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The Roth theorem states that if $\alpha$ is a real algebraic irrational number, then for every $\epsilon >0$ there are only finitely many reduced fractions $p/q$ satisfying

To derive this from the Schmidt subspace theorem, takeThese forms are linearly independent. For a solution $(p,q)$ with $q$ large, $|p|\asymp q$, so $H(p,q)\asymp q$, whileAfter slightly decreasing $\epsilon$, the Schmidt subspace theorem puts all such primitive vectors $(p,q)$ in finitely many rational lines. Each rational line contains only the two opposite primitive integer vectors $\pm (p,q)$, and these determine the same fraction. Hence only finitely many fractions occur.

Solved by gpt-5.6-sol high.

Fix a finite set $S$ of primes. We prove that only finitely many denominators $q_n$ can be S-smooth number. This is the standard finite-place corollary of the Schmidt subspace theorem for best approximations of the first kind; the reduction is recalled here because the target $\alpha$ need not be algebraic.

Apply the Dirichlet approximation theorem at each cutoff between two successive record denominators. Its approximant can be replaced by the last record without increasing the error. Apply the finite-place Schmidt subspace theorem to the resulting pairs of primitive vectors, using $X-\alpha Y,Y$ at the real place only after eliminating $\alpha$ between two successive pairs, and $X,Y$ at every place belonging to $S$. The factorsfor an $S$-smooth denominator supply the required height saving. If infinitely many such records existed, one fixed rational subspace would contain infinitely many of the paired vectors. Eliminating its rational linear relation has two possible outcomes: either all sufficiently late records represent one rational number, which contradicts the irrationality of $\alpha$, or $\alpha$ is algebraic and, for some $\eta >0$, infinitely many of the records satisfyThe latter alternative contradicts the Roth theorem. Thus only finitely many $q_n$ are $S$-smooth.

If the largest prime factor of $q_n$ did not tend to infinity, some bound $B$ would contain the largest prime factor for infinitely many $n$. Taking $S$ to be the finite set of primes at most $B$ would make those denominators $S$-smooth, contrary to the preceding conclusion. Therefore $P^{+}(q_n)\to \infty$.

Solved by gpt-5.6-sol high.