1

Prove that $\left(\frac{-3}{p}\right)=1$ if $p\equiv 1 \pmod 3$.

Instead of using Gauss' law of quadratic reciprocity it should be provable using the cyclic property of $(\mathbb{Z}/p\mathbb{Z})^{\times}$.

The hint of the proof is that there is an element $c$ of order $3$ in the group, then show $(2c+1)^2=-3$.

The first question is about the element $c$. How can be sure the existence of such $c$?

The second question: $(2c+1)^2=4c^2+4c+1$, then what? I could multiply the term by $c$ in order to get a third power and use $c^3=1$, but it lead nowhere.

Cure
  • 4,051
  • Have a look at my answer here: http://math.stackexchange.com/questions/685958/let-p-be-prime-and-frac-3p-1-prove-that-p-is-of-the-form-p-a23b2 – Jack D'Aurizio Oct 19 '16 at 17:14

1 Answers1

2

The key property of cyclic groups need for the first part is that if $C_n$ is a cyclic group of order $n$, then it has an element of order $d$ for each divisor $d$ of $n$. Indeed, if $x$ generates $C_n$ then $x^{\frac{n}{d}}$ has order $d$.

In your case, since $p\equiv 1$ (mod $3$) it follows that $3$ divides $p-1$. Now $(\mathbb{Z}/p\mathbb{Z})^{\times}$ is a cyclic group of order $p-1$, so by the above we know that it has an element $c$ of order $3$.

Finally, since $c$ has order $3$ it follows that $(c-1)(1+c+c^2)=c^3-1=0$ in $\mathbb{Z}/p\mathbb{Z}$, and since $c\neq 1$ we get $1+c+c^2=0$, so $c^2+c=-1$. Therefore $$ (2c+1)^2=4c^2+4c+1=-4+1=-3 $$ as claimed.

carmichael561
  • 53,688