OFFSET
1,3
COMMENTS
It is easy to show that ( a=x, b=x^3, c=x ) satisfies the constraint for any value of x, so the sequence is infinite. The more interesting values, such as ( 240, 3360, 6480, 46816 ), do not seem to have such an easy explanation.
This sequence follows from International Math Olympiad problem #6 1988.
If m is an integer such that 3*m^2+1 is square (i.e. m is in A001353), then n = 4*m*(2*m+sqrt(3*m^2+1)) is a term with a = 2*m, b = 4*m+2*sqrt(3*m^2+1) and c = 2, thus forming another infinite subsequence. The terms 240, 3360, 46816, 652080 correspond to these cases with m = 4, 15, 56, 209 respectively. Furthermore, for a, b, c corresponding to members of the sequence, a^2*c^4+4*(c^2-a^2) and b^2*c^4+4*(c^2-b^2) must be squares. - Chai Wah Wu, Apr 20 2016
LINKS
Giovanni Resta and Chai Wah Wu, Table of n, a(n) for n = 1..10445 a(n) for n = 1..346 from Giovanni Resta
Yimin Ge, The Method of Vieta Jumping.
EXAMPLE
6480 = 27 * 240. (27^2+ 240^2)/(1+6480) = 3^2.
MATHEMATICA
IntPairs[0] = {{0, 0}}; IntPairs[n_] := Union[Times @@ # & /@ Subsets[Flatten[ FactorInteger[n] /. {x_Integer, y_Integer} :> Table[x, {y}] ]][[2 ;; -1]]] /. {x_Integer :> {x , n/x}}; TestAB[n_] := Or @@ (IntegerQ /@ (IntPairs[n] /. {a_, b_} :> Divide[a^2 + b^2, 1 + a b])); b[n_] := DeleteCases[If[TestAB[#], #] & /@ Range[0, n], Null]; b[100]
ok[n_] := n == 0 || Block[{d = Divisors@n}, {} != Select[Take[d, Ceiling[ Length@d/2]], Mod[(#^2 + (n/#)^2), 1 + n] == 0 &, 1]]; Select[Range[0, 10^5], ok] (* faster, Giovanni Resta, Apr 17 2016 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Bradley Klee, Apr 16 2016
STATUS
approved