Monday, November 28, 2016

elementary number theory - Prove without induction that $3^{4n}-2^{4n}$ is divisible by $65$


My brother asked me this (for some reason).


My solution is:



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


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


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


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


$0\bmod{65}$



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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