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

Question

My brother asked me this (for some reason).

My solution is:

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

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

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

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

$0\bmod{65}$

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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 1

You can use the formula $$a^n-b^n = (a-b)\left(a^{n-1}+a^{n-2}b+a^{n-3}b^2+\ldots+ab^{n-2}+b^{n-1}\right)$$

Answer 2

You're almost there:

$$3^{4n}-2^{4n}=81^{n}-16^{n}=(65+16)^{n}-16^{n}=65a+16^{n}-16^{n}=65a$$

The key point is $(65+16)^{n}=65a+16^{n}$, which follows from the binomial theorem.