#  Reverse engineered Diophantine equations over $\mathbb{Q}$

Katerina Santicola  Mathematics Institute
University of Warwick
CV4 7AL
United Kingdom<EMAIL_ADDRESS>
###### Abstract.

Let $\mathscr{P}_{\mathbb{Q}}=\\{\alpha^{n}\;:\;\alpha\in\mathbb{Q},\;n\geq
2\\}$ be the set of rational perfect powers, and let
$S\subseteq\mathscr{P}_{\mathbb{Q}}$ be a finite subset. We prove the
existence of a polynomial $f_{S}\in\mathbb{Z}[X]$ such that
$f(\mathbb{Q})\cap\mathscr{P}_{\mathbb{Q}}=S$. This generalizes a recent
theorem of Gajović who recently proved a similar theorem for finite subsets of
integer perfect powers. Our approach makes use of the resolution of the
generalized Fermat equation of signature $(2,4,n)$ due to Ellenberg and
others, as well as the finiteness of perfect powers in non-degenerate binary
recurrence sequences, proved by Pethő and by Shorey and Stewart.

###### Key words and phrases:

Diophantine equations, rational points

###### 2010 Mathematics Subject Classification:

Primary 11D41, Secondary 11D61

The author’s research is funded by a doctoral studentship from the Heilbronn
Institute for Mathematical Research

##  1\. Introduction

By an _exponential Diophantine equation_ [14] we often mean an equation of the
form

(1) $f(X)=Y^{n}$

where $f$ is a polynomial with integer or rational coefficients, and the
unknowns $X$, $Y$ are taken to be either integral or rational. The exponent
$n$ is also an unknown, usually taken to run through the integers $n\geq 2$.
The epithet “exponential” simply refers to the fact that the exponent is a
variable (if the exponent is fixed the equation is called superelliptic). A
famous theorem of Schinzel and Tijdeman [12] concerning such equations asserts
that if $f\in\mathbb{Z}[X]$ as at least two distinct roots, then there are
finitely many solutions $(X,Y,n)$ to (1) with $X$, $Y$, $n\in\mathbb{Z}$,
$n\geq 3$ and $\lvert Y\rvert\neq 1$. One of the earliest examples of an
exponential Diophantine equation is

(2) $X^{2}+1=Y^{n},\qquad X,~{}Y,~{}n\in\mathbb{Z},\quad n\geq 2;$

In 1850 Victor Lebesgue [9] proved that the only solutions are $(X,Y)=(0,1)$
if $n$ is odd, and $(X,Y)=(0,\pm 1)$ if $n$ is even. Lebesgue’s argument is
elementary, and some version of this argument, with the help of the Primitive
Divisor Theorem of Bilu, Hanrot and Voutier [3] is capable of resolving many
equations of the form $X^{2}+C=Y^{n}$ (e.g. [6]). In contrast, the equation
$X^{2}+7=Y^{n}$ (which generalizes the famous Ramanujan–Nagell equation)
apparently cannot be tackled by elementary arguments, and was solved by
Bugeaud, Mignotte and Siksek [5] using a combination of tools from Diophantine
approximation and Galois representations of elliptic curves. We can think of
the Catalan equation

$X^{m}-Y^{n}=1,\qquad X,~{}Y,~{}m,~{}n\in\mathbb{Z},\quad m,~{}n\geq 2$

as an infinite family of exponential Diophantine equations of the form (1) by
rewriting it as $X^{m}+1=Y^{n}$, giving an exponential Diophantine equation
for each value of $m$. The famous Catalan conjecture, proved by Mihăilescu
[10] in 2004, states the only solution with $XY\neq 0$ is $3^{2}-2^{3}=1$.

We can view equation (1) as the question of which perfect powers belong to
$f(\mathbb{Z})$ (if the unknowns $X$, $Y$ belong to $\mathbb{Z}$) or to
$f(\mathbb{Q})$ (if the unknowns $X$, $Y$ belong to $\mathbb{Q}$). It is
therefore natural to ask which whether there is a restriction on the set of
perfect powers that can belong to $f(\mathbb{Z})$ or $f(\mathbb{Q})$. Indeed,
at the recent “Rational Points” conference (Schney, April 2022), Siksek posed
the following two questions.

######  Question 1.

Let

$\mathscr{P}_{\mathbb{Z}}\;=\;\\{a^{n}\;:\;a\in\mathbb{Z},\quad n\geq 2\\}$

be the set of perfect powers in $\mathbb{Z}$. Let $S$ be a finite subset of
$\mathscr{P}_{\mathbb{Z}}$. Is there a polynomial $f_{S}\in\mathbb{Z}[X]$ such
that $f_{S}(\mathbb{Z})\cap\mathscr{P}_{\mathbb{Z}}=S$?

######  Question 2.

Let

(3) $\mathscr{P}_{\mathbb{Q}}\;=\;\\{\alpha^{n}\;:\;\alpha\in\mathbb{Q},\quad
n\geq 2\\}$

be the set of perfect powers in $\mathbb{Q}$. Let $S$ be a finite subset of
$\mathscr{P}_{\mathbb{Q}}$. Is there a polynomial $f_{S}\in\mathbb{Q}[X]$ such
that $f_{S}(\mathbb{Q})\cap\mathscr{P}_{\mathbb{Q}}=S$?

Question 1 was answered affirmatively by Gajović [8, Theorem 3.1], and we
briefly recall (and slightly simplify) his elegant argument which yields an
explicit polynomial $f_{S}$ in terms of $S$. Let
$S=\\{b_{1},b_{2},\dotsc,b_{r}\\}\subseteq\mathscr{P}_{\mathbb{Z}}$. Let

$g(X)\;=\;\prod_{i=1}^{r}(X-b_{i})^{2}+1,\qquad h(X)\;=\;(X-1)\cdot
g(X)+1,\qquad f_{S}(X)\;=\;g(X)\cdot h(X).$

It is clear that $f_{S}(b_{i})=b_{i}$, so $S\subseteq
f_{S}(\mathbb{Z})\cap\mathscr{P}_{\mathbb{Z}}$. To prove the reverse inclusion
let $x\in\mathbb{Z}$ such that $f_{S}(x)=y^{n}$ for some $y\in\mathbb{Z}$ and
$n\geq 2$; it is enough to prove that $x=b_{i}$ for some $i$. Write
$c=\prod_{i=1}^{r}(x-b_{i})$. We consider two cases:

  * •

$y=0$. Then $h(x)=0$, so $x-1=-1/(c^{2}+1)$. Since $x\in\mathbb{Z}$, we have
$c=0$, and so $x=b_{i}$ for some $i$.

  * •

$y\neq 0$. Note that $g(x)>0$ and $h(x)$ are coprime integers whose product is
$y^{n}$. Thus $g(x)=z^{n}$ for some integer $z$, and so $c^{2}+1=z^{n}$.
Whence $c=0$, by the aforementioned theorem of Lebesgue concerning (2). Hence
$x=b_{i}$ for some $i$, completing the proof.

We point out that for the last step Gajović invokes Mihăilescu’s theorem [10]
(the Catalan conjecture), although it is enough to invoke the special (and
elementary) case due to Lebesgue.

The purpose this paper is to give an affirmative answer to Question 2.
Concisely, our theorem can be stated as follows.

######  Theorem 3.

Let $S$ be a finite subset of $\mathscr{P}_{\mathbb{Q}}$. Then there is a
polynomial $f_{S}\in\mathbb{Z}[X]$ such that
$f_{S}(\mathbb{Q})\cap\mathscr{P}_{\mathbb{Q}}=S$.

We give a recipe for $f_{S}$ in terms of a certain effectively computable
(though currently inexplicit) constant. Instead of Lebesgue’s theorem, or
Mihăilescu’s theorem, our proof that
$f_{S}(\mathbb{Q})\cap\mathscr{P}_{\mathbb{Q}}=S$ crucially relies on the
following theorem due to Ellenberg, to Bennett, Ellenberg and Ng, and to
Bruin.

######  Theorem 4 (Ellenberg et al.).

Let $n\geq 4$. Then the equation $a^{2}+b^{4}=c^{n}$ has no solutions in
coprime non-zero integers.

The theorem is due to Bruin [4] for $n=6$, who treats this case using an
elliptic Chabauty argument. It is due to Ellenberg [7] for prime $n\geq 211$,
and to Bennett, Ellenberg and Ng [2] for all other $n\geq 4$. These two papers
make use of deep results in theory of Galois representations of
$\mathbb{Q}$-curves as well as a careful study of critical values of Hecke
$L$-functions of modular forms.

We shall also need a famous theorem on perfect powers in non-degenerate binary
recurrence sequences, due to Pethő [11] and independently to Shorey and
Stewart [13]. We shall not need the theorem in its full generality, so we only
state a special case.

######  Theorem 5 (Pethő, Shorey and Stewart).

Let $a$, $b$, $\alpha$, $\beta$ be non-zero integers with
$\alpha\neq\pm\beta$. Let

$u_{t}\;=\;a\alpha^{t}+b\beta^{t},\qquad t\in\mathbb{Z},\qquad t\geq 0.$

Then there is an effectively computable constant $C(a,b,\alpha,\beta)$ such
that $u_{t}\notin\mathscr{P}_{\mathbb{Z}}$ for $t>C(a,b,\alpha,\beta)$.

The theorem is proved using the theory of lower bounds for linear forms in
logarithms of algebraic numbers.

##  2\. Preliminary results

We shall need the following consequence of Theorem 5.

######  Lemma 6.

Let $\gamma\in\mathbb{Q}$, $\gamma\neq 0$. Then there is an effectively
computable constant $D(\gamma)$ such for $t>D(\gamma)$,

$\gamma-2^{t}\notin\mathscr{P}_{\mathbb{Q}}.$

###### Proof.

Write $\gamma=u/v$ where $u$, $v\in\mathbb{Z}$, $u\neq 0$, $v\geq 1$ and
$\gcd(u,v)=1$. We let

$D(\gamma)\;=\;\max\\{0,\log_{2}{\lvert\gamma\rvert},\;C(u,-v,1,2)\\}$

where $C(a,b,\alpha,\beta)$ is as in Theorem 5. Note that since
$C(a,b,\alpha,\beta)$ is effectively computable, so is $D(\gamma)$. Let
$t>D(\gamma)$. We suppose $\gamma-2^{t}\in\mathscr{P}_{\mathbb{Q}}$ and derive
a contradiction.

Since $t>\log_{2}{|\gamma|}$, we have $\gamma-2^{t}\neq 0$. Observe that
$\gamma-2^{t}=(u-2^{t}v)/v$ where the numerator $u-2^{t}v$ and the denominator
$v$ are coprime. As $\gamma-2^{t}\in\mathscr{P}_{\mathbb{Q}}\setminus\\{0\\}$
we conclude that $u-2^{t}v\in\mathscr{P}_{\mathbb{Z}}$. We apply Theorem 5
with $a=u$, $b=-v$, $\alpha=1$, $\beta=2$. Since $t>C(u,-v,1,2)$, the theorem
tells us that $u-2^{t}v\notin\mathscr{P}_{\mathbb{Z}}$ giving a contradiction.
∎

We also need the following two corollaries to Theorem 4.

######  Corollary 7.

Let $n\geq 2$. Then the only solutions to the equation $A^{4}+B^{4}=2C^{n}$
with $A$, $B$, $C\in\mathbb{Z}$ and $\gcd(A,B)=1$ satisfy $A=\pm 1$, $B=\pm
1$.

###### Proof.

Note that $A$, $B$ are both odd. Write

$U\;=\;AB,\qquad V\;=\;\frac{A^{4}-B^{4}}{2}.$

Then $U$, $V$ are coprime integers and satisfy

$U^{4}+V^{2}\;=\;\left(\frac{A^{4}+B^{4}}{2}\right)^{2}\;=\;C^{2n}.$

By Theorem 4 we have $UVC=0$. However, $U$ is odd so $U\neq 0$. Thus $C\neq
0$. Hence $V=0$, so $A^{4}=B^{4}$. Since $A$, $B$ are coprime, $A=\pm 1$,
$B=\pm 1$ as required. ∎

######  Corollary 8.

Let $n\geq 2$. Then the only solutions to the equation $A^{4}+B^{4}=C^{n}$
with $A$, $B$, $C\in\mathbb{Z}$ and $\gcd(A,B)=1$ satisfy $A=0$, $B=\pm 1$, or
$B=0$, $A=\pm 1$.

###### Proof.

For $n=2$ this is a famous result of Fermat, proved by infinite descent. For
$n=3$ it is in fact a result of Lucas [1, Section 5]. Suppose $n\geq 4$. Then
we can rewrite the equation as $(A^{2})^{2}+B^{4}=C^{n}$ and conclude from
Theorem 4. ∎

##  3\. Proof of Theorem 3

Let $S$ be a finite subset of $\mathscr{P}_{\mathbb{Q}}$. We would like to
give a polynomial $f_{S}\in\mathbb{Z}[X]$ such that
$f_{S}(\mathbb{Q})\cap\mathscr{P}_{\mathbb{Q}}=S$. For $S=\emptyset$ we may
take any constant polynomial whose value is not a perfect power. Thus we may
suppose $S\neq\emptyset$. We write

$S\;=\;\\{\beta_{1},\dotsc,\beta_{r}\\}$

where $r\geq 1$, and $\beta_{i}\in\mathscr{P}_{\mathbb{Q}}$. As the
$\beta_{i}$ are rational numbers, we can write

$\beta_{i}\;=\;\frac{a_{i}}{c_{i}},\qquad a_{i},~{}c_{i}\in\mathbb{Z},\qquad
c_{i}\geq 1,\qquad\gcd(a_{i},c_{i})=1.$

Let $p_{1},p_{2},\dotsc,p_{t}$ be the distinct prime divisors of
$\prod_{i=1}^{n}c_{i}$, and let

(4) $k\;=\;\operatorname{lcm}(4,p_{1}-1,p_{2}-1,\dotsc,p_{t}-1).$

Let

(5) $F(X)\;=\;\prod_{i=1}^{r}(c_{i}X-a_{i})^{2}\;-1.$

Since $c_{i}\geq 1$ we note that $F$ has degree $2r$, and hence at most $2r$
roots. We are only interested in the non-zero rational roots of $F$, and we
let these be $\delta_{1},\delta_{2},\dotsc,\delta_{m}$. We let $s$ be an
integer satisfying the following:

(6) $\displaystyle\text{$s\,=\,2^{\kappa}-1$ for some $\kappa\geq 1$},$ (7)
$\displaystyle s\;\geq\;D(4\delta_{j}),\qquad j=1,\dotsc,m.$

Here $D(\cdot)$ is as in Lemma 6. It is clear that such an $s$ is effectively
computable. Let

(8) $\displaystyle g(X)\;=\;\prod_{i=1}^{r}(c_{i}X-a_{i})^{k}\;+\;1,\qquad
h(X)\;=\;(X-2^{s})g(X)\,+\,2^{s},$ (9) $\displaystyle f_{S}(X)\;=\;g(X)\cdot
h(X)\;\in\mathbb{Z}[X].$

To prove Theorem 3 we need to show that
$f_{S}(\mathbb{Q})\cap\mathscr{P}_{\mathbb{Q}}=S$. Observe that
$g(\beta_{i})=1$, $h(\beta_{i})=\beta_{i}$ and so
$f_{S}(\beta_{i})=\beta_{i}$. Thus $\beta_{i}\in
f_{S}(\mathbb{Q})\cap\mathscr{P}_{\mathbb{Q}}$, and therefore $S\subseteq
f_{S}(\mathbb{Q})\cap\mathscr{P}_{\mathbb{Q}}$. We would like to prove the
reverse inclusion. Thus let $x\in\mathbb{Q}$ and suppose that
$f_{S}(x)\in\mathscr{P}_{\mathbb{Q}}$. To prove Theorem 3 it will be enough to
show that $x=\beta_{i}$ for some $1\leq i\leq r$.

Write

(10) $x=\frac{u}{v},\qquad u,~{}v\in\mathbb{Z},\quad v\geq
1,\quad\gcd(u,v)=1.$

Let

(11) $A\;=\;\prod_{i=1}^{r}(c_{i}u-a_{i}v),\qquad
B\;=\;\frac{A}{\gcd(A,v^{r})},\qquad w\;=\;\frac{v^{r}}{\gcd(A,v^{r})},$

and note that

(12) $\gcd(B,w)=1,\qquad w\geq 1.$

Recall that we would like to show that $x=\beta_{i}$ (or equivalently
$u/v=a_{i}/c_{i}$) for some $1\leq i\leq r$. This is equivalent to $A=0$,
which is equivalent to $B=0$. Thus our proof will be complete on showing that
$B=0$.

We note that

(13)
$g(x)\;=\;\frac{A^{k}+v^{kr}}{v^{kr}}\;=\;\frac{B^{k}+w^{k}}{w^{k}},\qquad
h(x)\;=\;\frac{(u-2^{s}v)\cdot(B^{k}+w^{k})+2^{s}vw^{k}}{vw^{k}}.$

######  Lemma 9.

$\gcd(B^{k}+w^{k},v)$ is a power of $2$.

###### Proof.

Let $p$ be an odd prime dividing both $B^{k}+w^{k}$ and $v$. From (11) we see
that $(B^{k}+w^{k})$ divides $A^{k}+v^{kr}$. Thus $p\mid(A^{k}+v^{kr})$ and so
$p\mid A$. From the product definition of $A$ in (11), and since $\gcd(u,v)=1$
from (10), we see that $p\mid c_{i}$ for some $1\leq i\leq r$. It follows from
the definition of $k$ in (4) that $(p-1)\mid k$. Since $\gcd(B,w)=1$ (12), we
have that $p\nmid w$ and $p\nmid B$, so $B,w\in\mathbb{Z}_{p}^{*}$. Then we
have $B^{k}+w^{k}\equiv 0\mod p$, so $(Bw^{-1})^{k}\equiv-1\mod p$. This
contradicts Fermat’s little theorem. ∎

######  Lemma 10.

$B^{k}+w^{k}\neq 2$.

###### Proof.

Recall that $k$ is an even integer by (4), and that $w\geq 1$ from (12).
Suppose $B^{k}+w^{k}=2$. Then $\lvert B\rvert=w=1$. In particular, from (13)
we have $g(x)=2$. Thus

(14) $f_{S}(x)\;=\;g(x)\cdot h(x)\;=\;4x-2^{s+1}$

from (8) and (9). First we show that $x\neq 0$. Recall our assumption that
$f_{S}(x)\in\mathscr{P}_{\mathbb{Q}}$, and so $f_{S}(x)=y^{n}$ for some
rational $y$ and some integer $n\geq 2$. If $x=0$, then
$y^{n}=-2^{s+1}=-2^{2^{\kappa}}$ by (6). Thus $n\mid 2^{\kappa}$ and so $n$ is
even. Hence $y^{n}>0$ giving a contradiction. We conclude that $x\neq 0$.

Recall, from (10), that $v\geq 1$ and $x=u/v$. From (11), as $\lvert
B\rvert=w=1$, we see that

$\lvert A\rvert\;=\;\gcd(A,v^{r})\;=\;v^{r}.$

Thus $A=\pm v^{r}$. Dividing both sides of $A=\pm v^{r}$ by $v^{r}$ gives

$\prod_{i=1}^{r}(c_{i}x-a_{i})\;=\;\frac{A}{v^{r}}\;=\;\pm 1,$

from the product expression for $A$ in (11). Thus $x$ is a non-zero root of
the polynomial $F(X)$ given in (5). We have previously labelled the roots of
$F$ by $\delta_{1},\delta_{2},\dotsc,\delta_{m}$. Thus $x=\delta_{j}$ for some
$1\leq j\leq m$. By (7) we have $s+1>D(4\delta_{j})$. Hence, by Lemma 6 and
(14) we have
$f_{S}(x)=4x-2^{s+1}=4\delta_{j}-2^{s+1}\notin\mathscr{P}_{\mathbb{Q}}$,
giving a contradiction. ∎

######  Lemma 11.

If $f_{S}(x)=0$ then $B=0$.

###### Proof.

Suppose $f_{S}(x)=0$ and recall that $f_{S}(x)=g(x)h(x)$. Note that $k$ is
even by (4), and also $w\geq 1$ by (12). Thus $B^{k}+w^{k}$ is a positive
integer and hence, from (13) we have $g(x)\neq 0$. Hence $h(x)=0$, whence from
(13),

(15) $(u-2^{s}v)\cdot(B^{k}+w^{k})\;=\;-2^{s}vw^{k}.$

We claim that $B^{k}+w^{k}$ is a power of $2$. To prove this claim let
$p\mid(B^{k}+w^{k})$ be an odd prime. As $\gcd(B,w)=1$ by (12), we have
$p\nmid w$. From Lemma 9 we note that $p\nmid v$. However, from (15), we have
$p\mid 2^{s}vw^{k}$ giving a contradiction. Thus $B^{k}+w^{k}$ is indeed a
power of $2$. Since $k$ is even and $\gcd(B,w)=1$ we see that
$4\nmid(B^{k}+w^{k})$ and so $B^{k}+w^{k}=1$ or $2$. However, $B^{k}+w^{k}\neq
2$ from Lemma 10. We conclude that $B^{k}+w^{k}=1$. As $w\geq 1$, we obtain
$w=1$ and $B=0$ as required. ∎

Recall that we have supposed that $f_{S}(x)\in\mathscr{P}_{\mathbb{Q}}$ and to
complete the proof of Theorem 3, it is enough to show that $B=0$. Lemma 11
establishes this if $f_{S}(x)=0$. Thus we may suppose $f_{S}(x)\neq 0$. Hence
$f_{S}(x)=y^{n}$ where $y$ is a non-zero rational and $n\geq 2$. We claim that
$B^{k}+w^{k}=z^{n}$ or $2z^{n}$ for some odd positive integer $z$. To prove
this let $p$ be an odd prime dividing $B^{k}+w^{k}$. Then, as before $p\nmid
w$ since $\gcd(B,w)=1$, and $p\nmid v$ from Lemma 9. From the expression for
$h(x)$ in (13) we see that $p$ divides neither the numerator nor the
denominator of $h(x)$, and hence $\operatorname{ord}_{p}(h(x))=0$. However
$g(x)h(x)=f_{S}(x)=y^{n}$, so

$\operatorname{ord}_{p}(B^{k}+w^{k})\;=\;\operatorname{ord}_{p}\left(\frac{B^{k}+w^{k}}{w^{k}}\right)\;=\;\operatorname{ord}_{p}(g(x))\;=\;\operatorname{ord}_{p}(y^{n})\;\equiv\;0\pmod{n}.$

As this is true for every odd prime dividing $B^{k}+w^{k}$ we have
$B^{k}+w^{k}=2^{e}z^{n}$ for some $e\geq 0$ and some odd integer $z$. Now as
before $4\nmid(B^{k}+w^{k})$, since $k$ is even and $\gcd(B,w)=1$. Thus
$B^{k}+w^{k}=z^{n}$ or $B^{k}+w^{k}=2z^{n}$ where $z$ is odd.

Suppose first that $B^{k}+w^{k}=2z^{n}$. Recall that $4\mid k$ by 4. By
Corollary 7 we have $B=\pm 1$, $w=1$, contradicting Lemma 10.

Thus $B^{k}+w^{k}=z^{n}$. We apply Corollary 8 to conclude that $B=0$ or
$w=0$. However, $w\geq 1$, so $B=0$ completing the proof.

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