OFFSET
0,5
COMMENTS
Conjecture: (i) a(n) > 0 for any nonnegative integer n. Also, each n = 0,1,2,... can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that x^3 + ((y-z)/2)^3 is a square (or twice a square).
(ii) Let a and b be positive integers with gcd(a,b) squarefree. Then, every n = 0,1,2,... can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that a*x^3 + b*(y-z)^3 is a square, if and only if (a,b) is among the ordered pairs (1,1), (1,9), (2,18), (8,1), (9,5), (9,8), (9,40), (16,2), (18,16), (25,16), (72,1).
We have verified that a(n) > 0 for all n = 0..2*10^6.
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(56) = 1 since 56 = 0^2 + 6^2 + 2^2 + 4^2 with 72*0^3 + (6-2)^3 = 8^2.
a(120) = 1 since 120 = 4^2 + 2^2 + 10^2 + 0^2 with 72*4^3 + (2-10)^3 = 64^2.
a(159) = 1 since 159 = 2^2 + 3^2 + 11^2 + 5^2 with 72*2^3 + (3-11)^3 = 8^2.
a(1646) = 1 since 1646 = 5^2 + 10^2 + 0^2 + 39^2 with 72*5^3 + (10-0)^3 = 100^2.
a(1784) = 1 since 1784 = 12^2 + 22^2 + 30^2 + 16^2 with 72*12^3 + (22-30)^3 = 352^2.
a(3914) = 1 since 3914 = 2^2 + 45^2 + 21^2 + 38^2 with 72*2^3 + (45-21)^3 = 120^2.
a(5864) = 1 since 5864 = 50^2 + 0^2 + 0^2 + 58^2 with 72*50^3 + (0-0)^3 = 3000^2.
MATHEMATICA
SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
Do[r=0; Do[If[SQ[n-x^2-y^2-z^2]&&SQ[72x^3+(y-z)^3]&&Mod[y-z, 2]==0, r=r+1], {x, 0, Sqrt[n]}, {y, 0, Sqrt[n-x^2]}, {z, 0, Sqrt[n-x^2-y^2]}]; Print[n, " ", r], {n, 0, 80}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Mar 14 2017
STATUS
approved