# Deriving modular properties of factors of a polynomial based on the roots of the polynomial. [duplicate]

This is from an answer by Robert to this question When are $n!+1$ and $(n+2)!+1$ coprime?

"The roots of $$x^2 + 3x + 1$$ are $$(-3 \pm \sqrt{5})/2$$, so $$5$$ must be a square mod $$p$$, which says (if $$p > 5$$) $$p \equiv 1$$ or $$4 \mod 5$$."

Can anyone elaborate on how this is derived or point me somewhere where I can read up on it. Thanks.

• There are three assertions in a row in that quote; which one is the "this" you want information on? Dec 24, 2023 at 2:41

the discriminant of the binary quadratic form $$u^2 + 3uv + v^2$$ is $$9-4=5$$
If we have (positive) prime $$q$$ with Legendre symbol $$(5|q) = -1$$ and $$q | u^2 + 3uv + v^2, \; \;$$ then both $$q | u$$ and $$q | v.$$ Thus, no such prime $$q$$ can divide any $$x^2 + 3x + 1.$$ The number $$x^2 + 3x + 1$$ may be divisible by $$5$$ and it may be divisible by primes $$p \equiv \pm 1 \pmod 5$$
background material includes the Legendre symbol and quadratic reciprocity. Here, we used $$(q|5) = (5|q)$$ because $$5 \equiv 1 \pmod 4$$
• same thing. Except for $2$ itself, primes are odd, so $1,4 \pmod 5$ refer to $1,9 \pmod{10}$ Dec 24, 2023 at 3:02