Let $X,Y \geq 1$. I am interested in the number of solutions of the following diophantine equations:
$$S\_1\colon \, \, x\_1y\_1^3 = x\_2 y\_2^3 $$
Let $N\_1(X,Y) $ denote the number of solutions to $S\_1$ with $1\leq x\_i \leq X$ and $1\leq y\_i \leq Y$. There are the obvious solutions $x\_1=x\_2$ and $y\_1=y\_2$ which contribute $XY$. The true order of magnitude should be $XY \log(XY)^A$ with some small power of log. I would like to get $A$ as small as possible.


Also consider the equation $S\_2$ given by
$$ x\_1(y\_1^3 - y\_2^3) = x\_2 (z\_1^3 -z\_2^3)$$
with $1\leq x\_i\leq X$ and $1\leq y\_i,z\_i \leq Y$ with $y\_1\neq y\_2$ and $z\_1\neq z\_2$. Let $N\_2(X,Y)$ denote the corresponding number of solutions.
I expect something like $N\_2(X,Y) \ll XY^2\log(XY)^B$, with $B$ again hopefully small.


Is there some clever way to get the exponent of the logarithm small?



Diophantine equations and 'quasi-paucity'
For $S\_1$ you can take $A=0$.


Setting $g=\gcd(x\_1,x\_2)$ and $v\_i = x\_i/g$, and $h=\gcd(y\_1,y\_2)$ and $w\_i = y\_i/h$, one obtains $v\_1w\_1^3 = v\_2w\_2^3$ with $\gcd(v\_1,v\_2)=\gcd(w\_1,w\_2)=1$; this implies that $v\_1=w\_2^3$ and $v\_2=w\_1^3$. Therefore all solutions to $S\_1$ are of the form $(x\_1,x\_2,y\_1,y\_2) = (gw\_2^3,gw\_1^3,hw\_1,hw\_2)$ with $\gcd(w\_1,w\_2)=1$ and the four coordinates less than $X,X,Y,Y$, respectively.


Counting these solutions yields the sum
\begin{align\*}
\sum\_{g\le X} \sum\_{h\le Y} \sum\_{\substack{w\_1,w\_2 \le \min\{ (X/g)^{1/3},Y/h \} \\ \gcd(w\_1,w\_2)=1} } 1
&=
\sum\_{g\le X} \sum\_{h \le Y(g/X)^{1/3}} \sum\_{\substack{w\_1,w\_2 \le (X/g)^{1/3} \\ \gcd(w\_1,w\_2)=1}} 1
+
\sum\_{h\le Y} \sum\_{g \le X(h/Y)^3} \sum\_{\substack{w\_1,w\_2 \le Y/h \\ \gcd(w\_1,w\_2)=1}} 1 \\
&\le
\sum\_{g\le X} \sum\_{h \le Y(g/X)^{1/3}} \frac{X^{2/3}}{g^{2/3}}
+
\sum\_{h\le Y} \sum\_{g \le X(h/Y)^3} \frac{Y^2}{h^2} \\
&\le
\sum\_{g\le X} Y \frac{X^{1/3}}{g^{1/3}}
+
\sum\_{h\le Y} X\frac hY \ll YX^{1/3} X^{2/3} + \frac XY Y^2 \ll XY.
\end{align\*}


