Need a quick check on this proof.
Show that there are infinitely many primes congruent to 1 mod 3.
Proof: Suppose by contradiction finitely many primes $p≡1\pmod{3}$ say $p_1,…,p_k$.
Define $m=(2p_1\cdot\cdot\cdot p_k)^2+3$. So $m$ is odd therefore $m$ must be divisible by some odd prime $p$.
Then $(2p_1\cdot\cdot\cdot p_k)^2\equiv -3\pmod{p}$, so $-3\in Qp$ (quadratic residue of p) and hence $p\equiv 1\pmod{3}$.
So $p=p_i$ for some $i,…,k$. Then $m-(2p_1\cdot\cdot\cdotp_k)^2 =3$ so $p|3$.
Since the the only divisors of $3$ is $1$ or $3$, a contradiction since $p\equiv 1\pmod{3}$.
Best Answer
If you use the other bit of the hint that Dr H gave you, then the $-3 \in Q_p$ bit does a lot for you. (The proof looks correct to me though.)