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Can we find pairs $(x,y)$ of positive integers such that $x^2+3y$ and $y^2+3x$ are simultaneously perfect squares? Thanks a lot in advance. My progress is minimal.

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up vote 6 down vote accepted

Not a complete solution but an approach that seems like it will work.

Assume $y \gt x$

Then we have that

$(y+2)^2 \gt y^2 + 3y \gt y^2+3x \gt y^2$

If $y^2 + 3x$ was a perfect square, then we have that $y^2 + 3x = (y+1)^2$.

This gives us $3x = 2y+1$.

Substitute in the other expression, and form similar inequalities. This will narrow down to few small cases to consider.

The other case $y=x$ can be treated similarly.

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Well, it does work!However, can you please tell me the motivation behind the first step? – Eisen Nov 11 '11 at 18:37
Symmetry ($x^2+3y$ and $y^2+3x$) suggests that without loss of generality we can assume $y\ge x$. Also notice that if $x\sim y$, $y^2+3x$ is not much larger than $y^2$, so it may be equal to $(y+1)^2$ or $(y+2)^2$... To discuss this rigorously you need to assume $y\ge x$. – pharmine Nov 11 '11 at 19:22
@SabyasachiMukherjee: What pharmine said. The idea is similar to proving that $x^2 + K$ is a perfect square only 'few' times. Adding a linear multiple of $x$ is just transposing $x$ to $x+a$. – Aryabhata Nov 12 '11 at 0:15
How come this topic made it to the top of the list of asked questions?Does math.SE have some automatic system for bumping? – Eisen Dec 30 '11 at 8:45
@bgins: $x$, $y$ are positive integers, as stated in the problem. – Aryabhata Dec 30 '11 at 13:49

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