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

Bill Dubuque
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    Please drop the bit "or $\operatorname{lcm}(9,4)=36$". This is not the question, everyone knows that $\operatorname{lcm}(9,4)=36$. The question is: if $a$ is at the same time a $9$th power (of some integer. say $k$) and a $4$th power (of some integer, say $m$) then prove that $a$ is a $36$th power (of yet another integer). –  Jul 20 '23 at 11:35
  • And remember that you need to prove both sides of the equivalence. The $\impliedby$ part is trivial: if $a=x^{36}$ then $a=(x^4)^9=(x^9)^4=(x^6)^6$ so you can take $n=x^6, m=x^9, k=x^4$. The real onus is on proving the $\implies$ part. –  Jul 20 '23 at 11:39
  • This is only true when $k$ and $m$ also are integers. – Dominique Jul 20 '23 at 12:01
  • $a^9=m^{36}, a^6 = n^{36}, a^4 = k^{36}\Rightarrow a = a^9 a^4/a^{12} = (mk/n^2)^{36}$ and $,mk/n^2\in \Bbb Z$ by Rational Root Test. – Bill Dubuque Jul 20 '23 at 16:03
  • Same methods in the linked dupes work here. – Bill Dubuque Jul 20 '23 at 16:05

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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 $a=(b^6)^6=(b^4)^9=(b^9)^4$

hood09
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