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A256418
Congrua (possible solutions to the congruum problem): numbers n such that there are integers x, y and z with n = x^2-y^2 = z^2-x^2.
17
24, 96, 120, 216, 240, 336, 384, 480, 600, 720, 840, 864, 960, 1080, 1176, 1320, 1344, 1536, 1920, 1944, 2016, 2160, 2184, 2400, 2520, 2880, 2904, 3000, 3024, 3360, 3456, 3696, 3840, 3960, 4056, 4320, 4704, 4896, 5280, 5376, 5400, 5544
OFFSET
1,1
COMMENTS
n is a "congruum" iff n/4 is the area of a Pythagorean triangle, so these are the numbers 4*A009112.
Each congruum is a multiple of 24; it cannot be a square.
LINKS
Eric Weisstein's World of Mathematics, Congruum (but beware errors)
Wikipedia, Congruum (but beware errors).
EXAMPLE
a(11)=840 since 840=29^2-1^2=41^2-29^2 (indeed also 840=37^2-23^2=47^2-37^2).
MATHEMATICA
r[n_] := Reduce[0 < y < x && 0 < x < z && n == x^2 - y^2 == z^2 - x^2, {x, y, z}, Integers];
Reap[For[n = 24, n < 10^4, n += 24, rn = r[n]; If[rn =!= False, Print[n, " ", rn]; Sow[n]]]][[2, 1]] (* Jean-François Alcover, Feb 25 2019 *)
CROSSREFS
Cf. A004431 for possible values of x in definition. Cf. A057103, A055096 for triangles of all congrua and values of x.
Sequence in context: A283446 A208984 A103251 * A198387 A057102 A057103
KEYWORD
nonn
AUTHOR
N. J. A. Sloane, Apr 06 2015, following a suggestion from Robert Israel, Apr 03. 2015. This entry incorporates many comments that were originally in A057102. A057103 and A055096 need to be checked.
STATUS
approved