OFFSET
1,2
COMMENTS
Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 4^k*m (k = 0,1,2,... and m = 1, 7, 23, 31, 39, 47, 55, 71, 79, 119, 151, 191, 311, 671).
(ii) Any natural number can be written as x^2 + y^2 + z^2 + w^2 with (x+y+z)^2 + (4*(x+y-z))^2 a square, where x,y,z,w are nonnegative integers with x+y >= z.
(iii) For each tuple (a,b,c,d,e,f) = (1,1,1,3,6,-3), (1,1,1,4,12,-12), (1,1,2,1,1,-5), (1,1,2,1,8,-5), (1,1,2,3,3,-3), (1,1,2,4,4,-8), (1,3,11,12,4,4), (1,3,14,16,4,4), (1,3,14,18,4,2), (1,3,20,16,4,12), (1,4,11,6,3,3), (1,5,13,12,12,12), (1,5,14,15,12,21), (1,6,6,16,8,8), (1,6,14,12,8,8), (1,6,14,16,8,4), (1,6,17,20,8,4), (1,6,20,20,8,8), (1,7,8,4,2,6), (1,7,8,10,5,15), (1,7,9,10,5,12), (1,7,15,4,2,8), (1,7,15,10,5,20), any natural number can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that (a*x+b*y+c*z)^2 + (d*x+e*y+f*z)^2 is a square.
It was proved in arXiv:1604.06723 that any positive integer can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers and y > 0 such that x+4*y+4*z and 9*x+3*y+3*z are the two legs of a right triangle with positive integer sides.
LINKS
Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Zhi-Wei Sun, Refining Lagrange's four-square theorem, arXiv:1604.06723 [math.GM], 2016.
EXAMPLE
a(1) = 1 since 1 = 0^2 + 1^2 + 0^2 + 0^2 with 1 > 0 = 0 and (0+4*1+4*0)^2 + (9*0+3*1+3*0)^2 = 5^2.
a(7) = 1 since 7 = 2^2 + 1^2 + 1^2 + 1^2 with 0 < 1 = 1 = 1 and (2+4*1+4*1)^2 + (9*2+3*1+3*1)^2 = 26^2.
a(23) = 1 since 23 = 3^2 + 2^2 + 1^2 + 3^2 with 2 > 1 < 3 and (3+4*2+4*1)^2 + (9*3+3*2+3*1)^2 = 39^2.
a(31) = 1 since 31 = 2^2 + 1^2 + 1^2 + 5^2 with 0 < 1 = 1 < 5 and (2+4*1+4*1)^2 + (9*2+3*1+3*1)^2 = 26^2.
a(39) = 1 since 39 = 3^2 + 2^2 + 1^2 + 5^2 with 2 > 1 < 5 and (3+4*2+4*1)^2 + (9*3+3*2+3*1)^2 = 39^2.
a(47) = 1 since 47 = 5^2 + 3^2 + 2^2 + 3^2 with 3 > 2 < 3 and (5+4*3+4*2)^2 + (9*5+3*3+3*2)^2 = 65^2.
a(55) = 1 since 55 = 2^2 + 1^2 + 1^2 + 7^2 with 0 < 1 = 1 < 7 and (2+4*1+4*1)^2 + (9*2+3*1+3*1)^2 = 26^2.
a(71) = 1 since 71 = 6^2 + 5^2 + 1^2 + 3^2 with 5 > 1 < 3 and (6+4*5+4*1)^2 + (9*6+3*5+3*1)^2 = 78^2.
a(79) = 1 since 79 = 6^2 + 3^2 + 3^2 + 5^2 with 0 < 3 = 3 < 5 and (6+4*3+4*3)^2 + (9*6+3*3+3*3)^2 = 78^2.
a(119) = 1 since 119 = 5^2 + 3^2 + 2^2 + 9^2 with 3 > 2 < 9 and (5+4*3+4*2)^2 + (9*5+3*3+3*2)^2 = 65^2.
a(151) = 1 since 151 = 9^2 + 6^2 + 3^2 + 5^2 with 6 > 3 < 5 and (9+4*6+4*3)^2 + (9*9+3*6+3*3)^2 = 117^2.
a(191) = 1 since 191 = 10^2 + 9^2 + 1^2 + 3^2 with 9 > 1 < 3 and (10+4*9+4*1)^2 + (9*10+3*9+3*1)^2 = 130^2.
a(311) = 1 since 311 = 7^2 + 6^2 + 1^2 + 15^2 with 6 > 1 < 15 and (7+4*6+4*1)^2 + (9*7+3*6+3*1)^2 = 91^2.
a(671) = 1 since 671 = 17^2 + 11^2 + 6^2 + 15^2 with 11 > 6 < 15 and (17+4*11+4*6)^2 + (9*17+3*11+3*6)^2 = 221^2.
MATHEMATICA
SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
Do[r=0; Do[If[SQ[n-x^2-y^2-z^2]&&SQ[(x+4y+4z)^2+(9x+3y+3z)^2], r=r+1], {x, 0, Sqrt[n]}, {z, 0, Sqrt[(n-x^2)/3]}, {y, Max[1, z], Sqrt[n-x^2-2z^2]}]; Print[n, " ", r]; Continue, {n, 1, 80}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, May 15 2016
STATUS
approved