

A135789


Positive numbers of the form x^4  6 * x^2 * y^2 + y^4 (where x,y are integers).


9



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
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

1,1


COMMENTS

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 dn, 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Refers to A057102, which had an incorrect description and has been replaced by A256418.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 rechecked.  N. J. A. Sloane, Apr 06 2015


LINKS



MAPLE

isA135789 := proc(n) for d in numtheory[divisors](n) do dprime := n/d ; if abs(ddprime) mod 4 = 0 then for x from 1 to d/3 do y := (ddprime)/4/x ; if type(y, 'integer') and y< x and y> 0 then if n = (x^2y^2+2*x*y)*(x^2y^22*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


MATHEMATICA

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]


CROSSREFS



KEYWORD

nonn


AUTHOR



EXTENSIONS



STATUS

approved



