# Proving there exists such a polynomial

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!

• Hint: can you find a polynomial $g \in \mathbb Z_p[x]$, not equal to the zero polynomial, such that $g(x)$ nonetheless equals zero for all $x \in \mathbb Z_p$? If so, you should be able to modify $g(x)$ to produce a polynomial that is never 0. – Ravi Fernando Mar 29 '16 at 5:32
• Are you talking about the integers mod $p$ or the $p$-adic integers when you say $\mathbb Z_p$? – RKD Mar 29 '16 at 5:33
• The Fundamental Theorem of Algebra shouldn't be relevant here, because the statement you're proving says precisely that the FTA fails if you're working with polynomials over $\mathbb Z_p$ instead of $\mathbb C$. – Ravi Fernando Mar 29 '16 at 5:36
• @user1952009 You're thinking of the remainder theorem – RKD Mar 29 '16 at 6:16
• @Bill Do you know Fermat's little theorem ? – RKD Mar 29 '16 at 6:24

Write $f(x)=\Pi_{i\in Z_p}(X-i)+1$. It does not have a root.
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.
• typo aside, I think I have the right count, for example when $p=2$ there is only one irreducible polynomial of degree $2$, namely $x^2+x+1$ – mercio Mar 29 '16 at 13:31
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 $\mathbb Z_p$.
• I assume that $\mathbb Z_p = \mathbb Z/p\mathbb Z$. – lhf Mar 29 '16 at 10:44