2

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 "$(81-16)^{n}+65k$"?

Martin Sleziak
  • 53,687
barak manos
  • 43,109
  • @labbhattacharjee: This looks like a duplicate answer, not a duplicate question... don't you think? I mean, the answer to that other question may be the same as the answer to this question, but the questions themselves are different. If I had known that $a-b$ divides $a^n-b^n$, then I wouldn't have asked this to begin with. On top that, by the title, I am explicitly asking for any non-inductive solution to a completely different problem (although I do admit that in the question itself, I am implicitly focusing on binomial expansion, nevertheless, these two questions are not duplicates IMO). – barak manos Dec 30 '16 at 15:54

2 Answers2

7

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

timon92
  • 11,329
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.

lhf
  • 216,483