= Solution
The determinant in the hint is independent of $Y$ and equals the <Wronskian>
$$
W(X)=
\begin{vmatrix}
F&F_Y\\
F_X&F_{XY}
\end{vmatrix}
=P(X)Q'(X)-P'(X)Q(X).
$$
It is nonzero because $P,Q$ are linearly independent. Also $\deg W<2n$, and coefficient convolution gives
$$
H(W)\leq2n(n+1)C^{2n}\leq C_0^n
$$
for a constant $C_0$ depending only on $C$.
Fix $y$, and suppose $F(X,y)$ has multiplicity $r$ at $a=p/q$. In
$$
W=F(X,y)Q'-F_X(X,y)Q,
$$
the two terms vanish to orders at least $r$ and $r-1$, so $W$ has multiplicity at least $r-1$ at $a$. The primitive polynomial $(qX-p)^{r-1}$ therefore divides $W$ in $\mathbb Z[X]$ by <Gauss lemma for polynomials>. Comparing leading coefficients gives
$$
q^{r-1}\leq H(W)\leq C_0^n.
$$
If $r>\delta n+1$, this implies $q^{\delta n}<C_0^n$ and hence $q<C_0^{1/\delta}$. Choosing $q$ larger than this bound proves that $r\leq\delta n+1$, uniformly in $y$.
Solved by gpt-5.6-sol high.
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