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Consider the following transformation matrix that slides points on the curve along the curve $\ln(x)\ln(y)=1,$ with $s\in \Bbb R,$ and $(x,y)\in \Bbb R\cap(1,\infty).$
$$ T_s = \begin{bmatrix} e^{e^s} & 0 \\\ 0 & e^{e^{-s}} \end{bmatrix}.$$
We can relate $T_s$ to another transformation matrix $...
Suppose that $q_i$ is the $i$th odd prime, $q_1 = 3$.
Is it possible to simultaneously solve:
$x^{2^k} = 1 \pmod{q_i}, i=1..{k+1}$
whilst,
$x^{2^k} = $anything but $1 \pmod {q_i}$
for any $k \geq 1$? Can this be done using Chinese remaindering?
Doesn't this imply that $x^{2^k} - 1 = q_1 \c...
