So assume that $a^n|b^n$ then $b^n=a^n(x)$ for some integer $x$ taking the nth root of both sides you get $b=a(x^{\frac{1}{n}})$. Now I'm not sure if this is ok since $x^{\frac{1}{n}}$ is not always an integer. But it seems like it should always work so I"m not sure if this is the correct approach.
Asked
Active
Viewed 65 times
0
-
It could be made to work iff you first proved that $\sqrt[n]{|x|}$ is either an integer or irrational. – dxiv Jan 24 '17 at 03:34
-
Why either? I get showing its an integer but I don't think I can do that. Also my problem with the other proofs is that we have only covered divisibility not prime factorization yet. So I can't use that technique. – HighSchool15 Jan 24 '17 at 03:43
-
You only know $\sqrt[n]{|x|}$ is an integer after you proved $a|b,$. If you have some other proof in mind, please post it. – dxiv Jan 24 '17 at 03:49