A248194 - OEIS

%I #57 Nov 21 2014 09:17:30

%S 1,3,8,9,17,19,24,25,33,49,51,57,67,72,73,81,88,89,96,97,99,121,129,

%T 136,147,152,163,169,177,179,193,201,211,225,233,241,243,249,264,288,

%U 289,297,313,337,339,352,361,369,387,393,408,409,441,449,451,456,457

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

%C 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

%C 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

%C From _Jon Perry_, Nov 15 2014: (Start)

%C 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.

%C 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).

%C 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.

%C 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.

%C 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.

%C (End)

%H Lars Blomberg, <a href="/A248194/b248194.txt">Table of n, a(n) for n = 1..10000</a>

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

%Y Cf. A134419, A245226.

%K nonn

%O 1,2

%A _Colin Barker_, Oct 03 2014

%E More terms from _Lars Blomberg_, Nov 02 2014

%E "Positive" added to definition by _N. J. A. Sloane_, Nov 02 2014