Content deleted Content added
m →Introduction: -sp |
|||
Line 105:
If <math>q</math> turns out not to be a square modulo <math>l</math>, our assumption that <math>(x^{q^{2}}, y^{q^{2}}) = +q(x, y)</math> is rendered false, forcing us to consider the case where <math>(x^{q^{2}}, y^{q^{2}}) = - q(x, y)</math>, in which case our argument asserts that <math>\bar{t}</math> is <math>0 \, ( mod \, l)</math>.
As you might notice, it is possible for <math>q</math> to be a square <math>\pmod l</math> but <math>(x^{q^{2}}, y^{q^{2}}) = -q(x, y)</math>. In this case however, it suffices to check whether or not either square root <math>\pm w</math> satisfies <math>\phi(P) = \pm wP</math>, for <math>P \in E[l] \setminus \{O\}</math>, and this amounts to checking whether or not <math>\gcd(numerator(x^{q}-x_{w}),\phi_{l})=1</math>. If it is <math>1</math> then we are in the minus case and <math>t_{l}=0</math>; if <math>\neq 1</math> we proceed as in the plus case with <math>t_{l}=\pm 2w</math>.
===Additional Case: <math>l = 2</math>===
| |||