



 
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 2^{k}  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 2^{k}  1 factors nontrivially. (Experiment and look for a pattern!) Good Hunting!
If k = 2m then 2^{k}  1 = 2^{2m}  1 = (2^{m})^{2}  1 which has the form x^{2}  1; when could this be prime? Penny  


Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences. 