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Let $L$ be a fixed number field and let $V\subset L$ be a fixed rational linear subspace for which the associated norm form has no unit-family degeneracy. There are effective constants $A,C>0$ such that every $z\in V\cap {\mathcal{O}}_L$ with nonzero field norm $m$ satisfiesTo prove this, factor the principal ideal $(z)$, write its generators as a bounded factor times powers of fundamental units, use the linear relations defining $V$, and apply the Baker lower bound for a homogeneous linear form in logarithms to bound the unit exponents by $O(\log |m|)$. For $L=\mathbb{Q}(\alpha )K$ with $[\mathbb{Q}(\alpha ):\mathbb{Q}]=3$, $[K:\mathbb{Q}]=2$, and $V={\mathrm{span}}_{\mathbb{Q}}\{1,\alpha ,\omega \}$ for a basis $1,\omega$ of $K$, the coprime degrees rule out the degeneracy.

For the general result, choose logarithms of nonzero algebraic numbers ${\alpha }_1,\dots ,{\alpha }_s$ and algebraic coefficients ${\beta }_0,\dots ,{\beta }_s$, and putLetwhere $f_{{\alpha }_j}$ is the minimal polynomial and its $H$ is the naive polynomial height, and letThe general lower bound for a linear form in logarithms states that, if $\Lambda \ne 0$, thenwhere the effective constant $C$ depends only on $s$ and the degree of the number field generated by all the data.

For the improved homogeneous result, take $b_1,\dots ,b_s\in \mathbb{Z}$ andWith the same $A_j$, after ordering the terms setThe Baker lower bound for a homogeneous linear form in logarithms givesBoth constants are effective. The division by $\log A_s$ in $B^{\ast }$ is the improvement that matters when ${\alpha }_s$ has variable height.

Solved by gpt-5.6-sol high.

Put $\Lambda =n\log 2-m\log 3$. It is nonzero by unique prime factorization. If $3^m\le 2^{n-1}$ or $3^m\ge 2^{n+1}$, the claimed inequality follows immediately after increasing the effective constant. We may therefore assumewhich implies $m\asymp n$.

The Baker lower bound for a homogeneous linear form in logarithms, with the fixed algebraic numbers $2$ and $3$, givesfor an effective absolute constant $C_0$. If $|\Lambda |>1$, the desired conclusion is again immediate. Otherwise, the mean value theorem applied to the exponential function on $[-1,1]$ gives $|e^u-1|\ge c|u|$. HenceAbsorbing the fixed factor into a larger exponent proves $|2^n-3^m|\ge 2^n/n^C$.

Solved by gpt-5.6-sol high.

LetSuppose $F_n=x^k$ is a perfect power with $x\ge 2$. The finitely many small $n$ can be absorbed into the final effective constant. The Binet formula gives $x^k\asymp {\varphi }^n$ andThus, forthe local Lipschitz equivalence of $u$ and $e^u-1$ at zero yieldsThe form cannot vanish: applying the nontrivial field automorphism of $\mathbb{Q}(\sqrt{5})$ to ${\varphi }^n=\sqrt{5}{x}^k$ would give ${\psi }^n=-\sqrt{5}{x}^k$, whose absolute values are incompatible.

Apply the Baker lower bound for a homogeneous linear form in logarithms with the variable-height number $x$ placed last. The parameters belonging to $\varphi$ and $\sqrt{5}$ are absolute constants, while $\log A_x\asymp \log x$. Moreover,so $n/\log A_x\ll k$ and therefore $B^{\ast }\ll k$. The refined lower bound becomesComparison with the exponential upper bound gives $k\le C_3\log k$. Since $k/\log k$ tends to infinity, this bounds $k$ by an effective absolute constant. Enlarging it to cover the discarded small indices proves the claim.

Solved by gpt-5.6-sol high.

A suitable theorem is the Baker lower bound for a homogeneous linear form in logarithms. Let ${\alpha }_1,\dots ,{\alpha }_n$ be nonzero algebraic numbers with chosen logarithms and letChoose $A_i\ge 3$ to bound the degree-normalized Absolute logarithmic Weil height of ${\alpha }_i$ and $|\log {\alpha }_i|$, and put $B=\max (3,|b_1|,\dots ,|b_n|)$. If $\Lambda \ne 0$, thenwhere $C$ is an effectively computable constant depending only on $n$ and the degree of the number field generated by the ${\alpha }_i$.

Solved by gpt-5.6-sol high.

Let $f$ be the monic cubic minimal polynomial of the algebraic integer $\alpha$, let ${\alpha }_1=\alpha ,{\alpha }_2,{\alpha }_3$ be its conjugates, and let ${\beta }_1=\beta ,{\beta }_2$ be the two conjugates of a nonrational $\beta \in {\mathcal{O}}_K$. Because the degrees three and two are coprime, the fields are linearly disjoint, andis a nonzero integer.

AssumeFor large $H(\beta )$ this makes ${\beta }_1$ bounded. Since $\beta$ is a quadratic algebraic integer, the height-Mahler measure formula givesand therefore $|{\beta }_2|\asymp H(\beta {)}^2$. The two remaining factors with $j=1$ are bounded, while the three factors with $j=2$ are $O(|{\beta }_2|)$. Consequently

We now use the standard effective norm form consequence of Part (b). Forwhere $1,\omega$ is an integral basis of $K$, the effective norm-form height estimate supplies effective constants $A,C>0$, depending only on $\alpha$ and $K$, such thatfor every nonzero $z\in V\cap {\mathcal{O}}_L$. Its proof factors $(z)$, balances a generator using the Dirichlet unit theorem, and applies the Baker lower bound for a homogeneous linear form in logarithms to the linear relations defining $V$; the coprime degrees $3$ and $2$ exclude a unit-family degeneracy.

Apply this estimate to $z=\alpha -\beta$. Since $\beta =\alpha -z$, the height inequalities imply $H(\beta )\le 2H(\alpha )H(z)$. HenceChoose the effective value $\epsilon =1/(2A)$. The last inequality bounds $H(\beta )$ effectively. The rational integers $\beta \in \mathbb{Z}$ are already covered by the stronger degree-three Liouville approximation theorem, and the Northcott theorem leaves only finitely many remaining quadratic integers of bounded height. Taking the minimum ofover this effective finite set gives an effective $c>0$ and provesfor every $\beta \in {\mathcal{O}}_K$.

Solved by gpt-5.6-sol high.