# Thread: a^n | b^n => a | b ??

1. ## [Solved] a^n | b^n => a | b ??

Problem: Give a proof or a counterexample for the following.

- If a^n|b^n then a|b

Thought I had a solution in here, but it didn't work. I received help though, so I know what to do now. Thanks!

Since we can assume a^n|b^n, we can write x*a^n=b^n. Then we can rewrite as:

a^(n-1)*a*x=b^(n-1)*b
(a^(n-1)/b^(n-1))*x*a=b

From here I tried to say that a^(n-1)/b^(n-1) was an integer, using a^n|b^n. More evidence I have been doing too much math, and my brain is shutting down.

I know what to do now, I just had to correct this so it didn't look like I was a complete idiot.

2. Given a prime $\displaystyle p$ we'll define $\displaystyle v_p(n)$ to be the maximum integer such that $\displaystyle p^{v_p(n)}|n$ (it's 0 if p doesn't divide n)

First note that $\displaystyle a|b$ if and only if $\displaystyle v_p(b)\geq{v_p(a)}$ for all primes $\displaystyle p$ (use the prime descomposition)

But we also have $\displaystyle v_p(k^n)=n\cdot v_p(k)$ for all $\displaystyle n \in \mathbb{N}$. Using these 2 results you should not have problems in finding a proof.

3. PaulRS: Thanks for the reply! I reached my simple (incorrect) solution (and was debating the ettiquite of how to edit my post to reflect that, actually) just as you were posting your more advanced solution, it appears.

I just started Number Theory about two weeks ago, and we haven't covered the $\displaystyle v_p$ that you described in your solution, so I am wondering if it might be more sophisticated than he expects. I would not have come up with that approach on my own, anyways, so I suppose I am saying that I hope that is more than he expects. :)

Thank you for the reply though! It made sense how you did it.

Assuming you are around and reading this though - does my most recent attempt make sense and work well enough for a really simple proof? (Nevermind, I caught the obvious problem with my attempt.)