My solution: The desired result is equivalent to asserting the
is the so-called fundamental circulant matrix of order
, is nilpotent. Now using the Frobenius homomorphism (and induct on
), we have
, as claimed.
O.O whaaaaa? I'm not sure what most of this means.~ I looked up circulant matrices on wikipedia, but I have no clue how one can relate your first statement, to the original question. Hmmmmm I should learn algebra.
On a different note, I fail - I found out that there is a non-inductive solution (in addition to outermeasure
's solution). Well, it depends on whether or not you count a certain combinatorial identity as proven by induction or not.
Write your tuples of
and look at
--- a vector over
due to the ambiguity in the choice of
. The effect of your mapping on
. Hence the statement every tuple
will eventually end up as
is equivalent to saying
and large enough
--- which is the same as saying
is nilpotent in