I understand how to show that $2$ is a primitive root of $q$, but I am not sure how to use this fact to show that it is a primitive root of $p$ (or whether I should at all).
$p$ and $q$ are both primes where $p \equiv 1$ (mod 4) and $q = 2p + 1$. If $2^p \equiv -1$ (mod q), show that 2 is a primitive root modulo $p$.
2
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number-theory
2 Answers
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You're statement is false. Let $p=41, q=83$. Note that $$2^{41} \equiv \left( \frac{2}{83} \right) \equiv -1 \pmod {83}$$ By quadratic reciprocity and Euler's Criterion. But, $$2^{20} \equiv \left( \frac{2}{41}\right) \equiv 1 \pmod {41}$$ Thus $2$ is not a primitive root of $41$. You're wrong.
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0I am asking how to show that 2 is a primitive root of **p**, not q. – 2017-02-20
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0@alisa Then your statement is false. – 2017-02-20
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0Thank you. Must be an error in my book. – 2017-02-20
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0@aik Glad to be of assistance. I see you have changed you're profile name. Could you accept this answer? – 2017-02-20
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0@aik, Did you see my earlier comment? – 2017-02-20
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,By Fermat's Little Theorem $2^{q-1}\equiv1\,$mod $q$. The least number $m$ for which $2^m$ is 1 mod $q$ has to divide $q-1$. As $q-1$ has only $1,2,p,q-1$ as factors the statement follows.