A273134 Number of ordered ways to write n as x^2 + y^2 + z^2 + w^2 with (x+8*y+8*z+15*w)^2+(6*(x+y+z+w))^2 a square, where x,y,z,w are nonnegative integers with y < z.

1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 2, 2, 1, 1, 1, 2, 4, 2, 1, 1, 2, 1, 1, 3, 2, 3, 3, 1, 1, 2, 1, 3, 3, 1, 3, 3, 3, 1, 1, 2, 5, 3, 2, 3, 1, 2, 2, 3, 2, 2, 4, 2, 4, 3, 1, 3, 4, 2, 4, 3, 1, 3, 1, 2, 5, 4, 3, 2, 3, 1, 4, 5, 2, 3, 5, 3, 2, 2, 1

Offset: 1

Author

Zhi-Wei Sun, May 16 2016

Keywords

• nonn

Comments

Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 3, 7, 11, 15, 21, 23, 35, 39, 71, 95, 4^k*m (k = 0,1,2,... and m = 1, 2, 5, 6, 10, 14, 29, 30, 46, 62, 94, 110, 142, 190, 238, 334, 446).

(ii) For each polynomial P(x,y,z,w) = (x+3y+6z+17w)^2 + (20x+4y+8z+4w)^2, (x+3y+9z+17w)^2 + (20x+4y+12z+4w)^2, (x+3y+11z+15w)^2 + (12x+4y+4z+20w)^2, (3*(x+2y+3z+4w))^2 + (4*(x+4y+3z+2w))^2, (3*(x+2y+3z+4w))^2 + (4*(x+5y+3z+w))^2, any natural number can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that P(x,y,z,w) is a square.

Part (i) of this conjecture implies that any positive integer can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that x+8*y+8*z+15*w and 6*(x+y+z+w) are the two legs of a right triangle with positive integer sides. If a nonnegative integer n is not of the form 4^k*(16m+14) (k,m = 0,1,2,...), then n can be written as w^2+x^2+y^2+z^2 with w = x and hence (x+8y+8z+15w)^2 + (6(x+y+z+w))^2 = (8(2x+y+z))^2 + (6(2x+y+z))^2 = (10(2x+y+z))^2. Similar remarks apply to part (ii) of the conjecture.

See also A271714, A273107, A273108 and A273110 for similar conjectures related to Pythagorean triples. For more conjectural refinements of Lagrange's four-square theorem, one may consult arXiv:1604.06723.

Examples

			a(1) = 1 since 1 = 0^2 + 0^2 + 1^2 + 0^2 with 0 < 1 and (0+8*0+8*1+15*0)^2 + (6*(0+0+1+0))^2 = 10^2.
a(2) = 1 since 2 = 1^2 + 0^2 + 1^2 + 0^2 with 0 < 1 and (1+8*0+8*1+15*0)^2 + (6*(1+0+1+0))^2 = 15^2.
a(3) = 1 since 3 = 1^2 + 0^2 + 1^2 + 1^2 with 0 < 1 and (1+8*0+8*1+15*1)^2 + (6*(1+0+1+1))^2 = 30^2.
a(5) = 1 since 5 = 0^2 + 1^2 + 2^2 + 0^2 with 1 < 2 and (0+8*1+8*2+15*0)^2 + (6*(0+1+2+0))^2 = 30^2.
a(6) = 1 since 6 = 1^2 + 0^2 + 2^2 + 1^2 with 0 < 2 and (1+8*0+8*2+15*1)^2 + (6*(1+0+2+1))^2 = 40^2.
a(7) = 1 since 7 = 1^2 + 1^2 + 2^2 + 1^2 with 1 < 2 and (1+8*1+8*2+15*1)^2 + (6*(1+1+2+1))^2 = 50^2.
a(10) = 1 since 10 = 0^2 + 1^2 + 3^2 + 0^2 with 1 < 3 and (0+8*1+8*3+15*0)^2 + (6*(0+1+3+0))^2 = 40^2.
a(11) = 1 since 11 = 1^2 + 0^2 + 3^2 + 1^2 with 0 < 3 and (1+8*0+8*3+15*1)^2 + (6*(1+0+3+1))^2 = 50^2.
a(14) = 1 since 14 = 3^2 + 1^2 + 2^2 + 0^2 with 1 < 2 and (3+8*1+8*2+15*0)^2 + (6*(3+1+2+0))^2 = 45^2.
a(15) = 1 since 15 = 1^2 + 2^2 + 3^2 + 1^2 with 2 < 3 and (1+8*2+8*3+15*1)^2 + (6*(1+2+3+1))^2 = 70^2.
a(21) = 1 since 21 = 2^2 + 2^2 + 3^2 + 2^2 with 2 < 3 and (2+8*2+8*3+15*2)^2 + (6*(2+2+3+2))^2 = 90^2.
a(23) = 1 since 23 = 3^2 + 1^2 + 2^2 + 3^2 with 1 < 2 and (3+8*1+8*2+15*3)^2 + (6*(3+1+2+3))^2 = 90^2.
a(29) = 1 since 29 = 0^2 + 2^2 + 5^2 + 0^2 with 2 < 5 and (0+8*2+8*5+15*0)^2 + (6*(0+2+5+0))^2 = 70^2.
a(30) = 1 since 30 = 5^2 + 0^2 + 2^2 + 1^2 with 0 < 2 and (5+8*0+8*2+15*1)^2 + (6*(5+0+2+1))^2 = 60^2.
a(35) = 1 since 35 = 3^2 + 1^2 + 4^2 + 3^2 with 1 < 4 and (3+8*1+8*4+15*3)^2 + (6*(3+1+4+3))^2 = 110^2.
a(39) = 1 since 39 = 1^2 + 1^2 + 6^2 + 1^2 with 1 < 6 and (1+8*1+8*6+15*1)^2 + (6*(1+1+6+1))^2 = 90^2.
a(46) = 1 since 46 = 6^2 + 0^2 + 3^2 + 1^2 with 0 < 3 and (6+8*0+8*3+15*1)^2 + (6*(6+0+3+1))^2 = 75^2.
a(62) = 1 since 62 = 6^2 + 1^2 + 5^2 + 0^2 with 1 < 5 and (6+8*1+8*5+15*0)^2 + (6*(6+1+5+0))^2 = 90^2.
a(71) = 1 since 71 = 3^2 + 2^2 + 7^2 + 3^2 with 2 < 7 and (3+8*2+8*7+15*3)^2 + (6*(3+2+7+3))^2 = 150^2.
a(94) = 1 since 94 = 9^2 + 0^2 + 3^2 + 2^2 with 0 < 3 and (9+8*0+8*3+15*2)^2 + (6*(9+0+3+2))^2 = 105^2.
a(95) = 1 since 95 = 5^2 + 3^2 + 6^2 + 5^2 with 3 < 6 and (5+8*3+8*6+15*5)^2 + (6*(5+3+6+5))^2 = 190^2.
a(110) = 1 since 110 = 10^2 + 0^2 + 1^2 + 3^2 with 0 < 1 and (10+8*0+8*1+15*3)^2 + (6*(10+0+1+3))^2 = 105^2.
a(142) = 1 since 142 = 11^2 + 1^2 + 4^2 + 2^2 with 1 < 4 and (11+8*1+8*4+15*2)^2 + (6*(11+1+4+2))^2 = 135^2.
a(190) = 1 since 190 = 12^2 + 3^2 + 6^2 + 1^2 with 3 < 6 and (12+8*3+8*6+15*1)^2 + (6*(12+3+6+1))^2 = 165^2.
a(238) = 1 since 238 = 13^2 + 2^2 + 8^2 + 1^2 with 2 < 8 and (13+8*2+8*8+15*1)^2 + (6*(13+2+8+1))^2 = 180^2.
a(334) = 1 since 334 = 4^2 + 2^2 + 5^2 + 17^2 with 2 < 5 and
(4+8*2+8*5+15*17)^2 + (6*(4+2+5+17))^2 = 357^2.
a(446) = 1 since 446 = 17^2 + 6^2 + 11^2 + 0^2 with 6 < 11 and (17+8*6+8*11+15*0)^2 + (6*(17+6+11+0))^2 = 255^2.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
• Zhi-Wei Sun, Refining Lagrange's four-square theorem, arXiv:1604.06723 [math.GM], 2016.

Crossrefs

Cf. A000118, A000290, A260625, A261876, A262357, A267121, A268197, A268507, A269400, A270073, A271510, A271513, A271518, A271608, A271665, A271714, A271721, A271724, A271775, A271778, A271824, A272084, A272332, A272351, A272620, A272888, A272977, A273021, A273107, A273108, A273110.

Programs

Mathematica

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
Do[r=0;Do[If[SQ[n-x^2-y^2-z^2]&&SQ[(x+8*y+8*z+15*Sqrt[n-x^2-y^2-z^2])^2+36(x+y+z+Sqrt[n-x^2-y^2-z^2])^2],r=r+1],{x,0,Sqrt[n-1]},{y,0,(Sqrt[2(n-x^2)-1]-1)/2},{z,y+1,Sqrt[n-x^2-y^2]}];Print[n," ",r];Continue,{n,1,80}]