# Prove that the discriminant $D\equiv 0,\ 1 \pmod{4}$?

Let consider $P\in \mathbb{Z}[X]$ a monic polynomial of degree $3$ and of the form : $P=(X-x_1)(X-x_2)(X-x_3)$

First I will prove that $(x_1+x_2)(x_1+x_3)(x_2+x_3)\in \mathbb{Z}$.

I develop : $P=(X-x_1)(X-x_2)(X-x_3)=X^3-(x_1+x_2+x_3)X^2+(x_1x_2+x_1x_3+x_2x_3)X-x_1x_2x_3$

Using elementary symmetric functions I can write that : $P=X^3-\sigma_1X^2+\sigma_2X-\sigma_3$.

By developing and replacing : $(x_1+x_2)(x_1+x_3)(x_2+x_3)=\sigma_1\sigma_2-\sigma_3$.

As $P\in \mathbb{Z}[X]$ it means that $\sigma_{1,2,3}\in \mathbb{Z}$. That's why $(x_1+x_2)(x_1+x_3)(x_2+x_3)\in \mathbb{Z}$.

Now I must deduce that the discriminant $D$ of $P$ is $\equiv 0,\ 1 \pmod{4}$ but I don't see how to proceed.

Thanks in advance !

Likewise you proved it for the sums you have to prove that $D \in \Bbb Z$ where $D = [(x_1-x_2)(x_2-x_3)(x_3-x_1)]^2$. This is so because $D = \sigma_1^2\sigma_2^2-4\sigma_2^3-4\sigma_1^3\sigma_3-27\sigma_3^2+18\sigma_1\sigma_2\sigma_3$. We have $D \mod 4 = \sigma_1^2\sigma_2^2 + \sigma_3^2 +2\sigma_1\sigma_2\sigma_3 = (\sigma_1\sigma_2+\sigma_3)^2$.
• How do you know that $D=[(x_1-x_2)(x_2-x_3)(x_3-x_1)]^2$ ? By definition ? Nov 18, 2016 at 18:23
• Ok and why $D \not\equiv 2, 3 \pmod4$ ? There aren't square ? Nov 18, 2016 at 18:27
• Because it is a square of an integerr (mod 4): we have in $\Bbb Z_4$ : $0^2 = 0, 1^2 = 1, 2^2 = 0 , 3^2 = 1$ Nov 18, 2016 at 18:31