Problem: A triangle is called a \emph{parabolic} triangle
if its vertices lie on a parabola $y = x^2$.
Prove that for every nonnegative integer $n$,
there is an odd number $m$ and a parabolic triangle
with vertices at three distinct points
with integer coordinates with area $(2^nm)^2$.

Solution: For $n=0$, take instead $(a,b) = (1,0)$.

For $n > 0$, consider a triangle
with vertices at $(a,a^2)$, $(-a,a^2)$ and $(b, b^2)$.
Then the area of this triangle was equal to
\[ \half (2a) \left( b^2-a^2 \right) = a(b^2-a^2). \]

To make this equal $2^{2n} m^2$,
simply pick $a = 2^{2n}$,
and then pick $b$ such that $b^2-m^2 = 2^{4n}$,
for example $m = 2^{4n-2}-1$ and $b=2^{4n-2}+1$.