$\frac{xy^2}{x+y}=p \iff xy^2=xp+yp \iff (x+y)(y^2-p)=y^3$ First cases if $y$ prime , we get \item $y^2-p=y \iff y(y-1)=p$ and $x+y=y^2 \iff x=y(y-1)$ so $x=p$. Subtitute to the first equation and we get $(x,y)=(2,2)$ satisfied. \item $y^2-p=1 \iff y^2-1=p$ and $x+y=y^3 \iff x=y(y^2-1)$ so $x=py$. Subtitute to the first equation and we get $(x,y)=(2,3)$ satisfied. Second cases if $y$ not prime but i'm still stuck in this cases hehe

kentok wrote: ...and we get $(x,y)=(2,3)$ satisfied. This doesn't satisfy since $\frac{18}{5}$ is not even a positive integer. Solution Let $\frac{xy^2}{x+y}=p \Rightarrow xp+yp=xy^2$, $x=dm$ and $y=dn$ where $(m, n)=1$. Then we have $d^3mn^2=dmp+dnp \Rightarrow d^2n^2m=p(m+n)$. $p$ must divide $d$, $n$ or $m$. i) Let $p|n$. This implies $p^2|n^2|LHS=RHS=p(m+n) \Rightarrow p|m+n$. But we have $(m,n)=1 \Rightarrow (n,m+n)=1$, contradiction. ii) Let $p|m$. So the definition gives us $pk=m$. If we write it in the equation, it concludes with $pk+n=d^2n^2k \Rightarrow k|n$. We have $k|m$ and $k|n$, then since they are coprime $k=1 \Rightarrow p=m$. Writing in the equation gives us $p+n=d^2n^2 \Rightarrow n|p$. $n$ must be $1$. So $x=dp$ and $y=d$. If we put these into original equation, we have $$p+1=d^2 \Rightarrow p=(d-1)(d+1) \Rightarrow d-1=1 \Rightarrow d=2 \Rightarrow p=3 \; x=6 \; y=2$$which is indeed a solution. iii) Let $p|d^2 \Rightarrow p|d \Rightarrow pd_1=d$. If we put this into equation, we have $pd_{1}^{2}n^2m=m+n$. For $n \geq 2$ and $m \geq 1$ it obviously $LHS>RHS$. So either $m=1$ or $n=1$ must hold. Let $m=1$. We have $$n+1=pd_{1}^{2}n^2 \geq 2n \Rightarrow n \le 1\Rightarrow n=1 \Rightarrow x=y \Rightarrow \frac{x^3}{2x}=\frac{x^2}{2}=p \Rightarrow x=y=p=2$$which is indeed a solution. Let $n=1$. Similarly we have $m=1$ and the same solution. So the only solutions are $(x, y) \in \{(2, 2), (6, 2) \}$

Let $\gcd(x,y)=t$ and $(x,y)=(ta,tb)$. Notice $\frac{t^2ab^2}{a+b}=p$. Since $\gcd(ab^2,a+b)=1$ we must have $a+b\mid t^2$ and also $ab^2\mid p$. Since $b^2\mid p$ we have $b=1$. Indeed $$\frac{t^2a}{a+1}=p$$If $a=1$ we have $t^2=2p$ so $4\mid 2p$ i.e $t=2$ and $p=2$. If $a=p$ we have $\frac{t^2}{p+1}=1$ indeed $(t-1)(t+1)=p$ implies $t=2$ and $p=3$. So all the solutions are $(x,y)=(ta,tb)=(2\cdot 1,2\cdot 1),(2\cdot 3,2\cdot 1)$ which are $(2,2),(6,2)$.