8
$\begingroup$

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.

$\endgroup$
0
8
$\begingroup$

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.

To elaborate, given $3x = 2y+1$

$x^2 + 3y = x^2 + \frac{9x-3}{2} \lt (x+3)^2$

Thus $x^2 + 3y$ is either $(x+1)^2$ or $(x+2)^2$.

Thus we have that

$3y = 2x+1$ or $3y = 4x + 4$.

Substitute $y = \frac{3x -1}{2}$ and solve the linear equation in $x$ and compute $y$. In the end, don't forget to verify that both the expressions are indeed perfect squares.

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

$\endgroup$
10
  • $\begingroup$ Well, it does work!However, can you please tell me the motivation behind the first step? $\endgroup$ – Eisen Nov 11 '11 at 18:37
  • 2
    $\begingroup$ 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 $y\ge x$. $\endgroup$ – pharmine Nov 11 '11 at 19:22
  • 1
    $\begingroup$ @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 $x+a$. $\endgroup$ – Aryabhata Nov 12 '11 at 0:15
  • $\begingroup$ How come this topic made it to the top of the list of asked questions?Does math.SE have some automatic system for bumping? $\endgroup$ – Eisen Dec 30 '11 at 8:45
  • 1
    $\begingroup$ @bgins: $x$, $y$ are positive integers, as stated in the problem. $\endgroup$ – Aryabhata Dec 30 '11 at 13:49
1
$\begingroup$

$x^2+3y=a^2$ $y^2+3x=b^2$ $a,b \in \mathbb{N}$ assume wlog $x\ge y$ $a^2> x^2$ $a > x$ but since $x\ge y$ we have $x+2>a$

so $a=x+1 $

$3y=2x+1$

$x=3k+1 ,y=2k+1$

$4k^2+4k+1+9k+3=b^2$ solving for k we get

$16b^2+105=t^2$ is a perfect square

$b=1,2,4,13$

so $b=2,4,13$ and $x=y=1$ or $x=16 $, $y=11 $ and if $y\ge x$ we can get $x=11 $, $y=16$

Done!!!!

$\endgroup$
1

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.