OFFSET
0,3
COMMENTS
Conjecture: (i) a(n) > 0 for all n = 0,1,2,..., and a(n) = 1 only for n = 0, 1, 3, 43, 723, 1723, 3571, 3911 and 16^k*m (k = 0,1,2,... and m = 7, 23, 31, 71, 79, 143, 303, 1591).
(ii) Any nonnegative integer can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that both x and a*x^2 + b*x*y + c*y^2 are squares, whenever (a,b,c) is among the ordered triples (84,84,1), (16,144,9), (153,36,100), (177,214,9), (249,114,121).
The author has proved that any nonnegative integer can be expressed as the sum of a fourth power and three squares.
LINKS
Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
Zhi-Wei Sun, Refining Lagrange's four-square theorem, J. Number Theory 175(2017), 167-190.
Zhi-Wei Sun, Restricted sums of four squares, arXiv:1701.05868 [math.NT], 2017.
EXAMPLE
a(7) = 1 since 7 = 1^2 + 1^2 + 1^2 + 2^2 with 1 == 1 (mod 2), 1 = 1^2 and 1^2 + 62*1*1 + 1^2 = 8^2.
a(23) = 1 since 23 = 1^2 + 3^2 + 2^2 + 3^2 with 1 == 1 (mod 3), 1 = 1^2 and 1^2 + 62*1*3 + 3^2 = 14^2.
a(30) = 2 since 30 = 0^2 + 2^2 + 1^2 + 5^2 with 0 == 2 (mod 3), 0 = 0^2 and 0^2 + 62*0*2 + 2^2 = 2^2, and 30 = 1^2 + 3^2 + 2^2 + 4^2 with 1 == 3 (mod 2), 1 = 1^2 and 1^2 + 62*1*3 + 3^2 = 14^2.
a(79) = 1 since 79 = 1^2 + 7^2 + 2^2 + 5^2 with 1 == 7 (mod 2), 1 = 1^2 and 1^2 + 62*1*7 + 7^2 = 22^2.
a(143) = 1 since 143 = 9^2 + 3^2 + 2^2 + 7^2 with 9 == 3 (mod 2), 9 = 3^2 and 9^2 + 62*9*3 + 3^2 = 42^2.
a(303) = 1 since 303 = 1^2 + 3^2 + 2^2 + 17^2 with 1 == 3 (mod 2), 1 = 1^2 and 1^2 + 62*1*3 + 3^2 = 14^2.
a(723) = 1 since 723 = 1^2 + 7^2 + 12^2 + 23^2 with 1 == 7 (mod 2), 1 = 1^2 and 1^2 + 62*1*7 + 7^2 = 22^2.
a(1591) = 1 since 1591 = 9^2 + 9^2 + 23^2 + 30^2 with 9 == 9 (mod 2), 9 = 3^2 and 9^2 + 62*9*9 + 9^2 = 72^2.
a(1723) = 1 since 1723 = 1^2 + 1^2 + 11^2 + 40^2 with 1 == 1 (mod 2), 1 = 1^2 and 1^2 + 62*1*1 + 1^2 = 8^2.
a(3571) = 1 since 3571 = 9^2 + 3^2 + 0^2 + 59^2 with 9 == 3 (mod 2), 9 = 3^2 and 9^2 + 62*9*3 + 3^2 = 42^2.
a(3911) = 1 since 9^2 + 3^2 + 10^2 + 61^2 with 9 == 3 (mod 2), 9 = 3^2 and 9^2 + 62*9*3 + 3^2 = 42^2.
MATHEMATICA
SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
Do[r=0; Do[If[SQ[n-x^4-y^2-z^2]&&Mod[x-y, 2]==0&&SQ[x^4+62*x^2*y+y^2], r=r+1], {x, 0, n^(1/4)}, {y, 0, Sqrt[n-x^4]}, {z, 0, Sqrt[(n-x^4-y^2)/2]}]; Print[n, " ", r]; Continue, {n, 0, 80}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Feb 16 2017
STATUS
approved