Prove that $x^2 + y^2=3$ has no solutions in the rational numbers.

Question

I know that there are some solutions in the reel numbers, but how can I prove that there are none in the rational ones?

Answer 1

If you clear denominators, this is the same as proving that $x^2+y^2=3z^2$ has no integer solutions except $x=y=z=0$.

Now if you have a solution where $x$ and $y$ are both multiples of $3$, then the right-hand side is then a multiple of $9$, so $3\mid z$ and we can divide through by $3^2$. Thus, without loss of generality we can assume that one of $x$ and $y$ is not a multiple of $3$.

Now view the equation modulo 3 to get $$ x^2 + y^2 \equiv 0 \pmod 3 $$ The only squares modulo $3$ are $0$ and $1$, and we're assuming that they are not both $0$, so the left-hand side is one of $1+0$, $1+1$, or $0+1$ -- and neither of these are $0$ modulo $3$. So the equation is impossible.

Answer 2

Let $x,y,z\in\mathbb Z$ be the minimal solution of $x^2+y^2=3z^2$ in the sense that $|x|+|y|+|z|$ is minimal among the non zero solutions. Then $3|x^2+y^2$ and so $3|x$ and $3|y$, thus $9|3z^2\implies 3|z$. So $\frac{x}{3},\frac{y}{3},\frac{z}{3}$ is a solution to the equation with $|\frac{x}{3}|+|\frac{y}{3}|+|\frac{z}{3}|<|x|+|y|+|z|$, contradiction. So the only solution is $x=y=z=0$.