Does this prove that no sequential squares have a ratio of 2? The goal: 
Prove that there is no integer $k$ such that ${(k+1)^2\over{k^2}}=2$.
My proof:
If ${(k+1)^2\over{k^2}}=2$, then ${{k+1}\over{k}}=\sqrt2$, and if $k$ is an integer, $k+1$ is also an integer. This implies that $\sqrt2$ is a rational number, which is also provably false.

Just to be different:
$\frac {(k+1)^2}{k^2} = \frac {k^2 + 2k + 1}{k^2} = 1 + \frac {2k + 1}{k^2}$
So $\frac {(k+1)^2}{k^2} = 2\implies \frac {2k + 1}{k^2} = 1 \implies$
$2k + 1 =k^2$ and as $k \ne 0$ (otherwise $2 = 0$ which isn't true)
$2 + \frac 1k = k$.  But if $k \ne \pm 1$ then $\frac 1k$ is not an integer.
And if $k = \pm 1$ then $\frac 1k = k$ and $2 = k -\frac 1k = 0$ which is impossible.
....
But seriously...
$\frac {(k+1)^2}{k^2} = (\frac {k+1}{k})^2 = 2$ implies there is a rational number whose square is $2$ which is well-known to be false, if a perfect and irrefutable proof.  And probably the absolute easiest.
