Let $n \in \mathbb{Z}$. Prove $n^6 = k^9 = m^4$ and Let $a = n^6$ $\iff$ $a$ must be a perfect $36$-th power.

Question

Prove $n^6 = k^9 = m^4$ and Let $a = n^6$ $\iff$ $a$ must be a perfect $36$-th power.

Answer 1

It suffices to show that any prime dividing $a$, divides $a$ in powers of multiples of $36$ i.e if $p$ is a prime such that $p \mid a$, then, the exact power of $p$ in $a$ is a multiple of $36$. Let $p \mid a$. Then, $p \mid n, k, m$. Hence, the power of $p$ in $a$ is a multiple of $6,9,4$. Thus, it is a multiple of $36$.

Answer 2

If you know the p-adic valuation it can help you, all what you need is to prove that for all primes $p$ : $v_p(a)$ is divisible by $36$. that's clear because we have $v_p(a)=9v_p(k)=4v_p(m)$ holds for all prime, and using the fact that $4$ and $9$ are coprime, we can deduce that $36$ divide $v_p(a)$ for all primes.

if $a=b^{36}$ then $a=(b^6)^6=(b^4)^9=(b^9)^4$