We have giiven the Diophantine equation $(1) \;\; 2^3 \cdot 3^4(x^2 + y^2) = (xy)^3$. If $xy=0$, then $x=y=0$ by equation (1). Thus $(x,y)=(0,0)$ is a solution of equation (1). Next assume $(x,y) \neq (0,0)$. Then $xy \neq 0$. Clearly $xy>0$ by equation (3). Hence $(x,y)=(a,b)$ is a solution of equation (1) iff $(x,y)=(-a,-b)$ is a solution of equation (1). Consequently we may WLOG assume $x \geq y > 0$ (since $x$ and $y$ are symmetric in equation (1)). Let $d=GCD(x,y)$. Then there exists two coprime integers $s > t > 0$ s.t. $(x,y) = (sd,td)$, which inserted in equation (1) result in $(2) \;\; 2^3 \cdot 3^4(s^2 + t^2) = d^4(st)^3$. The fact that $GCD(s^2+t^2,st)=1$ combined with equation (2) give us $(st)^3 | 2^3 \cdot 3^4$, yielding $st | 6$. Therefore by equation (2) ${\textstyle (3) \;\; 3 (\frac{6}{st})^3(s^2 + t^2) = d^4}$. Thus we have $3 | d^4$ by equation (3), meaning $3 | d$. Hence $d=3k$, which inserted in equation (3) give us ${\textstyle (4) \;\; (\frac{6}{st})^3(s^2 + t^2) = 27k^4}$. Now $st | 6$ means that $3 | s$ and $3 | t$ is impossible, implying $3 \nmid s^2 + t^2$. Therefore ${\textstyle 3 | \frac{6}{st}}$ by equation (4), i.e. $st | 2$. Hence according to equation (4) we have ${\textstyle (4) \;\; (\frac{2}{st})^3(s^2 + t^2) = k^4}$. The only solution of equation (4) is $(s,t,k) = (1,1,2)$ (since $(s,t)=(2,1)$ yields $k^4=5$ by equation (4)). Thus $d = 3k = 3 \cdot 2 = 6$ and $(x,y) = (sd,td) = (1 \cdot 6,1 \cdot 6) = (6,6)$ is the only solution of equation (1). Conclusion: Equation (1) has exactly three integers solutions, namely $x=y=0$ and $x = y = \pm 6$.

First we notice the trivial solution $x=y=0$, and that both $x$ and $y$ have to have the same sign as $$LHS \geq 0 \Rightarrow x^3y^3 \geq 0$$We also notice that if $(x, y)$ is a solution, $(-x, -y)$ will also be a solution. In addition, as $x, y \in \mathbb{Z} \Rightarrow x^2+y^2 \in \mathbb{Z} \Rightarrow 2^33^4|x^3y^3$. Let’s start by assuming $y$ is not divisible by $2$ nor $3$, then: $x=2 \cdot 3^2 x’$, plugging it in the equation: $$3^42^2x’^2+y^2=3^2x’y$$Which is a contradiction therefore $3|x$ and $3|y$. Similarly assuming that $y$ is not divisible by $2$ we get a contradition that means that both $x$ and $y$ have to be divisible by $6$. Then we say $x=6x’$ and $y=6y’$. Plugging it in the equation we get this equality: $$2x’^3y’^3=x’^2+y’^2$$But for every case except $x’=y’=1$ $LHS \geq RHS$*. Therefore $x=y=6$ is a solution and $x=y=-6$ as well. Therefore the only solutions are $\boxed{(0, 0), (6, 6), (-6, -6)}$

arcf2x wrote: $LHS \geq RHS$* This can be proved like this: Assume $2xy<x^2+y^2$. Then: $$2xy < \frac{1}{x^2} + \frac{1}{y^2} \Rightarrow 2xy+ \frac{1}{2xy} < ( \frac{1}{x} + \frac{1}{y} )^2$$This means $$\frac{1}{x} + \frac{1}{y} > 2 \Rightarrow \frac{1}{x} + \frac{1}{y} < -2 $$As $x$ and $y$ have the same sign, we get that this last is impossible, therefore a contradiciton.

juanbri wrote: Find all pairs of integers $(x,y)$ that satisfy the equation $3^4 2^3(x^2+y^2)=x^3y^3$ Could you post your solution? I’m curious. @juanbri

juanbri wrote: Find all pairs of integers $(x,y)$ that satisfy the equation $3^4 2^3(x^2+y^2)=x^3y^3$ First note that $3|xy$, so, WLOG asume $3 |x$ and let $x=3a$. We get the following expresion $$3*2^3(9a^2+y^2)=a^3y^3$$ Then we know that $v_3(a^3y^3) = 3k$ for some $k$, and that $v_3(3*2^3(9a^2+y^2)) = v_3(3)+v_3(2^3) + v_3(9a^2+y^2)$ and so is clear that $9a^2+y^2 \equiv y^2 \equiv 0 \mod{9}$. This implies $3 | y$ and so we call $y = 3b$. We get the following expression $$2^3(a^2+b^2)=a^3b^3$$ Is now clear that $2|ab$ and so WLOG we asume $2|a$, and we call $a=2c$. The expression becomes $4c^2+b^2=c^3b^3$. From here we know that we can manipulate the expression as follows, $\frac{4c^2}{b^2}+1 = c^3b$ and $4+\frac{b^2}{c^2} = cb^3$. This implies that $b^2|4c^2$ and $c^2| b^2$ and so $v_p{(b)} \leq v_p(2)+v_p(c)$ and $v_p{(c)} \leq v_p{(b)}$. If $p$ is not 2 we get $v_p(b) = v_p(c)$. If $p=2$ then $v_2{(b)} \leq v_2(c) \leq v_2{(b)}+1$. This implies that either $b=c$ or $2b=c$. After this is trivial to check the solutions.