4

I'm stuck on a problem and I'm wondering if I'm missing something obvious.

Let $p \equiv 2 \mod 3$ be an odd prime. The problem is to show that every value is a cubic residue mod $p$.

So far I have reduced the problem to showing that for $a \not\equiv b \mod p$, it is never true that $a^2 + ab + b^2 = 0 \mod p$. I originally felt like there was some way to apply CRT, but I couldn't find a way. I see that $a^2 + ab + b^2$ is always $0$ or $1 \mod 3$, never $2$. And I'd like to apply that fact. I rewrote the original modular equation as $a^2 + ab + b^2 = pm = (6k - 1)m$ for some integers $k$ and $m$, but then I get an "$m$" in there that I don't want.

I also tried to look for a pattern in the values of $a$ and $b$ that make $a^2 + ab + b^2 \equiv 0 \mod p$ when $p \not\equiv 2 \mod 3$, but I couldn't see any obvious pattern in the small primes, and it didn't look promising enough for me to go to the trouble of generating a large table of values. Although, if I don't have any more ideas, that will be my next step.

Tac-Tics
  • 2,203

2 Answers2

3

Hint:

Since $\;|\Bbb F^*_{p}|=p-1\neq0\pmod 3\;$ , the map $\;x\mapsto x^3\;$ is an automorphism of the (cyclic) group $\;\Bbb F_p^*\;$ .

DonAntonio
  • 211,718
  • 17
  • 136
  • 287
3

A couple of ways to see this.

You can, indeed, show that $a^2+ab+b^2\equiv0$ cannot happen non-trivially. If you set $x=a/b$ here, you get $$ x^2+x+1=0 $$ with $x\in\Bbb{F}_p$. But the discriminant of that quadratic is $-3$ so it has no solutions unless $-3$ is a quadratic residue modulo $p$. That implies $p\equiv1\pmod3$.

A possibly easier way is to use the fact that $\Bbb{F}_p^*$ is cyclic (equivalently, there is a primitive root $g$ modulo $p$). If $3\nmid p-1$ then cubing in that group is injective, hence surjective.

Jyrki Lahtonen
  • 133,153
  • Ack. I had even looked at $1 + x + x^2$, but I was trying to show when it was 0 rather than when it can't be. I'll take a deeper look at this. The book I'm using, by the way, is Silverman's A Friendly Introduction to Number Theory. It does not assume any group theory or even knowledge of what a finite field is, so I'm also curious in finding a correspondingly "friendly" proof. – Tac-Tics Oct 21 '16 at 18:22
  • @Tac-Tics: Using a primitive root $g$ it would go as follows. Assume that $a=g^t$. Then either $t, t+(p-1)$ or $t+2(p-1)$ is divisible by three. So we can write $t+k(p-1)=3\ell$ for some integers $k,\ell$. Then, by Little Fermat, $$a\equiv g^t\equiv g^{t+k(p-1)}\equiv g^{3\ell}\equiv (g^\ell)^3,$$ so $a$ is a cubic residue. – Jyrki Lahtonen Oct 21 '16 at 18:36