Fermat's little theorem, humungously generalized

[Larry Hammick]

I haven't yet spelled out a complete proof but I'm almost certain of
this:

Let f(A,B,...,K) be a polynomial in any number of variables over Z,
and let n be a positive integer. Define a polynomial F by
F(A,...,K) = sum_{d|n} M(d) f(A^d, B^d, ..., K^d)^{n/d}
where M denotes the Mobius function.
Then all the coefficients of F are divisible by n.

Fermat's little theorem is the special case in which n is a prime and
f is a constant!

Anybody seen this before? I haven't.

[Chip Eastham]

No, that's interesting. My first thought
was about trying to invoke Mobius inversion
on your formula. My second was to try to
fit Euler's generalization of Fermat's
little theorem into this framework.

regards, chip

[Larry Hammick]

Here's the result for a constant f.
http://www.mathlinks.ro/Forum/viewtopic.php?t=156312
A reader supplies a proof, but I sketch a proof which uses certain
equivalence classes of mappings, in which each equivalence class has n
elements. The more general statement can be proved in an essentially
similar way -- using Mobius inversion to get a count of the aperiodic
functions on Z/nZ having values in a certain set, interpreting that
count as a coefficient of F, and noticing that the functions fall into
equivalence classes each of which has exactly n elements.
I'll write it all up in TeX when I get a bit of time, and copy it to
this group's webspace.