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

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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.

Answer 5

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 $\frac{x^2-x-2}2$.

Answer 6

let f(x) be a polynomial such that $$ f(x) = a_n x^n + a_{n-1} x^{n-1} + \ldots + a_2 x^2 + a_1 x + a_0 $$ let p be an integral root of f(x).

put x=0 $$ f(x) = a_n (0)^n + a_{n-1} (0)^{n-1} + \ldots + a_2 (0)^2 + a_1 (0) + a_0 $$
we get $$ f(x)= a_0 $$

f(0) is odd, and any integral root must divide f(0). so p divides f(0)

if p divides an odd number then p must be odd.

according to Rational root theorem, if f(x) has integral coefficients and a is an integer root of f(x) and m is any integer different from a then a-m divides f(m)

so (p-1) must divide f(1) which is odd

but p is odd so (p-1) is even.

(p-1) can't divide f(1) because it is odd