Show that there are no rational solutions on the ellipse $2x^2 + 3y^2 = 1$

Question

Show that there are no rational solutions on the ellipse $2x^2 + 3y^2 = 1$.

Want to make sure that my proof is correct.

Suppose there are rational solutions.

Note $2x^2 + 3y^2 = 1 \iff 2x^2 + 3y^2 = z^2\ |\ x,y.z \in \mathbb{Z}$

Since taking $(\frac{x}{z}, \frac{y}{z})$ as our rational solution, $2(\frac{x}{z})^2 + 3(\frac{y}{z})^2 = 1 \implies 2x^2 + 3y^2 = z^2$.

Now reducing $mod\ 3$, we have $2x^2 + 3y^2 \equiv 1 \ mod\ 3 \implies 2x^2 \equiv 1 \ mod\ 3 \implies -1x^2 \equiv 1\ mod\ 3 \implies x^2 \equiv -1\ mod\ 3 $, which is impossible.

Answer 1

You correctly arrived at the following: If there are rational points on this ellipse then the equation$$2x^2+3y^2=z^2\tag{1}$$ has integer solutions $\ne(0,0,0)$. But starting from here the argument is more intricate than you assumed.

If there are integer solutions of $(1)$ then after dividing them by a suitable power of $3$ we may assume that not all three of $x$, $y$, $z$ are divisible by $3$. Modulo $3$ we can rewrite $(1)$ as $$z^2+x^2\equiv0\qquad{\rm mod}\ 3\ .$$ Since all squares are $=0$ or $=1$ modulo $3$ this is only possible if both $x$ and $z$ are $\equiv0$ modulo $3$. But then $(1)$ would imply that $3y^2=z^2-2x^2$ is divisible by $9$, hence $y$ would be divisible by $3$ as well, which we have excluded.

Answer 2

As José Carlos Santos was pointing out in the comments, we have $$2x^2 \equiv z^2 \mod 3$$ But this does not imply your statement that $$2x^2 \equiv 1 \mod 3$$ If $z$ were divisible by $3$ and we would have $$2x^2 \equiv 0 \mod 3$$

Answer 3

One minor, easily fixed, problem with your proof is that you are using $x$ and $y$ first to denote rational numbers in the equation $2x^2+3y^2=1$, and then to denote the numerators of those rational numbers. It would be better to write $x=a/c$ and $y=b/c$ and then consider the equation $2a^2+3b^2=c^2$ with $a,b,c\in\mathbb{Z}$ (and $c\not=0$).

What's missing is the tacit assumption $\gcd(a,c)=1$. As José Carlos Santos and Isaac Browne point out, from $2a^2\equiv c^2$ mod $3$, it follows that $3\mid c$ implies $3\mid a$ (and vice versa), so you can assume $c\equiv\pm1$ mod $3$ (since those are the only other residue classes). It follows that $2a^2\equiv(\pm1)^2=1$, which now falls in line with the rest of your proof.

Just to give an alternative proof that there are no integer solutions to $2a^2+3b^2=c^2$ (with $c\not=0$), the (tacit) assumption $\gcd(b,c)=1$ tells us $b$ and $c$ are not both even, so $3b^2\equiv c^2$ mod $2$ tells us both are odd. Working mod $8$ now tell us $c^2\equiv1$ mod $8$, while

$$2a^2+3b^2\equiv \begin{cases} 3\text{ if }a\text{ is even} \\5\text{ if }a\text{ is odd}\end{cases}$$