(16n−16n)mod65=
0mod65
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:
34n−24n=
81n−16n=
(81−16)n+65k=
65n+65k=
65(65n−1+k)
How would I go from "81n−16n" to "(81−16)n+65k"?
Answer
You can use the formula an−bn=(a−b)(an−1+an−2b+an−3b2+…+abn−2+bn−1)
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