Here is my solution: $(x+y)(x^2+9y)=p^3$ Note that: $x+y<x^2+9y$ Then: $\boxed {x+y=p , x^2+9y=p^2}$ $(x+y)^2=p^2=x^2+9y$ $x^2+2xy+y^2=x^2+9y$ $y(2x+y-9)=0$; but $y$ it cannot be $0$ $2x+y-9=0$, then if $x=1$ and $y=7$ $\implies (1,7)$ but $x+y=8$ if $x=2$ and $y=5$ $\implies (2,5)$ if $x=3$ and $y=3$ $\implies (3,3)$ but $x+y=6$ if $x=4$ and $y=1$ $\implies (4,1)$ $\boxed {x+y=1 , x^2+9y=p^3}$ but $x\geq 1$ and $y\geq 1$ $x+y \geq 2$ Thus $(x,y)=(2,5)(4,1)$