= Solution
Suppose, for a contradiction, that suitable nonzero polynomials vanish at both $\alpha^k$ and $\alpha^{k+1}$. If $P(\gamma)=0$, then the primitive <minimal polynomial> $f_\gamma$ divides $P$. The multiplicativity of <Mahler measure> and the <Mahler measure bounded by polynomial length> give
$$
H(\gamma)^{[\mathbb Q(\gamma):\mathbb Q]}
=M(f_\gamma)
\leq M(P)
\leq\mathcal L(P)
<nh.
$$
If $[\mathbb Q(\alpha^k):\mathbb Q]=d$, then, using $H(\alpha^k)=H(\alpha)^k$, this inequality contradicts $H(\alpha)^{kd}>nh$. Hence $\mathbb Q(\alpha^k)$ is a proper intermediate field of $\mathbb Q(\alpha)/\mathbb Q$. Its degree divides the prime $d$ by the <tower law>, so $\alpha^k\in\mathbb Q$. Applying the same argument to $\alpha^{k+1}$ is even stronger and gives $\alpha^{k+1}\in\mathbb Q$. Since $\alpha\ne0$,
$$
\alpha=\frac{\alpha^{k+1}}{\alpha^k}\in\mathbb Q,
$$
contrary to $d\geq2$. At least one of the two proposed values of $\beta$ therefore has no such polynomial relation.
Solved by gpt-5.6-sol high.
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