# Thread: Determine solvablity

1. ## Determine solvablity

Prove that $\displaystyle x^{2}+2\equiv0 \mod p$ is unsolvable if $\displaystyle p\equiv 5,7\mod 8$ .
Thanks

2. Originally Posted by chipai
Prove that $\displaystyle x^{2}+2\equiv0 \mod p$ is unsolvable if $\displaystyle p\equiv 5,7\mod 8$ .
Thanks
Can we use Gauss's Lemma ?

To determine whether the given number $\displaystyle a \neq 0$ is quadratic residue , consider the list :

$\displaystyle a , 2a , 3a , ..., Pa ~,~ P = \frac{p-1}{2}$

If the number of the elements in the above list lying between $\displaystyle P+1$ and $\displaystyle 2P$ inclusively is even , then $\displaystyle a$ is quadratic residue , otherwise , it is quadratic non-residue .

In this case , $\displaystyle a \equiv -2 \bmod{p}$ and $\displaystyle P$ is either $\displaystyle 2\bmod{4}$ or $\displaystyle 3 \bmod{4}$

The elements lying between the range should be :

$\displaystyle -2 , -4 , .... , (-2) \frac{P}{2}$ for $\displaystyle P \equiv 2 \bmod{4}$ so the number is $\displaystyle \frac{P}{2}$ an odd number .

Or

$\displaystyle -2 , -4 , ..., (-2) \frac{P-1}{2}$ for $\displaystyle P \equiv 3 \bmod{4}$ so the number is $\displaystyle \frac{P-1}{2}$ also an odd .

Therefore , $\displaystyle - 2$ is not the quadratic residue of modulo $\displaystyle p \equiv 5,7 \bmod{8}$ .

Remarks : so we can see that $\displaystyle - 2$ is quadratic residue of modulo $\displaystyle p \equiv 1,3 \bmod{8}$ .