Quote from: PowerGlove on Today at 05:15:58 AM

But, after learning a bit more about finding roots in finite fields I can now appreciate how there are three solutions (given Y) to this equation (modulo P):

Equation B: X = (Y ** 2 - 7) ** 1/3

I'm still not sure what (if any) the corresponding operations are with private keys (i.e. what do you have to do to a private key to produce these "other" X co-ordinates?).

You could obtain the other private keys by multiplying by a non-trivial cubic root of 1 modulo n. Then y will be the same, and x will be multiplied by the corresponding non-trivial root of 1 modulo p.

Here are the two cube roots of 1 (besides the trivial 1) modulo n:
n2: 37718080363155996902926221483475020450927657555482586988616620542887997980018
n3: 78074008874160198520644763525212887401909906723592317393988542598630163514318
and their corresponding cube roots of 1 modulo p:
p2: 55594575648329892869085402983802832744385952214688224221778511981742606582254
p3: 60197513588986302554485582024885075108884032450952339817679072026166228089408

(X2, Y1) = n2 * (X1, Y1) = (p2*X1, Y1)
(X3, Y1) = n3 * (X1, Y1) = (p3*X1, Y1)
(X1, Y2) = -1 * (X1, Y1) = (X1, -Y1)
(X2, Y2) = -n2 * (X1, Y1) = (p2*X1, -Y1)
(X3, Y2) = -n3 * (X1, Y1) = (p3*X1, -Y1)