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If p is a prime and k is a positive number less than p, and $2^k$ is incongruent to 1, $2^{(n-1)}$ 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 1:13

It'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.

Firstly, no formula as simple as that always produces primes. Secondly, it's not clear what you mean by "congruent to 1" and "incongruent to 1" since you do not specify modulo what. Thirdly, you give no reason to believe this is true. –  Will Sawin May 28 '12 at 22:17
He probably means congruences modulo $n$, so one of his conditions is that $n$ is a base $2$ pseudoprime, if not prime. Here is a list of all such numbers up to $10^{15}$: cecm.sfu.ca/Pseudoprimes If there is no counterexample to his conjecture on this list, then it gets interesting. –  Felipe Voloch May 28 '12 at 23:13
Also posted to math.stackexchange, math.stackexchange.com/questions/150853/prime-of-the-form-kp21, without any mention of the post here. –  Gerry Myerson May 29 '12 at 1:06
@Felipe, I verified that no counterexamples exist in this list. –  Dan Brumleve May 29 '12 at 3:52
It's badly stated, unmotivated, rudely crossposted, and shows no sign of work on the part of OP --- but it is a real question! –  Gerry Myerson May 29 '12 at 5:50

1 Answer 1

I think that he omitted "modulo n". I tried to solve this problem with the fact : $ord_n{2}$ divides $n-1=kp^2$, and thought that his conjecture is true. But I couldn't proved it.

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