Finding $x$ such that $2x^2-2x+1=(2m+1)^2$ We need to find the smallest integral value of $x$ such that $2x^2-2x+1=(2m+1)^2$ and $x \ge 10^{15}$.
 A: HINT: $$2x^2 - 2x + 1 = (2m+1)^2 \iff x^2 + (x-1)^2 = (2m+1)^2$$
So the problem reduces to finding a specific Pythagorean triplet.
UPDATE:
We have that this is a primitive Pythagorean triplet, as $(x,x-1) = 1$. Therefore we either have $x = m^2 - n^2, x-1 = 2mn$ or $x=2mn, x-1 = m^2 - n^2$ for some positive integers $m,n$ which are coprime and one of them is odd.
Now going through the cases we have:
$$m^2-n^2 = 2mn+1 \iff (m-n)^2 = 2n^2 + 1$$
This is a Pell's Equation and has infinitely many solutions. So eventually you would be able to find a solution that will give you $x \ge 10^{15}$
Similarly for the other case we have:
$$m^2 - n^2 = 2mn - 1 \iff (m-n)^2 = 2n^2 - 1$$
Again this is a Pell's Equation, but this one is negative one, which doesn't change much.
A: According to Stefan4024, the solution for the Pythagorean triples are well-known:
$$\begin{pmatrix} x-1 \\ x \\ 2m+1 \end{pmatrix}=
\begin{pmatrix}
  \frac{(\sqrt{2}+1)^{2k+1}-(\sqrt{2}-1)^{2k+1}}{4}-\frac{1}{2} \\
  \frac{(\sqrt{2}+1)^{2k+1}-(\sqrt{2}-1)^{2k+1}}{4}+\frac{1}{2} \\
  \frac{(\sqrt{2}+1)^{2k+1}+(\sqrt{2}-1)^{2k+1}}{2\sqrt{2}}
\end{pmatrix}$$
Now $\dfrac{(\sqrt{2}+1)^{2k+1}}{4} \approx 10^{15} \implies 2k+1\gtrsim 40.76$
Put $2k+1=41$, then
$$\begin{pmatrix} x-1 \\ x \\ 2m+1 \end{pmatrix}=
\begin{pmatrix}
  1235216565974040 \\
  1235216565974041 \\
  1746860020068409
\end{pmatrix}$$
A: Thanks @Stefan4024 for the answer. However, on googling more, I found this site. This solves Quadratic Diophantine Equations of the type $$ a x^2 + b xy + c y^2 + dx + ey + f = 0$$ We need to provide values of coefficients $\{a,b,c,d,e,f\}$ and it gives us the output in recurrence form of the following type:
$$x_{n+1} = P x_n + Q y_n + K$$
$$y_{n+1} = R x_n + S y_n + L$$ where the coefficients $\{P,Q,R,S,K,L\}$ are also provided as an output. This is super awesome but I really don't know the mathematics behind it.
