# Prove that $f(x)\in \mathbb{Z}[x]$ such that $f(0)$ and $f(1)$ are odd has no integer roots

by Mohan   Last Updated July 12, 2019 10:20 AM - source

Suppose $f(x) \in \mathbb{Z}[x]$ is such that $f(0)$ and $f(1)$ are odd. How do I show that $f(x)$ has no integer roots?

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If $r$ is an integer root and $f(x)=\sum_{k=0}^na_kx^k$, $a_k\in\mathbb Z$ then $\sum_{k=0}^na_kr^k=0$.

• If $r$ is even, then reducing modulo $2$ we get that $a_0\equiv 0[2]$ hence $f(0)$ is even, which cannot be the case by hypothesis.

• If $r$ is odd, then $r^k\equiv 1[2]$ for each $k \geqslant 0$ hence $\sum_{k=0}^na_k\equiv 0[2]$. Thus $f(1)=\sum_{k=0}^na_k$ is even, and we get a contradiction.

Davide Giraudo
November 19, 2011 10:10 AM

Another proof can be done using the fact that if $$f$$ has integer coefficients and $$a\neq b$$ are integers, then $$a-b \mid f(a)-f(b)$$. Because $$f(0)$$ and $$f(1)$$ are odd, it follows that $$f(k)$$ is odd for every integer $$k$$, and therefore no integer can be a root.

Beni Bogosel
November 19, 2011 10:38 AM

Using the basic properties of congruences you can show easily that for any polynomial $f(x)$ with integer coefficients $$x \equiv y \pmod n \qquad \Rightarrow \qquad f(x) \equiv f(y) \pmod n.$$ See the proof at proofwiki.

In this case for $n=2$ you get:

• if $x$ is odd, i.e. $x \equiv 1 \pmod 2$, then $f(x)\equiv f(1)\equiv 1\pmod 2$;
• if $x$ is even, i.e. $x \equiv 0 \pmod 2$, then $f(x)\equiv f(0)\equiv 1\pmod 2$.

In both cases, $f(x)$ is odd integer, hence it is non-zero.

Note that this is basically the same answer as given by Beni Bogosel, but I thought that if you are familiar with congruences, this approach might be more clear for you.

Martin Sleziak
November 19, 2011 11:20 AM

I'm not being very original here, but reducing $x$ modulo 2 in the expression $f(x)$ gives $f(x)\equiv f(x\bmod 2)\pmod 2$. It may be interesting is to note that the result is false for polynomials with rational coefficients that take integer values on $\mathbf Z$, for instance $\frac{x^2-x-2}2$.

Marc van Leeuwen
November 19, 2011 11:47 AM

A slight (but cute, in my opinion) variation of the the other methods is this one:

If $f\in \Bbb Z[x]$ has an integer root $m$, then by Ruffini's rule $f(x)=(x-m)(b_n x^n+\cdots + b_0)=(x-m)g(x)$ for some $b_0,\cdots, b_m\in \Bbb Z$.

Then, $$\begin{cases} f(0)=-m\cdot g(0)\\ f(1)=(1-m)\cdot g(1)\end{cases}$$

But at least one between $-m$ and $(1-m)$ must be even.

user228113
July 01, 2016 13:25 PM