Equation with floor function How would one solve an equation with a floor function in it:
$$a - (2x + 1)\left\lfloor{\frac {a - 2x(x + 1)}{2x + 1}}\right\rfloor - 2x(x + 1) = 0$$
$a$ is a given and can be treated as a natural number, and all $x$ other than integers can be discarded. At least one non-trivial solution would be sufficient.
Maybe an algorithmic method could be used?

Rearranging we get, 
$$\frac{a - 2x(x + 1)}{(2x + 1) } =\left\lfloor{\frac {a - 2x(x + 1)}{2x + 1}}\right\rfloor $$
This means,$\frac{a - 2x(x + 1)}{(2x + 1) }$ is an integer. 
$$\frac{a - 2x(x + 1)}{(2x + 1) }=k$$
$$a=2xk+k+2x^2+2x$$
$$2x^2+2(k+1)x+k-a=0$$
Applying the quadratic formula, 
$$x = \frac{-2(k+1)\pm\sqrt{4(k+1)^2-8(k-a)}}{4}$$
$a$ is known, now we substitute integers in place of $k$ to obtain solutions.
