Math CentralQuandaries & Queries


Question from Chinonyerem, a student:

If p = 2^k - 1 is prime, show that k is an odd integer, except when k = 2.
[ Hint: 3/4^n - 1 for all n >= 1.]

The main difficulty here is the phrasing, I think. You are not meant to show that k is an integer, but to assume that it is (in fact, a natural number, as negative k are obviously irrelevant.) Also, the hint as given seems unhelpful - was it miscopied?

So try this: If k is a natural number greater than 2 and 2k - 1 is prime, show that k must be odd.

You might find it easier to think about the contrapositive: if k is even and greater than 2, show that 2k - 1 factors nontrivially. (Experiment and look for a pattern!)

Good Hunting!


If k = 2m then 2k - 1 = 22m - 1 = (2m)2 - 1 which has the form x2 - 1; when could this be prime?


About Math Central


Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.
Quandaries & Queries page Home page University of Regina PIMS