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.
 A: We are solving $2x^2-2x+1=K^2$ or equivalently $4x^2-4x+2=2K^2$ or equivalently $(2x-1)^2=2K^2-1$. For any explicit $K$, there are not many candidates!
If we are interested in the general theory, we might note that we are dealing here with the (negative) Pell Equation 
$$u^2-2v^2=-1.$$
There is a complete characterization of all the solutions. Let $n$ be an odd integer, and let $u_n+v_n\sqrt{2}=(1+\sqrt{2})^n$. Then $(u_n,v_n)$ is a solution, and we obtain all positive solutions in this way. 
One can also get a nice explicit recurrence, with integer coefficients, for the $u_n$ and for the $v_n$. This can be derived from the fact that 
$$u_{n+2}+v_{n+2}\sqrt{2}=(3+2\sqrt{2})(u_n+v_n\sqrt{2}).$$
Alternately, let $(s_k,t_k)$ be the $k$-th positive solution of $s^2-2t^2=-1$. A bit of playing shows that each of $(s_k)$ and $(t_k)$ satisfies the recurrence
$$w_{k+2}=6w_{k+1}-w_k.$$
The only difference is in the initial conditions. 
Remark: One can find bounds, indeed exact formulas, for the $k$-th positive solution of the equation $s^2-2t^2=-1$. For if we call this $(s_k,t_k)$, then $s_k+t_k\sqrt{2}=(1+\sqrt{2})^{2k-1}$. It follows that $s_k-t_k\sqrt{2}=(1-\sqrt{2})^{2k-1}$. By adding, we find that
$$2s_k=(1+\sqrt{2})^{2k-1}+(1-\sqrt{2})^{2k-1}.$$
One can do slightly better, by noting that $(1-\sqrt{2})^{2k-1}$ is negative, and small in absolute value. Thus
$$2s_k=\lfloor(1+\sqrt{2})^{2k-1}\rfloor.$$
One can in a similar way obtain an explicit formula for $t_k$.
A: Solving the quadratic gives $x = \frac{1 \pm \sqrt{2k^2-1}}{2}$. The bounds are clear from this.
