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 October 14th, 2009, 05:48 AM #1 Member   Joined: Jul 2009 Posts: 55 Thanks: 0 Odd primes as n²+2 Hello everybody Sloane A056899 says 3, 11, 83, 227, 443, 1091, 1523, 2027, 3251, 6563, 9803, 11027, 12323, 13691, 15131, 21611, 29243, 47963, 50627, 56171, 59051, 62003, 65027, 74531, 88211, 91811, 95483, 103043, 119027, 123203, 131771, 136163, 140627, 149771, 173891 COMMENT Note that all terms after the first are equal to 11 modulo 72 and that (a(n)-11)/72 is a triangular number, since they have to be 2 more than the square of an odd multiple of 3 to be prime and if k=6m+3 then a(n)=k^2+2=72m(m+1)/2+11. MY QUESTION Why they have to be an odd multiple of 3 ? Is it only a conjecture ? Euzenius
October 14th, 2009, 06:10 AM   #2
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Re: Odd primes as n²+2

Quote:
 Originally Posted by Euzenius MY QUESTION Why they have to be an odd multiple of 3 ? Is it only a conjecture ? Euzenius
My guess is:
If n is not divisible by 3, then n^2 = 1 mod 3, so 3 | n^2+2.
So n must be divisible by 3 for n^2+2 to be prime-- this also means they are 2 more than an odd multiple of 9.

They have to be two more than an odd multiple, because if they were 2 more than an even multiple, they would be even.

 October 14th, 2009, 11:14 AM #3 Member   Joined: Jul 2009 Posts: 55 Thanks: 0 Re: Odd primes as n²+2 Well done ! Euzenius
October 14th, 2009, 12:14 PM   #4
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Re: Odd primes as n²+2

Quote:
 Originally Posted by Euzenius Well done !
Thanks. My very little bit of experience with number theory problems suggests "primes of the form..." are always stated to have "known" constraints which have never-seen proofs. They tend to be easy to prove if you see the trick, but the trick is sometimes a bit hidden...

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