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A239064
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Positive integers whose square can be expressed as k*x^2+(k+1)*x+(k+2) with k and x positive integers.
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0
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3, 5, 6, 9, 10, 12, 15, 16, 18, 19, 21, 23, 24, 26, 27, 30, 33, 36, 37, 39, 40, 41, 42, 44, 45, 47, 48, 51, 52, 54, 57, 58, 60, 61, 63, 64, 65, 66, 67, 68, 69, 72, 75, 78, 79, 81, 82, 83, 84, 86, 87, 89, 90, 92, 93, 96, 99, 100, 102, 103, 105, 107, 108, 110, 111, 114, 117, 120, 121, 123, 124, 126, 128, 129, 130, 131, 132, 134, 135, 138, 141, 142, 144, 145, 147, 149, 150, 152, 153, 156, 159, 162, 163, 165, 166, 168, 170, 171, 173, 174, 176, 177, 180, 183, 184, 186, 187, 189, 191, 192, 194, 195, 196, 198, 201, 204
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OFFSET
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1,1
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COMMENTS
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Some can be expressed in more than one way. E.g., a(46) = 81 for x = 1, k = 26 and for x = 2 and k = 11.
x=1 for k=3*m^2-1 that is A080663, with a = 3*m.
The solution of this problem can be found as follows. Consider k*x^2 + (k+1)*x + (k+2) - a^2 = 0. Solve for x (positive) as function of k: x = (-(k+1) + sqrt((2*a)^2*k - c(k)))/(2*k), where c(k) = 3*k(k+2) - 1 is the sequence [8, 23, 44, 71, 104, 143, 188, 239, 296, 359,...]. A necessary condition for a solution is (2*a)^2*k - c(k) = b^2, or k*X^2 - Y^2 = c(k), with X = 2*a and Y = b. This is a binary indefinite quadratic form with discriminant D = (2^2)*k > 0. If k is a square then there is no solution with even X because (K*X)^2 - Y^2 = c(K^2) has only the solution with K=1, X=3, Y=1. For k not a square there are either no solutions for given k (e.g., k = 5, 6, 10, ...) or countable infinitely many ones of the representation k*X^2 - Y^2 = c(k). From the solutions one has to pick first the ones with even X (a = X/2) and then to test whether 2*k divides Y - (k+1). The solution for x is then x = (Y - (k+1))/(2*k).
For representations of integers by indefinite binary quadratic forms see, e.g., the D. A. Buell reference, and also a W. Lang link (with further references) with an on-line program.
(End)
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REFERENCES
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D. A. Buell, Binary Quadratic Forms. Springer-Verlag, NY, 1989.
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LINKS
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FORMULA
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Increasingly ordered set {a from the positive integers: a^2 = k*x^2 + (k+1)*x + (k+2), with k and x positive integers}. - Wolfdieter Lang, Apr 21 2014
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EXAMPLE
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a(10)=19 from 19^2=361=51*2^2+(51+1)*2+(51+2).
k = 2: There are two proper solutions of 2*X^2 - Y^2 = 23, namely [4, 3], [6, 7]. Both generate infinitely many new solutions, all with even X, namely [4, 3], [18, 25], [104, 147], [606, 857],[3532, 4995], [20586, 29113], ... and [6, 7], [32, 45], [186, 263], [1084, 1533],[6318, 8935], [36824, 52077], ... . Only every other solution has 2*k = 4 dividing Y - (2+1) = Y-3, giving the positive solutions for (a=X/2, b=Y; x), starting with the second proper solution and then alternating between the two sets of solutions (3, 7; 1), (52, 147; 36), (93, 263; 65), (1766, 4995; 1248), (3159, 8935; 2233), ... . Thus the positive x solutions for k = 2 are 1, 36, 65, 1248, 2233, ..., with a = 3, 52, 93, 1766, 3159, ... .
k = 3: the positive solutions for x are 2, 5, 38, 79, 538, 1109, ..., with a = 5, 10, 67, 138, 933, 1922, ..., coming from the even X solutions of 3*X^2 - Y^2 = 44, [4, 2], [10, 16], [36, 62], [134, 232], [500, 866], [1866, 3232], ... and [6, 8], [20, 34], [74, 128], [276, 478], [1030, 1784], [3844, 6658], ... . Then 2*k = 6 has to divide Y - 4, leaving every other of these solutions with (a = X/2, b=Y; x) given by (5, 16; 2), (10, 34; 5), (67, 232; 38), (138, 478; 79), (933, 3232; 538), (1922, 6658; 1109), ... .
k = 5: there are no solutions of 5*X^2 - Y2 = 104.
k = 6: there are no solutions of 6*X^2 - Y2 = 143.
(End)
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PROG
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(BASIC)
for k=1 to z
for x=1 to z
a=k*x*x+(k+1)*x+(k+2)
if sqr(a)-int(sqr(a))=0 then
j=j+1
a_n(j)=sqr(a)
endif
next x
next k
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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Edited, name specified, comment changed, reference and link added by Wolfdieter Lang, Apr 21 2014
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STATUS
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approved
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