cp's OEIS Frontend

A232530 Least square m^2 such that for all primes p where p and p-n are quadratic residues (mod 4*n), (m^2)*p can be written as x^2+n*y^2.

%I A232530 #16 Dec 16 2013 12:12:51
%S A232530 1,1,1,1,1,1,1,1,1,1,4,1,1,9,1,1,9,1,4,4,1,1,9,1,1,9,4,1,9,1,16,4,1,
%T A232530 25,4,4,1,9,16,1,81,1,4,9,1,25,64,1,25,9,4,4,9,9,16,9,1,1,36,1,25,81,
%U A232530 9,4,9,25,4,36,25,1,144,1,49,81,4,16,9,1,64,9,9,49,36,4,1,81,16,1,225,9,4,9,1,625,64,4,49
%N A232530 Least square m^2 such that for all primes p where p and p-n are quadratic residues (mod 4*n), (m^2)*p can be written as x^2+n*y^2.
%C A232530 If n is a convenient number (A000926) then a(n) = 1.
%C A232530 m^2 is also the smallest positive square that can be generated by using all the inequivalent primitive quadratic forms of discriminant = -4n.
%F A232530 a(n)=A232529(n)^2.
%e A232530 For n = 11, all primes p such that p such that p is a quadratic residue (mod 4*n) and p-n is a quadratic residue (mod 4*n) are either of the form x^2+11*y^2 or 3*x^2+2*x*y+4*y^2.
%e A232530 4*(x^2+11*y^2) = (2*x)^2+11*(2*y)^2, 4*(3*x^2+2*x*y+4*y^2) = (x+4*y)^2+11*x^2. Also, 4 is the smallest square to satisfy this condition. So, a(11) = 4.
%e A232530 For n = 14, all primes p such that p such that p is a quadratic residue (mod 4*n) and p-n is a quadratic residue (mod 4*n) are either of the form x^2+14*y^2 or 2*x^2+7*y^2.
%e A232530 9*(x^2+14*y^2) = (3*x)^2+14*(3*y)^2, 9*(2*x^2+7*y^2) = (2*x+7*y)^2+14*(x-y)^2 = (2*x-7*y)^2+14*(x+y)^2. Also, 9 is the smallest square to satisfy this condition. So, a(14) = 9.
%e A232530 For n = 17, all primes p such that p such that p is a quadratic residue (mod 4*n) and p-n is a quadratic residue (mod 4*n) are either of the form x^2+17*y^2 or 2*x^2+2*x*y+9*y^2.
%e A232530 9*(x^2+17*y^2) = (3*x)^2+17*(3*y)^2, 9*(2*x^2+2*x*y+9*y^2) = (x+9*y)^2+17*x^2. Also, 9 is the smallest square to satisfy this condition. So, a(17) = 9.
%e A232530 For n = 19, all primes p such that p such that p is a quadratic residue (mod 4*n) and p-n is a quadratic residue (mod 4*n) are either of the form x^2+19*y^2 or 4*x^2+2*x*y+5*y^2.
%e A232530 4*(x^2+19*y^2) = (2*x)^2+19*(2*y)^2, 4*(4*x^2+2*x*y+5*y^2) = (4*x+y)^2+19*y^2. Also, 4 is the smallest square to satisfy this condition. So, a(19) = 4.
%e A232530 For n = 20, all primes p such that p such that p is a quadratic residue (mod 4*n) and p-n is a quadratic residue (mod 4*n) are either of the form x^2+20*y^2 or 4*x^2+5*y^2.
%e A232530 4*(x^2+20*y^2) = (2*x)^2+20*(2*y)^2, 4*(4*x^2+5*y^2) = (4*x)^2+20*y^2. Also, 4 is the smallest square to satisfy this condition. So, a(20) = 4.
%Y A232530 Cf. A232529, A000926.
%K A232530 nonn,uned
%O A232530 1,11
%A A232530 _V. Raman_, Nov 25 2013