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Let $x$ and $y$ be nonzero integers where $xy$ is a cube. If $\gcd(x,y)=3$ and $3$ divides $y$ only once, then there are integers $r,s$, such that: $$3x=r^3$$ $$3^2y=s^3.$$

What is a simple method to prove this? Any input will be appreciated.

I put $$x=3^{t}pc^3$$ $$y=3qd^3$$ where $p,q$ are cube free integers.Then I argue that $3^{t+1}pq$ is a cube. Things get messy from there.

NumThcurious
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2 Answers2

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Since $(x,y)=3$, no other prime apart from $3$ divides both $x$ and $y$ and $xy$ is a cube. Also, $3$ divides $y$ only once. So, the prime factorisation of $y$ is \begin{align*} y=3p_1^{3i_1}p_2^{3i_2}\dots p_k^{3i_k} \end{align*} where $p_1,p_2,\dots ,p_k$ are distinct primes none of which equals $3$.

Now, $3$ divides $y$ only once but it divides $xy$, $3t$ times. So, $3$ must divide $x$, $3t-1$ times. So, prime factorisation of $x$ is \begin{align*} x=3^{3t-1}q_1^{3j_1}q_2^{3j_2}\dots q_{k^{\prime}}^{3j_k} \end{align*} where $q_1,q_2,\dots ,q_{k^{\prime}}$ are distinct primes none of which equals $3$ and $p_i\neq q_j \forall i,j$.

So, \begin{align*} 3x=3^{3t}q_1^{3j_1}q_2^{3j_2}\dots q_{k^{\prime}}^{3j_k} \end{align*} and \begin{align*} 3^2y=3^3p_1^{3i_1}p_2^{3i_2}\dots p_k^{3i_k} \end{align*} which completes the proof.

Does that help?

Sayan Dutta
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  • +1 : re your answer was nicely written. I don't know if your answer is going to be downvoted by other mathSE reviewers, who may regard your answer as premature, since the OP didn't really show work. Recently, there have been efforts made by (other) mathSE reviewers to discourage hint-answers. – user2661923 Jun 24 '21 at 07:00
  • @user2661923 Thanks for the information, but I'm new here, and I don't understand you completely. What is OP? And how is this a hint answer? I've done everything except the last two lines and some rigour, both of which are very easy. Can you please elaborate? – Sayan Dutta Jun 24 '21 at 07:46
  • @user2661923 I added the last two lines as well. – Sayan Dutta Jun 24 '21 at 07:49
  • OP refers to original poster, namely, the person who posted the question. The generally accepted standards for when it is okay to answer a mathSE question are based on this article. Personally, I used to avoid the issue, when the OP had not shown work, by providing Hint-Answers which always left the OP with work to do. In the last couple of months, there has been a concerted effort to discourage such Hint-Answers. ...see next comment – user2661923 Jun 24 '21 at 08:45
  • Two thinks to consider: [1] My (subjective perception) is that such efforts to discourage Hint-Answers are not always even-handed. [2] The issue of whether the OP has shown work is itself often subjective. In this particular situation, a case could be made either way. Personally, I'm unhappy with the recent Answer-Police activities, but it is not under my control. I have no idea how they will respond to your answer. – user2661923 Jun 24 '21 at 08:48
  • You bring up another issue. Actually, your answer is very close to a complete answer. In such circumstances, then (perhaps) the issue of whether the OP has shown work is more closely scrutinized. – user2661923 Jun 24 '21 at 08:52
  • @user2661923 Thanks for the comments. There are so many complicated policies here, it's really hard to follow all of them when answering posts. I'll try to keep these in mind from next time. – Sayan Dutta Jun 24 '21 at 09:38
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From the conditions given in the problem we can write $x=3X$ and $y=3Y$, where $\gcd(X,Y)=1$ and $3\nmid Y$. Then $$xy=m^3 \implies 9XY=m^3.$$ Thus $3 \mid m$. Thus we can say $m=3M$ for some $M \in \Bbb{Z}$. This implies $$XY=3M^3.$$ Now $3\nmid Y$ implies $3 \mid X$ (say $X=3T$). Thus we have $$TY=M^3.$$ But $\gcd(T,Y)=1$ as well, so $T$ and $Y$ must be cubes (from the comments it seems this is a result that you seem to be already aware of). So for some $u,v \in \Bbb{Z}$ we have $$T=u^3 \implies X=3u^3 \implies x=3^2u^3 \implies 3x=(3u)^3=r^3.$$ Likewise $$Y=v^3 \implies y=3v^3 \implies 3^2y=(3v)^3=s^3.$$

Anurag A
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