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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

2 Answers2

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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 $36$.

Vishnu Kadiri
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  • That is not correct. Why if $p|a,k,m$ we have that $u_{p}$ is a multiple of $6,9,4$? There is a gap there. You need to consider the largest power of $p$ that divides $a$ not just $p$, have a look at the other answer. – SotArmen Jul 20 '23 at 13:35
  • @SotArmen, yes you need to consider the largest power of $p$ dividing $k,m$ each, so that $a=p^{4x}x_1=p^{9y}y_1$ where $p \nmid x_1,y_1$. This should imply $p$ divides $a$ in multiples of $36$. Sorry, I thought this was trivial. – Vishnu Kadiri Jul 20 '23 at 14:44
  • Please strive not to post more (dupe) answers to dupes of FAQs (& PSQs) cf. recent site policy announcement here. – Bill Dubuque Jul 20 '23 at 16:06
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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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