<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] 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.
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