Prove that $f(x)\in \mathbb{Z}[x]$ such that $f(0)$ and $f(1)$ are odd 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.

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.