OFFSET
0,2
COMMENTS
Conjecture: a(n) = 0 only for n == 12 (mod 16).
This has been verified for n up to 10^8.
Now we show that a(n) = 0 whenever n == 12 (mod 16). If 16*q + 12 = 5*w^4 + x^4 + y^2 + z^2 with q,w,x,y,z integers, then the equality modulo 8 yields that w,x,y,z are all even, hence 4*q + 3 == 20*(w/2)^4 + 4*(x/2)^4 + (y/2)^2 + (z/2)^2 and thus (y/2)^2 + (z/2)^2 == 3 (mod 4) which is impossible.
It seems that a(n) = 1 only for n = 0, 3, 4, 27, 43, 48, 49, 63, 67, 72, 75, 192, 215, 303, 1092.
LINKS
Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
Zhi-Wei Sun, New conjectures on representations of integers (I), Nanjing Univ. J. Math. Biquarterly 34 (2017), no.2, 97-120.
Zhi-Wei Sun, Sums of four rational squares with certain restrictions, arXiv:2010.05775 [math.NT], 2020-2022.
EXAMPLE
a(192) = 1 with 192 = 5*1^4 + 3^4 + 5^2 + 9^2.
a(215) = 1 with 215 = 5*1^4 + 2^4 + 5^2 + 13^2.
a(303) = 1 with 303 = 5*1^4 + 0^4 + 3^2 + 17^2.
a(1092) = 1 with 1092 = 5*0^4 + 2^4 + 20^2 + 26^2.
MATHEMATICA
SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
tab={}; Do[r=0; Do[If[SQ[n-5x^4-y^4-z^2], r=r+1], {x, 0, (n/5)^(1/4)}, {y, 0, (n-5x^4)^(1/4)},
{z, 0, Sqrt[(n-5x^4-y^4)/2]}]; tab=Append[tab, r], {n, 0, 100}]; Print[tab]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Jan 28 2022
STATUS
approved