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A173200
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Solutions y of the Mordell equation y^2 = x^3 - 3a^2 - 1 for a = 0,1,2, ... (solutions x are given by A053755).
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2
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0, 11, 70, 225, 524, 1015, 1746, 2765, 4120, 5859, 8030, 10681, 13860, 17615, 21994, 27045, 32816, 39355, 46710, 54929, 64060, 74151, 85250, 97405, 110664, 125075, 140686, 157545, 175700, 195199, 216090, 238421, 262240, 287595, 314534, 343105
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OFFSET
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1,2
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COMMENTS
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For many values of k for the equation y^2 = x^3 + k, all the solutions are known. For example, we have solutions for k=-2: (x,y) = (3,-5) and (3,5). A complete resolution for all integers k is unknown. Theorem: Let k be < -1, free of square factors, with k == 2 or 3 (mod 4). Suppose that the number of classes h(Q(sqrt(k))) is not divisible by 3. Then the equation y^2 = x^3 + k admits integer solutions if and only if k = 1 - 3a^2 or 1 - 3a^2 where a is integer. In this case, the solutions are x = a^2 - k, y = a(a^2 + 3k) or -a(a^2 + 3k) (the first reference gives the proof of this theorem). With k = -1 - 3a^2, we obtain the solutions x = 4a^2 + 1, y = a(8a^2 + 3) or -a(8a^2 + 3). For the case k = 1 - 3a^2, we obtain the solution x = 4a^2 - 1 given by the sequence A000466.
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REFERENCES
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T. Apostol, Introduction to Analytic Number Theory, Springer, 1976
D. Duverney, Théorie des nombres (2e edition), Dunod, 2007, p.151
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LINKS
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FORMULA
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y = a*(8*a^2 + 3).
a(n) = 8*n^3 - 24*n^2 + 27*n - 11.
G.f.: x^2*(11 + 26*x + 11*x^2)/(1 - x)^4. (End)
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EXAMPLE
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With a=3, x =37 and y = 225, and then 225^2 = 37^2 - 28.
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MAPLE
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for a from 0 to 150 do : z := evalf(a*(8*a^2 + 3)) : print (z) :od :
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MATHEMATICA
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CoefficientList[Series[x*(11+26*x+11*x^2)/(1-x)^4, {x, 0, 40}], x] (* Vincenzo Librandi, Jul 02 2012 *)
LinearRecurrence[{4, -6, 4, -1}, {0, 11, 70, 225}, 40] (* Harvey P. Dale, Dec 21 2016 *)
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PROG
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(Magma) I:=[0, 11, 70, 225]; [n le 4 select I[n] else 4*Self(n-1)-6*Self(n-2)+4*Self(n-3)-Self(n-4): n in [1..40]]; // Vincenzo Librandi, Jul 02 2012
(Python) for n in range(1, 20): print(8*n**3 - 24*n**2 + 27*n - 11, end=', ') # Stefano Spezia, Dec 05 2018
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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