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# Show that 2 is a primitve root modulo 107

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"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
2. (Original post by Nazzy_HCrest)

"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.

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

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.
3. Yes there is a shortcut.
You only need to check that
(where q is prime factor- see below) is not congruent to .

Now, in you case, so and .

So check: and are not congruent to 1 (modulo 107).

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