# Proof for Perfect $n^{th}$ power.

A positive integer, $$k$$, is a perfect nth power if, there exist a positive integer a such that $$k=a^n$$, where $$n$$ is also a positive integer.

Question is:
How can I prove, for all positive integers $$k, m, n$$ (where $$m$$ and $$n$$ are coprime), if $$k$$ is a perfect $$m^{th}$$ power and a perfect $$n^{th}$$ power, then it is a perfect $$mn^{th}$$ power. This looks a lot like perfect squares, but it is more than perfect squares. So for example $$16 = 2^4 = 4^2$$

I'm not sure how to do this. I tried some patterns, is it possible to use gcd? I know that if $$gcd(a, b)=1$$, then $$gcd(a^n, b^n)=1$$. But I don't know how this can help either.

Hint: $$(m,n)=1 \Rightarrow \exists x, y \in \mathbb{N}, mx-ny=1.$$

Can you start from here?

If $$k=a^m=b^n$$, then $$a=a^{mx-ny}=\left(\frac{b^x}{a^y}\right)^n.$$

If you know the fundamental theorem of arithmetic, then you should be able to show the following lemma:

A positive integer $$k$$ is a perfect $$n$$-th power if and only if for every prime number $$p$$, the exponent of $$p$$ appearing in the factorization of $$k$$ is a multiple of $$n$$.

Now if $$k$$ is a perfect $$m$$-th power and a perfect $$n$$-th power, then by the lemma, for every prime $$p$$, the exponent of $$p$$ appearing in the factorization of $$k$$ is simultanously a multiple of $$m$$ and a multiple of $$n$$.

Since $$m$$ and $$n$$ are coprime, that exponent is a multiple of $$mn$$. This being true for every $$p$$, using the lemma again tells us that $$k$$ is a perfect $$mn$$-th power.