It is well known how to parameterize rational solutions to $x^2+y^2=a$ for $a\in \mathbb N^+$: first we know which $a$ admit integer solutions; our characterization is strong enough to tell us that theses are exactly the $a$ admitting rational solutions; then if we have one rational solution we can get all of them by drawing rational-slope lines through that one point.
I am interested in knowing if people understand well the algebraic solutions to $x^2+y^2=a$, i.e. the solutions $x,y\in \overline{\mathbb Q}$ to $x^2+y^2=a$.
In particular, I am interested in knowing if there exists an algebraic solution $(x_0,y_0)\in \overline{\mathbb Q} \times \overline{\mathbb Q}$ to $x^2+y^2=3$ such that the degree of the field extension $[\mathbb Q(x_0,y_0): \mathbb Q]$ is odd. (Indeed it has no solutions over $\mathbb Q$)
EDIT: a "famous" (in the sense of large number of upvotes/views) question on MSE in fact shows that $[\mathbb Q(x_0+iy_0):\mathbb Q]$ must be even. If $z_0:= x_0+iy_0$ was a root of unity (of say order $n$), then $z_0^{n-1} = \overline{z_0}$, and so $\mathbb Q(z_0) = \mathbb Q(z_0, \overline{z_0})= \mathbb Q(x_0,iy_0)$. Unfortunately, this does not give us what we want, and moreover, $z_0$ does not have to be a root of unity.
Theorem. Let $d \in \mathbf Q$ not be a square and $a$ be nonzero in $\mathbf Q$. If the equation $x^2 - dy^2 = a$ has a solution in an odd-degree extension of $\mathbf Q$ then it has a solution in $\mathbf Q$.
Therefore when $x^2 - dy^2 = a$ has no solution $(x,y)$ in $\mathbf Q$, it has no solution in any odd-degree extension of $\mathbf Q$.
Proof. This will be an application of transitivity of the norm map.
Suppose there's a number field $K$ with odd degree $n$ over $\mathbf Q$ such that $x^2 - dy^2 = a$ for some $x, y \in K$. Since $d$ is not a square in $\mathbf Q$ and $[K:\mathbf Q]$ is odd, $\sqrt{d} \not \in K$, so $[K(\sqrt{d}):K] = 2$. Then $$ a = x^2 - dy^2 = {\rm N}_{K(\sqrt{d})/K}(x + y\sqrt{d}). $$ Applying the norm map from $K$ down to $\mathbf Q$ to that equation: $$ a^n = {\rm N}_{K/\mathbf Q}({\rm N}_{K(\sqrt{d})/K}(x+y\sqrt{d})) = {\rm N}_{K(\sqrt{d})/\mathbf Q}(x+y\sqrt{d}). $$ We used transitivity of the norm map at the end of that calculation. Now going from $K(\sqrt{d})$ down to $\mathbf Q$ via the intermediate field $\mathbf Q(\sqrt{d})$ rather than the intermediate field $K$, $$ {\rm N}_{K(\sqrt{d})/\mathbf Q}(x+y\sqrt{d}) = {\rm N}_{\mathbf Q(\sqrt{d})/\mathbf Q}({\rm N}_{K(\sqrt{d})/\mathbf Q(\sqrt{d})}(x+y\sqrt{d})) . $$ Writing the norm ${\rm N}_{K(\sqrt{d})/\mathbf Q(\sqrt{d})}(x+y\sqrt{d})$ as $r + s\sqrt{d}$ where $r$ and $s$ are rational, $$ {\rm N}_{K(\sqrt{d})/\mathbf Q}(x+y\sqrt{d}) = {\rm N}_{\mathbf Q(\sqrt{d})/\mathbf Q}(r+s\sqrt{d}) = r^2 - ds^2, $$ so $a^n = r^2 - ds^2$. Since $n$ is odd, say $n = 2k+1$, we have $a = (r/a^k)^2 - d(s/a^k)^2$ since $a \not= 0$. Thus the equation $x^2 - dy^2 = a$ has a rational solution $(x,y) = (r/a^k,s/a^k)$. That proves the theorem.
While writing this up, I was pretty sure there is some general theorem about quadratic forms that has this result as a special case, and I thought Pete Clark may have written about it here or on MO once. A little searching led me to the desired result here. and Pete indeed posted an answer there. The general result about quadratic forms is called Springer's theorem.