Polynomial with no roots over the field $ \mathbb{F}_p $. How can I show that for any prime $p$ and any $d\ge 2$ there exists a polynomial of degree $d$ in $\mathbb{F}_p[X]$ with no roots? ($\mathbb{F}_p$ is the finite field with $p$ elements).
Thanks in advance for any idea.

Note that $f(x)=x^d-x$ has at least two roots in $\Bbb Z/p\Bbb Z$, namely $0$ and $1$. Thus, viewed as a map $f\colon\Bbb Z/p\Bbb Z\to \Bbb Z/p\Bbb Z$, it is not injective, hence also not surjective. Pick $a\in \Bbb Z/p\Bbb Z$ that is not in the image. Then $g(x):=x^d-x-a$ has no root in $\Bbb Z/p\Bbb Z$.
