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Can we find all rational numbers $x,y$ such that $\sqrt{x}+\sqrt{y}=\sqrt{2013}$?

Certainly possible answers are $(2013,0)$ and $(0,2013)$.

If we square the equation, we get $x+y+2\sqrt{xy}=2013$, so $\sqrt{xy}$ must be rational.

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  • $\begingroup$ Just an idea, I don't know if it goes anywhere, $x,\sqrt{xy},y$ make a geometric progression, so we can write $y=xq^2$ and $q$ is rational. $\endgroup$ – hhsaffar Nov 12 '13 at 20:09
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if $x\neq 0$ then $x,\sqrt{xy},y$ make a geometric progression, so we can write $y=xq^2$ and $q$ is rational.

$x+y+2\sqrt{xy} = 2013$

$x + xq^2+2xq = x(q+1)^2=2013$

$x= \frac{2013}{(q+1)^2}, y=xq^2, q\in \mathbb Q^+\cup\{0\}$

if $x=0$ then $y = 2013$.

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    $\begingroup$ You have a typo in the last line, $y = q^2x$, not $= qx$. Also, you should probably exclude $q = -1$. Rather, methinks we need $q \geqslant 0$. $\endgroup$ – Daniel Fischer Nov 12 '13 at 20:21
  • $\begingroup$ Thanks Daniel, you are right, just fixed it. $\endgroup$ – hhsaffar Nov 12 '13 at 20:25
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$$\sqrt x+\sqrt y=\sqrt a\iff\sqrt\frac xa+\sqrt\frac ya=1\iff\sqrt\frac xa=1-\sqrt\frac ya\iff\frac xa=1+\frac ya-2\sqrt\frac ya$$ $$\sqrt\frac ya=\frac{1+\frac ya-\frac xa}2\in\mathbb{Q}\iff y=a\cdot\left(\frac mn\right)^2$$ Can you take it from here ? :-)

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Subtracting $x + y$ from both sides and squaring makes the problem equivalent to "find rational points on a circle". $$4xy = 2013^2 + x^2 + y^2 - 4026x - 4026y - 2xy$$ $$x^2 + y^2 - 2xy - 4026x - 4026y = 2013^2$$

Rotate by 45 degrees and scale by $\sqrt{2}$; substitute $x' = x - y, \ y' = x + y$. This does not change if a point is rational.

This gives you a circle, I think (will compute when not on phone). From there it should be possible to use the radical points on the unit circle?

EDIT: Ack, it's an parabola. Well, perhaps that is also a manageable form?

$$x^2 - 4026y = 2013^2$$

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