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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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4 Answers 4

up vote 16 down vote accepted

Another proof can be done using the fact that if $f$ has integer coefficients and $a\neq b$ are integers, then $a-b | f(a)-f(b)$. 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.

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If $r$ is an integer root and $P(x)=\sum_{k=0}^na_kx^k$, $a_k\in\mathbb Z$ 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 $\sum_{k=0}^na_k\equiv 0[2]$, hence $f(1)$ is even, and we get a contradiction.

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

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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 odd, 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.

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