A126960 Primes p such that (3p)^2 + 2 is prime.

3, 5, 7, 11, 13, 19, 37, 41, 73, 79, 83, 101, 103, 107, 139, 149, 151, 167, 191, 227, 233, 251, 269, 311, 337, 443, 457, 479, 499, 503, 521, 541, 601, 613, 647, 673, 761, 811, 829, 863, 877, 883, 887, 907, 919, 941, 983, 997

Offset: 1

Author

Cino Hilliard, Mar 19 2007

Keywords

• easy
• nonn

Comments

A generalization of this would be primes p such that (kp)^2+2 is prime. Then k=3 is the only solution. This follows from the fact that k of the form 3m-1 or 3m+1 will give 9m^2 + 6m + 1 + 2, a multiple of 3.

References

• Garath A. Jones and Mary Jones, Elementary Number Theory, Springer - Verlag London, 1998; p. 35, Exercise 2.17.

Programs

Magma

[ p: p in PrimesUpTo(1000) | IsPrime((3*p)^2+2)] // Vincenzo Librandi, Jan 29 2011

Mathematica

lst={}; Do[p=Prime@n; If[PrimeQ@((3*p)^2+2),AppendTo[lst,p]],{n,6!}]; lst (* Vladimir Joseph Stephan Orlovsky, Aug 11 2009 *)

PARI

g(n) = forprime(x=0,n,y=9*x^2+2;if(isprime(y),print1(x",")))

Extensions

Entries confirmed by Zak Seidov, Mar 19 2007