Euler's proof of $3m+1\in \mathbb{P}\implies 3m+1=x^2+3y^2$?

Asked Feb 22 '15 at 15:50

Active Feb 22 '15 at 16:19

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as the title suggests I am looking for Euler's descent proof of the fact that all primes of the form $3m+1$ can be represented as $x^2+3y^2$.

Note: I am not interested in any other proofs but Euler's one.

Thanks!

EDIT: This isn't a duplicate, I am specifically asking for Euler's proof using Fermat's descent.

edited Feb 22 '15 at 16:19

asked Feb 22 '15 at 15:50

Is Ne

2

possible duplicate of Let $p$ be prime and $(\frac{-3}p)=1$. Prove that $p$ is of the form $p=a^2+3b^2$ – Jack D'Aurizio Feb 22 '15 at 16:17
That was Euler's proof (look at the descent part). – Jack D'Aurizio Feb 22 '15 at 16:22
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The basic step of the descent is: there exists some $c$ such that $c^2+3=k\cdot p$, so you first have to prove that $p\equiv 1\pmod{3}$ implies that $-3$ is a square $\pmod{p}$. I proved that in a completely elementary way: just the Cauchy's theorem for groups. – Jack D'Aurizio Feb 22 '15 at 16:36

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