#  Totally real Thue inequalities
over imaginary quadratic fields: an improvement

István Gaál,
University of Debrecen, Mathematical Institute
H–4002 Debrecen Pf.400., Hungary, e–mail<EMAIL_ADDRESS>

Borka Jadrijević
University of Split, Faculty of Science,
Ruđera Boškovića 33, 21000 Split, Croatia, e–mail<EMAIL_ADDRESS>

László Remete
University of Debrecen, Mathematical Institute
H–4002 Debrecen Pf.400., Hungary, e–mail<EMAIL_ADDRESS>Research supported in part by the EFOP-3.6.1-16-2016-00022 project. The
project is co-financed by the European Union and the European Social Fund.
Research supported by the ÚNKP-19-3 new national excellence program of the
Ministry of human capacities.

###### Abstract

We significantly improve our results of [2] reducing relative Thue
inequalities to absolute ones.

Mathematics Subject Classification: Primary 11D59; Secondary 11D57
Key words and phrases: relative Thue equations, Thue inequalities

##  1 Results

Let $F(x,y)$ be a binary form of degree $n\geq 3$ with rational integer
coefficients. Assume that $f(x)=F(x,1)$ has leading coefficient 1 and distinct
real roots $\alpha_{1},\ldots,\alpha_{n}$. Let $0<\varepsilon<1$ and let
$K\geq 1$. Let

$A=\min_{i\not=j}|\alpha_{i}-\alpha_{j}|,\;\;B=\min_{i}\prod_{j\not=i}|\alpha_{j}-\alpha_{i}|,\;\;C=\frac{K}{(1-\varepsilon)^{n-1}B},\;\;G=\frac{K^{1/n}}{\varepsilon
A}.$

Let $m\geq 1$ be a square-free positive integer, and set
$M={\mathbb{Q}}(i\sqrt{m})$. Consider the relative inequality

$|F(x,y)|\leq K\;\;{\rm in}\;\;x,y\in{\mathbb{Z}}_{M}.$ (1)

If $F$ is irreducible, then (1) is called a Thue inequality. We emphasize that
our statements are valid also if $F$ is reducible.

If $m\equiv 3\;(\bmod\;4)$, then $x,y\in{\mathbb{Z}}_{M}$ can be written as

$x=x_{1}+x_{2}\frac{1+i\sqrt{m}}{2}=\frac{(2x_{1}+x_{2})+x_{2}i\sqrt{m}}{2},\;\;y=y_{1}+y_{2}\frac{1+i\sqrt{m}}{2}=\frac{(2y_{1}+y_{2})+y_{2}i\sqrt{m}}{2},$

and if $m\equiv 1,2\;(\bmod\;4)$, then

$x=x_{1}+x_{2}i\sqrt{m},\;\;y=y_{1}+y_{2}i\sqrt{m},$

in both cases with $x_{1},x_{2},y_{1},y_{2}\in{\mathbb{Z}}$. Set $s=2$ if
$m\equiv 3\;(\bmod\;4)$ and $s=1$ if $m\equiv 1,2\;(\bmod\;4)$. In the
following Theorem we formulate our statements parallely in the two cases.

###### Theorem 1

Let $(x,y)\in{\mathbb{Z}}_{M}^{2}$ be a solution of (1). Then

$|F(sx_{1}+(s-1)x_{2},sy_{1}+(s-1)y_{2})|\leq
s^{n}K,\;\;|F(x_{2},y_{2})|\leq\frac{s^{n}K}{(\sqrt{m})^{n}},$ (2) ${\it
and}\;\;\;\;|F(sx_{1}+(s-1)x_{2},sy_{1}+(s-1)y_{2})|\cdot|F(x_{2},y_{2})|\leq\frac{s^{2n}K^{2}}{2^{n}\cdot(\sqrt{m})^{n}}.$
(3) ${\rm If}\;\;|y|>\max\left\\{G,\left(\frac{s\cdot
C}{\sqrt{m}}\right)^{\frac{1}{n-2}}\right\\},\;\;{\rm
then}\;\;x_{2}y_{1}=x_{1}y_{2}.$ (4) ${\rm If}\;\;|y|>\max\left\\{G,(s\cdot
C)^{\frac{1}{n-1}}\right\\}\;\;{\rm and}\;\;sy_{1}+(s-1)y_{2}=0,\;\;{\rm
then}\;\;sx_{1}+(s-1)x_{2}=0.$ (5) ${\rm
If}\;\;|y|>\max\left\\{G,\left(\frac{s\cdot
C}{\sqrt{m}}\right)^{\frac{1}{n-1}}\right\\}\;\;{\rm and}\;\;y_{2}=0,\;\;{\rm
then}\;\;x_{2}=0.$ (6)

Remark 1. The present inequality (2) is much sharper than the corresponding
inequalities of Theorem 2.1 of [2]. Moreover we obtain these inequalities
without any conditions on the variables. This makes the applications much
easier. If the values of $F$ are non-zero, then (3) yields further new
restrictions for the possible solutions of (1).

Proof of Theorem 1.
Let $(x,y)\in{\mathbb{Z}}_{M}^{2}$ be an arbitrary solution of (1). Let
$\beta_{j}=x-\alpha_{j}y,\;j=1,\ldots,n$, then inequality (1) can be written
as

$|\beta_{1}\cdots\beta_{n}|\leq K.$ (7)

We have

$\beta_{j}=\frac{1}{s}((sx_{1}+\left(s-1\right)x_{2})-\alpha_{j}(sy_{1}+\left(s-1\right)y_{2}))+\frac{i\sqrt{m}}{s}(x_{2}-\alpha_{j}y_{2}).$

Obviously,

$|{\rm Re}(\beta_{j})|\leq|\beta_{j}|,\;\;|{\rm
Im}(\beta_{j})|\leq|\beta_{j}|,\;1\leq j\leq n.$

Further,

$\prod_{j=1}^{n}|{\rm Re}(\beta_{j})|\leq\prod_{j=1}^{n}|\beta_{j}|\leq
K,\;\;{\rm and}\;\;\prod_{j=1}^{n}|{\rm
Im}(\beta_{j})|\leq\prod_{j=1}^{n}|\beta_{j}|\leq K,$

which imply (2). Moreover,

$\prod_{j=1}^{n}|{\rm Re}(\beta_{j})|\cdot\prod_{j=1}^{n}|{\rm
Im}(\beta_{j})|=\prod_{j=1}^{n}\left(|{\rm Re}(\beta_{j})|\cdot|{\rm
Im}(\beta_{j})|\right)\leq\prod_{j=1}^{n}\frac{|{\rm Re}(\beta_{j})|^{2}+|{\rm
Im}(\beta_{j})|^{2}}{2}=\prod_{j=1}^{n}\frac{|\beta_{j}|^{2}}{2}\leq\frac{K^{2}}{2^{n}},$

whence we obtain (3).

Assume now

$|y|\geq G.$ (8)

Let ${i_{0}}$ be the index with $|\beta_{i_{0}}|=\min_{j}|\beta_{j}|$. Then
$|\beta_{i_{0}}|\leq K^{\frac{1}{n}}$ and for $j\not={i_{0}}$

$|\beta_{j}|\geq|\beta_{j}-\beta_{i_{0}}|-|\beta_{i_{0}}|\geq|\alpha_{j}-\alpha_{i_{0}}|\cdot|y|-K^{\frac{1}{n}}\geq(1-\varepsilon)\cdot|\alpha_{j}-\alpha_{i_{0}}|\cdot|y|.$
(9)

From (7) and (9) we have

$|\beta_{i_{0}}|\leq\frac{K}{\prod_{j\not={i_{0}}}|\beta_{j}|}\leq\frac{C}{|y|^{n-1}}.$
(10)

Using that $\alpha_{i_{0}}|y|^{2}$ is real, by (10) we obtain

$|{\rm Im}(x\overline{y})|=|{\rm
Im}(\alpha_{i_{0}}|y|^{2}-x\overline{y})|\leq\left|\alpha_{i_{0}}|y|^{2}-x\overline{y}\right|=|y|^{2}\cdot\left|\alpha_{i_{0}}-\frac{x\overline{y}}{y\overline{y}}\right|=|y|^{2}\cdot\left|\alpha_{i_{0}}-\frac{x}{y}\right|\leq\frac{C}{|y|^{n-2}}.$

If

$|y|>\left(\frac{s\cdot C}{\sqrt{m}}\right)^{\frac{1}{n-2}},$

then this implies $x_{2}y_{1}=x_{1}y_{2}$.

Inequality (10) indicates that $|\beta_{i_{0}}|$ is small for sufficiently
large $|y|$ and so are its real and imaginary parts that can equal zero if we
impose some extra assumptions.

–If $|y|>(sC)^{\frac{1}{n-1}}$, then
$|(sx_{1}+\left(s-1\right)x_{2})-\alpha_{i_{0}}(sy_{1}+\left(s-1\right)y_{2})|<1$.
So, $sy_{1}+\left(s-1\right)y_{2}=0$ implies $sx_{1}+\left(s-1\right)x_{2}=0$.

–If $|y|>\displaystyle{\left(\frac{sC}{\sqrt{m}}\right)^{\frac{1}{n-1}}}$,
then $|x_{2}-\alpha_{i_{0}}y_{2}|<1$. So, $y_{2}=0$ implies $x_{2}=0$. $\Box$

##  2 How to apply Theorem 1?

Finally, we give useful hints for a practical application of Theorem 1. Using
the same notation let us consider again the relative inequality (1). We
describe our algorithm in case $m\equiv 3\;(\bmod\;4)$, since the other case
is completely similar.

First, we solve $F(x_{2},y_{2})=k_{1}$ for all $k_{1}\in{\mathbb{Z}}$ with
$|k_{1}|\leq 2^{n}K/(\sqrt{m})^{n}.$ Since the equation $F(x_{2},y_{2})=0$ can
also have non-trivial solutions if $F$ is reducible, we split our arguments
into two cases.

A. First suppose $F(x_{2},y_{2})=0$. This makes possible to determine
$x_{2},y_{2}$. If $F$ is irreducible, then $x_{2}=y_{2}=0$, if $F$ is
reducible, then $x_{2},y_{2}$ can be determined easily (if there are any). We
then determine the solutions $(a,b)\in{\mathbb{Z}}^{2}$ of $|F(a,b)|=k_{2}$
for all $k_{2}$ with $|k_{2}|\leq 2^{n}K$. Using all possible values of
$x_{2},y_{2}$ for each solution $(a,b)$ we determine
$x_{1}=(a-x_{2})/2,y_{1}=(b-y_{2})/2$ and check if these are integers. (Note
that if $F$ is irreducible, then $x_{2}=y_{2}=0$ implies $|F(x_{1},y_{1})|\leq
K$ and the procedure can be simplified.) Having all possible
$x_{1},x_{2},y_{1},y_{2}$ we test if $(x,y)\in{\mathbb{Z}}_{M}^{2}$ is a
solution of (1).

B. Assume now $F(x_{2},y_{2})=k_{1}\neq 0$ for some
$(x_{2},y_{2})\in{\mathbb{Z}}^{2}$. Then we solve $F(a,b)=k_{2}$ in
$(a,b)\in{\mathbb{Z}}^{2}$ for all $k_{2}\in{\mathbb{Z}}$ with
$|k_{1}k_{2}|\leq 2^{n}K^{2}/(\sqrt{m})^{n}$ (a part of this calculation was
already performed by solving $F(x_{2},y_{2})=k_{1}$). Having $a,b,x_{2},y_{2}$
we calculate $x_{1}=(a-x_{2})/2,y_{1}=(b-y_{2})/2$. For $x_{2},y_{2}$ and
integer values $x_{1},y_{1}$ we test if $(x,y)^{2}\in{\mathbb{Z}}_{M}$ is
indeed a solution of (1).

Remark 2. If $m$ is sufficiently large, then by (2) we have
$|F(x_{2},y_{2})|<1$. In case $F$ is irreducible, this implies
$x_{2}=y_{2}=0$, whence (1) reduces to an inequality in $x_{1},y_{1}$ over
${\mathbb{Z}}$.

Remark 3. Solving Thue equations over ${\mathbb{Z}}$ is no problem any more by
using well-known computer algebra packages. If $F$ is reducible, this task is
even easier.

## References

  * [1]
  * [2] I. Gaál, B. Jadrijević and L. Remete, Totally real Thue inequalities over imaginary quadratic fields, Glas. Mat., III. Ser. 53(2018), No. 2, 229-238.