A282463 Number of ways to write n as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers, x == y (mod 2) and z <= w such that both x and x^2 + 62*x*y + y^2 are squares.

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

Offset: 0

Author

Zhi-Wei Sun, Feb 16 2017

Keywords

• nonn

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.

Examples

			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.
		

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, A281976, A281977, A282013, A282014.

Programs

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}]