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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 $p\equiv 1\pmod{3}$.

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    $\begingroup$ Quadratic reciprocity seems too much here. $\endgroup$ – lhf Apr 28 '17 at 19:04
  • $\begingroup$ @lhf I guess but that was one of the hints given. I adapted this proof from a similar one from the book I'm using. $\endgroup$ – StillLearning Apr 28 '17 at 19:07
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    $\begingroup$ I believe quadratic reciprocity is actually needed for this particular residue class. Or at least the special case of which primes can be written in the form $x^2+3y^2$. (The parallel is to proving there are infinitely many primes $p\equiv1\pmod4$, where Fermat's theorem on the representation of integers in the form $x^2+y^2$ is used.) There seems to be a fundamental difference between $2\pmod3$ and $1\pmod3$ in this way (similarly, between $3\pmod4$ and $1\pmod4$). $\endgroup$ – Greg Martin Apr 28 '17 at 20:07
  • $\begingroup$ @GregMartin So in textbook I'm using it proves $p\equiv 1\pmod{4}$. I pretty much used that proof almost exactly except for the contradiction. Do I have to elaborate more on $-3\in Q_p$? $\endgroup$ – StillLearning Apr 28 '17 at 20:17
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Your proof by contradiction is correct.

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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.)

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