A282014 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 x and 121*x + 48*(y-z) are squares.

1, 2, 3, 3, 2, 2, 4, 2, 1, 5, 4, 3, 3, 2, 2, 3, 2, 4, 8, 4, 3, 3, 4, 2, 2, 6, 4, 7, 3, 1, 6, 1, 3, 7, 6, 5, 5, 3, 5, 4, 1, 4, 8, 5, 3, 4, 4, 2, 3, 5, 4, 9, 5, 3, 9, 4, 2, 7, 6, 2, 5, 2, 4, 4, 2, 5, 8, 8, 4, 4, 7, 2, 3, 6, 5, 9, 3, 2, 8, 2, 2

Offset: 0

Author

Zhi-Wei Sun, Feb 04 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 = 8, 29, 31, 40, 94, 104, 143, 319, 671).

The author has proved that any nonnegative integer can be written as the sum of a fourth power and three squares.

We have verified a(n) > 0 for all n = 0..10^7.

See also A281976, A281977 and A282013 for similar conjectures.

Qing-Hu Hou at Tianjin University verified a(n) > 0 for n up to 10^9. - Zhi-Wei Sun, Jun 02 2019

Examples

			a(8) = 1 since 8 = 0^2 + 2^2 + 2^2 + 0^2 with 0 = 0^2 and 121*0 + 48*(2-2) = 0^2.
a(29) = 1 since 29 = 0^2 + 5^2 + 2^2 + 0^2 with 0 = 0^2 and 121*0 + 48*(5-2) = 12^2.
a(31) = 1 since 31 = 1^2 + 2^2 + 1^2 + 5^2 with 1 = 1^2 and 121*1 + 48*(2-1) = 13^2.
a(40) = 1 since 40 = 4^2 + 2^2 + 2^2 + 4^2 with 4 = 2^2 and 121*4 + 48*(2-2) = 22^2.
a(94) = 1 since 94 = 0^2 + 6^2 + 3^2 + 7^2 with 0 = 0^2 and 121*0 + 48*(6-3) = 12^2.
a(104) = 1 since 104 = 4^2 + 6^2 + 6^2 + 4^2 with 4 = 2^2 and 121*4 + 48*(6-6) = 22^2.
a(143) = 1 since 143 = 1^2 + 6^2 + 5^2 + 9^2 with 1 = 1^2 and 121*1 + 48*(6-5) = 13^2.
a(319) = 1 since 319 = 1^2 + 17^2 + 2^2 + 5^2 with 1 = 1^2 and 121*1 + 48*(17-2) = 29^2.
a(671) = 1 since 671 = 9^2 + 5^2 + 23^2 + 6^2 with 9 = 3^2 and 121*9 + 48*(5-23) = 15^2.
		

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, A270969, A271518, A281939, A281941, A281975, A281976, A281977, A282013.

Programs

Mathematica

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
Do[r=0;Do[If[SQ[n-x^4-y^2-z^2]&&SQ[121x^2+48(y-z)],r=r+1],{x,0,n^(1/4)},{y,0,Sqrt[n-x^4]},{z,0,Sqrt[n-x^4-y^2]}];Print[n," ",r];Continue,{n,0,80}]