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{{fake heading|sub=3|Case 2: <math>(x^{q^{2}}, y^{q^{2}}) = \pm \bar{q}(x, y)</math>}}
We begin with the assumption that <math>(x^{q^{2}}, y^{q^{2}}) = \bar{q}(x, y)</math>. Since {{mvar|l}} is an odd prime it cannot be that <math>\bar{q}(x, y)=-\bar{q}(x, y)</math> and thus <math>\bar{t}\neq 0</math>. The characteristic equation yields that <math>\bar{t} \phi(P) = 2\bar{q} P</math>. And consequently that <math>\bar{t}^{2}\bar{q} \equiv (2q)^{2} \pmod l</math>
This implies that {{mvar|q}} is a square modulo {{mvar|l}}. Let <math>q \equiv w^{2} \pmod l</math>. Compute <math>w\phi(x,y)</math> in <math>\mathbb{F}_{q}[x,y]/(y^{2}-x^{3}-Ax-B, \psi_{l})</math> and check whether <math>
\bar{q}(x, y)=w\phi(x,y)</math>. If so, <math>t_{l}</math> is <math>\pm 2w \pmod l</math> depending on the y-coordinate.
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