OFFSET
0,4
COMMENTS
Conjecture: Let a and b be nonzero integers with gcd(a,b) squarefree. Then any natural number can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers and a*x^4 + b*y^3*z a square, if and only if (a,b) is among the ordered pairs (1,1), (1,15), (1,20), (1,36), (1,60), (1,1680) and (9,260).
If a natural number n is not of the form 4^k*(8m+7) (k,m = 0,1,...), then by the Gauss-Legendre theorem, there are nonnegative integers w,x,y such that n = w^2 + x^2 + y^2 + 0^2 and hence x^4 + 1680*y^3*0 is a square. Thus, the conjecture for (a,b) = (1,1680) has the following equivalent form: a(n) > 0 for all n = 0,1,...
See also A272336 for a similar conjecture.
Concerning the author's 1680-conjecture which states that each n = 0,1,... can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that x^4 + 1680*y^3*z is a square, Qing-Hu Hou at Tianjin Univ. has verified it for n up to 10^8. The author would like to offer 1680 RMB as the prize for the first complete solution of the 1680-conjecture. - Zhi-Wei Sun, Jun 22 2020
LINKS
Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
Zhi-Wei Sun, Refining Lagrange's four-square theorem, arXiv:1604.06723 [math.NT], 2016.
Zhi-Wei Sun, Refining Lagrange's four-square theorem, J. Number Theory 175 (2017), 167-190. (Cf. Conjecture 4.10(iv).)
EXAMPLE
a(0) = 1 since 8*0+7 = 1^2 + 1^2 + 1^2 + 2^2 with 1^4 + 1680*1^3*1 = 41^2.
a(11) = 1 since 8*11 + 7 = 95 = 6^2 + 3^2 + 1^2 + 7^2 with 6^4 + 1680*3^3*1 = 216^2.
a(244) = 1 since 8*244 + 7 = 1959 = 13^2 + 13^2 + 39^2 + 10^2 with 13^4 + 1680*13^3*39 = 11999^2.
a(289) = 1 since 8*289 + 7 = 2319 = 14^2 + 7^2 + 45^2 + 7^2 with 14^4 + 1680*7^3*45 = 5096^2.
a(664) = 1 since 8*664 + 7 = 5319 = 3^2 + 6^2 + 45^2 + 57^2 with 3^4 + 1680*6^3*45 = 4041^2.
a(749) = 1 since 8*749 + 7 = 5999 = 31^2 + 18^2 + 15^2 + 67^2
with 31^4 + 1680*18^3*15 = 12161^2.
a(983) = 1 since 8*983 + 7 = 7871 = 27^2 + 54^2 + 1^2 + 65^2 with 27^4 + 1680*54^3*1 = 16281^2.
a(1228) = 1 since 8*1228 + 7 = 9831 = 35^2 + 10^2 + 91^2 + 15^2 with 35^4 + 1680*10^3*91 = 12425^2.
a(1819) = 1 since 8*1819 + 7 = 14559 = 34^2 + 1^2 + 39^2 + 109^2 with 34^4 + 1680*1^3*39 = 1184^2.
a(2503) = 1 since 8*2503 + 7 = 20031 = 97^2 + 7^2 + 13^2 + 102^2 with 97^4 + 1680*7^3*13 = 9799^2.
a(2506) = 1 since 8*2506 + 7 = 20055 = 47^2 + 6^2 + 77^2 + 109^2 with 47^4 + 1680*6^3*77 = 5729^2.
MATHEMATICA
SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
Do[r=0; Do[If[SQ[8n+7-x^2-y^2-z^2]&&SQ[x^4+1680y^3*z], r=r+1], {x, 1, Sqrt[8n+6]}, {y, 1, Sqrt[8n+6-x^2]}, {z, 1, Sqrt[8n+6-x^2-y^2]}]; Print[n, " ", r]; Continue, {n, 0, 80}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Jan 08 2017
STATUS
approved