Solving $y^2(x^2+1) +x^2(y^2+16) =448$

Question

$y^2(x^2+1) +x^2(y^2+16) =448$

The task is to find all solutions in integers $(x,y)$. This is the fourth question of rmo 1st stage.The solution here is not complete. I have tried to solve unable to. enter link description here

Answer 1

First, this equation is symmetric under $x\to-x$ and $y\to-y$, independently. Therefore we can restrict attention to $x,y>0$ and such a solution will represent four actual solutions (we easily see that $x=0$ or $y=0$ are not possible integer solutions).

Solve with the quadratic formula to find $$y=4\frac{\sqrt{28-x^2}}{\sqrt{1+2x^2}}.$$ The allowed integer values of $x$ are $1$, $2$, $3$, $4$, $5$. Check these by hand: $x=1$ and $x=3$ give integer values of $y$, so there are eight solutions total: $$(\pm 1,\pm 12)$$ $$(\pm 3,\pm 4)$$

Answer 2

Note that all the terms on the left are positive or zero, so the way the problem is set up there can only be a finite number of solutions. Also if $(x,y)$ is a solution so are all of $(\pm x,\pm y)$ so we may as well restrict ourselves to $x,y\ge 0$.

Expanding the left-hand side we get $$2x^2y^2+y^2+16x^2=448$$

This means that $y$ must be even, say $2z$ so that $$8x^2z^2+4z^2+16x^2=448$$

or $$2x^2z^2+z^2+4x^2=112$$

Whence $z$ must be even, say $2w$ and $$2x^2w^2+w^2+x^2=28$$

So $x$ and $w$ are now both even, or both odd and no greater than $5$. $(1,3)$ works, and choosing for $w$ gives thge possibilities $(x,y)=(3,4), (1,12)$.