Polynomial with no roots over the field $ \mathbb{F}_p $.

Solution 1:

Even more than that, there exists an irreducible polynomial of degree $d$. The field $\mathbb{F}_{p^d}$ is a field extension of degree $d$ over $\mathbb{F_p}$. The multiplicative group a finite field is cyclic, so let $\alpha$ be a generator of $\mathbb{F}_{p^d}^{\times}$, and then we have $\mathbb{F}_{p^d}=\mathbb{F}_p(\alpha)$. Since $[\mathbb{F}_p(\alpha):\mathbb{F}_p]=d$ it follows that the minimal polynomial of $\alpha$ is an irreducible polynomial in $\mathbb{F}_p[x]$ of degree $d$.

Solution 2:

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 $\Bbb Z/p\Bbb Z$.