Lemma 4 ([Binary congruences]). Let z_1,…,z_r∈ℤ _2[G] . Given M≥1 , if b>r(2^M-1) , there is an x≠0 such that (z_j)_x≡(z_j)_02^M (1≤j≤r). The bound depends on the number of tests and their precision, and not on the group-ring coefficients.

Choose generators of G . Every character-value function z(x) has an expansion z(x)=a_∅+ ∑_ ∅≠I⊆1,…,b 2^ |I| a_I∏_ i∈I x_i, a_I∈ℤ _2. This follows by expanding each sign character as a product of factors 1-2x_i . The j th binary digit of a 2 -adic integer a is a 2^j 2 , equivalently the coefficient of Y^ 2^j in (1+Y)^a over 𝔽 _2 . On the Boolean cube, the contribution of the monomial x_I=∏_ i∈I x_i to this generating function is 1+x_I((1+Y)^ 2^ |I| a_I -1). Its nonconstant terms have Y -degree at least 2^ |I| . In a contribution to Y^ 2^j , the sum of the quantities 2^ |I| used is at most 2^j . Since |I|≤2^ |I| , the resulting Boolean polynomial has degree at most~ 2^j . These formulas for 2 -adic exponents follow either by continuity from nonnegative integers or in each finite coefficient quotient. Express all rM digit equalities as Boolean equations P_ ij (x)=0 . Their total degree is at most r(2^M-1) . The indicator of their common zero set is ∏_ i,j (1+P_ ij ) . The remaining count is the Chevalley--Warning degree argument . A Boolean polynomial of degree less than b has even sum over 𝔽 _2^b : every monomial omits a variable, and summing over that variable gives zero in 𝔽 _2 . Hence the number of common zeros is even. Since 0 is a common zero, another one exists.