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 ϕ(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.