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.

  • $\begingroup$ 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

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 $y = 2013$.

  • 1
    $\begingroup$ 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

$$\sqrt x+\sqrt y=\sqrt a\iff\sqrt\frac xa+\sqrt\frac ya=1\iff\sqrt\frac xa=1-\sqrt\frac ya\iff\frac xa=1+\frac ya-2\sqrt\frac ya$$ $$\sqrt\frac ya=\frac{1+\frac ya-\frac xa}2\in\mathbb{Q}\iff y=a\cdot\left(\frac mn\right)^2$$ Can you take it from here ? :-)


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?

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


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.