\section{The irrationality contradiction}\label{sec:contradiction}

We prove Proposition~\ref{prop:criterion}. Suppose that at most
$B-1$ of the beta values in \eqref{eq:forms} are irrational.
Choose $B-1$ distinct even indices
$\nu_1<\cdots<\nu_{B-1}$ containing all possible irrational
indices, adding other indices if necessary. There are enough
choices because $a>mDL\ge3B$.

The coefficients $c_j$ have already fixed the common multiplier
$W_n$ and its seed vector. We now choose weights on the $B$
forms to cancel the possible irrational values.
Following the coefficient-elimination framework
of~\cite[Section~5]{LY20}, choose fixed integer weights $w_b$
by the signed maximal minors of the matrix whose columns are
\[
 (1^{\nu_j},3^{\nu_j},\ldots,M^{\nu_j})^{\mathsf T},
 \qquad 1\le j\le B-1.
\]
They satisfy
\[
 \sum_{\substack{1\le b\le M\\b\text{ odd}}}w_b b^{\nu_j}=0,
 \qquad w_M\ne0.
\]
To see that the last minor is nonzero, regard it as a generalized
Vandermonde determinant at
$1^2,3^2,\ldots,(M-2)^2$ with exponents $\nu_j/2$.
A vanishing determinant would give a nonzero polynomial with
at most $B-1$ monomials and $B-1$ distinct positive roots.
A polynomial with $q$ monomials has at most $q-1$ such roots:
factor out its smallest power and apply Rolle's theorem
inductively to its derivative. This proves nonsingularity.
For $B=1$, take $w_1=1$.

Set
\[
 H_n=\sum_{\substack{1\le b\le M\\b\text{ odd}}}w_bF_{n,b},
 \qquad
 q_w(n)=\sum_{\substack{1\le b\le M\\b\text{ odd}}}w_bq_b(n).
\]
The possible irrational coefficients cancel at every $n$, so
$H_n$ is rational under the assumption. The seed choice
\eqref{eq:seed-target} gives $q_w(1)=w_M\eta\ne0$.
Let $D_0$ be a positive common denominator of the remaining
rational beta values. Proposition~\ref{prop:denominators} and
\eqref{eq:cleared-decay} give
\begin{equation}\label{eq:integers}
 D_0\delta_n^{a+k}H_n\in\Z,\qquad
 D_0\delta_n^{a+k}H_n\longrightarrow0
                 \quad(n\to\infty,\ n\text{ odd}).
\end{equation}

Exclude the finitely many primes dividing $D_0$ or the numerator
or denominator of $q_w(1)$. By Proposition~\ref{prop:dilation},
for every remaining sufficiently large prime,
\[
 p^\tau q_w(p)\equiv\ch(p)q_w(1)\not\equiv0\pmod p.
\]
Every remaining beta term of $p^\tau H_p$ is divisible by $p$
in $\Zp$, by \eqref{eq:rho-dilation} and the assumed rationality.
Hence $H_p\ne0$ for infinitely many primes.
Equation~\eqref{eq:integers} gives infinitely many nonzero
integers tending to zero, a contradiction.