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    Can someone guide me on answering this question please?

    "Show that 2 is a primitive root modulo 107" ?

    Is there an alternative way than computing all 106 powers of 2 which will take forever!

    Any help is much appreciated! Thanks
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    (Original post by Nazzy_HCrest)
    Can someone guide me on answering this question please?

    "Show that 2 is a primitive root modulo 107" ?

    Is there an alternative way than computing all 106 powers of 2 which will take forever!

    Any help is much appreciated! Thanks
    Number theory is definitely not my forté, so take this with a pinch of salt.

    From a quick google.

    The set of integers 1 to 106 which are coprime to 107 forms a multiplicative group mod 107. The order of this group is \phi(107) = 106

    Since the order of an element must divide the order of the group, and 106 = 2x53, it is sufficient to check 1,2,53 as possible orders. If it isn't one of these, then it must be 106, and we have a primative root.
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    Yes there is a shortcut.
    You only need to check that
    2^{\frac{p-1}{q}} (where q is prime factor- see below) is not congruent to 1.

    Now, in you case, p=107 so  p-1=106 and 106=2*53.

    So check: 2^{\frac{106}{2}}=2^{53} and 2^{\frac{106}{53}}=2^2 are not congruent to 1 (modulo 107).
 
 
 
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