I have been casually reading a set of notes (page 30 in reader) on number theory, but I am not certain on one of the steps in the reasoning. Here is a quote:
What can we say about $w^2 ≡ −3\mod 4n$? Let’s study the case when $n = p$ is an odd prime. By the Chinese Remainder Theorem, there is a solution to $w^2 ≡ −3\mod 4p$ iff there is a solution to:
$$w^2\equiv -3 \mod 4$$ $$w^2 \equiv -3 \mod p$$
There is clearly a solution to $w^2 ≡ −3\mod 4$, namely $w = 1$. So we can concentrate on $w^2 ≡ (−3)\mod p$. This has a solution iff $p=3$ or $\Big(\frac{-3}{p}\Big)=1$.
The author then simplifies the Legendre symbol to deduce that $p=3$ or $p \equiv 1 \mod 3$. However, this solution is not going to be $w=1$, as it was for the first congruence. As far as I can see, what the author proves is that there are solutions $u,v$ such that:
$$u^2\equiv -3 \mod 4$$ $$v^2 \equiv -3 \mod p$$
for $p$ as specified, but it is not clear to me how these separate existence results imply a solution to $w^2 ≡ −3\mod 4p$ even with the Chinese Remainder Theorem. Isn't it also necessary to show that at least one of the solutions $v$ to the congruence mod $p$ is congruent to $\pm 1 \mod 4$?
