A282562 Number of ways to write n as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that both 2*x - y and 9*z^2 + 666*z*w + w^2 are squares.

1, 2, 2, 2, 4, 4, 5, 2, 3, 5, 2, 2, 1, 5, 6, 2, 2, 5, 5, 1, 5, 7, 8, 1, 3, 4, 2, 4, 3, 9, 7, 5, 2, 5, 5, 4, 8, 3, 9, 3, 4, 4, 5, 3, 1, 7, 6, 3, 2, 13, 7, 5, 5, 5, 8, 3, 2, 3, 5, 2, 1, 6, 10, 6, 4, 8, 9, 1, 7, 8, 11, 3, 3, 6, 6, 5, 3, 11, 4, 4, 4

Offset: 0

Author

Zhi-Wei Sun, Feb 18 2017

Keywords

• nonn

Comments

Conjecture: a(n) > 0 for all n = 0,1,2,..., and a(n) = 1 only for n = 0, 16^k*m (k = 0,1,2,... and m = 12, 19, 23, 44, 60, 67, 139, 140, 248, 264, 347, 427, 499, 636, 1388, 1867, 1964, 4843).

By the linked JNT paper, any nonnegative integer can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers and z*(z-w) = 0. Whether z = 0 or z = w, the number 9*z^2 + 666*z*w + w^2 is definitely a square.

See also A282561 for a similar conjecture.

Examples

			a(12) = 1 since 12 = 2^2 + 0^2 + 2^2 + 2^2 with 2*2 - 0 = 2^2 and 9*2^2 + 666*2*2 + 2^2 = 52^2.
a(19) = 1 since 19 = 1^2 + 1^2 + 4^2 + 1^2 with 2*1 - 1 = 1^2 and 9*4^2 + 666*4*1 + 1^2 = 53^2.
a(44) = 1 since 44 = 5^2 + 1^2 + 3^2 + 3^2 with 2*5 - 1 = 3^2 and 9*3^2 + 666*3*3 + 3^2 = 78^2.
a(60) = 1 since 60 = 3^2 + 5^2 + 1^2 + 5^2 with 2*3 - 5 = 1^2 and 9*1^2 + 666*1*5 + 5^2 = 58^2.
a(67) = 1 since 67 = 4^2 + 7^2 + 1^2 + 1^2 with 2*4 - 7 = 1^2 and 9*1^2 + 666*1*1 + 1^2 = 26^2.
a(139) = 1 since 139 = 8^2 + 7^2 + 1^2 + 5^2 with 2*8 - 7 = 3^2 and 9*1^2 + 666*1*5 + 5^2 = 58^2.
a(140) = 1 since 140 = 3^2 + 5^2 + 5^2 + 9^2 with 2*3 - 5 = 1^2 and 9*5^2 + 666*5*9 + 9^2 = 174^2.
a(264) = 1 since 264 = 8^2 + 0^2 + 10^2 + 10^2 with 2*8 - 0 = 4^2 and 9*10^2 + 666*10*10 + 10^2 = 260^2.
a(499) = 1 since 499 = 7^2 + 5^2 + 20^2 + 5^2 with 2*7 - 5 = 3^2 and 9*20^2 + 666*20*5 + 5^2 = 265^2.
a(1388) = 1 since 1388 = 15^2 + 21^2 + 19^2 + 19^2 with 2*15 - 21 = 3^2 and 9*19^2 + 666*19*19 + 19^2 = 494^2.
a(1867) = 1 since 1867 = 16^2 + 31^2 + 5^2 + 25^2 with 2*16 - 31 = 1^2 and 9*5^2 + 666*5*25 + 25^2 = 290^2.
a(4843) = 1 since 4843 = 11^2 + 13^2 + 52^2 + 43^2 with 2*11 - 13 = 3^2 and 9*52^2 + 666*52*43 + 43^2 = 1231^2.
		

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.

Crossrefs

Cf. A000118, A000290, A271518, A281939, A281976, A282463, A282494, A282495, A282545, A282561.

Programs

Mathematica

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
Do[r=0;Do[If[SQ[2x-y],Do[If[SQ[n-x^2-y^2-z^2]&&SQ[9z^2+666z*Sqrt[n-x^2-y^2-z^2]+(n-x^2-y^2-z^2)],r=r+1],{z,0,Sqrt[n-x^2-y^2]}]],{y,0,Sqrt[4n/5]},{x,Ceiling[y/2],Sqrt[n-y^2]}];Print[n," ",r];Continue,{n,0,80}]