Final Round

We have $2xy=a^2$ for some nonnegative integer $a$, and $x^2+y^2=p$ for some prime number $p$. Since a prime number is never a perfect square, we see that $x,y\ne 0$. Since $2xy$ is a perfect square, it follows that $x$ and $y$ must both be positive, or both be negative. If $(x,y)$ is a solution, then so is $(-x,-y)$. Therefore, we may for now assume that $x$ and $y$ are positive, and at the end, add for each solution $(x,y)$ the pair $(-x,-y)$ to the list of solutions. Combining $2xy=a^2$ and $x^2+y^2=p$ yields $(x+y{)}^2=x^2+y^2+2xy=p+a^2$. By bringing $a^2$ to the other side, we find $$p=(x+y{)}^2-a^2=(x+y+a)(x+y-a).$$ Since $x+y+a$ is positive, also $x+y-a$ must be positive. The prime number $p$ can be written as a product of two positive integers in only two ways: $1\cdot p$ and $p\cdot 1$. Since $x+y+a\ge x+y-a$, we obtain $x+y+a=p$ and $x+y-a=1$.

Adding these two equations, we get $2x+2y=p+1$. We also know that $x^2+y^2+1=p+1$, so $2x+2y=x^2+y^2+1$. By bringing all terms to the right-hand side and adding 1 to both sides, we obtain $$1=x^2+y^2-2x-2y+2=(x-1{)}^2+(y-1{)}^2.$$ We now have two perfect squares that add up to 1. This implies that one of the squares is 0 and the other is 1. So $(x-1{)}^2=0$ and $(y-1{)}^2=1$, or $(x-1{)}^2=1$ and $(y-1{)}^2=0$. As $x$ and $y$ are positive, we find two possible solutions: $x=1$ and $y=2$, or $x=2$ and $y=1$. In both cases $2xy=4$ is a perfect square and $x^2+y^2=5$ is a prime number. It follows that both are indeed solutions. Adding the solutions obtained by replacing $(x,y)$ by $(-x,-y)$, we obtain a total of four solutions $(x,y)$, namely $$(1,2),(2,1),(-1,-2),(-2,-1).$$