A300219 Number of ways to write n^2 as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers and z <= w such that both x and 4*x - 3*y are powers of 4 (including 4^0 = 1).

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

Offset: 1

Author

Zhi-Wei Sun, Feb 28 2018

Keywords

• nonn

Comments

Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 4^k*m (k = 0,1,2,... and m = 1, 2, 3, 5, 15, 37, 83, 263). Also, for each n = 2,3,... we can write n^2 as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that both x and 4*x - 3*y lie in the set {2^(2k+1): k = 0,1,...}.

(ii) Let r be 0 or 1, and let n > r. Then n^2 can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that both x and x + 3*y belong to the set {2^(2k+r): k = 0,1,2,...}, unless n has the form 2^(2k+r)*81503 with k a nonnegative integer and hence n^2 = (2^(2k+r)*28^2)^2 + (2^(2k+r)*80)^2 + (2^(2k+r)*55937)^2 + (2^(2k+r)*59272)^2 with 2^(2k+r)*28^2 = 2^r*(2^k*28)^2 and 2^(2k+r)*28^2 + 3*(2^(2k+r)*80) = 2^(2(k+5)+r). So we always can write n^2 as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that x/2^r is a square and (x+3*y)/2^r is a power of 4.

In arXiv:1701.05868 the author proved that for each r = 0,1 and n > r we can write n^2 as (2^(2k+r))^2 + x^2 + y^2 + z^2 with k,x,y,z nonnegative integers.

We have verified both parts of the conjecture for n up to 10^7.

Examples

			a(2) = 1 since 2^2 = 1^2 + 1^2 + 1^2 + 1^2 with 1 = 4^0 and 4*1 - 3*1 = 4^0.
a(3) = 1 since 3^2 = 1^2 + 0^2 + 2^2 + 2^2 with 1 = 4^0 and 4*1 - 3*0 = 4^1.
a(5) = 1 since 5^2 = 4^2 + 0^2 + 0^2 + 3^2 with 4 = 4^1 and 4*4 - 3*0 = 4^2.
a(15) = 1 since 15^2 = 4^2 + 4^2 + 7^2 + 12^2 with 4 = 4^1 and 4*4 - 3*4 = 4^1.
a(37) = 1 since 37^2 = 16^2 + 16^2 + 4^2 + 29^2 with 16 = 4^2 and 4*16 - 3*16 = 4^2.
a(83) = 1 since 83^2 = 4^2 + 4^2 + 56^2 + 61^2 with 4 = 4^1 and 4*4 - 3*4 = 4^1.
a(263) = 1 since 263^2 = 4^2 + 5^2 + 22^2 + 262^2 with 4 = 4^1 and 4*4 - 3*5 = 4^0.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..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], 2017-2018.

Crossrefs

Cf. A000118, A271518, A279612, A281976, A299924, A300365, A300360.

Programs

Mathematica

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
Do[r=0;Do[If[SQ[n^2-16^k-((4^(k+1)-4^m)/3)^2-z^2],r=r+1],{k,0,Log[4,n]},{m,Ceiling[Log[4,Max[1,4^(k+1)-3*Sqrt[n^2-16^k]]]],k+1},{z,0,Sqrt[(n^2-16^k-((4^(k+1)-4^m)/3)^2)/2]}];Print[n," ",r];Label[aa],{n,1,80}]