Solution (outline). From a3-b3=(a-b)(a2+ab+b2) we have
Let and consider the number
Whenever 3 does not divide 2k+1 and 4k+2, the sum between the last parentheses is 1++2=0.
In those cases when 3 divides 2k+1 and 4k+2, the value of the last factor is 3. Such terms are integers, and the only odd value is the l;ast one when k=l=0 ann m=n. Hence, s is an integer and S3 (mod 6).
It is easy to check that and .
(E.g. these numbers are conjucates and
Therefore, the two latter terms in S are of absolute value less than 1.