OFFSET
1,2
COMMENTS
Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 4^k*m (k = 0,1,2,... and m = 1, 7, 15, 39, 119, 127, 159, 239, 359, 391, 527, 543, 863, 5791).
(ii) Any positive integer can be written as x^2 + y^2 + z^2 + w^2 with 2*x*y*(x+2y+z+2w) (or 2*x*y*(x+6y+z+2w), or x*y*(x+11y+z+2w)) a square, where x,y,z,w are nonnegative integers with z > 0 (or w > 0).
(iii) Any natural number can be written as w^2 + x^2 + y^2 + z^2 with w,x,y,z nonnegative integers such that w*(a*w+b*x+c*y+d*z) is a square, provided that (a,b,c,d) is among the following quadruples (1,1,2,3), (1,1,4,5), (1,1,6,9), (1,2,6,34), (1,3,6,m) (m = 12, 21, 27, 36), (1,3,9,18), (1,3,9,36), (1,3,18,27), (1,3,24,117), (1,3,90,99), (1,6,6,18), (1,6,6,30), (1,8,16,24).
For more refinements of Lagrange's four-square theorem, see arXiv:1604.06723.
LINKS
Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Zhi-Wei Sun, Refining Lagrange's four-square theorem, arXiv:1604.06723 [math.GM], 2016.
Zhi-Wei Sun, Refine Lagrange's four-square theorem, a message to Number Theory List, April 26, 2016.
EXAMPLE
a(4) = 1 since 4 = 2^2 + 0^2 + 0^2 + 0^2 with 2 > 0 and 2*0*0*(2+9*0+11*0+10*0) = 0^2.
a(7) = 1 since 7 = 2^2 + 1^2 + 1^2 + 1^2 with 2 > 0 and 2*1*1*(2+9*1+11*1+10*1) = 8^2.
a(15) = 1 since 15 = 2^2 + 1^2 + 3^2 + 1^2 with 2 > 0 and 2*1*3*(2+9*1+11*3+10*1) = 18^2.
a(39) = 1 since 39 = 1^2 + 1^2 + 1^2 + 6^2 with 1 > 0 and 1*1*1*(1+9*1+11*1+10*6) = 9^2.
a(119) = 1 since 119 = 1^2 + 1^2 + 9^2 + 6^2 with 1 > 0 and 1*1*9*(1+9*1+11*9+10*6) = 39^2.
a(127) = 1 since 127 = 5^2 + 1^2 + 1^2 + 10^2 with 5 > 0 and 5*1*1*(5+9*1+11*1+10*10) = 25^2.
a(159) = 1 since 159 = 3^2 + 1^2 + 7^2 + 10^2 with 3 > 0 and 3*1*7*(3+9*1+11*7+10*10) = 63^2.
a(239) = 1 since 239 = 3^2 + 3^2 + 10^2 + 11^2 with 3 > 0 and 3*3*10*(3+9*3+11*10+10*11) = 150^2.
a(359) = 1 since 359 = 9^2 + 11^2 + 11^2 + 6^2 with 9 > 0 and 9*11*11*(9+9*11+11*11+10*6) = 561^2.
a(391) = 1 since 391 = 19^2 + 5^2 + 1^2 + 2^2 with 19 > 0 and 19*5*1*(19+9*5+11*1+10*2) = 95^2.
a(527) = 1 since 527 = 21^2 + 6^2 + 7^2 + 1^2 with 21 > 0 and 21*6*7*(21+9*6+11*7+10*1) = 378^2.
a(543) = 1 since 543 = 15^2 + 13^2 + 10^2 + 7^2 with 15 > 0 and 15*13*10*(15+9*13+11*10+10*7) = 780^2.
a(863) = 1 since 863 = 9^2 + 9^2 + 5^2 + 26^2 with 9 > 0 and 9*9*5*(9+9*9+11*5+10*26) = 405^2.
a(5791) = 1 since 5791 = 57^2 + 38^2 + 33^2 + 3^2 with 57 > 0 and 57*38*33*(57+9*38+11*33+10*3) = 7524^2.
MATHEMATICA
SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
Do[r=0; Do[If[SQ[n-x^2-y^2-z^2]&&SQ[x*y*z(x+9y+11z+10*Sqrt[n-x^2-y^2-z^2])], r=r+1], {x, 1, Sqrt[n]}, {y, 0, Sqrt[n-x^2]}, {z, 0, Sqrt[n-x^2-y^2]}]; Print[n, " ", r]; Continue, {n, 1, 80}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, May 01 2016
STATUS
approved