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 $a-b | f(a)-f(b)$. 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:
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.
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.