The two bases #
The recursion needs a starting set at each residue of n modulo 8 that the construction
reaches. B_0 = {0} is the trivial square-difference-free subset of P_{3,0}, and
B_4 = { a T^3 + b T + c (1 - T^2) : a, b, c ∈ F_3 }
is a square-difference-free subset of P_{3,4} with 27 elements. Square-difference-freeness of
B_4 is the only place where the shape 1 - T^2 matters: a difference of two of its elements
has constant term c and T^2-coefficient -c, while a nonzero square of degree below 4 is
the square of a linear polynomial v + u T, with constant term v^2 and T^2-coefficient
u^2; so v^2 + u^2 = 0, which in F_3 forces u = v = 0.
The base B_0 = {0} ⊆ P_{3,0}.
Equations
Instances For
B_0 is square-difference-free.
One element of B_4, the polynomial a T^3 + b T + c (1 - T^2).
Equations
- NaslundCounterexample.base4Map t = Polynomial.C t.1 * Polynomial.X ^ 3 + Polynomial.C t.2.1 * Polynomial.X + Polynomial.C t.2.2 * (1 - Polynomial.X ^ 2)
Instances For
The base B_4 = { a T^3 + b T + c (1 - T^2) : a, b, c ∈ F_3 } ⊆ P_{3,4}.
Instances For
Distinct triples give distinct elements of B_4: the coefficients at T^3, T and 1 read
a, b and c back.
B_4 is square-difference-free. A difference of two of its elements has constant term c
and T^2-coefficient -c; a nonzero square of degree below 4 is the square of a linear
polynomial v + u T, whose constant term is v^2 and whose T^2-coefficient is u^2; so
v^2 + u^2 = 0, and in F_3 that forces u = v = 0.