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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 $p$.

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Tony
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1 Answers1

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HINT

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 $N =\frac{4\sqrt{3}}{\pi}p <3p$

S.C.B.
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  • $2p \equiv a^2 \pmod 3$ How did you get the $\equiv 2$? – Tony Mar 05 '16 at 16:26
  • @Tony because $p \equiv 1 \pmod 3$ – S.C.B. Mar 05 '16 at 16:27
  • So $p \equiv 1 \pmod 6$ implies $p \equiv 1 \pmod 3$? – Tony Mar 05 '16 at 16:28
  • @Tony Yes, that is true. Because $6|p-1$ implies $3|p-1$. – S.C.B. Mar 05 '16 at 16:29
  • Ok thanks. But how can I do this using Minkowski's theorem – Tony Mar 05 '16 at 16:34
  • @Tony I never read that proof by Minkowski's theorem thoroughly, but can't you just exchange $2$ for $3$(or so on)? I don't know that much about it. Why don't you see how they proved it for $-5$ http://math.stackexchange.com/questions/1069157/if-p-equiv-1-9-pmod20-is-a-prime-number-then-there-exist-a-b-in-mathbbz/1247186#1247186. – S.C.B. Mar 05 '16 at 16:45
  • How does the hint help also? Read my op (I have edited some info in) – Tony Mar 05 '16 at 16:48
  • @Tony Does this help? – S.C.B. Mar 05 '16 at 16:51
  • @Tony Since $x^2+3y^2 \equiv 0 \pmod p$, and it is between $0$ and $4p$ this implies $x^2+3y^2 =p$ or $3p$. – S.C.B. Mar 05 '16 at 16:57
  • @Tony But $3p=x^2+3y^2$ would imply $x=3t$. So, divide by $3$. – S.C.B. Mar 05 '16 at 16:58
  • The last bit should say less than $4p$? And (-3/p)=1 is the same as $a^2 \equiv -3 \pmod p$? – Tony Mar 05 '16 at 17:02
  • No, $\frac{4\sqrt{3}}{\pi} <3$. You can do this on a calculator, or with pencil and paper, using that $\sqrt 3$ is about $1.8$ and $\pi$ about $3.14$. This is because Minkowski is slightly more powerful than Pigeonhole. And yes, being a quadratic residue implies $a^2 \equiv -3 \pmod 3$. – S.C.B. Mar 05 '16 at 17:04
  • My argument was to use $N=2p$. I have area of $S = \frac{2p \pi}{\sqrt{3}} <2^2p=4p$ – Tony Mar 05 '16 at 17:05
  • Can't you say that Minkowski's theorem shows that $\Lambda$ contains a point $(x,y)$ with $0 < x^2+5y^2 \le \frac{4\sqrt{3}}{\pi}p <3p$. – S.C.B. Mar 05 '16 at 17:08
  • I am going to edit my OP with my answer for the case $p=a^2+2b^2$, because I don't see why I need the hint for this case. – Tony Mar 05 '16 at 17:09
  • Which Hint? (to reach comment length) – S.C.B. Mar 05 '16 at 17:09
  • @Tony My hint is for the problem you asked. Which case are you talking about? Please clarify. – S.C.B. Mar 05 '16 at 17:11
  • @Tony It is also right now 3 o'clock in the morning where I live, and I need to sleep. Can't you please tell me later? – S.C.B. Mar 05 '16 at 17:13
  • I have edited it in. – Tony Mar 05 '16 at 17:14
  • Why does the hint provided in the question help? – Tony Mar 05 '16 at 17:15
  • @Tony Ahh! So that's were you were asking! Because it is less than $x^2+3y^2 <3p$, there is also the possibility $x^2+3y^2=2p$. The hint is to remove this case. – S.C.B. Mar 05 '16 at 17:17
  • Ah I now see. Thanks a lot. – Tony Mar 05 '16 at 17:18