Problem: Prove that there exists an infinite set of points
\[ \dots, \; P_{-3}, \; P_{-2},\; P_{-1},\; P_0,\; P_1,\; P_2,\; P_3,\; \dots \]
in the plane with the following property:
For any three distinct integers $a$, $b$, and $c$,
points $P_a$, $P_b$, and $P_c$ are collinear if and only if $a+b+c=2014$.

Solution: The construction
\[ P_n = \left( n-\frac{2014}{3},
  \left( n-\frac{2014}{3} \right)^3 \right) \]
works fine, and follows from the following claim:
\begin{claim*}
  If $x$, $y$, $z$ are distinct real numbers
  then the points $(x,x^3)$, $(y,y^3)$, $(z,z^3)$
  are collinear if and only if $x+y+z=0$.
\end{claim*}
\begin{proof}
  Note that by the ``shoelace formula'',
  the collinearity is equivalent to
  \[
    0 =
    \det \begin{bmatrix}
      x & x^3 & 1 \\
      y & y^3 & 1 \\
      z & z^3 & 1 \\
    \end{bmatrix}
  \]
  But the determinant equals
  \[ \sum_{\text{cyc}} x(y^3-z^3)
    = (x-y)(y-z)(z-x)(x+y+z). \qedhere  \]
\end{proof}