proofs

98 Borsuk’s conjecture
On the other hand, if z ≢ 0, 1 (modq), then the numerator of (∗) contains
one factor that is divisible by q = p m . At the same time, the product has no
factors from two adjacent nonzero residue classes: one of them represents
numbers that have no p-factors at all, the other one has fewer p-factors
than q = p m . Hence there are more p-factors in the numerator than in the
denominator, and the quotient is divisible by p.
□
Now we consider an arbitrary subset Q ′ ⊆ Q that does not contain any
nearly-orthogonal vectors. We want to establish that Q ′ must be “small.”
Claim 1. If x, y are distinct vectors from Q, then 1 4
(〈x, y〉+2) is
an integer in the range
−(q − 2) ≤ 1 4
(〈x, y〉 + 2) ≤ q − 1.
Both x and y have an even number of (−1)-components, so the number of
components in which x and y differ is even, too. Thus
〈x, y〉 = (4q − 2) − 2#{i : x i ≠ y i } ≡ −2 (mod4)
for all x, y ∈ Q, that is, 1 4
(〈x, y〉 + 2) is an integer.
From x, y ∈ {+1, −1} 4q−2 we see that −(4q − 2) ≤ 〈x, y〉 ≤ 4q − 2,
that is, −(q − 1) ≤ 1 4
(〈x, y〉 + 2) ≤ q. The lower bound never holds with
equality, since x 1 = y 1 = 1 implies that x ≠ −y. The upper bound holds
with equality only if x = y.
Claim 2. For any y ∈ Q ′ , the polynomial in n variables x 1 , . . .,x n
of degree q − 2 given by
F y (x) := P ( ( 1
)
1
4 (〈x, y〉 + 2)) =
4
(〈x, y〉 + 2) − 2
q − 2
satisfies that F y (x) is divisible by p for every x ∈ Q ′ \{y}, but
not for x = y.
The representation by a binomial coefficient shows that F y (x) is an integervalued
polynomial. For x = y, we get F y (y) = 1. For x ≠ y, the
Lemma yields that F y (x) is not divisible by p if and only if 1 4
(〈x, y〉+2) is
congruent to 0 or 1 (modq). By Claim 1, this happens only if 1 4
(〈x, y〉+2)
is either 0 or 1, that is, if 〈x, y〉 ∈ {−2, +2}. So x and y must be nearlyorthogonal
for this, which contradicts the definition of Q ′ .
Claim 3. The same is true for the polynomials F y (x) in the n −1
variables x 2 , . . .,x n that are obtained as follows: Expand F y (x)
into monomials and remove the variable x 1 , and reduce all higher
powers of other variables, by substituting x 1 = 1, and x 2 i = 1 for
i > 1. The polynomials F y (x) have degree at most q − 2.
The vectors x ∈ Q ⊆ {+1, −1} n all satisfy x 1 = 1 and x 2 i = 1. Thus
the substitutions do not change the values of the polynomials on the set Q.
They also do not increase the degree, so F y (x) has degree at most q − 2.