OFFSET
1,5
COMMENTS
Four-square Conjecture: a(n) > 0 for all n > 1.
This is much stronger than Lagrange's four-square theorem. We have verified a(n) > 0 for all n = 2..10^9.
Note that 16265031 cannot be written as (2^a*3^b)^2 + (2^c*3^d)^2 + x^2 + y^2 with a,b,c,d,x,y nonnegative integers.
a(n) > 0 for 1 < n <= 10^10. - Giovanni Resta, Jun 28 2019
I promise to offer 2500 US dollars as the prize for the first correct proof of the Four-square Conjecture. - Zhi-Wei Sun, Jul 09 2019
Jiao-Min Lin (a student at Nanjing University) has verified a(n) > 0 for all 1 < n <= 1.6*10^11. - Zhi-Wei Sun, Jul 30 2022
LINKS
Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Soumyarup Banerjee, On a conjecture of Sun about sums of restricted squares, J. Number Theory 256 (2024), 253-289.
Zhi-Wei Sun, Refining Lagrange's four-square theorem, J. Number Theory 175 (2017), 167-190.
Zhi-Wei Sun, Restricted sums of four squares, Int. J. Number Theory 15 (2019), 1863-1893.
Zhi-Wei Sun, Various Refinements of Lagrange's Four-Square Theorem, Westlake Number Theory Symposium (Nanjing University, China, 2020).
Zhi-Wei Sun, New Conjectures in Number Theory and Combinatorics (in Chinese), Harbin Institute of Technology Press, 2021. (See Conjecture 5.16.)
EXAMPLE
a(2^(2k+1)) = 1 with 2^(2k+1) = (2^k*3^0)^2 + (2^k*5^0)^2 + 0^2 + 0^2.
a(2^(2k+2)) = 1 with 2^(2k+2) = (2^k*3^0)^2 + (2^k*5^0)^2 + (2^k)^2 + (2^k)^2.
a(3) = 1 with 3 = (2^0*3^0)^2 + (2^0*5^0)^2 + 0^2 + 1^2.
a(5) = 2 with 5 = (2^0*3^0)^2 + (2^1*5^0)^2 + 0^2 + 0^2 = (2^1*3^0)^2 + (2^0*5^0)^2 + 0^2 + 0^2.
a(11) = 2 with 11 = (2^0*3^0)^2 + (2^0*5^0)^2 + 0^2 + 3^2 = (2^0*3^1)^2 + (2^0*5^0)^2 + 0^2 + 1^2.
MATHEMATICA
SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
tab={}; Do[r=0; Do[If[SQ[n-4^a*9^b-4^c*25^d-x^2], r=r+1], {a, 0, Log[4, n]}, {b, 0, Ceiling[Log[9, n/4^a]]-1},
{c, 0, Log[4, n-4^a*9^b]}, {d, 0, Log[25, (n-4^a*9^b)/4^c]}, {x, 0, Sqrt[(n-4^a*9^b-4^c*25^d)/2]}]; tab=Append[tab, r], {n, 1, 80}]; Print[tab]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Jun 21 2019
STATUS
approved