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)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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