It is enough to prove the result for $0<\kappa<1$, since a construction for a smaller positive value gives the weaker vanishing requirement for any larger one. Apply part (b) with $\kappa/2$. For large $n$, part (c) gives linearly independent polynomials $P_0,Q_0$, and
$$
G(X,Y)=P_0(X)+YQ_0(X)
$$
satisfies $H(G)\leq C_1^n$ and has order at least
$$
\frac{(2-\kappa/2)n}{d}-2
$$
at $X=\alpha$ after setting $Y=\alpha$.

Set $\delta=\kappa/(4d)$. By the <rational multiplicity bound for a linear auxiliary polynomial>, once $q_1$ is sufficiently large, the one-variable polynomial $G(X,y)$ has multiplicity at most $\delta n+1$ at $p_1/q_1$. Consequently there is some integer $j\leq\delta n+1$ such that
$$
D_j^XG(p_1/q_1,y)\ne0.
$$
Put $F=D_j^XG$. The <normalized derivative of a polynomial> preserves integral coefficients and multiplies height by at most $2^n$, so $H(F)\leq C_2^n$. Differentiation lowers the vanishing order at $(\alpha,\alpha)$ by at most $j$; for sufficiently large $n$,
$$
\frac{(2-\kappa/2)n}{d}-2-j
\geq\frac{(2-\kappa)n}{d}.
$$
Finally write $F=P-YQ$ by replacing the coefficient of $Y$ by its negative. Then $F(p_1/q_1,y)\ne0$ and all the claimed bounds hold.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 166 / 3 / e / Solution

Let $F(X,Y)=P(X)+YQ(X)$ with $P,Q\in\mathbb Z[X]$ linearly independent, $\deg_XF\leq n$, and $H(F)\leq C^n$. For every $\delta>0$, if a reduced rational $p/q$ has sufficiently large denominator in terms of $C$ and $\delta$, then $F(X,y)$ has multiplicity at most $\delta n+1$ at $p/q$, for every fixed real $y$.

Indeed, the nonzero <Wronskian>
$$
W=PQ'-P'Q
$$
has degree below $2n$ and height at most $C_0^n$. A zero of multiplicity $r$ of $F(X,y)$ forces a zero of multiplicity at least $r-1$ of $W$. By <Gauss lemma for polynomials>, $(qX-p)^{r-1}$ then divides $W$ in $\mathbb Z[X]$, so $q^{r-1}$ divides the leading coefficient of $W$ and is at most $C_0^n$.

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