<s>[INST] 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 $x \ge 10^{15}$.
 [/INST] 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
$$\begin{pmatrix} x-1 \\ x \\ 2m+1 \end{pmatrix}=
\begin{pmatrix}
  1235216565974040 \\
  1235216565974041 \\
  1746860020068409
\end{pmatrix}$$
</s>