$p(x + y) = y^2x$.it follows that $p|y$ or $p|x$. Suppose that both of them are divisable by $p$. so there exist $a,b$ such that $p^{a}||y,p^{b}||x$. there are 2 cases: 1-$a\neq{b}$ then from the above equation we'll have $1 + \min{a,b} = 2a + b$ which is absurd. 2-$a = b$ then we'll have $(\frac {x}{p^{a}} + \frac {y}{p^{a}}) = p^{a - 1}(\frac {y^2}{p^{2a}})(\frac {x}{p^{a}})$ so if $\frac {y}{p^{a}} = m,\frac {x}{p^{a}} = n$ we'll have $m^2n\leq{m + n}$.we can easily find all the solutions. so assume that $p$ divides only one of them. u can see that in this case we must have $p|x$. if $px' = x$ u can see that $y |x'$ it follows $yt = x'$. then we'll have:$pt + 1 = y^2t$ the rest follows. i wish i didn't make any mistake I MUST GO TO BED!!

$p=\frac {y^2x}{x + y} \iff px+py=y^2x$ Taking mod $x$ then $x|py$ and taking mod $y$ then $y|px$ (1) $py=ax$ (2) $px=by$ Taking mod $p$ then $p|y^2x$ then either $p|x$ or $p|y$ If $p|x$ then $x=kp$ then in (1) $py=akp$ then $y=ak$ and in (2) $pkp=bak$ then $p^2=ab$ then three cases: $(a,b)=(1,p^2)$ or $(p^2,1)$ or $(p,p)$ If $p|y$ then $y=kp$ then in (2) $px=bkp$ then $x=bk$ and in (1) $pkp=abk$ then $p^2=ab$ then same three cases. Case 1) $py=x$ $p=\frac{y^3p}{y(p+1)} \iff p+1=y^2 \iff p=(y+1)(y-1) \Rightarrow y=2$ then $\boxed{(x,y,p)=(6,2,3)}$ Case 2) $y=px$ $p=\frac{p^2x^3}{x(p+1)} \iff p(p+1)=x^2 \Rightarrow p|x \Rightarrow x=cp \Rightarrow p+1=pc^2 \Rightarrow 1=p(c+1)(c-1)$ no solution Case 3) $y=x$ then $p=\frac{x^3}{2x}=\frac{x^2}{2}$ then $\boxed{(x,y,p)=(2,2,2)}$

lambruscokid wrote: Find all positive integers $(x,y)$ such that $\frac{y^2x}{x+y}$ is a prime number $(x,y)=d$ $x=ad , y=bd$ $\frac{d^{2}×b^{2}×a}{a+b}$ it must be prime $(a,a+b)=1$ and $(b,a+b)=1$ then $\frac{d^2}{a+b}$$=p$ it must be integer.$p×a×b^{2}$ is prime then $b=1$ $\frac{d^2}{a+1}$$=p$ then $p×a$ is prime $a=1$ or $a$ is prime and $p=1$ $p=1$ then $d^2=a+1$ $a=d^2-1$ and $a$ is prime then $a=3$ $d=2$ $\boxed{(x,y,q)=(2,2,2),(6,2,3)}$