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Suppose there is no rational points on $x^2 + y^2 = c$ where $c $ is a constant. If we plug in $x=0$, then there is no rational solution for $y^2=c$. Thus, $y=\sqrt{c}$ is irrational and we show that $\sqrt{c}$ is irrational.

This is the proof I saw in $x^2 + y^2 - 3 = 0 \implies 3^{1/2}$ is irrational..

Do we need to consider the case when both $x$ and $y$ are irrational? If not, why? Is it possible for both not be rational but $\sqrt{c}$ is rational? Thanks.

user665125
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  • By assumption $x^2+y^2=c$ has no rational solutions, this is defined to mean that there are no $x,y$ both rational that solve this equation. In other words if $(x,y)$ is a solution of this equation then at least one of $x,y$ is irrational. Since you pick $x=0$ which is rational then $y$ must be irrational. – Snaw Feb 16 '22 at 02:15
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    There’s no need to consider cases other than $(x, y) = (0, \sqrt{3})$; even cases where $x$ is rational and non-zero are irrelevant. If $\sqrt{3}$ were rational, this would be a rational point on $x^2 + y^2 = 3$, but (due to some other proof elsewhere), no rational points exist. By contradiction, $\sqrt{3}$ is irrational. – Theo Bendit Feb 16 '22 at 02:15
  • $x=y=\sqrt{2}, c=2$. – markvs Feb 16 '22 at 02:15

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