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Let $P,Q\in \mathbb{Z}[X_1,\dots ,X_k]$ have degree at most $n_j$ in $X_j$. Thenwhenever the quotient is defined. At each non-Archimedean place, the integral coefficients contribute at most one; at the Archimedean places, the triangle inequality contributes the polynomial length. Multiplying these local estimates and using the product formula gives the claim.

For every complex polynomial $P$,This follows from Jensen's formula, or directly by bounding $|P(z)|$ by its polynomial length on the unit circle and using the integral formula for the Mahler measure.

Write $\mathcal{L}(R)$ for the polynomial length. The height bound for a polynomial evaluation isprovided the denominator is nonzero. At non-Archimedean places the integral coefficients and ultrametric inequality give the local estimate without an extra constant; at Archimedean places the triangle inequality gives the polynomial length. Multiplication over every place of a number field and the product formula produce the displayed bound.

Taking $P(X,Y)=X+Y$ and $Q=1$ givesTaking $P(X)=X+1$ gives

Solved by gpt-5.6-sol high.

Take distinct $x,y\in A_n$. Their difference has the formwhere $P\in \mathbb{Z}[X]$ has degree at most $n-1$ and polynomial length at most $nh$. By the height bound for a polynomial evaluation,The algebraic number $x-y$ is nonzero and has degree at most $d$, so the Liouville height inequality gives the separation

All elements of $A_n$ lie in an interval of length at mostSince $H(1-\alpha )\le 2H(\alpha )$, another application of the Liouville height inequality givesThe number of points in an interval is at most one plus its length divided by their minimum separation. ConsequentlyThus the requested statement holds, for example, with the absolute constant $C=2$.

Solved by gpt-5.6-sol high.