Exercise 1.10 from Silverman "The Arithmetic of Elliptic Curves " I am having trouble with Silverman's exercise 1.10(b). The converse of (a) is easy because there is no integer solution to the equation when $p \equiv 3$ mod $4$. However, this method does not work for (b) because we always get empty set when intersecting whatever $V_p (p \equiv 3 $ mod $4)$ with $\mathbb{Q}$. I tried to intersect two varieties $V_p, V_q$ with something else, for example $\mathbb{Q}({\sqrt{p}})$ and try to tell the difference, but this is hard since it is very possible for $V_q$ to have nonempty intersection with $\mathbb{Q}({\sqrt{p}})$ and I think it is true that if so then $V_p, V_q$ are isomorphic over $\mathbb{Q}({\sqrt{p}})$. Are there some invariants I can use in this circumstance to show that two varieties are not isomorphic over $\mathbb{Q}$?
Using the Quadratic Reprocity Law, $\Big(\frac{p}{q}\Big)\Big(\frac{q}{p}\Big)=(-1)^\frac{(q-1)(p-1)}{4}=-1$, which implies that one of them (let's say p) is a square mod the other one, so $x^2+y^2=pz^2$ has solution in $\mathbf{F}_{q}$  because the equation $x^2+y^2=z^2$ has non trivial solution in $\mathbf{Z}$ but $x^2+y^2=qz^2$ has no solution in $\mathbf{F}_{q}$ because $q\equiv3\mod 4$ so they cannot be isomorphic over $\mathbf{Q}$.
You could calculate the Hilbert symbols $(\frac{p,-1}{q})_2$ for varying primes $q$.