OFFSET
1,2
COMMENTS
Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 7, 39, 47, 95, 191, 239, 327, 439, 871, 1167, 1199, 1367, 1487, 1727, 1751, 2063, 2351, 2471, 4647, 4^k*m (k = 0,1,2,... and m = 1, 3).
(ii) 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)*x*y*z is a square, whenever (a,b,c) is among the triples (1,3,7), (1,5,7), (1,5,11), (1,13,23), (2,4,6), (2,4,8), (2,6,8), (2,8,26), (3,5,21), (3,7,15), (3,9,43), (3,9,69), (3,9,141), (3,21,27), (3,27,39), (3,33,45), (3,39,123), (6,8,12), (6,8,18), (6,8,22), (6,8,28), (6,12,48), (6,18,132), (6,24,34), (6,24,36), (6,42,72), (7,13,29), (7,19,23), (12,18,24), (12,18,30), (12,26,48), (13,15,21), (13,17,19), (13,33,39), (14,28,58), (15,45,51), (16,22,62), (18,22,24), (21,27,33), (21,27,57), (23,37,61), (24,54,66), (33,57,79), (38,48,66), (42,58,84), (46,92,118).
For more refinements of Lagrange's four-square theorem, see arXiv:1604.06723.
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.
Zhi-Wei Sun, Refine Lagrange's four-square theorem, a message to Number Theory List, April 26, 2016.
EXAMPLE
a(3) = 1 since 3 = 1^2 + 1^2 + 0^2 + 1^2 with 1 > 0 and (1+3*1+13*0)*1*1*0 =0^2.
a(4) = 1 since 4 = 2^2 + 0^2 + 0^2 + 0^2 with 2 > 0, 0 = 0 and (2+3*0+13*0)*2*0*0 = 0^2.
a(7) = 1 since 7 = 2^2 + 1^2 + 1^2 + 1^2 with 2 > 0, 1 = 1 and
(2+3*1+13*1)*2*1*1 = 6^2.
a(39) = 1 since 39 = 2^2 + 3^2 + 1^2 + 5^2 with 2 > 0, 3 > 1 and (2+3*3+13*1)*2*3*1 = 12^2.
a(47) = 1 since 47 = 2^2 + 3^2 + 3^2 + 5^2 with 2 > 0, 3 = 3 and (2+3*3+13*3)*2*3*3 = 30^2.
a(95) = 1 since 95 = 2^2 + 3^2 + 1^2 + 9^2 with 2 > 0, 3 > 1 and (2+3*3+13*1)*2*3*1 = 12^2.
a(191) = 1 since 191 = 2^2 + 3^2 + 3^2 + 13^2 with 2 > 0, 3 = 3 and (2+3*3+13*3)*2*3*3 = 30^2.
a(239) = 1 since 239 = 2^2 + 3^2 + 1^2 + 15^2 with 2 > 0, 3 > 1 and (2+3*3+13*1)*2*3*1 = 12^2.
a(327) = 1 since 327 = 11^2 + 3^2 + 1^2 + 14^2 with 11 > 0, 3 > 1 and (11+3*3+13*1)*11*3*1 = 33^2.
a(439) = 1 since 439 = 10^2 + 5^2 + 5^2 + 17^2 with 10 > 0, 5 = 5 and (10+3*5+13*5)*10*5*5 = 150^2.
a(871) = 1 since 871 = 21^2 + 15^2 + 3^2 + 14^2 with 21 > 0, 15 > 3 and (21+3*15+13*3)*21*15*3 = 315^2.
a(1167) = 1 since 1167 = 22^2 + 11^2 + 11^2 + 21^2 with 22 > 0, 11 = 11 and (22+3*11+13*11)*22*11*11 = 726^2.
a(1199) = 1 since 1199 = 14^2 + 21^2 + 21^2 + 11^2 with 14 > 0, 21 = 21 and (14+3*21+13*21)*14*21*21 = 1470^2.
a(1367) = 1 since 1367 = 14^2 + 21^2 + 21^2 + 17^2 with 14 > 0, 21 = 21 and (14+3*21+13*21)*14*21*21 = 1470^2.
a(1487) = 1 since 1487 = 9^2 + 29^2 + 6^2 + 23^2 with 9 > 0, 29 > 6 and (9+3*29+13*6)*9*29*6 = 522^2.
a(1727) = 1 since 1727 = 2^2 + 21^2 + 21^2 + 29^2 with 2 > 0, 21 = 21 and (2+3*21+13*21)*2*21*21 = 546^2.
a(1751) = 1 since 1751 = 9^2 + 17^2 + 15^2 + 34^2 with 9 > 0, 17 > 15 and (9+3*17+13*15)*9*17*15 = 765^2.
a(2063) = 1 since 2063 = 18^2 + 19^2 + 3^2 + 37^2 with 18 > 0, 19 > 3 and (18+3*19+13*3)*18*19*3 = 342^2.
a(2351) = 1 since 2351 = 15^2 + 35^2 + 15^2 + 26^2 with 15 > 0, 35 > 15 and (15+3*35+13*15)*15*35*15 = 1575^2.
a(2471) = 1 since 2471 = 1^2 + 18^2 + 11^2 + 45^2 with 1 > 0, 18 > 11 and (1+3*18+13*11)*1*18*11 = 198^2.
a(4647) = 1 since 4647 = 10^2 + 45^2 + 29^2 + 41^2 with 10 > 0, 45 > 29 and (10+3*45+13*29)*10*45*29 = 2610^2.
MATHEMATICA
SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
Do[r=0; Do[If[SQ[n-x^2-y^2-z^2]&&SQ[(x+3y+13z)x*y*z], r=r+1], {x, 1, Sqrt[n]}, {z, 0, Sqrt[(n-x^2)/2]}, {y, z, Sqrt[n-x^2-z^2]}]; Print[n, " ", r]; Label[aa]; Continue, {n, 1, 80}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Apr 30 2016
STATUS
approved