OFFSET
1,3
COMMENTS
Conjecture: a(n) > 0 for all n > 1; in other words, for any integer n > 1 there is a nonnegative integer k such that either n^2 - 2*4^k or n^2 - 5*4^k can be written as the sum of two squares. Moreover, a(n) = 1 only for n = 2^k with k > 0.
This conjecture is stronger than the first conjecture in A300448. We have verified that a(n) > 0 for all n = 2..5*10^7.
Consider positive integers c not divisible by 4 such that for any integer n > 1 there is a nonnegative integer k for which n^2 - 2*4^k or n^2 - c*4^k can be written as the sum of two squares. Our computation for n up to 3*10^7 shows that the only candidates for values of c smaller than 160 are 5, 17, 18, 26, 29, 41, 45, 65, 74, 89, 98, 101, 113, 122, 125, 146, 149, 153. These numbers have the form 9^a*(3*b+2) with a and b nonnegative integers and the p-adic order of 3*b+2 is even for any prime p == 3 (mod 4). For n = 42211965 there is no nonnegative integer k such that n^2 - 2*4^k or n^2 - 162*4^k can be written as the sum of two squares.
Qing-Hu Hou at Tianjin Univ. reported that he had verified a(n) > 0 for n up to 10^9. - Zhi-Wei Sun, Mar 14 2018
Qing-Hu Hou found that 29, 65, 113 should be excluded from the candidates. In fact, for c = 29, 65, 113 there is no nonnegative integer k such that N(c)^2 - 2*4^k or N(c)^2 - c*4^k can be written as the sum of two squares, where N(29) = 51883659, N(65) = 56173837 and N(113) = 65525725. - Zhi-Wei Sun, Mar 23 2018
a(n) > 0 for 1 < n < 6*10^9. - Giovanni Resta, Jun 14 2019
LINKS
Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Zhi-Wei Sun, Refining Lagrange's four-square theorem, J. Number Theory 175(2017), 167-190.
Zhi-Wei Sun, Restricted sums of four squares, arXiv:1701.05868 [math.NT], 2017-2018.
EXAMPLE
a(1) = 0 since 1^2 - 4^k*(m^2+1) < 0 for k = 0,1,2,... and m = 1, 2.
a(2) = 1 since 2^2 = 4^0*(1^2+1) + 1^2 + 1^2.
a(3) = 2 since 3^2 = 4^0*(2^2+1) + 0^2 + 2^2 = 4^1*(1^2+1) + 0^2 + 1^2.
MATHEMATICA
f[n_]:=f[n]=FactorInteger[n];
g[n_]:=g[n]=Sum[Boole[Mod[Part[Part[f[n], i], 1]-3, 4]==0&&Mod[Part[Part[f[n], i], 2], 2]==1], {i, 1, Length[f[n]]}]==0;
QQ[n_]:=QQ[n]=n==0||(n>0&&g[n]);
SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
tab={}; Do[r=0; Do[If[QQ[n^2-4^k*(m^2+1)], Do[If[SQ[n^2-4^k(m^2+1)-x^2], r=r+1], {x, 0, Sqrt[(n^2-4^k(m^2+1))/2]}]], {m, 1, 2}, {k, 0, Log[4, n^2/(m^2+1)]}]; tab=Append[tab, r], {n, 1, 80}]; Print[tab]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Mar 07 2018
STATUS
approved