Tuesday, 6 June 2017

elementary number theory - Prove without induction that 34n24n is divisible by 65




My brother asked me this (for some reason).



My solution is:



(34n24n)mod65=



(81n16n)mod65=




((81mod65)n16n)mod65=



(16n16n)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 (8116)n.



In other words, something like:



34n24n=



81n16n=



(8116)n+65k=




65n+65k=



65(65n1+k)



How would I go from "81n16n" to "(8116)n+65k"?


Answer



You can use the formula anbn=(ab)(an1+an2b+an3b2++abn2+bn1)


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