A283269 Number of ways to write n as x^2 + y^2 + z^2 with x,y,z integers such that x + 3*y + 5*z is a square.

1, 1, 2, 2, 0, 3, 5, 0, 3, 4, 4, 1, 0, 3, 7, 0, 1, 5, 3, 3, 0, 5, 3, 0, 6, 2, 8, 2, 0, 8, 3, 0, 2, 3, 6, 7, 0, 2, 6, 0, 6, 8, 4, 1, 0, 4, 3, 0, 2, 3, 7, 5, 0, 4, 13, 0, 8, 5, 2, 3, 0, 6, 6, 0, 0, 7, 13, 2, 0, 7, 3, 0, 5, 4, 9, 1, 0, 5, 3, 0, 3

Offset: 0

Author

Zhi-Wei Sun, Mar 04 2017

Keywords

• nonn

Comments

Conjecture: (i) a(2^k*m) > 0 for any positive odd integers k and m. Also, a(4*n+1) = 0 only for n = 63.

(ii) For any positive odd integers k and m, we can write 2^k*m as x^2 + y^2 + z^2 with x,y,z integers such that x + 3*y + 5*z is twice a square.

(iii) For any positive odd integer n not congruent to 7 modulo 8, we can write n as x^2 + y^2 + z^2 with x,y,z integers such that x + 3*y + 4*z is a square.

(iv) Let n be any nonnegative integer. Then we can write 8*n + 1 as x^2 + y^2 + z^2 with x + 3*y a square, where x and y are integers, and z is a positive integer. Also, except for n = 2255, 4100 we can write 4*n + 2 as x^2 + y^2 + z^2 with x + 3*y a square, where x,y,z are integers.

(v) For each n = 0,1,2,..., we can write 8*n + 6 as x^2 + y^2 + z^2 with x + 2*y a square, where x,y,z are integers with y nonnegative and z odd.

The Gauss-Legendre theorem asserts that a nonnegative integer can be written as the sum of three squares if and only if it is not of the form 4^k*(8m+7) with k and m nonnegative integers. Thus a(4^k*(8m+7)) = 0 for all k,m = 0,1,2,....

See also A283273 for a similar conjecture.

Examples

			a(11) = 1 since 11 = 3^2 + 1^2 + (-1)^2 with 3 + 3*1 + 5*(-1) = 1^2.
a(43) = 1 since 43 = (-5)^2 + (-3)^2 + 3^2 with (-5) + 3*(-3) + 5*3 = 1^2.
a(75) = 1 since 75 = (-1)^2 + 5^2 + 7^2 with (-1) + 3*5 + 5*7 = 7^2.
a(262) = 1 since 262 = 1^2 + 15^2 + (-6)^2 with 1 + 3*15 + 5*(-6) = 4^2.
a(277) = 1 since 277 = (-6)^2 + 4^2 + 15^2 with (-6) + 3*4 + 5*15 = 9^2.
a(617) = 1 since 617 = 17^2 + 18^2 + 2^2 with 17 + 3*18 + 5*2 = 9^2.
a(1430) = 1 since 1430 = (-13)^2 + (-6)^2 + 35^2 with (-13) + 3*(-6) + 5*35 = 12^2.
a(5272) = 1 since 5272 = (-66)^2 + 30^2 + (-4)^2 with (-66) + 3*30 + 5*(-4) = 2^2.
a(7630) = 1 since 7630 = (-78)^2 + 39^2 + 5^2 with (-78) + 3*39 + 5*5 = 8^2.
a(7933) = 1 since 7933 = (-56)^2 + 69^2 + (-6)^2 with (-56) + 3*69 + 5*(-6) = 11^2.
a(14193) = 1 since 14193 = (-7)^2 + 112^2 + 40^2 with (-7) + 3*112 + 5*40 = 23^2.
		

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. A000290, A005875, A271518, A283170, A283196, A283204, A283205, A283239, A283273.

Programs

Mathematica

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
Do[r=0;Do[If[SQ[n-x^2-y^2]&&SQ[(-1)^i*x+(-1)^j*3y+(-1)^k*5*Sqrt[n-x^2-y^2]],r=r+1],{x,0,Sqrt[n]},{y,0,Sqrt[n-x^2]},{i,0,Min[x,1]},{j,0,Min[y,1]},{k,0,Min[Sqrt[n-x^2-y^2],1]}];Print[n," ",r];Continue,{n,0,80}]