1. ## Questions about polynomials over Z mod 5

How many irreducible quadratic polynomials are there over Z mod 5?

Thanks!

2. Originally Posted by xxie

How many irreducible quadratic polynomials are there over Z mod 5?

Thanks!
let's solve the problem for the general case, i.e. over $\mathbb{Z}/p,$ where $p$ is any prime number. a quadratic polynomial is irreducible iff it has no root. this number is obviously $n=p^2(p-1) - k,$ where

$k$ is the number of those polynomials of degree 2 which are reducible, i.e. they are in the form $a(x-b)(x-c).$ clearly $a$ can be any element of $\mathbb{Z}/p - \{0 \}.$ so there are $p-1$ possibilities for $a.$

we also have $\binom{p}{2}$ distinct pairs $a,b$ in $\mathbb{Z}/p.$ it is also possible to have $a=b,$ which gives us $p$ possibilities. thus $k=(p-1) \left(\binom{p}{2} + p \right)=\frac{p(p^2-1)}{2}.$ thus: $n=p^2(p-1)-k=\frac{p(p-1)^2}{2}.$

if you're looking for the number of monic irreducible polynomials of degree 2, then $a=1$ would be the only possibility for $a$ and thus in this case $n=\frac{p(p-1)}{2}.$