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Guest
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Another virtuoso étude from the Sortov Institute:
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.
Hints for provers:
Attack first the case
f = A + B + ... + K.
The coefficient of A^a B^b ... K^k, in the polynomial F, turns out to be the
number of mappings g of the cyclic group of n elements into the set
{A,B,...,K}
1) g takes the value A, a times, the value B b times, etc.
2) if g(x) = g(b+x) for all x then b is 0 (mod n).
Since such mapping fall into equivalence classes each with n elements, the
number of those mappings is a multiple of n.
To pass to the general case, add a multiple of n to every coefficient, so
the new coefficients are all positive, and then replace each monomial with a
sum of indeterminates.
Fermat's Little Theorem is the special case in which n is a prime and f is a
constant! Roll over, Fermat.
LH |
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