A248194 Positive integers n such that the equation x^2 - n*y^2 = n*(n+1)/2 has integer solutions.

1, 3, 8, 9, 17, 19, 24, 25, 33, 49, 51, 57, 67, 72, 73, 81, 88, 89, 96, 97, 99, 121, 129, 136, 147, 152, 163, 169, 177, 179, 193, 201, 211, 225, 233, 241, 243, 249, 264, 288, 289, 297, 313, 337, 339, 352, 361, 369, 387, 393, 408, 409, 441, 449, 451, 456, 457

Offset

1,2

Comments

All odd squares are in this sequence. Proof: Set n = (2k + 1)^2, then we have x^2 - (2k + 1)^2 * y^2 = (2k + 1)^2 * (2k^2 + 2k + 1). Rearranging gives x^2 = (2k + 1)^2 * (y^2 + 2k^2 + 2k + 1). As 2k^2 + 2k + 1 is odd, a careful selection of y makes the RHS square. So [(2k+1) * (k(k + 1) + 1), k(k + 1)]. E.g., if k=2, then (5*7)^2 - 25*6^2 = 1225 - 900 = 325 = 25*26/2. - Jon Perry, Nov 07 2014

No even squares are in the sequence. Proof: Rearrange the equation to read x^2 = n(n + 1 + 2y^2)/2, with n = 4k^2. n + 1 + 2y^2 is always odd and so the RHS contains an odd exponent of 2, and therefore cannot be square. - Jon Perry, Nov 15 2014

From Jon Perry, Nov 15 2014: (Start)

Odd squares + 8 are always in this sequence. Proof: Let m = 4k^2 + 4k + 9 and let n = (m+1)/2 = 2k^2 + 2k + 5.

Rearranging the equation x^2 - m*y^2 = m(m + 1)/2, we get x^2 = m(m + 1 + 2y^2)/2, and so x^2 = m(n + y^2) = (4k^2 + 4k + 9)(2k^2 + 2k + 5 + y^2).

We aim to find a y such that the last bracket on the RHS is z^2 * (4k^2 + 4k + 9), so that x equals z*m. We claim that if we let Y = ((n-3)/2)^2*m - n, then Y is a square, and letting Y = y^2, we have y^2 + n = Y + n = z^2 * m as required, with z = (n-3)/2 = k^2 + k + 1.

To prove that Y is a square, Y = [(n^2 - 6n + 9)*(2n - 1) - 4n]/4 = [2n^3 - 13n^2 + 20n - 9]/4 = [(n-1)^2*(2n-9)]/4, and with n as it is, 2n - 9 = 4k^2 + 4k + 1 = (2k + 1)^2, and so we arrive at Y = [(n-1)^2*(2k+1)^2]/4 = [(n-1)(2k+1)/2]^2 = [(k^2 + k + 2)(2k + 1)]^2, a square as required, with y = (k^2 + k + 2)(2k + 1). Also GCD(n-3,2n-1)=1 as required.

This gives a solution as [(k^2 + k + 1)*(4k^2 + 4k + 9), (k^2 + k + 2)*(2k + 1)]. E.g., if k=4, n=45 and a solution is [21*89, 22*9] = [1869, 198]. To validate, 1869^2 - 89*198^2 = 3493161 - 3489156 = 4005 = 89*45.

(End)

Links

Lars Blomberg, Table of n, a(n) for n = 1..10000

Example

3 is in the sequence because x^2 - 3*y^2 = 6 has integer solutions, including (x, y) = (3, 1) and (9, 5).

Crossrefs

Cf. A134419, A245226.

Sequence in context: A152411 A080517 A264898 * A295289 A212849 A191487

Adjacent sequences: A248191 A248192 A248193 * A248195 A248196 A248197

Keyword

nonn

Author

Colin Barker, Oct 03 2014

Extensions

More terms from Lars Blomberg, Nov 02 2014

"Positive" added to definition by N. J. A. Sloane, Nov 02 2014

Status

approved