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?


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.


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.

  • $\begingroup$ Question: For the second case -How is $\sum_{k=0}^n a_k\equiv 0$?If $\deg f(x)=2\implies f(r)=a_0+a_1+a_2\mod 2$ ;why is that zero?Can you explain $\endgroup$ – Learnmore Jan 18 '17 at 6:32
  • $\begingroup$ @user407186 $a_kr^k\equiv a_k\mod 2$ for any $k$ and $\sum_{k=0}^na_kr^k=0$. $\endgroup$ – Davide Giraudo Jan 18 '17 at 9:50
  • $\begingroup$ oh yes i got that $\endgroup$ – Learnmore Jan 18 '17 at 9:53

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.

  • 1
    $\begingroup$ Although I am not the OP and this question is 4 years old,I would like to thank you as this answer was really helpfull for me !!(Btw there's a slight typo :in the second case you mean $x$ is even). $\endgroup$ – Mr. Y Dec 7 '15 at 16:37
  • $\begingroup$ Thanks for noticing! Corrected now. (It was really an odd typo.) $\endgroup$ – Martin Sleziak Apr 7 '16 at 7:23

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


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.


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