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</math>
Here, it seems not right, we throw away <math>x^q^2-x_{\bar{q}}</math>?
Now if <math>X \equiv x^{q} _ {\bar{t}}\bmod \psi_l(x)</math> for one <math>\bar{t}\in [0,(l-1)/2]</math> then <math>\bar{t}</math> satisfies
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If you recall, our initial considerations omit the case of <math>l = 2</math>.
Since we assume {{mvar|q}} to be odd, <math>q + 1 - t \equiv t \pmod 2</math> and in particular, <math>t_{2} \equiv 0 \pmod 2</math> if and only if <math>E(\mathbb{F}_{q})</math> has an element of order 2. By definition of addition in the group, any element of order 2 must be of the form <math>(x_{0}, 0)</math>. Thus <math>t_{2} \equiv 0 \pmod 2</math> if and only if the polynomial <math>x^{3} + Ax + B</math> has a root in <math>\mathbb{F}_{q}</math>, if and only if <math>\gcd(x^{q}-x, x^{3} + Ax + B)\neq 1</math>.
==The algorithm==
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