My try $(2n^2+5m-5n-mn)^2=m^3n\Longrightarrow$ either $m,n\le0$, or $m,n\ge0$ 1) Case with $m=n$ Replacing $m$ with $n\Longrightarrow n^4=n^4\Rightarrow$ infinite solutions $(n,m)=(t,t), (-t,-t)$ 2) Case with $m\neq n$ The LHS is a perfect square, $m^3n$ should also be a square. This occurs if $mn$ is a square. Since $m\neq n$, assuming, in the simplest case that $m=nk^2\Rightarrow mn=n^2k^2$ Subst. $m=nk^2$ in the title equation $\Longrightarrow (2n^2+5nk^2-5n-n^2k^2)^2=n^4k^6 \Longrightarrow 2n^2+5nk^2-5n-n^2k^2=\pm n^2k^3$ 2.1) Case with "$+n^2k^3$" in the RHS $\Longrightarrow n^2(k^3+k^2-2)=5n(k^2-1)\Longrightarrow n=\frac{5k^2-5}{k^3+k^2-2}$ $\Longrightarrow \frac{\mid 5k^2-5\mid}{\mid k^3+k^2-2\mid}\ge 1\Longrightarrow -5\le k\le 3$ $k=-5\Longrightarrow n=\frac{5k^2-5}{k^3+k^2-2}=-\frac{60}{51}$ $k=-4\Longrightarrow n=\frac{5k^2-5}{k^3+k^2-2}=-\frac{75}{50}$ $k=-3\Longrightarrow n=\frac{5k^2-5}{k^3+k^2-2}=-2, m=nk^2=(-2)(-3)^2=-18$ $k=-2\Longrightarrow n=\frac{5k^2-5}{k^3+k^2-2}=-\frac{15}{6}$ $k=-1\Longrightarrow n=\frac{5k^2-5}{k^3+k^2-2}=0$ $k=0\Longrightarrow n=\frac{5k^2-5}{k^3+k^2-2}=\frac{5}{2}$ $k=1\Longrightarrow n=\frac{5k^2-5}{k^3+k^2-2}=0$ $k=2\Longrightarrow n=\frac{5k^2-5}{k^3+k^2-2}=\frac{15}{10}$ $k=3\Longrightarrow n=\frac{5k^2-5}{k^3+k^2-2}=\frac{40}{34}$ In conclusion, one solution $(n,m)=(-2,-18)$ 2.2) Case with "$-n^2k^3$" in the RHS $\Longrightarrow n^2(k^2-k^3-2)=5n(k^2-1)\Longrightarrow n=\frac{5k^2-5}{k^2-k^3-2}$ $\Longrightarrow \frac{\mid 5k^2-5\mid}{\mid k^2-k^3-2\mid}\ge 1\Longrightarrow -5\le k\le 5$ Same solution as above $(n,m)=(-2,-18)$ for $k= -3, k=3$ @ below: fixed it

iniffur wrote: (WLOG $m>n),\Longrightarrow m=nk^2\Rightarrow mn=n^2k^2$ Subst. $m=nk^2$ in the title equation 1) you can not say "WLOG $m>n$" since there is no symetrie 2) $mn$ perfect square does not imply $m=nk^2$ (look for example at $(m,n)=(18,8)$)

$(2n^2+5m-5n-mn)^2=m^3n$ $(2n^2+5m-5n-mn)^2-n^4=m^3n-n^4$ $(n-m)(n-5)(3n^2+5m-5n-mn)=n(m-n)(m^2+mn+n^2)$ $m=n$ is solution Let $m \neq n$ $(n-5)(3n^2+5m-5n-mn)+n(m^2+mn+n^2)=0$ $m^2 n + 10 m n - 25 m + 4 n^3 - 20 n^2 + 25 n=0$ Solve it for $m$ $D=(10n-25)^2-4n(4n^3-20n^2+25n)=(10n-25)^2-(4n^2-10n)^2=-(2n-5)^2(2n+5)$ $D \geq 0 \to 25-4n^2 \geq 0 \to -2 \leq n \leq 2$ and as $25-4n^2$ should be square than $n^2=0,4$ $n=0 \to m=0$ $n=-2 \to m=-18$ $n=2 \to m=2$ Answer: $ (m,n)=(-18,-2),(t,t)$ for any integer $t$