Proof verification: $x^2 + y^2 - 3 = 0$ has no rational points.

Question

This is clearly true, but I got this proof wrong. My proposed proof by contradiction: Suppose $$x^2 + y^2 = 3$$ has rational solutions. Let $$x = \frac{q}{r},\quad y = \frac{p}{n}$$ So, $$\frac{q^2n^2+p^2r^2}{n^2r^2} = 3$$ Therefore, $a^2 + b^2 = 3c^2$, where $a=qn, b = pr, c = nr$. We get $3 \mid a^2+b^2$. At this point, I wrote a short lemma how $3 \mid a^2+b^2 \implies 3 \mid a^2 \land 3 \mid b^2$. I am not including it because it was the prof's proof. I claim that a, b, and c are relatively prime, as if there was an integer that divided all 3, we would STILL have $a^2 + b^2 = 3c^2$, for some different integers a, b, c. The contradiction arises here, since $3 \mid a^2 \implies 3 \mid a$, so $a$ can be written as $3f$, and $b$ can be written as $3g$. Therefore, $3c^2 = 9f^2 + 9g^2, c^2 = 3(f^2 + g^2)$. So, $3 \mid c$, but a, b, c are relatively prime, so this is a contradiction. Apparently it was incorrect to assume a, b, c are relatively prime, why?

Answer 1

Let $x = \frac{q}{r}, y = \frac{p}{n}, \gcd(q,r)=1, \gcd(p,n)=1$.

At first, there is no integer solution, so $r > 1, n > 1$.

If $r=n$, then $\frac{q^2+p^2}{r^2}=3\Rightarrow q^2+p^2=3r^2$.

• If $r$ is odd, the right side is $3\text{odd}^2 \bmod 4 = 3$, so it's impossible.

• If $r$ is even, because $\gcd(q,r)=1,\gcd(p,n=r)=1$, so the left side is $\text{odd}^2+\text{odd}^2\bmod 4 = 2$, so it's also impossible.

If $r\ne n$, let $t = \gcd(r,n)$, so that $r=r't,n=n't,\gcd(r',n')=1$.

Then, $\frac{q^2}{r'^2t^2}+\frac{p^2}{n'^2t^2}=3$.

Hence, $\frac{q^2n'^2+p^2r'^2}{r'^2n'^2t^2}=3$.

Since $r\ne n$, at least one of $r'$ and $n'$ is larger than 1. Assume that $r' > 1$.

As we know, if $\frac{b}{a} \in \mathbb{Z}$, it means $\forall d | a \Rightarrow d| b$.

But $r' | r'^2n'^2t^2$, and $r' \nmid q^2n'^2+p^2r'^2$ because of $\gcd(q,r't)=1,\gcd(n',r')=1$.

So $\frac{q^2n'^2+p^2r'^2}{r'^2n'^2t^2} \not \in \mathbb{Z}$