<s>[INST] 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.
 [/INST] you want a solution to the equation:
$$x^2+y^2=3$$
you could try proving that if this is a point on a circle, it cannot have both coordinates be integer, so say:
$$x=\sqrt{3}\cos\theta,y=\sqrt{3}\sin\theta$$
for $x$ to even be rational $\cos\theta$ must be be a rational multiple of $\sqrt3$ and similar for $y$. Now show there exists no value $0\le\theta< 2\pi$ where both $\cos,\sin$ are multiples of $\sqrt3$
</s>