Problem A. 465. (November 2008)
A. 465. Show that 6 does not divide for any positive integer n.
(5 pont)
Deadline expired on 15 December 2008.
Solution (outline). From a^{3}b^{3}=(ab)(a^{2}+ab+b^{2}) we have
and
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.
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