Problem: Prove that if $f \colon \ZZ \to \ZZ$ is any function,
then there are infinitely many integers $c$ such that the function
$g(x) = f(x)+cx$ is not a bijection.

Solution: Assume for contradiction that there exists a finite ``bad'' set $S$
such that $f(x)+cx$ is bijective for all $c \notin S$.

The first observation is basically that given just $f(0)$ and $f(1)$,
or any two consecutive $f$-values, we can already find bad values of $c$.
\begin{claim*}
  The numbers $f(0)-f(1)$, $f(1)-f(2)$, $f(2)-f(3)$, \dots\ are all in $S$.
  That is, consecutive differences of $f$ only take finitely many values.
\end{claim*}
\begin{proof}
  To see that $f(0) - f(1) \in S$, just note that
  \[ x \mapsto f(x) + \left( f(0) - f(1) \right) \cdot x  \]
  is obviously not a bijective function,
  since it takes the same values at $x=0$ and $x=1$, namely $f(0)$.
  The same is true for other elements.
\end{proof}

In particular, suppose $M > \max_{s \in S} |s|$.
Then the function
\[ g(x) = f(x) + (M+100) x \]
is supposed to be a bijection (because $M+100 \notin S$),
and yet $g(x+1) - g(x) > 100$ for all $x$, which is a contradiction.