Задачи и решения 44й международной математической олимпиады (Токио, 2003)
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where the sum is over the set of all vectors (x
1
, . . . , x
r
2
+1
) in the vector space F
r
2
+1
p
over the
finite field F
p
. By Fermat’s little theorem, this claim evidently implies that the number of
solutions to the equation (∗) is divisible by p.
We prove the claim. In each monomial in F
r
, there are at most r
2
variables, and there-
fore at least one of the variables is absent. Suppose that the monomial is of the form
bx
α
1
i
1
x
α
2
i
2
···x
α
k
i
k
, where 1 ≤ k ≤ r
2
. Then
P
bx
α
1
i
1
x
α
2
i
2
···x
α
k
i
k
, where the sum is over the same
set as above, is equal to p
r
2
+1−k
P
x
i
1
,...,x
i
k
bx
α
1
i
1
x
α
2
i
2
···x
α
k
i
k
, which is divisible by p. This proves
the claim.
Comment. In general, if we replace p − 1 in (i) with any positive integer d, the answer is
(p − 1)d. In fact, if k > (p − 1)d, then the constant term of the element (1 − g
1
) ···(1 − g
k
)
of the group algebra Q
p
(ζ
p
)
£
(Z/pZ)
d
¤
can be evaluated p-adically so we see that it is not
equal to 1. Here g
1
, . . . , g
k
∈ (Z/pZ)
d
, Q
p
is the p-adic number field, and ζ
p
is a primitive
p-th root of 1. This also gives an alternative solution to the problem.