OFFSET
0,2
COMMENTS
Theorem: a(n) > 0 for all n = 0,1,2,.... In other words, any nonnegative integer can be written as the sum of a fourth power and three squares.
This is stronger than Lagrange's four-square theorem, and it can be proved by induction on n. It is easy to check that a(n) > 0 for all n = 0..16. Now let n be an integer greater than 16, and assume that a(m) > 0 for all m = 0..n-1. If 16|n, then n/16 can be written as w^4+x^2+y^2+z^2 with w,x,y,z integers, and hence n = (2w)^4+(4x)^2+(4y)^2+(4z)^2. If n == 8 (mod 16), then n is not of the form 4^k*(8q+7) and hence n = 0^4+x^2+y^2+z^2 for some integers x,y,z. If n == 4 (mod 8), then n-1^4 can be written as the sum of three squares. If n == 2 (mod 4), then n-0^4 is a sum of three squares. If n == 7 (mod 8), then n-1^4 can be written as the sum of three squares. If n is odd but not congruent to 7 modulo 8, then n-0^4 can be expressed as the sum of three squares.
We have a(n) = 1 if n has the form 16^k*q with k a nonnegative integer and q among 7, 8, 15, 23, 31, 47, 71, 79. In fact, if n = 16*m with m > 0, and 16*m = w^4+x^2+y^2+z^2 with w,x,y,z integers, then w,x,y,z are all even and hence m = (w/2)^4+(x/2)^2+(y/2)^2+(z/2)^2. Therefore a(16*m) = a(m) for all m > 0. It is easy to check that a(q) = 1 for every q = 7, 8, 15, 23, 31, 47, 71, 79.
For (a,b,c) = (1,1,2),(1,1,3),(1,1,4),(1,1,6),(1,2,2),(1,2,3),(1,2,4),(1,2,5), we are also able to show that any natural number can be written as w^4+a*x^2+b*y^2+c*z^2 with w,x,y,z integers.
Conjecture: For each triple (a,b,c) = (1,2,11),(1,2,12),(1,2,13),(2,3,5), any natural number can be written as w^4+a*x^2+b*y^2+c*z^2 with w,x,y,z integers.
LINKS
Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
Zhi-Wei Sun, A result similar to Lagrange's theorem, J. Number Theory 162(2016), 190-211.
EXAMPLE
a(7) = 1 since 7 = 1^4 + 1^2 + 1^2 + 2^2.
a(8) = 1 since 8 = 0^4 + 0^2 + 2^2 + 2^2.
a(15) = 1 since 15 = 1^4 + 1^2 + 2^2 + 3^2.
a(23) = 1 since 23 = 1^4 + 2^2 + 3^2 + 3^2.
a(31) = 1 since 31 = 1^4 + 1^2 + 2^2 + 5^2.
a(47) = 1 since 47 = 1^4 + 1^2 + 3^2 + 6^2.
a(71) = 1 since 71 = 1^4 + 3^2 + 5^2 + 6^2.
a(79) = 1 since 79 = 1^4 + 2^2 + 5^2 + 7^2.
MATHEMATICA
SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
Do[r=0; Do[If[SQ[n-w^4-x^2-y^2], r=r+1], {w, 0, n^(1/4)}, {x, 0, Sqrt[(n-w^4)/3]}, {y, x, Sqrt[(n-w^4-x^2)/2]}]; Print[n, " ", r]; Continue, {n, 0, 80}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Mar 27 2016
STATUS
approved