How to prove if $xy$ is a cube, $\gcd(x,y)=3$ and $3$ divides $y$ only once, then $3x=r^3$ and $3^2y=s^3$ for some integers $r,s$?

Question

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.

Answer 1

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?

Answer 2

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 $$Y=v^3 \implies y=3v^3 \implies 3^2y=(3v)^3=s^3.$$