OFFSET
1,1
COMMENTS
Conjecture: For any integer x with gcd(x,6) = 1, the number x^3 + 6^3 is never a sum of two squares.
We have verified this for x up to 5*10^6.
Note also that 6^3 + (-2)^3 = 8^2 + 12^2.
Hao Pan at Nanjing Univ. confirmed the conjecture on Jan. 3, 2016. - Zhi-Wei Sun, Jan 06 2016
LINKS
Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
EXAMPLE
a(1) = 14 since 14^3 + 6^3 = 16^2 + 52^2.
a(7) = 237 since 237^3 + 6^3 = 162^2 + 3645^2.
MATHEMATICA
f[n_]:=f[n]=FactorInteger[n]
Le[n_]:=Le[n]=Length[f[n]]
n=0; Do[Do[If[Mod[Part[Part[f[x^3+6^3], i], 1], 4]==3&&Mod[Part[Part[f[x^3+6^3], i], 2], 2]==1, Goto[aa]], {i, 1, Le[216+x^3]}]; n=n+1; Print[n, " ", x]; Label[aa]; Continue, {x, 0, 2750}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Jan 02 2016
STATUS
approved