Talk:Cocks IBE scheme

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== I cannot follow the correctness. You write the following:

\begin{align} r^2 &= \left(a^{(n+5-p-q)/8}\right)^2 \\ &= \left(a^{(n+5-p-q - \Phi(n))/8}\right)^2 \\ &= \left(a^{(n+5-p-q - (p-1)(q-1))/8}\right)^2 \\ &= \left(a^{(n+5-p-q - n+p+q-1)/8}\right)^2 \\ &= \left(a^{4/8}\right)^2  \\ &= \pm a \end{align} $$ In one-to-last line let us denote by $$b$$ one of the square roots of $$a$$. So you proved:

\begin{align} r^2 &= b^2  \\ &= \pm a \end{align} $$ Square root of $$a$$ is by definition an integer $$b$$ such that the $$b^2=a$$.

How can also be that $$a=b^2= -a$$? — Preceding unsigned comment added by 85.10.106.82 (talk) 09:14, 28 October 2019 (UTC)