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12 July, 01:12

Show that if 2^m is an odd prime then m = 2^n for some nonnegative integer n

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  1. 12 July, 01:13
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    Okay, im not sure if this is what you needed but if we are saying its (2^m) + 1, we know that m={1,2} will all result in an odd prime. You can use log₂ (m) = n and you will find that for the sequence m={1,2}, the corresponding n values are n={0,1} which are nonnegative integers.

    I do not know if you need to show proof but both m={1,2} and the corresponding n values show that this works.
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