Condition on integers $a,b$ such that $a^{1/b}$ is rational number

Question

If $a, b\in \mathbb Z$ then what is condition on them such that $a^{1/b}\in \mathbb Q$?

I wanted to show above. My attempt:

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Let $a^{1/b}=p/q$ then $a=p^b/q^b$ that is $aq^b=p^b$...

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I am not able to proceed further, any help will be appreciated.

Answer 1

Hint: $aq^b=p^b$ implies that $v_\pi(a)$ is a multiple of $b$ for every prime $\pi$.

($v_\pi(a)$ is the exponent of $\pi$ in the prime factorization of $a$).

Answer 2

It's not enough to assume just $a^{1/b}=\frac{p}{q}$, you still need to assume (which makes sense) that $\frac{p}{q}$ is irreducible or $\gcd(p,q)=1$. But then $\gcd\left(p^b,q^b\right)=1$ (left as an exercise) and from Bezout, $\exists m,n \in\mathbb{Z}: mp^b+nq^b=1$. Combine this with $aq^b=p^b$ and the result is $$maq^b+nq^b=1 \Rightarrow q^b(ma+n)=1 \Rightarrow q^b \mid 1$$ which is only possible when $q=1$ or $q=-1$. Let's assume $q=1$, then $a^{1/b}=\frac{p}{q}=p \Rightarrow \color{red}{a=p^b}$.

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A few notes on why assuming $\frac{p}{q}$ is irreducible makes sense. If it's not irreducible, it can be made irreducible, via $$\gcd(p,q)=d>1 \Rightarrow p=p_1\cdot d, q=q_1\cdot d, \color{red}{\gcd(p_1,q_1)=1} \Rightarrow\\ \frac{p}{q}=\frac{p_1\cdot d}{q_1\cdot d}=\frac{p_1}{q_1}$$ and we can use $\frac{p_1}{q_1}$ instead.