1. Introduction
Given
$a_1, \dots, a_r\in \mathbb{Q}^*\setminus\{\pm1\}$
, we say that the Schinzel–Wójcik problem has an affirmative answer if there exist infinitely many primes p such that they have the same multiplicative order modulo p.
In 1992, Schinzel–Wójcik proved that
Theorem 1·1 [
Reference Schinzel and Wójcik9
]. Given any
$a,b \in \mathbb Q^*\setminus \{\pm 1\}$
, there exist infinitely many primes p such that
$\operatorname{ord}_p a = \operatorname{ord}_p b$
.
The proof of Schinzel and Wójcik’s result is very ingenious and uses Dirichlet’s Theorem for primes in arithmetic progressions. In the last line of their paper, Schinzel and Wójcik concluded by stating the following problem:
Given
$a,b,c \in \mathbb Q^*\setminus\{\pm 1\}$
, do there exist infinitely many primes p with the property that
$\operatorname{ord}_p a = \operatorname{ord}_p b = \operatorname{ord}_p c$
?
In 1996, Wójcik produced a family of triplets of rational numbers
$\left( a,b,c \right)$
for which the above property is not satisfied for any odd prime p (see [
Reference Wójcik10
]). Here is another family of non-rational triplets in cyclotomic extensions, let
$a_1=\zeta_n a \ ,a_2= a \ ,a_3= a^n$
such that
$a\in \mathbb{Q}(\zeta_n)^*$
is not a root of unity and
$n\gt2$
. Let
$\mathfrak{p}$
be a prime ideal in
$\mathbb{Q}(\zeta_n)$
such that
$v_{\mathfrak{p}}(a_j)=0$
for all
$j=1,2,3$
. If
$d=\text{ord}_{\mathfrak{p}}(a_1)=\text{ord}_{\mathfrak{p}}(a_2)=\text{ord}_{\mathfrak{p}}(a_3)$
, then
$\zeta_n^d= (a_1/a_2)^d \equiv 1(\!\operatorname{mod} \mathfrak{p})$
. Therefore n divides d. Hence
$a^d=a_3^{d/n} \not\equiv 1 (\!\operatorname{mod} \mathfrak{p})$
but also
$a^d=a_2^d\equiv 1(\!\operatorname{mod} \mathfrak{p})$
which is a contradiction. Finally,
$(a_1,a_2,a_3)$
is an example of a triplet of non-rational numbers for which the Schinzel–Wójcik problem has negative answer. However, we have the following result proved by Wójcik [
Reference Wójcik10
] under the assumption of Schinzel’s Hypothesis H as in [
Reference Schinzel and Sierpiński8
]:
Hypothesis 1·2 (Schinzel’s Hypothesis H, 1958). Let
$f_1,\ldots,f_k \in\mathbb{Z}[x]$
be irreducible polynomials with positive leading coefficients and such that
$\gcd(f_1(n)\cdots f_k(n)\ |\, n\in\mathbb{N})=1$
. Then there are infinitely many
$t\in\mathbb{N}$
such that
$f_1(t),\ldots,f_k(t)$
are all prime.
Theorem 1·3 [
Reference Wójcik10
]. Let
$K/\mathbb{Q}$
be a finite extension and
$a_1, \dots, a_r \in K\setminus\{0, 1\}$
be such that the multiplicative group
$\langle a_1,\dots, a_r\rangle\subset K$
is torsion-free. Then, Schinzel’s Hypothesis H implies that there exist infinitely many prime ideals
$\mathfrak{p}$
of degree one such that
$\operatorname{ord}_\mathfrak{p} a_1= \dots = \operatorname{ord}_\mathfrak{p} a_r$
.
Also, in 2009, F. Pappalardi and A. Susa proved that:
Proposition 1·4 [
Reference Pappalardi and Susa7
]. Let
$\{a_1, \ldots, a_r\}\subset\mathbb{Q}^*\setminus\{\pm1\}$
be such that both the following properties are satisfied:
-
(i) there exist
$\omega_1,\ldots,\omega_r\in\mathbb{Z}$
with
$a_1^{\omega_1}\cdots a_r^{\omega_r}=-1$
; and -
(ii) there exist
$v_1,\ldots,v_r\in\mathbb{Z}$
with
$v_1+\cdots+v_r$
odd and
$a_1^{v_1}\cdots a_r^{v_r}=1$
.
Then the Schinzel–Wójcik problem for
$a_1,\ldots, a_r$
has a negative answer.
In the case where
$S=\{ a_1, \ldots, a_r\} \subset\mathbb{Z}$
, under the assumption of the Generalised Riemann Hypothesis for suitable number fields, in 1976, K. Matthews proved [
Reference Matthews5
] that
$\mathcal P_S=\left\{p\text{ prime } \left|\ \forall a\in S, a \text{ is a primitive root modulo }p\right. \right\}$
is finite if and only if at least one of the two following conditions is satisfied:
-
(1) there exist
$1\leq i_1\lt\cdots\lt i_{2s+1}\leq r$
such that
$a_{i_1}\cdots a_{i_{2s+1}}\in{\mathbb{Q}^*}^2;$
-
(2) there exist
$1\leq i_1\lt\cdots\lt i_{2s}\leq r$
such that
$a_{i_1}\cdots a_{i_{2s}}\in -3{\mathbb{Q}^*}^2,$
and for all primes
$\ell\equiv 1\, \textrm{mod}\, 3$
there exists at least one element of S which is a cube modulo
$\ell$
.
In 2017, F. Pappalardi and M. Anwar proved [
Reference Anwar and Pappalardi2
] that the same characterisation holds in the case of
$\{a_1, \ldots, a_r\} \subset\mathbb{Q}$
, under the assumption of Hypothesis H.
The aim of this paper is to prove a sufficient condition of the Schinzel–Wójcik problem under Hypothesis H. We shall prove:
Theorem 1·5. Let
$\{a_1, \dots, a_r\}\subset \mathbb{Q}^*\setminus\{\pm 1\}$
be such that:
-
(i) there exist
$v_1, \ldots, v_r \in \mathbb{Z}$
such that
$-1= a_{1}^{v_{1}} \cdots a_{r}^{v_{r}}$
with
$\sum^{'}{v_{i}} \equiv 0 \, (\!\operatorname{mod} 2),$
where
$\sum^{'}$
means that the sum runs over the positive
$a_{i},$
and -
(ii) for every
$v_1,\dots,v_r\in\mathbb{Z}$
, if
$a_1^{v_1}\cdots \ a_r^{v_r}= 1$
, then
$v_1+\dots+v_r \equiv 0 \, (\!\operatorname{mod} 2).$
Then Hypothesis H implies the existence of infinitely many prime numbers q such that
\begin{equation*}\operatorname{ord}_q a_1 = \operatorname{ord}_q a_2= \dots= \operatorname{ord}_q a_r.\end{equation*}
Throughout this paper, the notation
$\sum^{'}$
denotes that the summation is taken only over the positive values of
$a_i$
, and this convention will be used consistently.
The conditions (i) and (ii) in Theorem 1·5 are satisfied by many families of rational numbers, for example,
$\{-a^2, a^{-1}, b^2, b^{-2}\}$
and
$\{-a^2, a^{-1}b^{-1}, b\}$
$\subset \mathbb{Q}^*\setminus\{\pm 1\},$
where a, b are coprime rational numbers except
$0,1,-1.$
Note that the new contribution of this work is solving the case
$-1= a_{1}^{v_{1}} \cdots a_{r}^{v_{r}}$
for some
$v_1, \ldots, v_r \in \mathbb{Z}$
with
$\sum^{'}{v_{i}} \equiv 0 (\!\operatorname{mod} 2),$
and that the only remaining case is when
$\sum^{'}{v_{i}} \equiv 1 (\!\operatorname{mod} 2)$
for any such identity.
In light of Theorem 1·3 and Proposition 1·4, Theorem 1·5, under Hypothesis H, completes a characterisation of the Schinzel–Wójcik r-rational tuples (i.e. r-tuples of rational numbers for which the Schinzel–Wójcik problem has an affirmative answer) which satisfy that the sum of the exponents of the positive elements
$a_i$
in the representation of
$-1$
in terms of the
$a_i$
’s in the multiplicative group
$\langle a_1,\dots, a_r\rangle\subset \mathbb{Q}^*$
is even whenever
$-1 \in \langle a_1,\dots, a_r\rangle$
. (Note that if
$-1 = \prod a_i^{v_i}$
for some
$v_1, \ldots, v_r$
with
$\sum' v_i$
even, then the sum of the exponents of the positive
$a_i$
’s in any representation of
$-1$
has to be even.) More precisely,
Theorem 1·6. Let
$\{a_1, \dots, a_r\}\subset \mathbb{Q}^*\setminus\{\pm 1\}$
such that if
$-1= a_{1}^{v_1}\cdots a_{r}^{v_{r}}$
for some
$v_{1}, \ldots , v_{r} \in \mathbb{Z}$
, then the sum of the exponents of the positive
$a_{i}$
’s is even. Assuming Hypothesis H, then the Schinzel–Wójcik problem has an affirmative answer for
$\{a_1, \dots, a_r\}$
if and only if at least one of the following two conditions is satisfied:
-
(i)
$-1\notin\langle a_1,\dots, a_r\rangle$
; -
(ii) for every
$v_1,\dots,v_r\in\mathbb{Z}$
, if
$a_1^{v_1}\cdots a_r^{v_r}= 1$
, then
$v_1+\dots+v_r \equiv0 \, (\!\operatorname{mod} 2)$
.
This characterisation was originally introduced by M. Anwar in his Ph.D. thesis [ Reference Anwar1 ] in 2018 for all r-tuples of odd rational numbers—that is, rational numbers whose denominators are supported only at odd primes—for which the Schinzel–Wójcik problem admits an affirmative answer. However, a gap was later identified in the proof. In 2021, Olli Järviniemi established the same characterisation for all r-rational tuples, under the assumption of the generalised Riemann hypothesis [ Reference Järviniemi4 ].
Remark 1·7. The proof of Theorem 1·3 relies on Artin’s reciprocity law, more precisely on the following consequence of the latter for power-residue symbols ([
Reference Hasse3
], page 51). Let K be an algebraic number field and
$\mathfrak{m}$
,
$\mathfrak{p}$
,
$\mathfrak{q}$
ideals of K. We write
$\mathfrak{p} \sim \mathfrak{q} \,(\!\operatorname{mod} \mathfrak{m})$
if
$(\mathfrak{p}, \mathfrak{q}, \mathfrak{m}) = 1$
and there exist totally positive integers
$\alpha$
,
$\beta$
of K such that
$(\alpha \beta, \mathfrak{m}) = 1$
,
$\alpha\equiv\beta \, (\!\operatorname{mod} \mathfrak{m})$
and
$(\alpha) \mathfrak{p} = (\beta) \mathfrak{q}.$
Assume now that
$k\in\mathbb{N}$
,
$\zeta_k\in K$
,
$\gamma \in K$
and let
$\mathfrak{f}$
be the conductor of the extension
$K(\sqrt[k]{\gamma})/ K$
denoted further by
$\mathfrak{f}\left(K(\sqrt[k]{\gamma})/ K\right).$
If ideals
$\mathfrak{p}$
,
$\mathfrak{q}$
of K satisfy
$\mathfrak{p} \sim \mathfrak{q} \, (\!\operatorname{mod} \,\mathfrak{f})$
, then
\begin{equation*}\left[\frac{\gamma}{\mathfrak{p}}\right]_k=\left[\frac{\gamma}{\mathfrak{q}}\right]_k \end{equation*}
provided the
$k\textrm{th}$
-power-residue symbols are defined.
Let us conclude the introduction by giving a brief sketch of the proof of the main theorem. To do so, we first set some notation. Let
$\{a_1, \ldots, a_r\}\subset \mathbb{Q}^*\setminus\{\pm 1\}$
be such that
$-1= a_{1}^{v_{1}} \cdots a_{r}^{v_{r}}$
for some
$v_1, \ldots, v_r \in \mathbb{Z}$
with
$\sum^{'}{v_{i}} \equiv 0 \, (\!\operatorname{mod} \, 2).$
Let
$$\mathcal L=\{\ell\ \textrm{prime}\ |\ v_\ell(a_{j})\neq0\ \textrm{for some} \ j\in \{1,\ldots,r\}\}.$$
Write
$$\mathcal L=\{\ell_1,\ldots,\ell_m\}$$
and for all
$j=1,\ldots, r$
,
\begin{equation} a_j=(\!-\!1)^{e_{0j}}\ell_1^{e_{1j}}\ell_2^{e_{2j}}\cdots \ell_{m}^{e_{mj}},\end{equation}
where
$ e_{0j}=0 \ \textrm{or} \ 1$
and
$e_{ij}\in\mathbb{Z}$
for all
$1\leq i\leq m, \, 1\leq j\leq r.$
We furthermore decompose the
$\ell_i$
’s in
$(\mathbb{Q}^{mc})^*,$
the cyclotomic closure of
$\mathbb{Q}$
,
\begin{equation} \ell_i=\zeta_w^{b_i}\pi^{a_{i1}}_1\cdots \pi^{a_{in}}_n \quad \left( 1 \leq i \leq m \right)\end{equation}
with
$w\in\mathbb{N}$
,
$b_i$
,
$a_{ij}\in\mathbb{Z}$
,
$(b_1, \dots, b_m,w)=1$
, where
$\pi_1, \pi_2,\dots$
is a multiplicative basis of
$(\mathbb{Q}^{mc})^*$
(see Lemma 2·1) and
$\pi_1,\ldots , \pi_n\in\mathbb{Q}(\zeta_g)$
for some
$g\in\mathbb{Z}^+$
.
For some suitable positive integer k, the main points in the proof are the following:
-
(1) there exists a prime ideal
$\mathfrak{q}_0$
in
$\mathbb{Q}( \zeta_k)$
such that
\begin{equation*} \left[\frac{a_i}{\mathfrak{q}_0}\right]_k=-1 \quad \text{for all} \ \ i=1, \ldots, r;\end{equation*}
-
(2) there exist infinitely many pairs of primes (p,q) satisfying
$q = kp + 1$
and Hence,
\begin{equation*} a^p_j \equiv -1 \, (\!\operatorname{mod} q) \quad \mbox{for all} \, j=1, 2, \dots,r.\end{equation*}
\begin{equation*}\operatorname{ord}_q a_1 = \operatorname{ord}_q a_2= \dots= \operatorname{ord}_q a_r=2p.\end{equation*}
Using Lemma 2·2, we will be able to prove that for any integers
$x_1, \ldots, x_n$
and every
$t \in \mathbb{Z}$
such that
$(t,F)=1$
, there exists a prime ideal
$\mathfrak{q}_0$
in
$\mathbb{Q}( \zeta_k)$
such that
\begin{equation*} \left[\frac{\pi_i}{\mathfrak{q}_0}\right]_k = \zeta^{x_i}_{k} \ \ (1\leq i\leq n), \quad N\mathfrak{q}_0 \equiv t \,(\!\operatorname{mod} F),\end{equation*}
where
$F \in \mathbb{Z}$
such that
$k \mid F$
and
$N=N_{\mathbb{Q}\left(\zeta_k\right)/\mathbb{Q}}\left(\cdot\right)$
. A main ingredient in the proof is to choose
$\zeta_k$
suitably so that this implies 1 above. This corresponds to proving the solvability of a system of linear congruences. We show in Lemma 2·4 that a solution exists. Consequently, the second claim can be achieved using Lemma 2·3 which guarantees the existence of a polynomial
$f_1(x)$
such that
\begin{equation*} (p,q)= \left( f_1(x), \frac{f_1(x)-1}{k} \right)\end{equation*}
is a pair of primes for infinitely many
$x \in \mathbb{Z},$
using Hypothesis H. Furthermore, if
$q=f_{1}(x)$
is prime for some positive integer x, then
$q=N\mathfrak{q}^{'}$
,
$\mathfrak{q}^{'} \sim \mathfrak{q}_{0}^{-1} \ (\!\operatorname{mod} F)$
. Using Remark 1·7 and the first point, one can easily see that
\begin{equation*} a_j^{\frac{q-1}{k}}\equiv\left[\frac{a_j}{\mathfrak{q}^{'}}\right]_k=\left[\frac{a_j}{\mathfrak{q}_{0}}\right]_k ^{-1} =-1 \, (\!\operatorname{mod} \mathfrak{q}^{'}) \quad (1 \leq j \leq r).\end{equation*}
Consequently,
\begin{equation*} a^p_j\equiv-1 \, (\!\operatorname{mod} \mathfrak{q}^{'} ) \quad \mbox{for all} \, j=1, 2, \dots, r,\end{equation*}
which completes the second claim.
The paper is organised as follows. Section 2 includes some prior lemmas. Section 3 presents the proof of the main Theorem 1·5. Finally, Section 4 presents a family of examples that shows that the applied strategy is not applicable in the case that the sum of the exponents of the positive elements
$a_i$
in the representation of
$-1$
in terms of the
$a_i$
’s in the multiplicative group
$\langle a_1,\dots, a_r\rangle\subset \mathbb{Q}^*$
is odd whenever
$-1 \in \langle a_1,\dots, a_r\rangle.$
2. Lemmata
In this section, we introduce a series of lemmata crucial to the proof of Theorem 1·5. The first lemma is due to W. May. Its proof follows from the corollary to theorem 2 in [ Reference May6 ]. The second and third lemmas are due to J. Wojcik [ Reference Wójcik10 ].
Lemma 2·1. Let K be an algebraic number field. Let
$K^{mc}$
denote the cyclotomic closure of the field K. In the field
$K^{mc}$
there is a multiplicative basis i.e. such sequence
$\pi_1, \pi_2, \ldots ,$
that any non-zero element of
$K^{mc}$
is represented uniquely as
$\zeta \prod\limits_{s=1}^{t}\pi_{s}^{x_s}$
, where
$x_s$
are rational integers and
$\zeta$
is a root of unity.
Lemma 2·2. Let K,
$K_1$
be algebraic number fields such that
$K\subset K_1\subset K^{mc}$
and
$k, F \in \mathbb{Z}$
such that
$k|F$
. Assume that
$\pi_1,\ldots , \pi_n ,\zeta_k\in K_1$
, where
$\pi_1,\ldots , \pi_n$
are elements of the multiplicative basis described in Lemma 2·1. Let G be the subgroup of the multiplicative group of residue classes
$ \mbox{mod}\, F$
corresponding to the field
$K_1\cap\mathbb{Q}(\zeta_F)$
. Finally let
$\overline{t}\in G$
denote the residue class
$\mbox{mod} \, F$
containing
$t\in\mathbb{Z}.$
For all rational integers
$x_1,\ldots, x_n$
and every
$\overline{t}\in G$
there exist infinitely many prime ideals
$\mathfrak{q}$
of degree one of
$K_1$
such that
\begin{equation*} \left[\frac{\pi_i}{\mathfrak{q}}\right]_k = \zeta^{x_i}_{k} \ \ (1\leq i\leq n), \quad N\mathfrak{q}\equiv t \, (\!\operatorname{mod} F). \end{equation*}
Lemma 2·3. Let
$K_2$
be a normal field and k a positive integer. Assume that
$n_2$
denotes the degree of
$K_2$
and that g(x) is the minimal polynomial of an algebraic integer
$\theta$
such that
$K_2=\mathbb{Q}(\theta)$
. If a prime ideal
$\mathfrak{q}_0$
of degree one of
$K_2$
and a positive integer F satisfy the condition
\begin{equation*} F\equiv 0 \, (\!\operatorname{mod} k(2n_2)! \ \operatorname{disc}(g)), \quad N\mathfrak{q}_0 \equiv 1 \, (\!\operatorname{mod} k), \ ( \mathfrak{q}_{0},F)=1, \quad \left(\frac{N \mathfrak{q}_{0}-1}{k},F\right)=1 \end{equation*}
then there exists a polynomial
$f_1(x)$
such that the polynomials
$f_1(x)$
and
$f_2(x)=(\,{f_1(x)-1})/{k}$
satisfy the assumptions of Hypothesis H. Moreover, this
$f_{1}(x)$
is such that if
$q=f_1(x)$
is prime for some positive integer x, then
$q=N\mathfrak{q}^{'}$
,
$\mathfrak{q}^{'} \sim \mathfrak{q}_0 ^{-1} \, (\!\operatorname{mod} F)$
, where
$\mathfrak{q}^{'}$
is a prime ideal of degree one of
$K_2$
.
Define the matrices
\begin{equation*} E_{r\times (m+1)}\;:\!=\; \begin{pmatrix} e_{01} & e_{11} & e_{21} & \cdots & e_{m1}\\[5pt] \vdots & \vdots & \vdots & &\vdots\\[5pt] e_{0r} & e_{1r} & e_{2r} & \cdots & e_{mr}\\[5pt] \end{pmatrix} \quad \mbox{and} \quad A_{(m+1)\times n}\;:\!=\; \begin{pmatrix} 0 & 0 &\cdots &0 \\[5pt] a_{11} &a_{12} &\cdots &a_{1n}\\[5pt] \vdots & \vdots & &\vdots\\[5pt] a_{m1} &a_{m2} &\cdots &a_{mn}\\[5pt] \end{pmatrix}.\end{equation*}
As defined in Section 1, the entries of the matrices E and A are associated with the rational tuple
$a_1, \ldots, a_r.$
Lemma 2·4. Let
$k, w, u \in \mathbb{Z}^{+}$
be such that
$w \mid k$
and
$b_1, b_2, \ldots , b_m \in \mathbb{Z}$
where
$m \in \mathbb{Z}^{+}.$
Let
$\{a_1, \dots, a_r\}\subset \mathbb{Q}^*\setminus\{\pm 1\}$
such that:
-
(i) there exist
$v_1, \ldots, v_r \in \mathbb{Z}$
such that
$-1= a_{1}^{v_{1}} \cdots a_{r}^{v_{r}}$
with
$\sum^{'}{v_{i}} \equiv 0 \ (\!\operatorname{mod} 2),$
and -
(ii) for every
$v_1,\dots,v_r\in\mathbb{Z}$
, if
$a_1^{v_1}\cdots \ a_r^{v_r}= 1$
, then
$v_1+\dots+v_r \equiv 0 \, (\!\operatorname{mod} 2).$
Then the system
\begin{equation}E A \begin{pmatrix} x_1 \\[2pt] \vdots\\[2pt] x_n \end{pmatrix}\equiv \frac{k}{2} \begin{pmatrix} 1-e_{01} \\[2pt] \vdots \\[2pt] 1-e_{0r} \\[2pt] \end{pmatrix} - \frac{k}{w} u \ E \begin{pmatrix} 0 \\[2pt] b_{1} \\[2pt] \vdots \\[2pt] b_{m} \\[2pt] \end{pmatrix}(\!\operatorname{mod} k)\end{equation}
is solvable.
Proof. The first condition implies that there exists
$v_i$
for some
$i=1,2, \ldots, r$
such that
$v_i$
is odd and
$a_i\lt0.$
So, without loss of generality we may assume that
$v_1$
is odd and
$a_1\lt0$
(i.e.,
$e_{01}=1$
). Note that a solution exists if and only if a solution exists modulo
$p^{v_p(k)}$
for each prime
$p\mid k.$
We will show how to prove the existence of a solution for
$p=2;$
the case
$p\gt2$
is easier, since in this case the column
\begin{equation*} \frac{k}{2} \begin{pmatrix} 1-e_{01} \\[2pt] \vdots \\[2pt] 1-e_{0r} \\[2pt] \end{pmatrix}\end{equation*}
is
$\textbf{0}$
modulo
$p^{v_p(k)}$
.
Consider the system (2·1) modulo
$2^{\alpha}$
, where
$\alpha \;:\!=\; v_2(k)$
is assumed to be sufficiently large. The system (2·1) takes the form:
\begin{equation} E A \begin{pmatrix} x_1 \\[2pt] \vdots\\[2pt] x_n \end{pmatrix}\equiv \begin{pmatrix} 0 \\[2pt] k_{2} \\[2pt] \vdots \\[2pt] k_{r} \\[2pt] \end{pmatrix} - \frac{k}{w} u \ E \begin{pmatrix} 0 \\[2pt] b_{1} \\[2pt] \vdots \\[2pt] b_{m} \\[2pt] \end{pmatrix}(\!\operatorname{mod} 2^{\alpha}),\end{equation}
where for all
$i=1, 2,\dots, r,$
\begin{equation*} k_i=\frac{k}{2}(1-e_{0i}) \equiv \left\{ \begin{array}{ll} 2^{\alpha -1} \, (\!\operatorname{mod} 2^\alpha) & \text{if} \ a_i\gt 0, \\[2pt] \ \ \ 0 \ \ \ \ (\!\operatorname{mod} 2^\alpha) & \text{if} \ a_i\lt0. \end{array} \right.\end{equation*}
The first condition of Lemma 2·4 implies that
\begin{equation}\begin{array}{rcl} \mu_0\;:\!=\;e_{01} v_{1} + \cdots + e_{0r} v_{r} & \equiv 1 & (\!\operatorname{mod} 2), \\[2pt] \mu_j\;:\!=\;e_{j1} v_{1} + \cdots + e_{jr} v_{r} & = 0 & \text{for all} \ j=1, \ldots ,m,\\[2pt] \sum\limits_{i=1}^{r} v_i k_i= 2^{\alpha -1} \sum^{'} v_{i} & \equiv 0& (\!\operatorname{mod} 2^\alpha). \end{array}\end{equation}
Therefore, replacing Row
$_1$
by
$\sum_i v_i \text{Row}_i$
in the system (2·2) and using equations (2·3), we get the following
\begin{equation} E_1 A \begin{pmatrix} x_1 \\[2pt] x_2\\[2pt] \vdots\\[2pt] x_n \end{pmatrix}\equiv \begin{pmatrix} 0\\[2pt] k_{2} \\[2pt] \vdots \\[2pt] k_{r} \\[2pt] \end{pmatrix} - \frac{k}{w} u \ E_1 \begin{pmatrix} 0 \\[2pt] b_{1} \\[2pt] \vdots \\[2pt] b_{m} \\[2pt] \end{pmatrix}(\!\operatorname{mod} 2^{\alpha}),\end{equation}
where
\begin{equation*} E_{1}= \begin{pmatrix} \mu_0 & 0 & 0 & \cdots & 0\\[2pt] e_{02} & e_{12} & e_{22} & \cdots & e_{m2}\\[2pt] \vdots & \vdots &\vdots & &\vdots\\[2pt] e_{0r} & e_{1r} & e_{2r} & \cdots & e_{mr} \end{pmatrix}. \end{equation*}
Now, replacing Row
$_1$
by
$ \mu_0^{-1} \cdot \, \text{Row}_1$
in the system (2·4), we get
\begin{equation} E_2 A \begin{pmatrix} x_1 \\[2pt] x_2\\[2pt] \vdots\\[2pt] x_n \end{pmatrix}\equiv \begin{pmatrix} 0\\[2pt] k_{2} \\[2pt] \vdots \\[2pt] k_{r} \\[2pt] \end{pmatrix} - \frac{k}{w} u \ E_{2} \begin{pmatrix} 0 \\[2pt] b_{1} \\[2pt] \vdots \\[2pt] b_{m} \\[2pt] \end{pmatrix}(\!\operatorname{mod} 2^{\alpha}),\end{equation}
where
\begin{equation*} E_{2}= \begin{pmatrix} 1 & 0 & 0 & \cdots & 0\\[2pt] e_{02} & e_{12} & e_{22} & \cdots & e_{m2}\\[2pt] \vdots & \vdots &\vdots & &\vdots\\[2pt] e_{0r} & e_{1r} & e_{2r} & \cdots & e_{mr} \end{pmatrix}. \end{equation*}
Replacing Row
$_i$
by
$ \left( -e_{0i} \cdot\text{Row}_1+\text{Row}_i \right)$
for
$i=2,3, \ldots,r$
in the system (2·5), we get
\begin{equation} E_3 A \begin{pmatrix} x_1 \\[2pt] x_2\\[2pt] \vdots\\[2pt] x_n \end{pmatrix}\equiv \begin{pmatrix} 0\\[2pt] k_{2} \\[2pt] \vdots \\[2pt] k_{r} \\[2pt] \end{pmatrix} - \frac{k}{w} u \ E_{3} \begin{pmatrix} 0 \\[2pt] b_{1} \\[2pt] \vdots \\[2pt] b_{m} \\[2pt] \end{pmatrix}(\!\operatorname{mod} 2^{\alpha}),\end{equation}
where
\begin{equation*} E_{3}= \begin{pmatrix} 1 & 0 & 0 & \cdots & 0\\[2pt] 0 & e_{12} & e_{22} & \cdots & e_{m2}\\[2pt] \vdots & \vdots &\vdots & &\vdots\\[2pt] 0 & e_{1r} & e_{2r} & \cdots & e_{mr} \end{pmatrix}. \end{equation*}
Hence, using row operations, the system (2·6) can be written as follows
\begin{equation} D A \begin{pmatrix} x_1 \\[2pt] \vdots\\[2pt] x_n\\[2pt] \end{pmatrix}\equiv \begin{pmatrix} 0 \\[2pt] k^{'}_2 \\[2pt] \vdots\\[2pt] k^{'}_s \\[2pt] k^{'}_{s+1} \\[2pt] \vdots\\[2pt] k^{'}_r\\[2pt] \end{pmatrix} - \frac{k}{w} u \begin{pmatrix} 0 \\[2pt] b_2^{'} \\[2pt] \vdots \\[2pt] b_{s}^{'}\\[2pt] 0 \\[2pt] \vdots\\[2pt] 0\\[2pt] \end{pmatrix}(\!\operatorname{mod} 2^{\alpha} ),\end{equation}
where
\begin{equation*} D= \begin{pmatrix} 1 & 0 & 0 & \cdots & 0\\[5pt] 0 &d_{21} &d_{22} &\cdots &d_{2m}\\[5pt] \vdots &\vdots & \vdots & &\vdots\\[5pt] 0 &d_{s1} &d_{s2} &\cdots &d_{sm}\\[5pt] 0 &0 &0 &\cdots &0\\[5pt] \vdots\\[5pt] 0 &0 &0 &\cdots &0 \end{pmatrix}\end{equation*}
is a row echelon form of E with
$s \leq r $
independent rows,
$b_{i}^{'}= d_{i1} b_1 + d_{i2} b_2 + \cdots + d_{im} b_m$
and
$k^{'}_i=0$
or
$2^{\alpha-1}$
modulo
$2^{\alpha}$
for all
$i=2,\dots, r$
. Obtaining a zero row, say row j, through row operations on E corresponds to replacing
$a_j$
with a product of powers of the
$a_i$
’s that equals 1. Hence, by the second condition of Lemma 2·4, we deduce that any sequence of row operations on E resulting in a zero row in D must replace the initial row with an even sum of rows of E. Hence, we get
$k_{i}^{'}=0$
modulo
$2^{\alpha}$
for all
$i=s+1,s+2,\dots, r$
. Note that
$\alpha$
is chosen sufficiently large such that the non-zero rows of D are not zero modulo
$2^{\alpha}$
.
The rows of the matrix DA are linear combinations of the rows of the matrix A. More precisely,
\begin{equation*} DA= \begin{pmatrix} \textbf{0} \\[5pt] d_{21} \text{Row}_2 (A) + d_{22} \text{Row}_3 (A)+ \cdots + d_{2m} \text{Row}_{m+1} (A) \\[5pt] \vdots\\[5pt] d_{s1} \text{Row}_2 (A) + d_{s2} \text{Row}_3 (A)+ \cdots + d_{sm} \text{Row}_{m+1} (A) \\[5pt] \textbf{0} \\[5pt] \vdots\\[5pt] \textbf{0}\\[5pt] \end{pmatrix}.\end{equation*}
In fact, the rows
$\text{Row}_2 (DA), \ldots , \text{Row}_{s} (DA)$
are linearly independent, otherwise there are
$c_2, \ldots, c_{s}$
not all zeros such that
\begin{equation*} c_2 \text{Row}_2 (DA)+ c_3 \text{Row}_3 (DA)+ \cdots + c_s \text{Row}_s (DA)= \textbf{0}.\end{equation*}
Hence,
\begin{equation*} f_2 \text{Row}_2 (A)+ f_3 \text{Row}_3 (A)+ \cdots + f_{m+1} \text{Row}_{m+1} (A)= \textbf{0},\end{equation*}
where
\begin{equation*}f_i =c_2 d_{2,i-1}+ \cdots +c_s d_{s,i-1}, \ \ \textrm{for all} \ i=2,3, \ldots, m+1.\end{equation*}
Equivalently, we have
\begin{eqnarray*} f_2 a_{11} + f_3 a_{21} + \cdots + f_{m+1} a_{m1} &=& 0,\\[5pt] f_2 a_{12} + f_3 a_{22} + \cdots + f_{m+1} a_{m2} &=& 0,\\[5pt] \cdots \ \ \ \ \cdots \ \ \ \ \cdots \ \ \ \ \ && \\[5pt] f_2 a_{1n} + f_3 a_{2n}+ \cdots + f_{m+1} a_{mn} &=& 0.\end{eqnarray*}
Consequently, using the factorisation of the
$\ell_i$
’s in equation (1·2), we get
$$\ell_1 ^{f_2} \ell_2 ^{f_3} \cdots \ell_m ^{f_{m+1}}=1.$$
Since
$\ell_1, \ell_2, \ldots, \ell_m$
are multiplicatively independent, therefore
$$f_2 =f_3= \cdots = f_{m+1} =0.$$
Equivalently,
\begin{equation*} c_2 \begin{pmatrix} d_{21} \\[5pt] \vdots\\[5pt] d_{2m}\\[5pt] \end{pmatrix} + c_3 \begin{pmatrix} d_{31} \\[5pt] \vdots\\[5pt] d_{3m}\\[5pt] \end{pmatrix} +\cdots + c_s \begin{pmatrix} d_{s1} \\[5pt] \vdots\\[5pt] d_{sm}\\[5pt] \end{pmatrix}= \begin{pmatrix} 0 \\[5pt] \vdots\\[5pt] 0\\[5pt] \end{pmatrix}.\end{equation*}
Since the rows
$\operatorname{Row}_2 (D), \ldots , \operatorname{Row}_s (D)$
are linearly independent, therefore
$$c_2 = c_3 = \cdots = c_s =0$$
which contradicts the assumption that not all the
$c_i$
are zeros. Therefore, the rows
$\operatorname{Row}_2 (DA), \ldots , \operatorname{Row}_{s} (DA)$
are linearly independent.
By applying row and column operations to the system (2·7) and using the method described in [
Reference Anwar1
, chapter 2, section 5, p. 17], while noting that the matrix DA has rank
$s-1$
, we obtain the following comparable system:
\begin{equation} \begin{pmatrix} g_1 x^{'}_1 \\[5pt] \vdots\\[5pt] g_{s-1} x^{'}_{s-1}\\[5pt] \end{pmatrix}\equiv \begin{pmatrix} k^{''}_2 - \frac{k}{w}u \, b_2^{''}\\[5pt] \vdots\\[5pt] k^{''}_s -\frac{k}{w}u \, b_s^{''}\\[5pt] \end{pmatrix}(\!\operatorname{mod} 2^{\alpha} ),\end{equation}
where
$g_i=2^{\beta_i}$
for some positive integer
$\beta_i, k^{''}_j\in\{0,2^{\alpha-1}\}$
and
$b_i^{''} \in \mathbb{Z}$
for all
$ i=1,\dots,s-1, j=2,\dots,s$
. For sufficiently large
$\alpha$
, satisfying
$\beta_i \lt \alpha$
and
$\beta_i \lt v_{2}\left({k}/{w}\right)$
for all
$i=1,2, \dots, s-1,$
the system (2·8) is solvable and hence the system (2·2) is so. Therefore, the system (2·1) is solvable.
3. Proof of Theorem 1·5
Let
$k_0=gw, \ k=2k_0$
and choose
$K_1=K_2=\mathbb{Q}(\zeta_k),$
which is a normal extension. The integers g,w are defined as in equation (1·2).
Put
\begin{equation} F=k(2 \phi(k))! \ \operatorname{disc}(\Phi_k) \prod\limits^{r}_{i=1}N\left(\mathfrak{f}\left(\frac{K_2\left(\sqrt[k]{a_i}\right)}{K_2}\right)\right),\end{equation}
where
$\phi$
is the Euler totient function,
$\Phi_k$
is the
$k\textrm{th}$
cyclotomic polynomial and
$N(\cdot)=N_{K_2/\mathbb{Q}}(\cdot).$
Further, put
$\overline{F}=kF$
,
$F=F_1F_2$
, where
$F_1$
contains prime factors dividing k and
$(F_2,k)=1$
. Note that
$F_2$
is odd since k is even. Let t satisfy the system of congruences
\begin{equation*} t\equiv\left\{ \begin{array}{ll} k+1 & (\hbox{$\operatorname{mod} k^2$),} \\[5pt] 2 & (\hbox{$ \operatorname{mod} F_2)$.} \end{array} \right.\end{equation*}
Consequently,
\begin{equation*} \left\{ \begin{array}{ll} t & \equiv 1 \ (\hbox{$\operatorname{mod} k)$,} \\[5pt] t-1 & \equiv 1 \ (\hbox{$ \operatorname{mod} F_2)$,}\\[5pt] \frac{t-1}{k} & \equiv 1 \ (\hbox{$\operatorname{mod} k)$.} \\[5pt] \end{array} \right.\end{equation*}
Therefore,
\begin{equation*} (t,kF_2)=1, \quad \left(t-1,F_2\right)=1, \quad \left(\frac{t-1}{k},F_1\right)=1.\end{equation*}
Hence,
\begin{equation} (t,\overline{F})=1, \quad t\equiv1\, (\!\operatorname{mod} k), \quad \left(\frac{t-1}{k},F\right)=1.\end{equation}
Lemma 2·4 guarantees the existence of a solution of the system (2·1) for the integers
$e_{ij}, a_{ij}, b_i$
introduced in equations (1·1), (1·2) and
$u=({t-1})/{k}$
. Let
$x_1, x_2, \dots, x_n \in{\mathbb{Z}/k\mathbb{Z}}$
be such a solution. Using Lemma 2·2, there exists a prime ideal
$\mathfrak{q}_0$
in
$K_1=\mathbb{Q}(\zeta_k)$
of degree one with the properties:
\begin{equation}\left[\frac{\pi_i}{\mathfrak{q}_0}\right]_k=\zeta_{k}^{x_i} \quad (1\leq i\leq n), \quad N\mathfrak{q}_0\equiv t \, (\!\operatorname{mod} \overline{F}),\end{equation}
where the
$\pi_i$
’s are defined as in equation (1·2). Since equations (3·2) and (3·3) imply that
$({t-1})/{k}$
is odd and
$({N\mathfrak{q}_0 - 1})/{k}\equiv ({t-1})/{k} \, (\!\operatorname{mod} k),$
therefore
\begin{equation*}\begin{split} \left[\frac{-1}{\mathfrak{q}_0}\right]_k=(\!-\!1)^{\frac{N\mathfrak{q}_0-1}{k}}=(\!-\!1)^{\frac{t-1}{k}}=-1 \ \ \ \ \mbox{and}\ \ \ \ \ & \left[\frac{\zeta_w}{\mathfrak{q}_0}\right]_k=\zeta_w^{\frac{N\mathfrak{q}_0-1}{k}}=\zeta^{\frac{k}{w} \frac{N\mathfrak{q}_0-1}{k}}_k=\zeta^{\frac{k}{w} u}_k.\end{split}\end{equation*}
Consequently, with the aid of equations (1·1), (1·2) and (3·3), we get
\begin{eqnarray*}\left[\frac{a_j}{\mathfrak q_{0}}\right]_k &=& \left[\frac{-1}{\mathfrak q_{0}}\right]_k^{e_{0j}}\cdot \prod_{y=1}^{m}\left[\frac{\ell_y}{\mathfrak q_{0}}\right]_k^{e_{yj}} = (\!-\!1)^{e_{0j}} \prod_{y=1}^{m} \left( \left[\frac{\zeta_w}{\mathfrak q_{0}}\right]_k^{b_y} \cdot \prod_{i=1}^{n} \left[\frac{\pi_i}{\mathfrak q_{0}}\right]_k^{a_{yi} }\right)^{e_{yj}}\\[5pt] &=& \zeta_k^{\frac{k}{2}e_{0j}} \cdot \zeta_k^{\left(\sum\limits_{y=1}^{m}b_ye_{yj}\right)\frac{k}{w}u} \cdot \zeta_k^{\sum\limits_{y=1}^{m}\left( \sum\limits_{i=1}^{n}a_{yi} x_i\right) e_{yj}}.\end{eqnarray*}
The sum of the three exponents in the previous equation is congruent to
$ k/2$
modulo k by Lemma 2·4, and hence we obtain
\begin{equation} \left[\frac{a_1}{\mathfrak{q}_0}\right]_k=-1, \dots, \left[\frac{a_r}{{\mathfrak{q}}_0}\right]_k=-1.\end{equation}
Hence, by equations (3·1), (3·2) and (3·3), we have
\begin{equation*} F\equiv 0 \,\left(\operatorname{mod} k(2 \phi(k))! \ \mbox{disc}(\Phi_k)\right), \quad N\mathfrak{q}_0\equiv 1 \, (\!\operatorname{mod} k), \ (\mathfrak{q}_0,F)=1, \quad \left(\frac{N\mathfrak{q}_0-1}{k},F\right)=1.\end{equation*}
Using Lemma 2·3, there exists a polynomial
$f_1(x)$
such that the polynomials
$f_1(x)$
and
$f_2(x)=({f_1(x)-1})/{k}$
satisfy the assumptions of Hypothesis H. By this Hypothesis, there exist infinitely many
$x \in \mathbb{N}$
such that
$q=f_1(x)$
,
$p=f_2(x)$
are primes. Again by Lemma 2·3, if
$q=f_1(x)$
is a prime number for some
$x \in \mathbb{N}$
, then
\begin{equation*}q=N\mathfrak{q}^{'} \ \ \ \ \ \ \text{and} \ \ \ \ \ \mathfrak{q}_0 ^{-1} \sim \mathfrak{q}^{'} \, (\!\operatorname{mod} F),\end{equation*}
where
$\mathfrak{q}^{'}$
is a prime ideal of degree one of
$K_2$
. Consequently, equation (3·1) implies
$$\mathfrak{q}_0 ^{-1} \sim \mathfrak{q}^{'} \, \operatorname{mod} \mathfrak{f}\left(\frac{K_2\left(\sqrt[k]{a_i}\right)}{K_2}\right) \quad \mbox{for all} \,1 \leq i \leq r.$$
By equations (3·4), then Remark 1·7, the consequence of Artin’s reciprocity law, and
$({q-1})/{k}=p$
guarantee
\begin{equation*} a^p_j= a_j^{\frac{q-1}{k}}\equiv\left[\frac{a_j}{\mathfrak{q}^{'}}\right]_k=\left[\frac{a_j}{\mathfrak{q}_0}\right]_k ^{-1} =-1 \, (\!\operatorname{mod} \mathfrak{q}^{'}) \quad (1 \leq j \leq r).\end{equation*}
Hence,
\begin{equation*} a^{2p}_j\equiv 1 \, (\!\operatorname{mod} q) \quad \mbox{for all} \, j=1, 2, \dots, r.\end{equation*}
For all sufficiently large
$q=kp+1$
, we have
\begin{equation*}\operatorname{ord}_q a_1 = \operatorname{ord}_q a_2= \dots= \operatorname{ord}_q a_r=2p.\end{equation*}
4. Worked example
In this section, we give a family of examples that shows that the system (2·1) is not solvable in the case that
$-1= a_{1}^{v_{1}} \cdots a_{r}^{v_{r}}$
for some
$v_1, \ldots, v_r \in \mathbb{Z}$
with
$\sum^{'}{v_{i}} \equiv 1 (\!\operatorname{mod} 2).$
Consider
\begin{equation*}a_1 = - \ell_1^{-2} \ell_2 ^{-2}, \quad \quad a_2= - \ell_1 \ell_2 \ell_3 ^{-1}\quad \quad \textrm{and} \quad a_3= \ell_3^{2},\end{equation*}
where
$\ell_1,\ell_2$
and
$\ell_3$
are prime numbers. Clearly, we have
\begin{equation*}a_1^{v_1} a_2^{v_2} a_3^{v_3}= -1 \quad \textrm{if and only if} \quad v_1=v_3 \equiv \ 1 \ (\!\operatorname{mod} 2)\, \textrm{and}\ v_2 =2v_1 \equiv \ 0 \ (\!\operatorname{mod} 2).\end{equation*}
In this case, the sum of the exponents of the positive
$a_i$
’s in the representation of
$-1$
in the multiplicative group
$\langle a_1,a_2, a_3\rangle\subset \mathbb{Q}^*$
is odd. Now, we will show the insolvability of the system (2·1) modulo
$2^{\alpha}$
, which is
\begin{equation*} \! \begin{pmatrix} 1 & -2 & -2 & 0 \\[5pt] 1 & 1 & 1 & -1 \\[5pt] 0 & 0 & 0 & 2 \\[5pt] \end{pmatrix} \!\! \begin{pmatrix} 0 & 0 &\cdots &0 \\[5pt] a_{11} &a_{12} &\cdots &a_{1n}\\[5pt] a_{21} &a_{22} &\cdots &a_{2n}\\[5pt] a_{31} &a_{32} &\cdots &a_{3n}\\[5pt] \end{pmatrix} \!\! \begin{pmatrix} x_1 \\[5pt] \vdots\\[5pt] x_n \end{pmatrix}\equiv 2^{\alpha- 1} \! \begin{pmatrix} 0 \\[5pt] 0 \\[5pt] 1 \\[5pt] \end{pmatrix} - \frac{k}{w} u \! \begin{pmatrix} 1 & -2 & -2 & 0 \\[5pt] 1 & 1 & 1 & -1 \\[5pt] 0 & 0 & 0 & 2 \\[5pt] \end{pmatrix}\!\! \begin{pmatrix} 0 \\[5pt] b_{1} \\[5pt] b_{2}\\[5pt] b_{3} \\[5pt] \end{pmatrix}\!\!(\!\operatorname{mod} 2^{\alpha}).\end{equation*}
By applying the row operations mentioned in Lemma 2·4 on E and then multiplying by A, we get the following equivalent system
\begin{equation*} \begin{pmatrix} 0 & 0 &\cdots &0 \\[5pt] c_{11} &c_{12} &\cdots &c_{1n}\\[5pt] c_{21} &c_{22} &\cdots &c_{2n}\\[5pt] \end{pmatrix} \begin{pmatrix} x_1 \\[5pt] \vdots\\[5pt] x_n \end{pmatrix} \equiv 2^{\alpha- 1} \begin{pmatrix} 1 \\[5pt] 1 \\[5pt] 1 \\[5pt] \end{pmatrix} - \frac{k}{w} u \begin{pmatrix} 0 \\[5pt] b_{1}+b_{2}-b_{3} \\[5pt] 2b_{3} \\[5pt] \end{pmatrix} (\!\operatorname{mod} 2^{\alpha}),\end{equation*}
which is not solvable.
Note that
$a_1,a_2$
and
$a_3$
satisfy the proved characterisation in [
Reference Järviniemi4
] by Olli Järviniemi, under the assumption of the generalised Riemann hypothesis.
Acknowledgements
The author would like to thank Prof. Francesco Pappalardi for his useful comments. The author also gratefully acknowledges the anonymous reviewers and the editors for their valuable suggestions and constructive feedback, which helped improve the quality and clarity of this work.
This research was financially supported by the Fulbright U.S. Scholar Program, which is sponsored by the U.S. Department of State and the Egyptian-American Fulbright Commission, and was conducted at Pennsylvania State University. Its contents are solely the responsibility of the author and do not necessarily represent the official views of the Fulbright Program, the Government of the United States, or the Egyptian-American Fulbright Commission.
