:sad: The the condition of the original problem is $x^2|y^3+1,y|x^2+1$....... It cost me too much time....... :sad: For the orginal one. Case 1.y=x ,then y=x=1 Case 2,y<=x-1.because $x^2y|(x^2+1)(y^3+1)$ ,$x^2y|x^2+y^3+1$ $0<=x^2+y^3+1-x^2y=(1-x^2)y+y^3+1<=(1-(y-1)^2)y+y^3+1$ then $y<=2$ Case 3: y>=x+1,let $y_1=\frac{x^2+1}{y},x_1=x ,=>$Case2.

You have a mistake, $x^2+y^3+1 - x^2y = (1-x^2)y + y^3 + 1$ ?

vulalach wrote: You have a mistake, $x^2+y^3+1 - x^2y = (1-x^2)y + y^3 + 1$ ? That's fixable. $0 \le x^2+y^3+1 - x^2y = y^3+1 - x^2(y-1) \le y^3+1 - (y+1)^2(y-1) = y-y^2$, therefore $y=1$.

The problem is actually a transformation of problem no. 78 which means, $y \vert x^2 +1$ => $x^2+1=Ky$ .......(1) => $x^2=(Ky-1)|(y^3+1)$, where K is positive integer. This is nothing but problem no. 78 that is the question find all ordered pairs (m,n) such that $\frac{n^3+1}{mn-1}$ is integer where m,n are positive integers, which can be solved easily and the solution was given there anyone can see that and from the solution there we get $(K,y)= (2,2),(1,2),(2,1),(1,3),(3,1),(2,5),(5,2),(3,5),(5,3)$ are the only solutions and (1) gives us when (K,y)=(1,2) or (2,1) we have x=1 and when (K,y)=(2,5) or (5,2) we get x=3. Hence all solutions are (x,y)= (1,1),(1,2),(3,2),(3,5).

which also satisfy the conditions given