Positive integers $(x,y)$ their sum and product is a perfect square Is there any genuine approach to find pair of positive integers $(x,y)$ such that both their sum and product is a perfect square?
One pair is $(5,20)$ but it looks to me that this question can be solved by hit and trial only which is very difficult thing to do.
 A: Let $\gcd(x,y)=G$. 
Then, $x=aG, y=bG$ where $\gcd(a,b)=1$. Note that $xy=abG^2=t^2$
This implies that $a$ and $b$ are both squares. 
Let $a=X^2, b=Y^2$. 
Then, $G(X^2+Y^2)=k^2$. 
I) If $G$ is a square
This implies that $x,y$ are both squares, and that $X=(p^2-q^2), Y=2pq$ where $p$, $q$ are coprime integers and $p \not \equiv q \pmod 2$($\because$Pythagorean Triples).
II)If $G$ is not a square
Let $G=n^2m$ where $m$ is a squarefree integer.
This also implies that $X^2+Y^2=mS^2$
Note that if $X^2+Y^2=a^2b$, $b$ is a square free integer, and $b$ has only prime divisors of the form $4l+1$ (from Fermat`s Theorem on the Sum of two Squares) as $X,Y$ are coprime.
This implies that $m$ also has only prime divisors that are of the form $4l+1$, as $(X^2+Y^2)m$ is a square.
Since $m$ can be represented as a sum of two squares, let $m=a_1^2+a_2^2$
By multiplying $a_1^2+a_2^2$ to both sides, note that we must prove that $(a_1X-a_2Y)^2+(a_2X+a_1Y)^2=(mS)^2$
This implies that $(a_1X-a_2Y,a_2X+a_1Y,mS)$ are Pythagorean triples. 
I suggest you proceed with calculations from here. 
