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Let $n=2^{2^{\alpha}}-1.$ We know that $$n=\frac{2^{2^{\alpha}}-1}{(2+1)}= (2^{2^{\alpha-1}}+1)(2^{2^{\alpha-2}}+1)\dots(2^{2^2}+1)(2^2+1).$$

My question is are the numbers $ 2^{2^1}+1, 2^{2^2}+1, \dots$ relatively prime to each other? Since I want to apply Chinease remainder theorem and find out a number $X$ such that $x^2\equiv -1 \mod n.$

Mike
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Raheel
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1 Answers1

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The very factorization you give implies that $2^{2^i} + 1$ divides $2^{2^j}-1$ for all $j>i$. So if $2^{2^i}+1$ and $2^{2^j}+1$ have a common divisor $d$, then $d$ divides both $2^{2^j}-1$ and $2^{2^j}+1$, so it can only be $1$ or $2$. But wait, it can't be $2$ either, because $2^{2^j}+1$ is odd.

Misha Lavrov
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