Proving there exists such a polynomial

Question

I'm having trouble proving the following statement:

For all primes $p$, there exists a non-constant polynomial $f(x)\in \mathbb Z_p[x]$ such that f(x) does not have a root in $\mathbb Z_p$

What I have tried so far is using the Fundamental Theorem of Algebra, which states that all polynomials $f(x)$ with $deg(f(x)) \ge 1$, there exists $x_0\in \mathbb C$ such that $f(x_0)=0$ assuming $f(z)$ has 0 roots, but that did not get me anywhere. Does anyone know of a way to solve this? Thanks!

Answer 1

Write $f(x)=\Pi_{i\in Z_p}(X-i)+1$. It does not have a root.

Answer 2

By Fermat's theorem, all elements of $\mathbb Z_p$ are roots of $x^p-x$.

Therefore, $f(x)=x^p-x+1$ has no roots in $\mathbb Z_p$.

Answer 3

Over $\Bbb Z/p\Bbb Z$, there are $p^2$ polynomials of the form $x^2+ax+b$ but there are only $p(p+1)/2$ polynomials of the form $(x+a)(x+b)$, which leaves $p(p-1)/2$ irreducible degree $2$ polynomials.