Finding bound for the equation $\sqrt{1+2x(x-1)} = K$ In this equation: $\sqrt{1+2x(x-1)} = K$, $K$ is an integer constant, I need to check if there exists integer $x$ satisfying this equation. Can anyone prove some lower and upper bound for $x$. Here is how I proceeded(and failed):
Obviously: $\sqrt{2x(x-1)} \approx \sqrt{1+2x(x-1)}$
$\Rightarrow \sqrt{2x(x-1)} \approx K$
$\Rightarrow \sqrt{x(x-1)} \approx K/\sqrt{2}$
Also obvious is: $x - 1 \approx x$
$\Rightarrow x(x-1) \approx x \times x$
$\Rightarrow \sqrt{x(x-1)} \approx x$
Combining $\sqrt{x(x-1)} \approx K / \sqrt{2}$ and $\sqrt{x(x-1)} \approx x$ implies $x \approx K / \sqrt{2}$. But that doesn't give any specifig upper/lower bound.

Solving the quadratic gives $x = \frac{1 \pm \sqrt{2k^2-1}}{2}$. The bounds are clear from this.
