# Prove that no points on a circle of radius $\sqrt{3}$ can have both $x$ and $y$ coordinates rational

The problem is stated in the title.

$$x^2+y^2=3$$

Assume one coordinate is rational, i.e. $$y=\frac{n}{m}$$. Then $$x^2+\frac{n^2}{m^2}=3$$ , which implies : $$x=\sqrt{3-\frac{n^2}{m^2}}$$ $$\ \ \ \ \ =\frac{\sqrt{3m^2-n^2}}{m}$$ So for $$x$$ to be rational,

$$3n^2-m^2$$ has to be a perfect square $$p^2$$.

$$3n^2=p^2+m^2$$.

I need to prove that there are no integer solutions for $$n$$ , $$m$$ and $$p$$ in order to show that $$x$$ cannot be rational. I don't know how to do this as I have not mastered number theory yet. I found a discussion of this question in the post "When are $$x$$ and $$y$$ both rational..." but I don't understand the answer.

Suppose, for the sake of contradiction, there was such an integer solution $$p,m,n$$, and so there must exist some coprime solution because if they share a common factor it can just cancel out. Then in particular $$p^{2} + m^{2} \equiv 0$$ modulo $$3$$, or in other words $$p^{2} \equiv - m^{2}$$ modulo $$3$$. Now since $$-1$$ is not a square modulo $$3$$, it follows that $$p,m \equiv 0$$ modulo $$3$$, and so we can write $$p = 3p'$$, $$m = 3m'$$ for some integers $$p'$$, $$m'$$, and then $$n^{2} = 3({p'}^{2} + {m'}^{2})$$ and so $$n \equiv 0$$ modulo $$3$$ which contradicts that $$n,p,m$$ are coprime (unless of course $$p,n,m = 0$$).
Dividing $$n,p,m$$ by their gcd, you can suppose that $$n,p,m$$ are coprime.
But then $$p^2 + m^2 \in \{\bar 1, \bar 2\} \subseteq \mathbb Z_3$$ in contradiction with $$p^2+m^2=3n^2 \equiv \bar 0$$. This follows from the fact that the square of an integer is congruent to $$0$$ or $$1$$ modulo $$3$$.
• How does it follow from n,p and m being coprime that $p^2+m^2\in \{1,2\}$? Jul 30, 2020 at 10:22
• The only option for $p^2+m^2$ to be congruent to $0$ module $3$ would be to have both $p,m$ dividable by $3$. Which can't be if those are coprime. Jul 30, 2020 at 10:24