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Find all rational solutions to $x^2+y^2=2$

I rewrote the equation to $x= \sqrt{2-y^2}$ and thought that $x$ is rational if and only if $2-y^2$ is a square. So the only solution to the first problem is $x=1$ or $-1$ and $y=1$ or $-1$.

Is there another approach? Thank you.

Edit: Another question: Is there maybe a way to show this using elliptic curves?

Ng Chung Tak
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deavor
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    There are indeed no rational solutions to $x^2+y^2=3$. But you need to prove it. For $x^2+y^2=2$ you have found the integer solutions. There are infinitely many rational solutions, and you need to show how to generate them. – André Nicolas Jul 31 '16 at 20:18
  • I don't think this should be closed, as mentioned above in the comments this curve has rational solutions, so the methods for finding them are different than proving the other curve has no rational solutions. – Josh B. Jul 31 '16 at 21:13
  • @JoshB. I agree and voted for reopening – egreg Jul 31 '16 at 21:17
  • http://www.artofproblemsolving.com/community/c7t177f7h1277169_find_all_integer_solutions_of_a2__b2__2_c2 – individ Aug 01 '16 at 04:11

3 Answers3

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Given any rational $$ u^2 + v^2 = 1,$$ we find $$ (u-v)^2 + (u+v)^2 = 2 $$

Meanwhile, given any integer Pythagorean triple $$ a^2 + b^2 = c^2, $$ we get rational $$ \left( \frac{a}{c} \right)^2 + \left( \frac{b}{c} \right)^2 = 1 $$

From the comments above, I guess I should add that all primitive solutions in integers to $x^2 + y^2 = 2 z^2,$ since this means $x \equiv y \pmod 2$ and $z$ odd, so we get $$ \left( \frac{x-y}{2} \right)^2 + \left( \frac{x+y}{2} \right)^2 = z^2 $$ in integers and primitive. We can find all of these with the tandard parametrization for primitive Pythagorean triples.

Will Jagy
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  • Conversely, given rational $x,y$ with $x^2+y^2=2$, set $u=(x+y)/2$ and $v=(x-y)/2$ and $u^2+v^2=1$. – egreg Jul 31 '16 at 21:15
  • @egreg yes. Indeed, given integers $r^2 + s^2 = 2 t^2,$ we must have $r \equiv s \pmod 2,$ so we get integers $\left( \frac{r-s}{2} \right)^2 +\left( \frac{r+s}{2} \right)^2 = t^2, $ so this one comes out quite neatly over the integers. Given typical indefinite ternary quadratic form $F(x,y,z) = 0,$ we expect a finite number of integer parametrizations to get all integer triples...for Pythagorean triple we just need the one to get all primitive, $e^2 - f^2, 2ef, e^2 + f^2.$ Usually more than one is needed. – Will Jagy Jul 31 '16 at 21:26
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Since you already have a rational solution, one strategy is to find a line with rational slope which passes through that point. The line will also intersect the curve in a second spot, which gives another solution (unless the line is tangent to the curve.) This is similar to in elliptic curves when we find the line through 2 points to find a thrid point.

Let $m$ be a rational number, find the line with slope $m$ which passes through the point $(1,1)$:
$y=m*x+(1-m)$.
Plug this back into the curve to get
$x^2+(m*x+(1-m))^2=2$
Solve for $x$, to find $x=\frac{m^2-2m-1}{m^2+1}$.
Plug this back into the equation for the line to find $y=\frac{m^2+2m-1}{m^2+1}$
So, we get a rational solution for any rational number $m$.
Convince yourself that this generates every rational solution (except for the one with a vertical line, $(1,-1)$).

Josh B.
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2

Let $\cal P$ be the set of rational points in the circle $\Gamma:x^2+y^2=1$. It is well known that $\cal P$ is dense in $\Gamma$ and that $$(a,b)\in \cal P \iff \left(\frac{a+b}{\sqrt 2}, \frac{a-b}{\sqrt 2}\right)\in \Gamma$$ The second fact is because of the parametrization of the Pythagorean triples and the simple verification of $$\left(\frac{2st}{s^2+t^2}, \frac{ s^2-t^2}{ s^2+t^2}\right)\in \cal P \iff \left(\frac{s(2t+s)-t^2}{\sqrt 2(s^2+t^2)}\right)^2+\left( \frac{t(2s+t)-s^2}{\sqrt 2(s^2+t^2)}\right)^2=1$$ This gives a parametrization of the solutions and shows that the set of these is dense in the circle $\Gamma$ also (because to each element of $\cal P$ it is associated a solution and reciprocally). The set of solutions is for $t,s$ rational with $st\ne 0$

$$\begin{cases}x=\frac{s(2t+s)-t^2}{(r^2+t^2)}\\y=\frac{t(2s+t)-s^2}{(r^2+t^2)}\end{cases}$$

Piquito
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