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A134419
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Numbers n for which the generalized Pell equation x^2 - n*y^2 = n(n-1)(n+1)/3 has an integer solution for x and y.
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12
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1, 2, 4, 11, 16, 23, 24, 25, 26, 33, 47, 49, 50, 52, 59, 64, 73, 74, 88, 96, 97, 100, 107, 121, 122, 146, 148, 169, 177, 184, 191, 193, 194, 196, 218, 239, 241, 242, 244, 249, 256, 276, 289, 292, 297, 299, 311, 312, 313, 337, 338, 347, 352, 361, 362, 376, 383
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OFFSET
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1,2
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COMMENTS
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This generalized Pell equation appears in the solution of problems posed in A001032 (and A001033): numbers n such that the sum of squares of n consecutive (odd) positive integers is a square. This sequence is the union of A001032, A001033 and the number 4, which is not a solution to either problem. When n is a square > 1 and not divisible by 3, then the equation has only a finite number of solutions; otherwise it has an infinite number of solutions.
For an n in this sequence, consider solutions with x>0 and y>n. (For n=4, there will be no such solutions.) If y-n+1 is even, then n is in A001032, the n consecutive positive integers begin with (y-n+1)/2 and the sum of the squares is x/2. If y-n+1 is odd, then the n is in A001033, the n consecutive odd positive integers begin with y-n+1 and the sum of the squares is x. For some n, such as 33, there are solutions y1 and y2 such that y1-n+1 is even and y2-n+1 is odd. In this case, n is in both A001032 and A001033.
The reason that 4 is not in A001033 is that there is no sequence of 4 consecutive positive odd squares that add to a square. However, there is a sequence of 4 consecutive odd integers whose squares add up to a square: (-1)^2 + 1^2 + 3^2 + 5^2 = 6^2. - Thomas Andrews, Feb 22 2011
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LINKS
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MATHEMATICA
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t={}; n=0; While[Length[t]<200, n++; If[Reduce[x^2-n*y^2==n(n^2-1)/3, {x, y}, Integers] =!= False, AppendTo[t, n]]]; t
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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