Baker lower bound for a homogeneous linear form in logarithms

This quantitative refinement belongs to the theory of Baker's theorem; its dependence on the scaled coefficient bound is more specific than that broad theorem.

Choose logarithms of nonzero algebraic numbers ${\alpha }_1,\dots ,{\alpha }_n$, putand choose $A_i\ge 10$ large enough to bound the naive polynomial height of the algebraic minimal polynomial of ${\alpha }_i$ and $\exp (|\log {\alpha }_i|)$. Relabel so that $A_1\le \cdots \le A_n$, and setThere is an effectively computable constant $C$, depending only on $n$ and the degree of the number field generated by the ${\alpha }_i$, such that

The division by $\log A_n$ is the useful refinement over the general lower bound for a linear form in logarithms when one algebraic number has a large height correlated with a coefficient.

Fix positive integers $a,b$ and a nonzero integer $c$. In positive integer solutions of $a{x}^n-b{y}^n=c$ with $n\ge 1$ and $Y=\max (x,y)>1$, the exponent $n$ is bounded effectively in terms of $a,b,c$.

When $x\ne y$ and the larger-variable power is sufficiently large, put $\Lambda =\log (a/b)+n\log (x/y)$. The equation makes $\Lambda$ nonzero and givesA Baker lower bound for a homogeneous linear form in logarithms gives $\log |\Lambda |\ge -C(a,b)\log Y\log (2n)$: the algebraic numbers are the rationals $a/b$ and $x/y$, the latter has height at most $Y$, and the coefficients are $1,n$. Dividing by $\log Y\ge \log 2$ yields $n\le C\log (2n)+O(1)$, which bounds $n$. If the larger-variable power is bounded, so is $n$. For $x=y>1$, the equation $(a-b)x^n=c$ gives the bound directly.

This is a bound on the exponent. Fixed exponents one or two can still admit infinitely many pairs $(x,y)$; for instance, $2x^2-y^2=1$ is a negative Pell equation after exchanging variables. The Thue theorem must not be applied to this quadratic case. The constant solution $x=y=1$, possible when $a-b=c$, is excluded by $Y>1$.

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.