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Consider the prime numbers of the form :

$p=4 \cdot k^2+1~$ , where $~k~$ is an odd prime number .

For the first $~1200000~$ primes of this form except when $~p=4 \cdot 193^2+1~$

$2~$ is a primitive root modulo $~p$ .

Note that :

Euler's totient function is given by :

$\phi(p)=4 \cdot k^2$

One can show that :

$\operatorname{ord}_p(2) \neq k~ ,\operatorname{ord}_p(2) \neq 2k~ ,\operatorname{ord}_p(2) \neq k^2~ ,\operatorname{ord}_p(2) \neq 2k^2~ $ , hence :

$\operatorname{ord}_p(2) = 4k~ \text {or}~\operatorname{ord}_p(2) = 4k^2~ $ since it cannot be $2 ~\text{or}~ 4$ .

My question :

Is there some special reason why $~2~$ isn't primitive root modulo $~p~$ in case of number

$~p=4 \cdot 193^2+1~$ ?

  • 0
    There are some typos in the text : "For the first $~291422~$ primes of this form only in case when $~p=4 \cdot 193^2+1~$ $2~$ isn't primitive root modulo $~p$" should be "For the first $~291422~$ primes of this form except when $~p=4 \cdot 193^2+1~$ $2~$ is a primitive root modulo $~p$" .2012-02-18
  • 0
    @EwanDelanoy,Thanks,English isn't my native language so....2012-02-18
  • 0
    Note that you can get proper formatting for things like $\operatorname{ord}$ by using `\operatorname{ord}`. If you just write `ord`, $\TeX$ interprets this as juxtaposed variable names and italicizes them.2012-02-19
  • 0
    @joriki,thanks...fixed2012-02-19
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    Note that the answer to this question had already been accepted and was apparently unaccepted for non-mathematical reasons (at least no mathematical reasons have been given) after a dispute about [this question](http://math.stackexchange.com/questions/111228).2012-02-20

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