How do I show $x^2+1$ is reducible or irreducible in $\mathbb{Z}_7$?

Question

How do I show $x^2 + 1$ is reducible or irreducible in $\mathbb{Z}_7$? Is there a standardized procedure to check ?

Thanks

Answer 1

Since $a^6\equiv 1\pmod 7$ for any $a$ not divisible by $7$, then if $-1$ is a square modulo $7$ then $(-1)^3\equiv 1\pmod 7$. We can see it is not.

More generally, the answer is to use various forms of quadratic reciprocity to determine if $x^2+n$ is irreducible modulo $p$.

Answer 2

Alternatively, as mentioned in comments, the first way I would think:

We will look at $x^2+1$ over $\Bbb F_7$. It is irreducible if it has no roots in the field $\Bbb F_7$. Now we can see that a root will have $x^2=-1\equiv 6\pmod 7$

$1^2=1$, $2^2=4$, $3^2\equiv 2\pmod 7$, $4^2\equiv 2\pmod 7$, $5^2 \equiv 4 \pmod 7$, $6^2\equiv 1\pmod 7$, $7^2 \equiv 0 \pmod 7$. Hence $x^2$ cannot equal $6\pmod 7$, and no root is present in $\Bbb F_7$.