A232528 Numbers n such that for all primes p where p and p-n are quadratic residues (mod 4*n), 4*p can be written as x^2 + n*y^2.

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 15, 16, 18, 19, 20, 21, 22, 24, 25, 27, 28, 30, 32, 33, 35, 36, 37, 40, 42, 43, 45, 48, 51, 52, 57, 58, 60, 64, 67, 70, 72, 75, 78, 84, 85, 88, 91, 93, 96, 99, 100, 102, 105, 112, 115, 120, 123, 130, 132, 133, 147, 148, 160, 163, 165, 168, 177, 180, 187, 190, 192, 195

Offset

1,2

Comments

Convenient numbers (A000926) are numbers n such that for all primes p where p and p-n are quadratic residues (mod 4*n), p can be written as x^2 + n*y^2, so convenient numbers are a subsequence.

All non-convenient numbers which are members of this sequence are either congruent to 0 (mod 4) or 3 (mod 4).

The equation 4*p = x^2 + n*y^2 is important because if n is a squarefree integer congruent to 3 (mod 4), then the ring of integers Q[sqrt(-n)] will be all integers of form (x/2) + (y/2)*sqrt(-n) for x and y of the same parity, whose norm is (x/2)^2 + n*(y/2)^2. If prime p = (x/2)^2 + n*(y/2)^2, then 4*p = x^2 + n*y^2.

Is this sequence finite?

Is 7392 the largest term of this sequence?

There are no further terms up to 10^6. - Andrew Howroyd, Jun 08 2018

Links

V. Raman, Table of n, a(n) for n = 1..139

Thomas R. Hagedorn, Primes of the form x^2+ny^2 and the geometry of (convenient) numbers

Example

n = 14 is not a member of this sequence because for prime p = 71, 4*p = 284 cannot be written as x^2 + 14*y^2.

Prog

(PARI) ok(n)=!#select(k->k<>2, quadclassunit(-n*if((-n)%4>1, 4, 1)).cyc) \\ Andrew Howroyd, Jun 08 2018

Crossrefs

Cf. A000926.

Sequence in context: A348960 A367613 A095392 * A254075 A377916 A140401

Adjacent sequences: A232525 A232526 A232527 * A232529 A232530 A232531

Keyword

nonn

Author

V. Raman, Nov 25 2013

Status

approved