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This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A280831 Number of ways to write 8*n+7 as x^2 + y^2 + z^2 + w^2 with x^4 + 1680*y^3*z a square, where x,y,z,w are positive integers.

Original entry on oeis.org

1, 1, 1, 2, 1, 3, 2, 1, 2, 1, 1, 1, 4, 2, 1, 4, 5, 3, 3, 1, 3, 2, 3, 2, 6, 5, 3, 4, 4, 3, 12, 6, 2, 7, 5, 3, 10, 4, 5, 2, 7, 5, 4, 5, 3, 8, 2, 2, 3, 4, 6, 7, 8, 1, 5, 2, 6, 9, 6, 5, 9, 9, 4, 6, 1, 4, 14, 5, 4, 12, 3, 11, 12, 1, 4, 8, 6, 7, 4, 6, 7
Offset: 0

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Author

Zhi-Wei Sun, Jan 08 2017

Keywords

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

Examples

			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.
		

Crossrefs

Programs

  • 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}]