$ 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 $a=3$ $d=2$ $\boxed{(x,y,q)=(2,2,2),(6,2,3)}$