Show that if $p \equiv 1 \pmod 6$ then there exist integers $a$ and $b$ such that $p = a^2 + 3b^2$.

Let $p$ be a prime such that $p \equiv 1,3 \pmod 8$.

There exists $c \in \mathbb{Z}$ such that $c^2 \equiv −2 \pmod p$.

Define $L=\{(x,y) \in \mathbb{Z^2} \mid x\equiv cy \pmod p\}$ and we know that $(a,b) \in L \implies a^2+2b^2 \equiv 0 \pmod p$.

Let $S = \{(x,y) \in \mathbb{R^2} \mid x^2+2y^2<2p\}$. S has area $\pi p \sqrt{2}$ and there exist integers $a$ and $b$ such that $p = a^2 + 2b^2$ using Minkowski's theorem.

I have to use a similar argument to the above and use the hint below.

Hint: first prove that $2p = a^2 + 3b^2$ has no solutions for such primes $p$.

  • $\begingroup$ Do you know how Minkowski's theorem is applied to primes of the form $a^2+b^2$ or $a^2+2b^2$? $\endgroup$ – Wojowu Mar 5 '16 at 16:20
  • $\begingroup$ Yes I do know they exist. $\endgroup$ – Tony Mar 5 '16 at 16:21
  • $\begingroup$ So, what do you think? $\endgroup$ – S.C.B. Mar 5 '16 at 17:01
  • $\begingroup$ Maybe you can use quadratic forms... $\endgroup$ – Maman Mar 5 '16 at 19:11


The hint can be proven so: $2p \equiv 2 \equiv a^2 \pmod 3$, a contradiction.

Pigeonhole Principle

Also, note that since $-3$ is a quadratic residue, this implies there exists such $a$ that $a^2 \equiv -3 \pmod p$.

By Thue's Lemma, we get that there exists such $-\sqrt{p}< x,y < \sqrt{p}$ that $x \equiv ay \pmod {p}$.

This would imply there exists such $x,y$ that $0<x^2+3y^2<4p$ and $x^2+3y^2 \equiv 0 \pmod p$.

By Minkowski's theorem

$p \equiv 1$ $\bmod 6$, then $\left( \frac{-3}{p} \right) = 1$. Let $p$ be in one of these residue classes and let $a$ be a square root of $-3$ modulo $p$.

Let $\Lambda \subset \mathbb{Z}^2$ be the lattice of pairs $(x,y)$ such that $x \equiv ay \bmod p$.

For any $(x,y) \in \Lambda$, we have $x^2+3 y^2 \equiv (ay)^2 + 3 y^2 \equiv (-3+3) y^2 \equiv 0 \bmod p$. Since $\Lambda$ has index $p$ in $\mathbb{Z}^2$, the fundamental domain of $\Lambda$ has area $p$.

The ellipse $\{ (x,y) : x^2+3y^2 < N \}$ has area $\frac{\pi N}{\sqrt{3}}$. Now take $N =\frac{4\sqrt{3}}{\pi}p <3p$

  • $\begingroup$ $2p \equiv a^2 \pmod 3$ How did you get the $\equiv 2$? $\endgroup$ – Tony Mar 5 '16 at 16:26
  • $\begingroup$ @Tony because $p \equiv 1 \pmod 3$ $\endgroup$ – S.C.B. Mar 5 '16 at 16:27
  • $\begingroup$ So $p \equiv 1 \pmod 6$ implies $p \equiv 1 \pmod 3$? $\endgroup$ – Tony Mar 5 '16 at 16:28
  • $\begingroup$ @Tony Yes, that is true. Because $6|p-1$ implies $3|p-1$. $\endgroup$ – S.C.B. Mar 5 '16 at 16:29
  • $\begingroup$ Ok thanks. But how can I do this using Minkowski's theorem $\endgroup$ – Tony Mar 5 '16 at 16:34

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