On the finiteness of solutions for polynomial-factorial Diophantine equations

TL;DR

This paper proves the finiteness of solutions for certain polynomial and factorial Diophantine equations, linking number theory conjectures and special factorial functions, with some results unconditional and others conditional on conjectures.

Contribution

It establishes finiteness results for polynomial-factorial Diophantine equations, including cases involving norm forms and Bhargava factorials, and connects these to the Oesterlé-Masser conjecture.

Findings

Finiteness of solutions for equations involving norm forms and factorials.

Conditional finiteness results assuming the Oesterlé-Masser conjecture.

Unconditional finiteness for specific infinite subsets of integers.

Abstract

We study the Diophantine equations obtained by equating a polynomial and the factorial function, and prove the finiteness of integer solutions under certain conditions. For example, we show that there exists only finitely many $l$ such that $l!$ is represented {by} $N_A(x)$, where $N_A$ is a norm form constructed from the field norm of a field extension $K/\mathbf Q$. We also deal with the equation $N_A(x)=l!_S$, where $l!_S$ is the Bhargava factorial. In this paper, we also show that the Oesterl\'e-Masser conjecture implies that for any infinite subset $S$ of $\mathbf Z$ and for any polynomial $P(x)\in\mathbf Z[x]$ of degree $2$ or more the equation $P(x)=l!_S$ has only finitely many solutions $(x,l)$. For some special infinite subsets $S$ of $\mathbf Z$, we can show the finiteness of solutions for the equation $P(x)=l!_S$ unconditionally.

Equations

$$P(x)=l!,$$

$$\mathrm{rad}(n)=\prod _{p|n}p.$$

$$\max \{|a|,|b|,|c|\}<\beta (\epsilon )\mathrm{rad}(abc{)}^{1+\epsilon }$$

$$|abc|<\mathrm{rad}(abc{)}^s$$

$$\underset{n\to \infty }{\lim }\frac{|\{l\le n|\text{there exists x∈Z such that P(x)=l!}\}|}{n}=0.$$

$$\left(\genfrac{}{}{0ex}{}{al}{l,\dots ,l}\right)=\frac{(al)!}{(l!{)}^a}$$

$$\sum _{i=0}^n{a}_i{x}^i{y}^{n-i}=l!.$$

$$\sum _{i=0}^n{a}_i{x}^i{y}^{n-i}={\Pi }_K(l),$$

$${\Pi }_K(l)=\prod _{\begin{array}{c}\mathfrak{a}:\text{ideal}\\ \mathfrak{Na}\le l\end{array}}\mathfrak{Na},$$

$$F(x,y)=\prod _{j=1}^u{F}_j(x,y).$$

$$\bigcap _i\mathbf{P}({\mathcal{C}}_{F_i})\supset \{p:\text{prime}|p\equiv a\mathrm{mod}b\}.$$

$$F(x,y)=\prod _{j=1}^u{F}_j(x,y{)}^{m_j}.$$

$$S=\{\sigma |\text{there exists i such that σ(i)=i}\}.$$

$$|S|\le \sum _{i=1}^n|H_i|-(n-1)=|H|-n+1<|H|.$$

$${\Delta }_{F,mod}=\frac{{\Delta }_F}{\gcd (a_n,\dots ,a_0{)}^{2n-2}},$$

$$N=g{p}_1\cdots p_s{q}_1^{l_1}\cdots q_t^{l_t},$$

$$a{F}_1(x,y)\cdots F_r(x,y)\mathrm{mod}q,$$

$$F(x,y)=\prod _{j=1}^u{F}_j(x,y)$$

$$N=g{p}_1\cdots p_s{q}_1^{l_1}\cdots q_t^{l_t},$$

$$aN=2^{l}g{p}_1\cdots p_s{q}_1\cdots q_t{r}_1^{l_1}\cdots r_u^{l_u},$$

$${\mathfrak{p}}^{\ell }{\mathfrak{p}}_1\cdots {\mathfrak{p}}_s{\mathfrak{q}}_1\cdots {\mathfrak{q}}_t(r_1{)}^{\frac{l_1}{2}}\cdots (r_u{)}^{\frac{l_u}{2}}\in P_{\mathcal{O}},$$

$$aN$$

$$\equiv a^2{x}^2+ax+\frac{b^2-{\Delta }_F}{4}\mathrm{mod}2.$$

$$aN$$

$$\equiv 1\mathrm{mod}2.$$

$$aN$$

$$=\left(ax+\frac{b+\sqrt{{\Delta }_F}}{2}y\right)\left(ax+\frac{b-\sqrt{{\Delta }_F}}{2}y\right)$$

$$=\left(ax+\frac{b-{\Delta }_F}{2}y+\frac{{\Delta }_F+\sqrt{{\Delta }_F}}{2}y\right)\left(ax+\frac{b-{\Delta }_F}{2}y+\frac{{\Delta }_F-\sqrt{{\Delta }_F}}{2}y\right).$$

$$\mathfrak{b}={\mathfrak{p}}^{\ell }{\mathfrak{p}}_1\cdots {\mathfrak{p}}_s{\mathfrak{q}}_1\cdots {\mathfrak{q}}_t(r_1{)}^{\frac{l_1}{2}}\cdots (r_u{)}^{\frac{l_u}{2}}\in P_{\mathcal{O}}.$$

$$\mathfrak{b}=\left(x+\frac{{\Delta }_F+\sqrt{{\Delta }_F}}{2}y\right)\mathcal{O}$$

$$x^2+{\Delta }_{F}xy+\frac{{\Delta }_F^2-b^2}{4}{y}^2$$

$$\left(x+\frac{{\Delta }_F+b}{2}y\right)\left(x+\frac{{\Delta }_F-b}{2}y\right)$$

$$F(x,y)=\prod _{j=1}^u{F}_j(x,y).$$

Full text

1. Introduction

The Brocard-Ramanujan problem, which is an unsolved problem in number theory, is to find integer solutions $(x,l)$ of $x^{2}-1=l!$. Brocard and Ramanujan independently considered this problem and conjectured that the only solutions are $(x,l)=(5,4),(11,5)$ and $(71,7)$ [Br76, Br85, Ra13]. As one of its generalizations, it has been proposed that there are only finitely many solutions of the polynomial-factorial Diophantine equation

[TABLE]

where $P(x)$ is a polynomial of degree $2$ or more with integer coefficients. The generalized Brocard-Ramanujan problem excludes the case $\deg P=1$. In this case, we can observe that if $a_{1}|a_{0}$ the equation $a_{1}x+a_{0}=l!$ has infinitely many solutions $(x,l)$, and otherwise has only finitely many solutions.

The Oesterlé-Masser conjecture, also known as the ABC-conjecture, implies that polynomial-factorial equations (1.1) have only finitely many solutions. To explain the statement of the Oesterlé-Masser conjecture, we define the algebraic radical. For any non-zero integer $n$, the algebraic radical $\mathrm{rad}(n)$ is defined by

[TABLE]

The Oesterlé-Masser conjecture states that for any $\varepsilon>0$ there exists a positive constant $\beta(\varepsilon)$ such that

[TABLE]

for any triples $(a,b,c)$ of non-zero coprime integers with $a+b=c$ [Ma85, Os88]. In the following, we summarize results applying this conjecture to polynomial-factorial Diophantine equations.

First, Overholt showed that if the weak form of Szpiro’s conjecture is true, then $x^{2}-1=l!$ has only finitely many solutions [Ov93]. The weak form of Szpiro’s conjecture states that there exists a positive constant ${s>6}$ such that

[TABLE]

for any triples $(a,b,c)$ of non-zero coprime integers with $a+b=c$. One can check that the Oesterlé-Masser conjecture implies this conjecture. More generally, Dąbrowski showed that if the weak form of Szpiro’s conjecture is true, then for any integer $A$ the equation $x^{2}-A=l!$ has only finitely many solutions $(x,l)$ [Da96]. He also showed that when $A$ is not square of an integer this result becomes unconditional. As a generalization of these results, Luca showed that for any polynomial $P(x)$ with integer coefficients of degree $\geq 2$, the Oesterlé-Masser conjecture implies that the equation $P(x)=l!$ has only finitely many solutions $(x,l)$ [Lu02].

There are many unconditional results. It is known that for $m\geq 3$ the equation $x^{m}+y^{m}=l!$ has no solution with $\gcd(x,y)=1$ except for $(x,y,l)=(1,1,2)$ and $x^{m}-y^{m}=l!$ has no solution with $\gcd(x,y)=1$ except for $m=4$ [EO37]. In 1973, Pollack and Shapiro showed that $x^{4}-1=l!$ also has no solution [PS73]. If we remove the assumption $\gcd(x,y)=1$, there are infinitely many solutions $(x,y,l)$ of $x^{2}-y^{2}=l!$. In fact, for any $a\geq 4$, $(x,y,l)=(\frac{a!}{4}+1,\frac{a!}{4}-1,a)$ are solutions of $x^{2}-y^{2}=l!$.

Dąbrowski and Ulas showed that for all positive integer $B$ there exist infinitely many integers $A$ such that $x^{2}-A=B\cdot l!$ has at least three solutions in positive integer $x$ [DU13].

Berend and Osgood dealt with all polynomial $P(x)\in\mathbf{Z}[x]$ and showed that for any polynomial $P$ of degree $2$ or more with integer coefficients, the equation $P(x)=l!$ has only a density [math] set of solutions $l$ [BO92], that is,

[TABLE]

In 2006, Berend and Harmse considered several related problems. They showed that for any polynomial $P$ which is an irreducible polynomial or satisfies certain technical conditions, there exist only finitely many solutions of $P(x)=H_{l}$, where $H_{l}$ is a highly divisible sequence [BH06]. They chose the following three sequences as highly divisible sequences $H_{l}$,

• •

$H_{l}=l!$.

• •

$H_{l}=[1,2,\ldots,l]$,

• where $[1,2,\ldots,l]$ is the least common multiple of all positive integers less than or equal to $l$.
• •

$H_{l}=p_{1}p_{2}\cdots p_{l}$,

• where $2=p_{1}<p_{2}<\cdots<p_{l}<\cdots$ is the sequence of all primes.

They also consider $H_{l}$ defined by multinomial coefficients

[TABLE]

for a fixed integer $a$.

In this paper, we use algebraic number theoretical approaches to consider the number of solutions for equation (1.1). Therefore, it is appropriate for our approach to deal with the equation

[TABLE]

More generally, we focus on the equation

[TABLE]

where $a_{i}\in\mathbf{Z}$ and $\Pi_{K}$ is a generalized factorial function over number fields. Let $K$ be a number field and $\mathcal{O}_{K}$ be its ring of integers. Then the function $\Pi_{K}(l)$ is defined by

[TABLE]

where $\mathfrak{Na}=\#\mathcal{O}_{K}/\mathfrak{a}$.

Instead of studying the number of $(x,y,l)$, we study the number of $l$ for which there exists a pair $(x,y)$ such that $a_{n}x^{n}+\cdots+a_{0}y^{n}=\Pi_{K}(l)$ by the following reasons. It is known that when $d$ is not a square integer $x^{2}-dy^{2}=1$ has infinitely many solutions from the theory of Pell’s equation. Therefore, we can find $x^{2}-dy^{2}=l!$ has infinitely many solutions $(x,y,l)$ easily. To consider the relation between integers represented by polynomial and those of factorial, we consider the number of $l$ for which there exists a pair $(x,y)$ such that $a_{n}x^{n}+\cdots+a_{0}y^{n}=\Pi_{K}(l)$. In the case $K=\mathbf{Q}$ and $y=1$, equation (1.2) is reduced to the generalized Brocard-Ramanujan’s equation (1.1). Therefore, it is expected that our results give some improvement of the generalized Brocard-Ramanujan problem. For the equation $a_{n}x^{n}+\cdots+a_{0}y^{n}=l!$, we show that

Theorem 1.3**.**

Let $F(x,y)$ be a homogeneous irreducible polynomial with $\deg F\geq 2$, then there exist only finitely many $l$ such that $l!$ is represented by $F(x,y)$.

As a corollary of this theorem, we obtain the result of Berend and Harmse [BH06] for irreducible polynomials. For reducible polynomials, we prove the following result.

Theorem 1.4**.**

Let $F(x,y)$ a polynomial in $\mathbf{Z}[x,y]$ whose irreducible factorization is

[TABLE]

Assume that there exist positive integers $a$ and $b>1$ such that

[TABLE]

Then there exist only finitely many $l$ such that $l!$ is represented by $F(x,y)$.

We explain the definition of $\mathbf{P}(\mathcal{C}_{F_{i}})$ in Section 4, roughly speaking, $\mathbf{P}(\mathcal{C}_{F_{i}})$ is the set of all primes $p$ such that $F_{i}(x,1)\equiv 0\mod p$ has no solution in $\mathbf{Z}/p\mathbf{Z}$.

Taking $y=1$ in Theorem 1.4, we get the result of Berend and Harmse for reducible polynomials partially. As in their results, we can replace $l!$ with the above highly divisible sequences $H_{l}$. Theorem 1.4 is one of corollaries of the folloing result for equation (1.2).

Theorem 1.5**.**

Let $K$ be a Galois extension field over $\mathbf{Q}$ and $F(x,y)$ a polynomial in $\mathbf{Z}[x,y]$ whose irreducible factorization is

[TABLE]

Assume that there exist a conjugacy class $C$ of ${\rm Gal}(K/\mathbf{Q})$, positive integers $a$ and $b>1$ such that $\mathbf{P}(C)\cap\bigcap_{i}\mathbf{P}(\mathcal{C}_{F_{i}})\supset\{p:\text{prime}~{}|~{}p\equiv a\mod b\}$. If $\deg F$ does not divide $[K:\mathbf{Q}]$ then there exist only finitely many $l$ such that $\Pi_{K}(l)$ is represented by $F(x,y)$.

Moreover, for special quadratic forms $F(x,y)$ there exist infinitely many $l$ with $\Pi_{K}(l)$ represented by $F$.

In Section 2, we review some of the standard facts on algebraic number theory. First, we check that for any Galois group $G\not=\{1\}$ of splitting field of polynomial $P$, there exists an element such that it fixes no roots of $P$. This fact is very important to prove our theorems. Second, we review the Frobenius map of prime $p$. This map controls the irreducible factorization modulo $p$ of the polynomial corresponding to itself.

In Section 3, we introduce two auxiliary lemmas. The first one characterizes the prime factorization of integers which can be written as $F(x,y)$. This is one of the generalizations of Cho’s results [Ch14, Ch16]. Cho characterized the prime factorization of integers written as $x^{2}+ny^{2}$ or $x^{2}+xy+ny^{2}$ with congruent condition for $n\geq 1$. The second one gives a Bertrand type estimate for prime ideals corresponding to a conjugacy class of Galois group such that their ideal norm is of the form $p^{f}$. In a previous paper [Ta19], we considered a Bertrand type estimate for primes splitting completely, which is the simplest case of the second auxiliary lemma. This result plays a crucial role in proving our main theorem.

In Section 4, we give a necessary condition for the existence of infinitely many solutions of $F(x,y)=\Pi_{K}(l)$. As a corollary, we obtain the result of Berend and Harmse for irreducible polynomials. Also, we gave their result for reducible polynomials partially. Moreover, for special quadratic forms $F(x,y)$ there exists infinitely many $l$ such that $\Pi_{K}(l)$ is represented by $F$.

In Section 5, we consider a generalization of the result in Section 4 to multi-variable homogeneous polynomials. However, irreducible polynomial $x^{2}+y^{2}+z^{2}+w^{2}$ represents any positive integers, that is, there exists infinitely many $l$ such that $H_{l}$ is represented by $x^{2}+y^{2}+z^{2}+w^{2}$. Therefore, the simplest generalization of the result of Section 4 does not hold. In place of multi-variable homogeneous polynomials, we deal with norm forms of number fields. We can apply the same argument as in the proof of the result in Section 4 and obtain the finiteness of $l$ such that $H_{l}$ is represented by a norm form $N_{\alpha_{1},\ldots,\alpha_{n}}(x_{1},\ldots,x_{n})$ except for the case $n=1$. Since $N_{\alpha_{1}}(x)=x$, the equation $N_{\alpha_{1}}(x)=H_{l}$ has infinitely many solutions $(x,l)$. After that, we consider the equation $N_{\alpha_{1},\ldots,\alpha_{n}}(x_{1},\ldots,x_{n})=l!_{S}$, where $l!_{S}$ is the Bhargava factorial for $S\subset\mathbf{Z}$. Since the Bhargava factorial $l!_{\mathbf{Z}}$ is the ordinary factorial $l!$, we can regard this equation as one of the generalizations of the Brocard-Ramanujan problem. We point out that we can generalize Luca’s result, which states that the Oesterlé-Masser conjecture implies the finiteness of solution of the equation $P(x)=l!$, to the equation $P(x)=l!_{S}$ by following Luca’s proof. Also, we show the finiteness of solution of the equation $N_{\alpha_{1},\ldots,\alpha_{n}}(x_{1},\ldots,x_{n})=l!_{S(a,b,c)}$ for $S(a,b,c)=\{an^{2}+bn+c~{}|~{}n\in\mathbf{Z}\}$ with $(a,b,c)\in\mathbf{Z}^{3}$ and $ab\neq 0$.

2. Preliminaries on algebraic number theory

We recall some basic definitions and some propositions of algebraic number theory in this section. Let $P(x)=a_{n}x^{n}+\cdots+a_{0}\in\mathbf{Z}[x]$ be an irreducible polynomial with $\deg P=n$ and discriminant $\Delta_{P}$. When $\alpha_{1},\ldots,\alpha_{n}$ are roots of $P$, the splitting field $K_{P}$ of $P$ is $\mathbf{Q}(\alpha_{1},\ldots,\alpha_{n})$. It is known that $K_{P}/\mathbf{Q}$ is a Galois extension and the Galois closure of $\mathbf{Q}(\alpha_{i})/\mathbf{Q}$ for all $i=1,\ldots,n$. The Galois group $G_{P}$ of $K_{P}/\mathbf{Q}$ plays a crucial role for splitting of primes in $\mathbf{Q}(\alpha_{i})$ as follows.

We call a subgroup $H$ of $S_{n}$ transitive if the group orbit $H(i)=\{\sigma(i)~{}|~{}\sigma\in H\}$ is equal to $\{1,\ldots,n\}$ for $1\leq i\leq n$. From Galois Theory, the Galois group $G_{P}$ can be identified with a transitive subgroup $H$ of the symmetric group $S_{n}$ of degree $n$. For $\sigma\in H$, the cycle type of $\sigma$ is defined as the ascending ordered list $[f_{1},\ldots,f_{r}]$ of the sizes of the cycles in the cycle decomposition of $\sigma$. For example, the cycle type of $(1\ 2)(3\ 4)(6\ 7\ 8)=(1\ 2)(3\ 4)(5)(6\ 7\ 8)(9)\in H\subset S_{9}$ is $[1,1,2,2,3]$. Since if two permutations are conjugate in $H$ then they have the same cycle type, we can define the cycle type of conjugacy class $C=[\sigma]$ of $H$ by the cycle type of a representative $\sigma$. We introduce a lemma for transitive subgroups of the symmetric group $S_{n}$.

Lemma 2.1**.**

Let $H$ be a transitive subgroup of the symmetric group $S_{n}$ of degree $n\geq 2$. Then there exists an element $\sigma\in H$ such that $\sigma(i)\not=i$ for all $i=1,\ldots,n$.

Proof.

Let $H_{i}$ be the stabilizer subgroup of $H$ with respect to $i$. Then the orbit-stabilizer theorem and transitivity of $H$ leads to $|H_{i}|=|H|/n$. Now we consider the number of elements of the set

[TABLE]

Since the identity element belongs to all stabilizer $H_{i}$, we have

[TABLE]

Therefore, there exists an element $\sigma\in H$ such that $\sigma(i)\not=i$ for all $i=1,\ldots,n$. ∎

This lemma ensures that for any Galois group $G\not=\{1\}$ of an irreducible polynomial of degree $\geq 2$ contains an element that does not fix any of the roots.

Next, we review the definitions and properties of the Frobenius map. Let $p$ be a prime and $\mathfrak{P}$ a prime ideal of $\mathcal{O}_{K_{P}}$ lying above $p$. For a prime ideal $\mathfrak{P}$ in $\mathcal{O}_{K_{P}}$, we define the decomposition group $D_{\mathfrak{P}}$ of $\mathfrak{P}$ by $\{\sigma\in G_{P}~{}|~{}\sigma(\mathfrak{P})=\mathfrak{P}\}$. Since $\sigma(\mathfrak{P})=\mathfrak{P}$ and $\sigma(\mathcal{O}_{K_{P}})=\mathcal{O}_{K_{P}}$ for $\sigma\in D_{\mathfrak{P}}$, $\sigma$ induces an automorphism $\overline{\sigma}$ of $\mathcal{O}_{K_{P}}/\mathfrak{P}$ over $\mathbf{Z}/p\mathbf{Z}$. Now we consider the Galois group ${\rm Gal}((\mathcal{O}_{K_{P}}/\mathfrak{P})/(\mathbf{Z}/p\mathbf{Z}))$. It is known that this group is cyclic and there exists an unique automorphism $\sigma:x\rightarrow x^{p}$ which generates it. Then the Frobenius map $(p,K_{P}/\mathbf{Q})$ of $p$ is the image of $\sigma$ in Galois group $G_{P}$. If the Frobenius map $(p,K_{P}/\mathbf{Q})$ of $p$ belongs to a conjugacy class $C$ of $G_{P}$, then we say that $p$ corresponds to $C$. We denote the set of primes corresponding to $C\in\mathcal{C}$ by $\mathbf{P}(\mathcal{C})$. The following theorem gives a relation between the cycle type of the Frobenius map of $p$ and the monic irreducible factorization of $P(x)\mod p$, where $p$ does not divide $a_{n}\Delta_{P}$.

Theorem 2.2** (Frobenius).**

Let $p$ be a prime such that $p$ does not divide $a_{n}\Delta_{P}$. We denote the cycle type of the Frobenius map $(p,K_{P}/\mathbf{Q})$ of $p$ by $[f_{1},\ldots,f_{r}]$. Then the monic irreducible factorization of $P(x)\mod p$ is $P(x)\equiv a_{n}P_{1}(x)\cdots P_{r}(x)\mod p$, where $P_{i}(x)$ are distinct and $f_{i}=\deg P_{i}(x)$.

3. Auxiliary lemmas

Let $F(x,y)=a_{n}x^{n}+a_{n-1}x^{n-1}y+\cdots+a_{0}y^{n}$ be an irreducible homogeneous polynomial (i.e. binary form) and $K_{F}$ the splitting field of $F(x,1)$. Also, we define the modified discriminant $\Delta_{F,mod}$ by

[TABLE]

where $\Delta_{F}$ is the discriminant of $F(x,1)$. Cho studied the representation of integers as $x^{2}+ny^{2}$ or $x^{2}+xy+ny^{2}$ for $n\geq 1$ under the congruence conditions $x\equiv 1\mod m$ and $y\equiv 0\mod m$ [Ch14, Ch16]. We remove this congruence conditions and consider all polynomials with integer coefficients. Let $\mathcal{C}_{F}$ be the set of conjugacy classes $C$ of the Galois group $G_{F}={\rm Gal}(K_{F}/\mathbf{Q})$ whose cycle type $[f_{1},\ldots,f_{r}]$ satisfies $f_{i}\geq 2$ for all $i=1,\ldots,r$. This classification is very important to characterize integers represented by $F(x,y)$.

Lemma 3.1**.**

Let $F(x,y)=a_{n}x^{n}+a_{n-1}x^{n-1}y+\cdots+a_{0}y^{n}$ be an irreducible homogeneous polynomial with $g=\gcd(a_{n},a_{n-1},\ldots,a_{0})$. Let $N$ be an integer with

[TABLE]

where $q_{i}$ are primes corresponding to a conjugacy class $C\in\mathcal{C}_{F}$ with

$\gcd(q_{i},a_{n}\Delta_{F,mod})=1$ or $\gcd(q_{i},a_{0}\Delta_{F,mod})=1$ and $p_{i}$ are the other primes. If $N$ is represented by $F(x,y)$ then $n|l_{i}$ for all $i$.

Proof.

An integer $N$ is represented by $a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{0}y^{n}$ with $\gcd(a_{n},\ldots,a_{0})=g$ if and only if $\frac{N}{g}$ is represented by $\frac{a_{n}}{g}x^{n}+\cdots+\frac{a_{0}}{g}y^{n}$ with $\gcd(\frac{a_{n}}{g},\ldots,\frac{a_{0}}{g})=1$. Hence, we may assume without loss of generality that $\gcd(a_{n},a_{n-1},\ldots,a_{0})=1$ and $\Delta_{F}=\Delta_{F,mod}$ .

Let $q$ be a prime corresponding to a conjugacy class $C\in\mathcal{C}_{F}$ with

$\gcd(q_{i},$ $a\Delta_{F,mod})=1$, where $a\in\{a_{n},a_{0}\}$. By the definition of $\mathcal{C}_{F}$, the cycle type $[f_{1},\ldots,f_{r}]$ of $C$ satisfies $f_{i}\geq 2$ for all $i=1,\ldots,r$. As we seen in Lemma 2.2, the monic irreducible factorization of $F(x,y)\mod q$ is

[TABLE]

where the $F_{i}(x,y)$ are distinct and $f_{i}=\deg F_{i}(x,y)$. Since $\deg F_{i}(x,y)\geq 2$ for all $i=1,\ldots,r$, it follows that $F(x,y)\mod q$ has no solution expect for $(0,0)$ in $\mathbf{F}_{q}\times\mathbf{F}_{q}$. Therefore, if $q|N$ then $q$ divides both of $x$ and $y$. Thus $q^{n}|N$. We can apply the same argument repeatedly to $F(\frac{x}{q^{n}},\frac{y}{q^{n}})=\frac{N}{q^{n}}$, which leads to $n|l_{i}$.

This proves the lemma. ∎

As one of the corollaries of Lemma 3.1, we obtain the following theorem.

Theorem 3.2**.**

Let $F(x,y)=a_{n}x^{n}+a_{n-1}x^{n-1}y+\cdots+a_{0}y^{n}$ be a homogeneous polynomial whose irreducible factorization is

[TABLE]

and $g=\gcd(a_{n},a_{n-1},\ldots,a_{0})$. Let $N$ be an integer with

[TABLE]

where $q_{i}\in\cap_{j=1}^{u}\mathcal{C}_{F_{j}}$ with $\gcd(q_{i},a_{n}\Delta_{F,mod})=1$ or $\gcd(q_{i},a_{0}\Delta_{F,mod})=1$ and $p_{i}$ are the other primes. If $N$ is represented by $F(x,y)$ then $n|l_{i}$ for all $i$.

Proof.

By assumption, there exists a pair $(x_{0},y_{0})\in\mathbf{Z}^{2}$ such that $N=F(x_{0},y_{0})$. Since $q_{i}|N=F(x_{0},y_{0})$, there exists a polynomial $F_{j}(x,y)=a_{j,n_{j}}x^{n_{j}}+\cdots+a_{j,0}y^{n_{j}}$ such that $q_{i}|F_{j}(x_{0},y_{0})$. By the definition of discriminant of polynomial, we have $\Delta_{F_{j},mod}|\Delta_{F,mod}$, that is, $\gcd(q_{i},a\Delta_{F,mod})=1$ implies $\gcd(q_{i},a^{(j)}\Delta_{F_{j},mod})=1$, where $a\in\{a_{n},a_{0}\}$ and $a^{(j)}\in\{a_{j,n_{j}},a_{j,0}\}$. It follows that $x_{0}\equiv y_{0}\equiv 0\mod q$ by Lemma 3.1. Therefore we obtain $q^{n}|N=F(x_{0},y_{0})$ and $n|l_{i}$. This is the desired conclusion. ∎

Next we change the assumption of Lemma 3.1 and give a necessary and sufficient condition for integers to be represented by $F(x,y)$. We call a discriminant $\Delta$ fundamental, if one of the following statements holds

• •

$\Delta\equiv 1\mod 4$ and is square-free,

• •

$\Delta=4m$, where $m\equiv 2$ or $3\mod 4$ and $m$ is square-free.

In the following, we assume that one of $a$ and $c$ is a prime number or $1$ and the discriminant $\Delta_{F}$ is fundamental. We characterize the prime factorization of integers which are expressed by $ax^{2}+bxy+cy^{2}$.

Theorem 3.3**.**

Let $F(x,y)=ax^{2}+bxy+cy^{2}$ be a positive definite quadratic form with fundamental modified discriminant $\Delta_{F,mod}$ and $g=\gcd(a,b,c)$. We denote the corresponding order to $F$ by $\mathcal{O}$ and the set of principal ideals of $\mathcal{O}$ by $P_{\mathcal{O}}$. Let $N$ be an integer with

[TABLE]

where $p_{i}$ ramifies in $\mathbf{Q}(\sqrt{\Delta_{F}})$, $q_{i}$ splits completely in $\mathbf{Q}(\sqrt{\Delta_{F}})$ and $r_{i}$ are distinct odd inert primes in $\mathbf{Q}(\sqrt{\Delta_{F}})$. If $a$ is a prime number or $1$, then $N$ is represented by $F(x,y)$ if and only if

$l$ is even if $\Delta_{F,mod}\equiv 5\mod 8$. 2. 2.

$l_{i}$ are even numbers. 3. 3.

There exist prime ideals $\mathfrak{p},\mathfrak{p}_{1},\ldots,\mathfrak{p}_{s},\mathfrak{q}_{1},\ldots,\mathfrak{q}_{t}$ lying above $2$, $p_{1},\ldots,p_{s},$

$q_{1},\ldots,q_{t}$ respectively such that

[TABLE]

where $\ell$ is $\frac{l}{2}$ if $2$ is inert in $\mathbf{Q}(\sqrt{\Delta_{F}})$, $l$ otherwise.

Proof.

As in the proof of Lemma 3.1, we can assume $g=1$ without loss of generality. We assume that there exists a pair $(x,y)\in\mathbf{Z}^{2}$ such that $N=ax^{2}+bxy+cy^{2}$. We can obtain the second assertion by Lemma 3.1. Therefore, it suffices to prove the first and the third assertion.

First, we prove the first assertion. If $\Delta_{F}\equiv 5\mod 8$, $2|aN$ and $y$ is odd, then $\Delta_{F}=b^{2}-4ac$ is odd, that is, $b$ is. Therefore,

[TABLE]

This contradicts $2|aN$ and leads to $2|y$. From the identity $aN=a^{2}x^{2}+abxy+acy^{2}$ and this result, $ax$ is even. Therefore, $4|aN$ holds. The same argument repeatedly applied to $\frac{aN}{4}$ leads to $2|l$.

Next we show the third assertion. Since $\Delta_{F}=b^{2}-4ac$, we have

[TABLE]

Since $b\equiv\Delta_{F}\mod 2$, $\frac{b-\Delta_{F}}{2}$ is an integer. Therefore, $ax+\frac{b-\Delta_{F}}{2}y+\frac{\Delta_{F}+\sqrt{\Delta_{F}}}{2}y\in\mathcal{O}$ and $\left(ax+\frac{b-\Delta_{F}}{2}y+\frac{\Delta_{F}+\sqrt{\Delta_{F}}}{2}y\right)\mathcal{O}\in P_{\mathcal{O}}$. Now we denote this ideal by $\mathfrak{b}$ then we have $(aN)\mathcal{O}=\mathfrak{b\overline{b}}$. Since $p_{i}$ ramifies in $Q(\sqrt{\Delta_{F}})$, $p_{i}\mathcal{O}=\mathfrak{p}_{i}^{2}$ and $\mathfrak{p}_{i}|\mathfrak{b}$. Also, $r_{i}\mathcal{O}$ is a prime ideal in $Q(\sqrt{\Delta_{F}})$ and $(r_{i}\mathcal{O})^{\frac{l_{i}}{2}}|\mathfrak{b}$. Also, $q_{i}$ splits completely as $q_{i}\mathcal{O}=\mathfrak{q}_{i}\overline{\mathfrak{q}}_{i}$ in $Q(\sqrt{\Delta_{F}})$. Hence, there exist prime ideals $\mathfrak{q}_{i}$ such that

[TABLE]

Conversely, let $\mathfrak{b}$ be $\mathfrak{p}_{1}\cdots\mathfrak{p}_{s}\mathfrak{q}_{1}\cdots\mathfrak{q}_{t}(r_{1})^{\frac{l_{1}}{2}}\cdots(r_{u})^{\frac{l_{u}}{2}}\in P_{\mathcal{O}}$. This ideal can be expressed as

[TABLE]

and $(aN)\mathcal{O}=\mathfrak{b\overline{b}}$. Since $aN>0$, we obtain $aN=x^{2}+\Delta_{F}xy+\frac{\Delta_{F}^{2}-\Delta_{F}}{4}y^{2}$. Therefore we get

[TABLE]

Now we assume that $a$ is a prime number or $1$, so $a|(x+\frac{\Delta_{F}+b}{2}y)$ or $a|(x+\frac{\Delta_{F}-b}{2}y)$. Without loss of generality, we assume that $a\alpha=x+\frac{\Delta_{F}-b}{2}y$ for some integer $\alpha$. If we take $y=\beta$, then $aN=x^{2}+\Delta_{F}xy+\frac{\Delta_{F}^{2}-d}{4}y^{2}=a^{2}\alpha^{2}+ab\alpha\beta+ac\beta^{2}$. This proves the theorem. ∎

Remark 3.4**.**

By swapping $x$ and $y$ in the binary form $ax^{2}+bxy+cy^{2}$ , we can replace $a$ by $c$ in Theorem 3.3

In the following, we consider a Bertrand type estimate for primes corresponding to a conjugacy class $C$ of Galois group $G$ by following the method of Hulse and Murty. They gave one of the generalizations of Bertrand’s postulate, or Chebyshev’s theorem, to number fields [HM17]. We can obtain the following lemma, which gives a Bertrand type estimate for primes corresponding to $C$ of $G$, by following the argument of Hulse and Murty [HM17].

Lemma 3.5** (cf. [HM17, Ta19]).**

Let $L/K$ be a Galois extension with $[L:\mathbf{Q}]=n$ and $D_{L}$ the absolute value of the discriminant of $L$. For any $A>1$ there exists an effectively computable constant $c(A)>0$ such that for $x>\exp(c(A)n(\log D_{L})^{2})$ there is a prime corresponding to a conjugacy class $C$ of ${\rm Gal}(L/K)$ with $p\in(x,Ax)$.

Next, we consider the distribution of prime ideals $\mathfrak{p}$ corresponding to a conjugacy class $C$ such that their ideal norm is of the form $p^{f}$.

Theorem 3.6**.**

Let $L$ be the Galois closure of $K/\mathbf{Q}$ with $k=[L:\mathbf{Q}]$ and $p$ a prime corresponding to a conjugacy class $C$ of ${\rm Gal}(L/\mathbf{Q})$. For any $A>1$ there exists an effectively computable constant $c(A)>0$ such that for $p^{f_{i}}>\exp(c(A)k(\log D_{L})^{2})$ there exists a prime ideal $\mathfrak{q}$ with $\mathfrak{Nq}=q^{f_{i}}\in(p^{f_{i}},Ap^{f_{i}})$, where $q\in\mathbf{P}(C)$.

Proof.

Let $\mathfrak{p}$ be a prime ideal with $\mathfrak{Np}=p^{f}$ and $\mathfrak{P}$ a prime ideal lying above $\mathfrak{p}$. We denote the order of the decomposition group $D_{\mathfrak{P}/\mathfrak{p}}$ by $l$, then $\mathfrak{N}\mathfrak{P}=p^{fl}$. From Lemma 3.5, any $A>1$ there exists a constant $c(A)>0$ such that for $p^{fl}=\mathfrak{N}\mathfrak{P}>\exp(c(A)k(\log D_{L})^{2})$ there is a prime $\mathfrak{Q}$ corresponding to $C_{\mathfrak{P}}$ with $q^{fl}=\mathfrak{N}\mathfrak{Q}\in(p^{fl},Ap^{fl})$.

Since $p,q\in\mathbf{P}(C_{\mathfrak{P}})$, there exists a prime ideal $\mathfrak{Q}$ with $\mathfrak{N}\left(\mathfrak{Q}\cap\mathcal{O}_{K}\right)=q^{f}$. We denote $\mathfrak{q}=\mathfrak{Q}\cap\mathcal{O}_{K}$. Thus for $p^{f}>\exp(c(A)k(\log D_{L})^{2})$ there is a prime ideal $\mathfrak{q}$ with $\mathfrak{Nq}=q^{f}\in(p^{f},Ap^{f})$. ∎

4. Main theorems

In this section, we give a necessary condition for the existence of infinitely many solutions for the equation $F(x,y)=\Pi_{K}(l)$.

First, we consider the equation $F(x,y)=l!$, where $F$ is an irreducible homogeneous polynomial.

Theorem 4.1**.**

Let $F(x,y)=a_{n}x^{n}+a_{n-1}x^{n-1}y+\cdots+a_{0}y^{n}$ be a homogeneous irreducible polynomial with $\deg F\geq 2$, then there exist only finitely many $l$ such that $l!$ is represented by $F(x,y)$.

Proof.

Lemma 2.1 provides that $\mathcal{C}_{F}\not=\emptyset$. Let $C\in\mathcal{C}_{F}$ be a fixed conjugacy class of $G_{F}$. The assumption $\deg F\geq 2$ and Lemma 3.1 imply that if $N$ is represented by $F(x,y)$ and $p|N$ for prime $p$ corresponding to $C$ with $\gcd(q_{i},a_{n}\Delta_{F,mod})=1$ or $\gcd(q_{i},a_{0}\Delta_{F,mod})=1$, then $N$ is divisible by $p^{2}$ at least. In particular, $F(x,y)=p!$ has no integer solution $(x,y)$. Moreover, since the second smallest positive integer divisible by $p$ is $2p$, $l!$ is not of the form in Lemma 3.1 for $p\leq l<2p$, that is, there exists no pair $(x,y)\in\mathbf{Z}^{2}$ such that $F(x,y)=l!$ for $p\leq l<2p$.

Let $\alpha$ be a root of $F(x,1)$ and let $k$ be the extension degree of $K_{F}/\mathbf{Q}$. We denote the ring of integers of $\mathbf{Q}(\alpha)$ by $\mathcal{O}_{\alpha}$. Theorem 3.6 states that there exists $c>0$ such that for $x>\exp(ck(\log D_{K_{F}})^{2})$ there is a prime ideal $\mathfrak{p}$ of $\mathcal{O}_{\alpha}$ corresponding to $C$ with $\mathfrak{Np}=p^{f}\in(x,2x)$. Let $\mathfrak{p}$ be a prime ideal of $\mathcal{O}_{\alpha}$ corresponding to $C$ with $\mathfrak{Np}=p^{f}>\max\{\exp(ck(\log D_{K_{F}})^{2}),(a_{n}\Delta_{F,mod})^{f},$ $(a_{0}\Delta_{F,mod})^{f}\}$. Since we have $p^{f}>\exp(ck(\log D_{K_{F}})^{2})$, there exists $\mathfrak{q}$ corresponding to $C$ with $\mathfrak{Nq}=q^{f}\in(p^{f},2p^{f})$, that is, there exists a prime $q$ corresponding to $C$ with $q\in(p,2p)$.

As above, $l!$ is not of the form in Lemma 3.1 for $q\leq l<2q$ and there exists a prime $q_{1}$ corresponding to $C$ with $q_{1}\in(q,2q)$. By induction, $l!$ is not of the form in Lemma 3.1 for $p\leq l$. This shows the finiteness of $l$ such that $l!$ is represented by $F(x,y)$.

∎

As a corollary of this theorem, we obtain the result of Berend and Harmse for irreducible polynomial.

Theorem 4.2** (Theorem 3.1. of [BH06]).**

For any irreducible polynomial $P(x)\in\mathbf{Z}[x]$ with $\deg P\geq 2$, the equation $P(x)=H_{l}$ has only finitely many solutions $(x,l)$.

Next we consider the general case $F(x,y)=\Pi_{K}(l)$. For a prime $p$ and its Frobenius map $(p,K_{P}/\mathbf{Q})$ with cycle type $[f_{1},\ldots,f_{r}]$, we define $G_{p}(l;K)$ as the number of $f_{i}$ such that $f_{i}=l$. If $K/\mathbf{Q}$ is a Galois extension with extension degree $k$, then $fG_{p}(f;K)=k$ for all primes $p$ unramified in $K$, where $f$ is the inertia degree of $p$ in $K$. Therefore, we obtain the following theorem.

Theorem 4.3**.**

Let $K$ be a Galois extension of $\mathbf{Q}$ and $F(x,y)$ a polynomial in $\mathbf{Z}[x,y]$ whose irreducible factorization is

[TABLE]

Assume that there exist a conjugacy class $C$ of ${\rm Gal}(K/\mathbf{Q})$, positive integers $a$ and $b>1$ such that $\mathbf{P}(C)\cap\bigcap_{i}\mathbf{P}(\mathcal{C}_{F_{i}})\supset\{p:\text{prime}~{}|~{}p\equiv a\mod b\}$. If $\deg F$ does not divide $[K:\mathbf{Q}]$ then there exist only finitely many $l$ such that $\Pi_{K}(l)$ is represented by $F(x,y)$.

Proof.

We denote the extension degree $[K:\mathbf{Q}]$ by $k$. As we remarked above, $fG_{p}(f;K)=k$ for all primes $p$ unramified in $K$, where $f$ is the inertia degree of $p$ in $K$. Then for all primes $p$ unramified in $K$, $p^{fa(p^{f})}=p^{k}$. Since $\deg F$ does not divide $[K:\mathbf{Q}]$, Lemma 3.2 implies that there is no pair $(x,y)$ such that $F(x,y)=\Pi_{K}(p^{f})$ for $p\in\{p:\text{prime}~{}|~{}p\equiv a\mod b\}$.

By a Bertrand type estimate for primes in an arithmetic progression $bm+a$, there exists $c>0$ such that for $x>c$ there is a prime $p\in\{p:\text{prime}~{}|~{}p\equiv a\mod b\}$ with $p\in(x,2^{\frac{1}{f}}x)$. Let $\mathfrak{p}$ be a prime ideal of $K$ corresponding to $C$ with $\mathfrak{Np}=p^{f}>\max\{c^{f},(a_{n}\Delta_{F,mod})^{f},(a_{0}\Delta_{F,mod})^{f}\}$. Since $p>c$, there exist a prime $q$ satisfying $q=bm_{q}+a\in(p,2^{\frac{1}{f}}p)$. By the assumption, we have $q\in\mathbf{P}(C)$ and there exist $\mathfrak{q}$ lying above $q$ with $\mathfrak{Nq}=q^{f}\in(p^{f},2p^{f})$.

As above, $\Pi_{K}(l)$ is not of the form in Lemma 3.2 for $q^{f}\leq l<2q^{f}$ and there exists a prime ideal $\mathfrak{q}_{1}$ corresponding to $C$ with $\mathfrak{Nq}_{1}\in(q^{f},2q^{f})$. By induction, $\Pi_{K}(l)$ is not of the form in Lemma 3.2 for $p^{f}\leq l$. This shows the finiteness of $l$ such that $\Pi_{K}(l)$ is represented by $F(x,y)$. ∎

Since the $p$-factor of the above highly divisible sequences $H_{l}$ appears with regularity, we can replace $\Pi_{K}(l)$ with $H_{l}$ in Theorem 4.3. When $K=\mathbf{Q}$, the conjugacy class $C$ in Theorem 4.3 is equal to $\{1\}$ and $\deg F$ does not divide $[K:\mathbf{Q}]$. Therefore, we obtain the following corollary.

Corollary 4.4**.**

Let $F(x,y)$ a polynomial in $\mathbf{Z}[x,y]$ whose irreducible factorization is

[TABLE]

Assume that there exist positive integers $a$ and $b>1$ such that

[TABLE]

Then there exist only finitely many $l$ such that $H_{l}$ is represented by $F(x,y)$.

Taking $y=1$ in Corollary 4.4, we get the result of Berend and Harmse for reducible polynomials partially. To explain their result, we introduce the natural density $d(S)$ for a subset $S$ of the set of all primes defined by

[TABLE]

where $\pi(x)$ is the number of primes $p\leq x$ and $\pi(x,S)$ is the number of those belonging to $S$.

Theorem 4.5** (Theorem 4.1. of [BH06]).**

Consider the equation

[TABLE]

Let $Q(x)\in\mathbf{Z}[x]$ be any factor (irreducible or not) of $P$. Denote by $S(Q)\subset P$ the set of all primes $p$ for which the congruence $Q(x)\equiv 0\mod p$ has a solution. If $d(S(Q))<\frac{\deg Q}{\deg P}$, then (4.6) has only finitely many solutions.

The assumption $\bigcap_{i}\mathbf{P}(\mathcal{C}_{F_{i}})\supset\{p:\text{prime}~{}|~{}p\equiv a\mod b\}$ in Corollary 4.4 leads to $d(S(F(x,1)))<1$. Thus, Theorem 4.5 implies Corollary 4.4 with $y=1$.

If $\deg F$ is even then we can remove the assumption that $K/\mathbf{Q}$ is a Galois extension.

Theorem 4.7**.**

Let $K$ be a number field and $F(x,y)$ a polynomial of even degree in $\mathbf{Z}[x,y]$ whose irreducible factorization is

[TABLE]

Let $p$ be a prime whose Frobenius map $(p,K_{P}/\mathbf{Q})$ has the cycle type $[f_{1},\ldots,f_{r}]$ such that $G_{p}(m;K)$ is odd for some odd $m$ and $C$ the conjugacy class of $(p,K/\mathbf{Q})$. If there exist positive integers $a$ and $b>1$ such that $\mathbf{P}(C)\cap\bigcap_{i}\mathbf{P}(\mathcal{C}_{F_{i}})\supset\{p:\text{prime}~{}|~{}p\equiv a\mod b\}$, then there exist only finitely many $l$ such that $\Pi_{K}(l)$ is represented by $F(x,y)$.

Proof.

We assume that $G_{p}(i;K)$ is odd for some prime $p$ and some odd $i$, and the Frobenius map $(p,K_{P}/\mathbf{Q})$ has the cycle type $[f_{1},\ldots,f_{r}]$ in $K$. Let $m=\min\{i~{}|~{}G_{p}(i;K)\text{ is odd}\}$. For all odd $i<m$, the number $G_{p}(i;K)$ is even. Let $a(n)$ be the number of ideals of $\mathcal{O}_{K}$ with $\mathfrak{Na}=n$ It follows from the Chinese Remainder Theorem that the function $a(n)$ satisfies the multiplicative property

[TABLE]

The ideals $\mathfrak{a}$ such that $\mathfrak{Na}=p^{m}$ is expressed by product of prime ideals $\mathfrak{p}$ with $\mathfrak{Np}=p^{k}(k\leq m)$. If $\mathfrak{a}$ is expressed as $\mathfrak{p}_{1}\cdots\mathfrak{p}_{s}$ and the number of $\mathfrak{p}_{t}$ with $\mathfrak{Np}_{t}=p^{i}$ equals $a_{i}$, then we have $a_{1}+\cdots+ma_{m}=m$. By considering the number of combinations with repetition, we get

[TABLE]

Since $m$ is odd, there exists an odd $i$ such that $a_{i}$ is odd in each product. For this odd $i$

[TABLE]

is even, since binomial coefficients $\binom{e}{o}$, where $e$ is an even number and $o$ is an odd number, are always even. Therefore

[TABLE]

Accordingly, $\Pi_{K}(p^{m})$ is not of the form in Theorem 3.2. Since $p^{m}$-factor does not appear in $(p^{m},2p^{m})$, for all $l\in(p^{m},2p^{m})$ the left hand side $\Pi_{K}(l)$ is not of the form in Theorem 3.2.

On the other hand, a Bertrand type estimate for primes in an arithmetic progression $bt+a$ leads to the conclusion that there exists $c>0$ such that for $x>c$ there is a prime $p\in\{p:\text{prime}~{}|~{}p\equiv a\mod b\}$ with $p\in(x,2^{\frac{1}{m}}x)$. Let $\mathfrak{p}$ be a prime ideal of $K$ corresponding to $C$ with $\mathfrak{Np}=p^{m}>\max\{c^{m},(a_{n}\Delta_{F,mod})^{m},(a_{0}\Delta_{F,mod})^{m}\}$. Since $p>c$, there exists a prime $q$ satisfying $q=bt_{q}+a\in(p,2^{\frac{1}{m}}p)$. By the assumption, we have $q\in\mathbf{P}(C)$ and there exist $\mathfrak{q}$ lying above $q$ with $\mathfrak{Nq}=q^{m}\in(p^{m},2p^{m})$.

As above, $\Pi_{K}(l)$ is not of the form in Lemma 3.2 for $q^{m}\leq l<2q^{m}$ and there exists a prime ideal $\mathfrak{q}_{1}$ corresponding to $C$ with $\mathfrak{Nq}_{1}\in(q^{m},2q^{m})$.

By induction, $\Pi_{K}(l)$ does not satisfy the condition in Theorem 3.2 for all $l\geq p^{m}$. This shows the theorem. ∎

From Lemma 3.5, we observe that for a number field $K$ there exists an effectively computable constant $c(2)>0$ such that for $l>\exp(c(2)k(\log D)^{2})$ there is a prime $p$ such that $p^{n}$ divides $\Pi_{K}(l)$ but $p^{n+1}$ does not divide $\Pi_{K}(l)$. Therefore, we obtain the following result.

Theorem 4.8**.**

Let $K$ be a number field and $F(x,y)$ a polynomial in $\mathbf{Z}[x,y]$ whose irreducible factorization is

[TABLE]

where $F_{j}$ are distinct irreducible polynomials. If $\min\{e_{1},\ldots,e_{u}\}>[K:\mathbf{Q}]$, then there exist only finitely many $l$ such that $\Pi_{K}(l)$ is represented by $F(x,y)$.

For special quadratic forms, we give a sufficient condition for existence of infinitely many solutions. We denote the set of primes which is inert in $\mathbf{Q}(\sqrt{\Delta})$ by $P_{\Delta}$.

Theorem 4.9**.**

Let $K$ be a number field with $n=[K:\mathbf{Q}]$ and $D_{K}$ its discriminant. Let $F(x,y)=ax^{2}+bxy+cy^{2}$ be a positive definite quadratic form with fundamental modified discriminant $\Delta_{F,mod}$, where one of $a$ and $c$ is a prime number or $1$. We denote $P_{\Delta_{F},D_{K}}=P_{\Delta_{F}}\setminus\{p|D_{K}\}$. We assume that the class number of $\mathbf{Q}(\sqrt{\Delta_{F}})$ equals $1$. If for all $p\in P_{\Delta_{F,mod},D_{K}}$ and for odd $i$, $G_{p}(i;K)$ is even, then there exist infinitely many $l$ such that $\Pi_{K}(l)$ is represented by $F(x,y)$.

Proof.

We assume for all $p\in P_{\Delta_{F},D_{K}}$ and odd $i$, $G_{p}(i;K)$ is even. It suffices to show that the prime factorization of $\Pi_{K}(l)$ contains no prime $p\in P_{\Delta_{F},D_{K}}$ raised to an odd power for infinitely many $l$. From the multiplicative property of $a(n)$ we show $a(p^{m})$ is even for all primes $p\in P_{\Delta_{F},D_{K}}$ and odd $m$ in the following. As in the proof of Theorem 4.7, we get

[TABLE]

Now we assume $G_{p}(i;K)$ is even for all odd $i$. Since $m$ is odd, there exists an odd $i$ such that $a_{i}$ is odd in each product. As we remarked above, one of the binomial coefficients in the above product are even. Therefore, $a(p^{m})$ is a sum of even numbers, $a(p^{m})$ is also even.

If $G_{p}(i;K)$ is odd for some $p\in P_{\Delta_{F}}\setminus P_{\Delta_{F},D_{K}}$ and some odd $i$, we denote $m=\min\{i:\text{odd}~{}|~{}G_{p}(i,K)\text{ is odd}\}$. As we mentioned above, $a(p^{m})$ is odd. Chebotarev’s density theorem says that for any number fields $K$ there exist infinitely many primes splitting completely in $K$. Let $q$ be a prime splitting completely in $K$. Then we have $a(q^{k})=\binom{n+k-1}{n-1}$. One can see easily that $\binom{n+k-1}{n-1}$ takes odd values infinitely many times and $a(p^{m}q^{k})$ does. Since $P_{\Delta_{F}}\setminus P_{\Delta_{F},D_{K}}$ is a finite set, $\Pi_{K}(l)$ satisfies the first and second conditions in Theorem 3.3 infinitely many times. By assumption, the third condition in Theorem 3.3 is trivial. This shows the theorem. ∎

5. Some generalizations

In the previous sections, we deal with two variables homogeneous polynomial. Naturally, we have an interest in the Brocard-Ramanujan problem for multi-variable homogeneous polynomial. In this section, we consider polynomials in more variables.

It is known that a positive integer $n$ is the sum of three squares of integers if and only if it is not of the form $4^{l}(8k+7)$, where $k,l$ are non-negative integers. Using this criterion, we check that there are infinitely many $l$ such that $l!$ is represented by $x^{2}+y^{2}+z^{2}$. Therefore, irreducibility of polynomials $f(x_{1},\ldots,x_{n})$ is not important for the finiteness of the solutions of $f(x_{1},\ldots,x_{n})=l!$.

In this paper, we consider the equations involving norm forms and factorial functions. Let $\mathcal{O}$ be an order of number field $K$ and $\{\alpha_{1},\ldots,\alpha_{n}\}$ be their basis over $\mathbf{Z}$. Then the norm form $N_{\alpha_{1},\ldots,\alpha_{n}}$ is defined by

[TABLE]

There exists the matrix $A$ converting the basis $\{\alpha_{1},\ldots,\alpha_{n}\}$ to the basis $\{1,\alpha_{2}^{\prime},\ldots,\alpha_{n}^{\prime}\}$ of $\mathcal{O}$. Also, since $A\in{\rm SL}_{n}(\mathbf{Z})$, an integer $N$ is represented by $N_{\alpha_{1},\ldots,\alpha_{n}}$ if and only if $N$ is also represented by $N_{1,\alpha_{2}^{\prime},\ldots,\alpha_{n}^{\prime}}$. Therefore, it suffices to consider the case $\alpha_{1}=1$.

As a corollary of Theorem 4.1 we have

Corollary 5.1**.**

For any norm form $N_{\alpha_{1},\alpha_{2}}$ of quadratic fields, there exists only finitely many $l$ such that $H_{l}$ is represented by $N_{\alpha_{1},\alpha_{2}}(x_{1},x_{2})$.

We generalize this corollary to all norm forms by following the proof of Theorem 4.1 as follows.

Theorem 5.2**.**

For any order $\mathcal{O}\not=\mathbf{Z}$ of a number field and their basis $\{\alpha_{1},\ldots,\alpha_{n}\}$ over $\mathbf{Z}$, there exists only finitely many $l$ such that $H_{l}$ is represented by $N_{\alpha_{1},\ldots,\alpha_{n}}(x_{1},\ldots,x_{n})$.

Using the following lemma, we can show Theorem 5.2 by the same argument with the proof of Theorem 4.1.

Lemma 5.3**.**

Let $\mathcal{O}$ be an order of a number field $K\neq\mathbf{Q}$ and $A=\{1,\alpha_{2},\ldots,\alpha_{n}\}$ be a basis of $\mathcal{O}$ over $\mathbf{Z}$. Let $F(x_{1},x_{2})$ be the homogeneous polynomial obtained by substituting $x_{i}=0$ for $i\geq 3$ into $N_{A}(x_{1},\ldots,x_{n})$. Let $N$ be an integer with

[TABLE]

where $q_{i}$ are primes corresponding to a conjugacy class $C\in\mathcal{C}_{F}$ with

$\gcd(q_{i},F(0,1)\Delta_{F})=1$ and $p_{i}$ are the other primes. If $N$ is represented by $N_{A}$ then $l_{i}\geq 2$ for all $i$.

Remark 5.4**.**

By considering other intermediate fields $\mathbf{Q}(\alpha_{i})$ of $K/\mathbf{Q}$, we can characterize the prime factorization of integers represented by $N_{A}$ more specifically.

Next, we replace the right hand side of the equation $N_{A}(x)=l!$ with the Bhargava factorial. Bhargava introduced a generalization of the factorial function to generalize classical results in $\mathbf{Z}$ to Dedekind domains and unify them [Bh97, Bh00]. In this section, we consider the number of solutions for equations involving polynomials and the Bhargava factorial. Since the ordinary factorial $l!$ is one of examples of the Bhargava factorial, we regard this equation as one of the generalizations of the Brocard-Ramanujan problem. The Bhargava factorial is defined as follows.

Let $S$ be an infinite subset of $\mathbf{Z}$. First, we define $p$-ordering of $S$. A $p$-ordering of $S$ is any sequence $\{a_{n}\}$ of elements of $S$ that is formed as follows:

• •

Choose any element $a_{0}\in S$;

• •

For $k\geq 1$ choose an element $a_{n}\in S$ such that

[TABLE]

where $v_{p}$ is the $p$-adic valuation defined by $v_{p}(p^{v}a)=v$ with an integer $a$ relatively prime to $p$.

For a $p$-ordering of $S$, we construct the $p$-sequence $\{v_{p}(n;S)\}$ as

[TABLE]

It is known that the associated $p$-sequence of $S$ is independent of the choice of $p$-ordering of $S$ [Bh00].

With these settings, we define the Bhargava factorial $l!_{S}$ by

[TABLE]

We give some examples of the Bhargava factorial. When $S=\mathbf{Z}$, we can choose the natural ordering $0,1,2,3,\ldots$ as $p$-ordering of $l!_{S}$ for all primes $p$ and find $l!_{S}$ is the ordinary factorial $l!$. This is why we can regard the equation $P(x)=l!_{S}$ as one of the generalizations of the Brocard-Ramanujan problem $x^{2}-1=l!$. Also, when $S(a,b)=\{an+b~{}|~{}n\in\mathbf{Z}\}$ for some $a,b\in\mathbf{Z}$ then $l!_{S(a,b)}=a^{l}l!$. Since the arguments in the proof of Theorem 4.1 also works for the equation $P(x)=l!_{S(a,b)}$, we obtain the finiteness of solutions $(x,l)$ for the equation $P(x)=l!_{S(a,b)}$.

In this section, we point out that we can generalize Luca’s result by following his proof. Luca showed that the Oesterlé-Masser conjecture implies that the equation $P(x)=l!$ has only finitely many solutions $(x,l)$ [Lu02]. In the proof of this result, Luca used the facts that ${\rm rad}(l!)<4^{l}$ and the Stirling formula $\log l!\sim l\log l$ as $l\rightarrow\infty$ and he estimate ${\rm rad}(l!)$ and $l!$. Therefore, if we can estimate $l!_{S}$ and ${\rm rad}(l!_{S})$, we can argue as in the proof of Luca’s result. We summarize this more general form.

Theorem 5.5** (cf. Luca. [Lu02]).**

Let $P(x)\in\mathbf{Z}[x]$ be a polynomial of $\deg P\geq 2$ and $F(l)$ be a function satisfying ${\rm rad}(F(l))=o(F(l))$ as $l\rightarrow\infty$. Then the Oesterlé-Masser conjecture implies that the equation $P(x)=F(l)$ has only finitely many solutions $(x,l)$.

We check that the Bhargava factorial for an infinite subset $S\subset\mathbf{Z}$ satisfies the condition of Theorem 5.5. Since $l!|l!_{S}$, for all primes $p$, the $p$-adic valuation $v_{p}(l!_{S})$ tends to infinity as $l\rightarrow\infty$. Therefore, as $l\rightarrow\infty$, ${\rm rad}(l!_{S})=o(l!_{S})$.

Corollary 5.6**.**

Let $P(x)\in\mathbf{Z}[x]$ be a polynomial of $\deg P\geq 2$ and $S$ be an infinite subset of $\mathbf{Z}$. Then the Oesterlé-Masser conjecture implies that the equation $P(x)=l!_{S}$ has only finitely many solutions $(x,l)$.

For some special case, we can show the finiteness of solutions for the equation $P(x)=l!_{S}$ unconditionally.

Theorem 5.7**.**

Let $N_{\alpha_{1},\ldots,\alpha_{n}}(x_{1},\ldots,x_{n})$ be a norm form of number field $K\neq\mathbf{Q}$. For a polynomial $f(x)=ax^{2}+bx+c$ with $(a,b,c)\in\mathbf{Z}^{3}-\{(0,0,c)~{}|~{}c\in\mathbf{Z}\}$ we denote $S=f(\mathbf{Z})$. Then there exist only finitely many $l$ such that $l!_{S}$ is represented by $N_{\alpha_{1},\ldots,\alpha_{n}}(x_{1},\ldots,x_{n})$.

Proof.

As we remarked above, when $a=0$, we find that $l!_{S}=b^{l}l!$ and the same argument as in the proof of Theorem 4.1 also works for the equation $P(x)=l!_{S}$. Therefore, it suffices to show the case $a\neq 0$. Let $p$ be an odd prime not dividing $a$. Then we have

[TABLE]

and we can choose an ordering $f(n_{0}),\ldots,f(n_{p-1}),\ldots$ satisfying the following three conditions:

1. (1)

$\{n_{0},\ldots,n_{p-1}\}=[0,p-1]\cap\mathbf{Z}$; 2. (2)

$f^{\prime}(n_{0})\equiv 0\mod p$; 3. (3)

For $0\leq i<j\leq\frac{p-1}{2}$, $f(n_{i})\not\equiv f(n_{j})\mod p$.

This ordering forms a $p$-ordering of $S$ and we can estimate $v_{p}(l!_{S})$ as

[TABLE]

Following the notation in Lemma 5.3, we denote

[TABLE]

Lemma 5.3 leads to the conclusion that if $N$ is represented by $N_{\alpha_{1},\ldots,\alpha_{n}}$ and $p|N$ for odd prime $p$ corresponding to $C\in\mathcal{C}_{F}$ with $\gcd(q_{i},F(0,1)\Delta_{F})$ $=1$, then $N$ is divisible by $p^{2}$ at least. In particular, $N_{\alpha_{1},\ldots,\alpha_{n}}(x_{1},\ldots,x_{n})$ $=\frac{p+1}{2}!_{S}$ has no integer solution $(x_{1},\ldots,x_{n})$. Moreover, since the second smallest $p$-factor appears at $l=p$ by estimate (5.8), $l!_{S}$ is not of the form in Lemma 5.3 for $\frac{p+1}{2}\leq l<p$, that is, there exists no $n$-tuple $(x_{1},\ldots,x_{n})\in\mathbf{Z}^{n}$ such that $N_{\alpha_{1},\ldots,\alpha_{n}}(x_{1},\ldots,x_{n})=l!_{S}$ for $\frac{p+1}{2}\leq l<p$.

As in the proof of Theorem 4.1, for sufficient large prime $p$, there exists a prime $q$ corresponding to $C$ with $\frac{q+1}{2}\in(\frac{p+1}{2},p)$ and $l!_{S}$ is not of the form in Lemma 5.3 for $\frac{q+1}{2}\leq l<q$. By induction, $l!_{S}$ is not of the form in Lemma 5.3 for $\frac{p+1}{2}\leq l$. This shows the finiteness of $l$ such that $l!_{S}$ is represented by $N_{\alpha_{1},\ldots,\alpha_{n}}(x_{1},\ldots,x_{n})$. ∎

The case $\deg f\geq 3$, it depends on the base field $K$. For example, when $f(x)=x^{3}$ then we find

[TABLE]

If $K/\mathbf{Q}$ is an abelian extension, then there exists a positive integer $D$ which characterizes the set of primes corresponding to a conjugacy class $C\subset{\rm Gal}(K/\mathbf{Q})$. Therefore, for any norm form $N_{A}$ of $K$, we can show the finiteness of solutions for $N_{A}(x)=l!_{S}$. On the other hand, if $K/\mathbf{Q}$ is not an abelian extension, then we cannot characterize the set of primes corresponding to a conjugacy class $C\subset{\rm Gal}(K/\mathbf{Q})$ by any modulus and it is difficult to show the finiteness of solutions in general.

6. acknowledgement

The author deeply express their sincere gratitude to Professor M. Ram Murty and Professor Andrzej Dąbrowski for fruitful discussions. The author also deeply thanks Professor Kohji Matsumoto and Professor Masatoshi Suzuki for their precious advice. This work was supported by Grant-in-Aid for JSPS Research Fellow (Grant Number: 19J10705).