Proving a curve has no curve point I am trying to show that the curve $x^2 + y^2 - 3 = 0$ has no rational point, where a rational point is defined as a solution $(x,y) \in \mathbb{Q}^2$ to this equation.
My attempt was to proceed by contradiction. Assume there is such a point $(x,y) \in \mathbb{Q}^2$ and let $x = \frac{a}{b}$, $y = \frac{c}{d}$ for $a,b,c,d \in \mathbb{Z}$, $b,d \neq 0$ where we assume without loss of generality that these fractions are fully reduced. Then we substitute:
\begin{align*}
\frac{a^2}{b^2} + \frac{c^2}{d^2} - 3 = 0.
\end{align*}
Rearranging and finding a common denominator of $b^2 d^2$ on the left:
\begin{align*}
\frac{a^2 d^2 + b^2 c^2}{b^2 d^2} = 3.
\end{align*}
We multiply through by $b^2 d^2$:
\begin{align*}
a^2 d^2 + b^2 c^2 = 3b^2 d^2.
\end{align*}
At this point, I'm stuck. I could try to argue that $3$ divides the left-hand side and apply Euclid's lemma, but I can't figure out how to get a contradiction out of the assumption that $\frac{a}{b}$ and $\frac{c}{d}$ have no common factors.

You can rearrange the equality to get $a^2 d^2 = 3 b^2 d^2 - b^2 c^2 = b^2(3 d^2 - c^2)$ which means that $b^2 | a^2 d^2 \implies b | ad$, but $\gcd(a, b) = \gcd(c, d) = 1$ means that this implies $b | d$, and by a similar argument $d | b$, but that only happens if $b = d$.
From there, you can reduce your equation to $a^2 + c^2 = 3b^2$. Then you can prove that this requires both $a$ and $c$ to be multiples of 3 and you're just about home.
