If p is a prime and k is a positive number less than p, and $2^k$ is incongruent to 1, $2^{(n1)}$ is congruent to 1, then $kp^2+1$ would be a prime number?
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closed as not a real question by Will Sawin, Will Jagy, George Lowther, Yemon Choi, Igor Rivin May 29 '12 at 1:13It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center. If this question can be reworded to fit the rules in the help center, please edit the question. 


I think that he omitted "modulo n". I tried to solve this problem with the fact : $ord_n{2}$ divides $n1=kp^2$, and thought that his conjecture is true. But I couldn't proved it. 

