Assume p triplebar 1 mod 3 is prime Show that 3 is a quadrat

Assume p triplebar 1 mod 3 is prime. Show that -3 is a quadratic residue mod p. Show that x^3 -1 = 0 in integernumberset_p. Conclude that z^2 + z + 1 =0. Let tau = 2z + 1. By direct calculation, show that tau^2 + 3 = 0 in integernumberset_p. Conclude.

Solution

Solution :

First, by quadratic reciprocity,

(p/3) (3/p) = (-1)^{[(p-1)/2][(3-1)/2]} = (-1)^(p-1)/2, so
(-1)^(p-1)/2(3/p)=(p/3).

Then, by the identities (ab/P) = (a/P) (b/P) and (-1/P) = (-1)^(p-1)/2,

(-3/p) = (-1/p)(3/p)=(-1)^(p-1)/2(3/p)=(p/3).

Now, to find the primes, p, for which -3 is a quadratic residue (mod p) -- which are also the primes, p, that are a quadratic residue (mod 3),

(p/3) = p^(3-1)/2 (mod 3) = p (mod 3)
if p = 6k+1, then p (mod 3)=1,

and

- 3 is a quadratic residue (mod p).