A282972 Number of ways to write n as 4*x^4 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that 79*y^2 - 220*y*z + 205*z^2 is a square.

1, 1, 1, 1, 2, 2, 3, 1, 2, 4, 3, 2, 2, 4, 4, 2, 2, 3, 5, 2, 2, 4, 4, 2, 3, 3, 3, 2, 2, 3, 2, 2, 1, 4, 2, 1, 4, 2, 3, 1, 4, 3, 2, 1, 3, 5, 2, 1, 3, 6, 3, 2, 2, 5, 5, 2, 4, 3, 4, 2, 3, 5, 2, 2, 2, 6, 5, 2, 4, 5, 6, 1, 5, 6, 5, 4, 5, 5, 6, 2, 4

Offset

0,5

Comments

Conjecture: (i) a(n) > 0 for all n = 0,1,2,....

(ii) Any positive integer n can be written as 4*x^4 + y^2 + z^2 + w^2 with x,y,z nonnegative integers and w a positive integer such that 169*y^2 - 444*y*z + 396*z^2 (or 289*y^2 - 654*y*z + 401*z^2) is a square.

This is much stronger than Lagrange's four-square theorem, and we have verified parts (i) and (ii) of the conjecture for n up to 10^7 and 10^6 respectively.

By the linked JNT paper, any nonnegative integer n can be written as 4*x^4 + y^2 + z^2 + w^2 with x,y,z,w integers, and we can also write n as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers and (y-z)*(y-2*z) = 0. Whether y = z or y = 2*z, the number 79*y^2 - 220*y*z + 205*z^2 is definitely a square.

See also A282933 for a similar conjecture.

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.

Example

a(2) = 1 since 2 = 4*0^4 + 1^2 + 1^2 + 0^2 with 79*1^2 - 220*1*1 + 205*1^2 = 8^2.
a(35) = 1 since 35 = 4*0^4 + 3^2 + 1^2 + 5^2 with 79*3^2 - 220*3*1 + 205*1^2 = 16^2.
a(119) = 1 since 119 = 4*1^4 + 9^2 + 3^2 + 5^2 with 79*9^2 - 220*9*3 + 205*3^2 = 48^2.
a(124) = 1 since 124 = 4*1^4 + 4^2 + 2^2 + 10^2 with 79*4^2 - 220*4*2 + 205*2^2 = 18^2.
a(1564) = 1 since 1564 = 4*3^4 + 14^2 + 30^2 + 12^2 with 79*14^2 - 220*14*30 + 205*30^2 = 328^2.
a(4619) = 1 since 4619 = 4*2^4 + 51^2 + 27^2 + 35^2 with 79*51^2 - 220*51*27 + 205*27^2 = 228^2.
a(6127) = 1 since 6127 = 4*5^4 + 49^2 + 35^2 + 1^2 with 79*49^2 - 220*49*35 + 205*35^2 = 252^2.
a(7119) = 1 since 7119 = 4*1^4 + 51^2 + 17^2 + 65^2 with 79*51^2 - 220*51*17 + 205*17^2 = 272^2.
a(9087) = 1 since 9087 = 4*3^4 + 61^2 + 71^2 + 1^2 with 79*61^2 - 220*61*71 + 205*71^2 = 612^2.

Mathematica

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
Do[r=0; Do[If[SQ[n-4x^4-y^2-z^2]&&SQ[79y^2-220*y*z+205z^2], r=r+1], {x, 0, (n/4)^(1/4)}, {y, 0, Sqrt[n-4x^4]}, {z, 0, Sqrt[n-4x^4-y^2]}]; Print[n, " ", r]; Continue, {n, 0, 80}]

Crossrefs

Cf. A000118, A000290, A000583, A270969, A271518, A281976, A282933.

Sequence in context: A079688 A346264 A261789 * A354584 A088598 A200751

Adjacent sequences: A282969 A282970 A282971 * A282973 A282974 A282975

Keyword

nonn

Author

Zhi-Wei Sun, Feb 25 2017

Status

approved