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A135789
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Positive numbers of the form x^4 - 6 * x^2 * y^2 + y^4 (where x,y are integers).
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9
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28, 41, 161, 448, 476, 656, 721, 956, 1081, 1241, 1393, 2108, 2268, 2576, 3281, 3321, 3713, 3836, 4633, 4681, 5593, 6076, 7168, 7616, 8188, 9401, 9641, 10496, 11536, 11753, 12121, 12593, 13041, 13916, 15296, 16828, 17296, 17500, 19516, 19856
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OFFSET
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1,1
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COMMENTS
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Squares of these numbers are of the form N^4 - M^2 (where N belongs to A135786 and M to A057102). Proof is based on the identity (x^4 - 6x^2 * y^2 + y^4)^2 = (x^2 + y^2)^4 - (4(x^3y - xy^3))^2.
Since x^4 - 6x^2 * y^2 + y^4 = d*d' where d = x^2 - y^2 + 2xy and d' = x^2 - y^2 - 2xy, and d - d' = 4xy, the computational technique is to consider the divisors d|n, d'=n/d, to check that the difference is a multiple of 4, and to check x in the range 1..d/3. - R. J. Mathar, Sep 18 2009
Refers to A057102, which had an incorrect description and has been replaced by A256418. As a result the present sequence should be re-checked. - N. J. A. Sloane, Apr 06 2015
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LINKS
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MAPLE
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isA135789 := proc(n) for d in numtheory[divisors](n) do dprime := n/d ; if abs(d-dprime) mod 4 = 0 then for x from 1 to d/3 do y := (d-dprime)/4/x ; if type(y, 'integer') and y< x and y> 0 then if n = (x^2-y^2+2*x*y)*(x^2-y^2-2*x*y) then RETURN(true); fi; fi; od: fi: od: RETURN(false) ; end: for n from 1 do if isA135789(n) then printf("%d, \n", n) ; fi; od: # R. J. Mathar, Sep 18 2009
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MATHEMATICA
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a = {}; Do[Do[w = x^4 - 6x^2 y^2 + y^4; If[w > 0&&w<10000, AppendTo[a, w]], {x, y, 2000}], {y, 1, 2000}]; Union[a]
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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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STATUS
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approved
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