The problem is stated in the title.


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 $p^2$.


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.


2 Answers 2


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 $3$.

  • $\begingroup$ How does it follow from n,p and m being coprime that $p^2+m^2\in \{1,2\}$? $\endgroup$
    – Anna Naden
    Jul 30, 2020 at 10:22
  • $\begingroup$ 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. $\endgroup$ Jul 30, 2020 at 10:24

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