<figure><img src="image_1.png" /></figure>

BLAISE PASCAL

YASUHIRO KISHI

On $D_5$-polynomials with integer coefficients

Volume 16, n° 1 (2009), p. 113-125.

<http://ambp.cedram.org/item?id=AMBP_2009__16_1_113_0>

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Publication éditée par le laboratoire de mathématiques
de l'université Blaise-Pascal, UMR 6620 du CNRS
Clermont-Ferrand — France
# On $D_5$-polynomials with integer coefficients

YASUHIRO KISHI

## Abstract

We give a family of $D_5$-polynomials with integer coefficients whose splitting fields over $\mathbb{Q}$ are unramified cyclic quintic extensions of quadratic fields. Our polynomials are constructed by using Fibonacci, Lucas numbers and units of certain cyclic quartic fields.

## 1. Introduction

The following is a fundamental problem in the theory of quadratic fields: For a given positive integer $N$, find quadratic fields whose class number is divisible by $N$. Several authors (for example, T. Nagell [6], N. C. Ankeny and S. Chowla [1], Y. Yamamoto [12], P. J. Weinberger [11] and H. Ichimura [3]) gave an infinite family of quadratic fields whose class number is divisible by arbitrary given integer $N$. If limited to the case $N = 5$, C. J. Parry [8], J.-F. Mestre [5], M. Sase [10] and D. Byeon [2] gave a family of quadratic fields whose class number is divisible by 5. In particular, Sase [10] gave a family of polynomials whose splitting field is a $D_5$-extension of $\mathbb{Q}$ and an unramified $C_5$-extension of containing the quadratic field. In the present paper, we will give other such polynomials by the use of the result of our previous result [4]. Then we get a new family of quadratic fields whose class number is divisible by 5. As a consequence, the following conjecture arises:

**Conjecture 1.1.** Let $F_n$ denote the $n$-th number in the Fibonacci sequence:

1, 1, 2, 3, 5, 8, 13, ...

Then the class number of the quadratic field $\mathbb{Q}(\sqrt{-F_{50s+25}})$ ($s \ge 0$) is divisible by 5.

<footnote>Keywords: class number, Fibonacci number, polynomial.</footnote>

<footnote>Math. classification: 11R29.</footnote>
The organization of this paper is as follows. In Section 1, we state the main theorem that the above conjecture is true under some conditions. In Section 2, we give a parametric $D_5$-polynomial with integer coefficients. It is a review of [4]. In Section 3, we study the Fibonacci and the Luca sequences. In Section 4, we give a proof of the main theorem. We give a numerical example in the last Section 5.

We list here those symbols which will be used throughout this article.

Let $\mathbf{Q}$ denote the field of rational numbers and $\mathbf{Z}$ denote the ring of rational integers.

For an integer $n$, let $C_n$ and $D_n$ denote the cyclic group of order $n$ and the dihedral group of order $2n$, respectively.

For an extension $L/K$, denote the norm map and the trace map of $L/K$ by $N_{L/K}$ and by $\mathrm{Tr}_{L/K}$, respectively. For simplicity, we denote $N_L$ and $\mathrm{Tr}_L$ if the base field $K = \mathbf{Q}$. For a Galois extension $L/K$, we denote the Galois group of $L/K$ by $\mathrm{Gal}(L/K)$.

For a polynomial $f(X)$ and a field $K$, we denote the minimal splitting field of $f(X)$ over $K$ by $\mathrm{Spl}_K(f)$.

## 2. The main Theorem

To state the main theorem, we prepare some notations.

Let $(F_n)$ and $(L_n)$ be the Fibonacci and the Luca sequences, respectively, defined as follows:

$$F_1 = 1, \quad F_2 = 1, \quad F_{n+2} = F_{n+1} + F_n \quad (n \ge 1),$$

$$L_1 = 1, \quad L_2 = 3, \quad L_{n+2} = L_{n+1} + L_n \quad (n \ge 1).$$

Let $\zeta := e^{2\pi i/5}$ be a primitive fifth root of unity. For a non-negative integer $m$, we set $k_m := \mathbf{Q}(\sqrt{-F_{2m+1}})$. Moreover we define a cyclic quartic field $M_m$ as follows: $M_m$ is the proper subextension of $k_m(\zeta)/\mathbf{Q}(\sqrt{5})$ other than $k_m(\sqrt{5})$ and $\mathbf{Q}(\zeta)$. Then we can express $M_m = \mathbf{Q}(\sqrt{-F_{2m+1}}(\zeta - \zeta^{-1}))$. We define an element $\delta_m$ of $M_m$ by

$$\delta_m := 1 + \epsilon^m \sqrt{-F_{2m+1}}(\zeta - \zeta^{-1}),$$

where $\epsilon := (1 + \sqrt{5})/2$ is a fundamental unit of $\mathbf{Q}(\sqrt{5})$. As we will see in Section 4, $\delta_m$ is a unit in $M_m$. Let $\tau$ be a generator of $\mathrm{Gal}(k_m(\zeta)/k_m)$ ($\cong C_4$) with $\zeta^\tau = \zeta^2$.
ON $D_5$-POLYNOMIALS

For a non-negative integer $m$, we define a polynomial $g_m(X)$ of degree 5 with integer coefficients by

$$g_m(X) := X^5 - 10X^3 - 20X^2 + 5(20F_{2m+1}^2 - 3)X + 40F_{2m+1}^2((-1)^m L_{2m+1} + 1) - 4.$$

Under the above notations and assumptions, we have the following.

**Theorem 2.1.** Let $m$ be a non-negative integer. If $\delta_m^{2+3\tau+\tau^2}$ is not a fifth power in $M_m$, then $\text{Spl}_Q(g_m)$ is a $D_5$-extension of $Q$ containing $k_m$. If, moreover, $m \equiv 12 \pmod{5^2}$, then $\text{Spl}_Q(g_m)$ is an unramified cyclic quintic extension of $k_m$. Hence, by putting $m = 25s+12$, the class number of the quadratic field $Q(\sqrt{-F_{50s+25}})$ is divisible by 5.

*Remarks 2.2.* The author think the assumption “$\delta_m^{2+3\tau+\tau^2}$ is not a fifth power in $M_m$” can be excluded (cf. Remark 6.2).

### 3. Construction of $D_5$-polynomials

This section is a review of [4] in the case $p = 5$ and the base field $Q$. Let $\zeta$ be a primitive fifth root of unity, and let $k = Q(\sqrt{D})$ be a quadratic field which does not coincide with $Q(\sqrt{5})$. Then there exists a unique proper subextension of the bicyclic biquadratic extension $k(\zeta)/Q(\sqrt{5})$ other than $k(\sqrt{5})$ and $Q(\zeta)$. We denote it by $M$. Then $M$ is a cyclic quartic field. Let us call $M$ the associated field with $k$. Fix the generator $\tau$ of $\text{Gal}(k(\zeta)/k)$ with $\zeta^\tau = \zeta^2$, and define a subset $\mathcal{M}(k)$ of $k(\zeta)^\times$ as follows:

$$\mathcal{M}(k) := \{\gamma \in k(\zeta)^\times \mid \gamma^{3+4\tau+2\tau^2+\tau^3} \notin k(\zeta)^5\}.$$

For an element $\gamma \in M$, we define a polynomial $f_\gamma(X)$ by

$$f_\gamma(X) := X^5 - 10N_M(\gamma)X^3 - 5N_M(\gamma)NT(\gamma)X^2 + 5N_M(\gamma)\{N_M(\gamma) - NT(\gamma^{1+\tau})\}X - N_M(\gamma)NT(\gamma^{2+\tau}),$$

where $NT = N_{Q(\sqrt{5})} \text{Tr}_{M/Q(\sqrt{5})}$.

Applying [4, Theorem 2.1, Corollary 2.6] to the case $p = 5$, we get the following proposition.

**Proposition 3.1.** Let the notation be as above. Then for $\delta \in \mathcal{M}(k) \cap M$, $\text{Spl}_Q(f_\delta)$ is a $D_5$-extension of $Q$ containing $k$. Putting $E := \text{Spl}_Q(f_\delta)$,
moreover, we have

$$E = k \left( \mathrm{Tr}_{E(\zeta)/E} \left( \sqrt[5]{\delta^{3+4\tau+2\tau^2+\tau^3}} \right) \right).$$

*Remarks 3.2.* In [4], we can see that every $D_5$-extension E of Q containing k is given as $E = \mathrm{SpI}_Q(f_\delta)$ for some $\delta \in \mathcal{M}(k) \cap M$.

## 4. Properties of Fibonacci and Lucas numbers

There are many relations between Fibonacci and Lucas numbers. (See for example [9] and [7].) In this section, we show the following four properties which we need in the next section.

(A) The power of $\varepsilon = (1 + \sqrt{5})/2$ is expressed by

$$\varepsilon^m = \frac{L_m + F_m \sqrt{5}}{2}.$$

(B) For positive integer $m$, we have

$$F_{m+1} = \frac{L_m + F_m}{2}, \qquad (4.1)$$

$$L_{m+1} = \frac{L_m + 5F_m}{2}. \qquad (4.2)$$

(C) Let $m$ be a positive integer. If $m$ is divisible by $5^2$, then so is $F_m$.

(D) Let $n$ and $m$ be positive integers. If $d = \mathrm{gcd}(n, m)$, then we have

$$\mathrm{gcd}(F_n, F_m) = F_d.$$

We easily get the property (A) by mathematical induction on $m$, using

$$\begin{aligned} \varepsilon^{m+1} &= \frac{L_m + F_m \sqrt{5}}{2} \cdot \frac{1 + \sqrt{5}}{2} \\ &= \frac{(L_m + 5F_m)/2 + (L_m + F_m)/2 \cdot \sqrt{5}}{2}, \end{aligned}$$

and the property (B).

We will prove the property (B) by mathematical induction on $m$. The equations (4.1) and (4.2) hold clearly for $m = 1$. Assume that (4.1) and
(4.2) hold for $m$. Then we have

$$ 
\begin{aligned}
\frac{L_{m+1} + F_{m+1}}{2} &= \frac{(L_m + 5F_m)/2 + F_{m+1}}{2} = \frac{L_m + 5F_m + 2F_{m+1}}{4} \\
&= \frac{(2F_{m+1} - F_m) + 5F_m + 2F_{m+1}}{4} = F_{m+1} + F_m = F_{m+2}
\end{aligned}
$$

and

$$ 
\begin{aligned}
\frac{L_{m+1} + 5F_{m+1}}{2} &= \frac{L_{m+1} + 5(L_m + F_m)/2}{2} = \frac{2L_{m+1} + 5L_m + 5F_m}{4} \\
&= \frac{2L_{m+1} + 5L_m + (2L_{m+1} - L_m)}{4} = L_{m+1} + L_m = L_{m+2}.
\end{aligned}
$$

Hence (4.1) and (4.2) hold for $m+1$.

Before proving the property (C), we show that the relation

$$ F_{n+m} = F_m F_{n+1} + F_{m-1} F_n \quad (4.3) $$

holds for positive integers $n, m$. For any $n$, we have

$$ 
\begin{aligned}
F_{n+1} &= 1 \cdot F_{n+1} + 0 \cdot F_n = F_1 F_{n+1} + F_0 F_n, \\
F_{n+2} &= 1 \cdot F_{n+1} + 1 \cdot F_n = F_2 F_{n+1} + F_1 F_n.
\end{aligned}
$$

(For convenience we define $F_0 = 0$.) Then (4.3) holds for $m=1, 2$. Assume that (4.3) holds for $m=k, k-1$:

$$ 
\begin{aligned}
F_{n+k} &= F_k F_{n+1} + F_{k-1} F_n, \\
F_{n+(k-1)} &= F_{k-1} F_{n+1} + F_{k-2} F_n.
\end{aligned}
$$

Then we have

$$ 
\begin{aligned}
F_{n+(k+1)} &= F_{n+k} + F_{n+(k-1)} \\
&= (F_k F_{n+1} + F_{k-1} F_n) + (F_{k-1} F_{n+1} + F_{k-2} F_n) \\
&= (F_k + F_{k-1}) F_{n+1} + (F_{k-1} + F_{k-2}) F_n = F_{k+1} F_{n+1} + F_k F_n.
\end{aligned}
$$

Hence (4.3) holds for $m=k+1$.

Next we prove

$$ m \equiv 0 \pmod{n} \implies F_m \equiv 0 \pmod{F_n}, \quad (4.4) $$

namely,

$$ F_{nk} \equiv 0 \pmod{F_n} $$
for each $k \in \mathbb{Z}$, $k \ge 1$. It holds clearly for $k = 1$. Assume that $F_{nk} \equiv 0 \pmod{F_n}$. Then by (4.3), we have

$$F_{n(k+1)} = F_{nk+n} = F_n F_{nk+1} + F_{n-1} F_{nk} \equiv 0 \pmod{F_n}.$$

Hence (4.4) is proved.

The property (C) follows from (4.4) and $F_{25} = 75025 \equiv 0 \pmod{5^2}$.

For positive integers $n$ and $m$, express $n = qm + r$, $0 \le r \le m$. Then we claim

$$\gcd(F_n, F_m) = \gcd(F_r, F_m). \qquad (4.5)$$

Using (4.3) and (4.4), we have

$$\begin{aligned} \gcd(F_n, F_m) &= \gcd(F_{qm+r}, F_m) \\ &= \gcd(F_{qm}F_{r+1} + F_{qm-1}F_r, F_m) \\ &= \gcd(F_{qm-1}F_r, F_m). \end{aligned}$$

Here we have

$$\begin{aligned} \gcd(F_{qm-1}, F_{qm}) &= \gcd(F_{qm-1}, F_{qm-1} + F_{qm-2}) = \gcd(F_{qm-1}, F_{qm-2}) \\ &= \dots = \gcd(F_2, F_1) = \gcd(1, 1) = 1. \end{aligned}$$

From this together with $F_m \mid F_{qm}$, we have $\gcd(F_{qm-1}, F_m) = 1$. Then we get (4.5). Hence by using the Euclidean algorithm, the property (D) follows.

*Remarks 4.1.* The inverse of the property (C) also holds true.

## 5. Proof of the main theorem

The goal of this section is to give a proof of our main theorem.

By the definition, $M_m$ is the associated field with $k_m$. Hence we can apply Proposition 3.1. Now let us calculate $f_{\delta_m}(X)$;

$$f_{\delta_m}(X) = X^5 - 10N_M(\delta_m)X^3 - 5N_M(\delta_m)NT(\delta_m)X^2 \\ + 5N_M(\delta_m)\{N_M(\delta_m) - NT(\delta_m^{1+\tau})\}X - N_M(\delta_m)NT(\delta_m^{2+\tau}).$$

We note that $\tau$ satisfies the following:

$$\zeta^{\tau} = \zeta^2, \quad (\sqrt{5})^{\tau} = -\sqrt{5}, \quad (\sqrt{-F_{2m+1}})^{\tau} = \sqrt{-F_{2m+1}}.$$
Put $\bar{\epsilon} := \epsilon^\tau$; then we have

$$
\begin{align*}
N_M(\delta_m) &= (1 + \epsilon^m \sqrt{-F_{2m+1}}(\zeta - \zeta^{-1}))(1 + \bar{\epsilon}^m \sqrt{-F_{2m+1}}(\zeta^2 - \zeta^{-2})) \\
&\quad \times (1 - \epsilon^m \sqrt{-F_{2m+1}}(\zeta - \zeta^{-1}))(1 - \bar{\epsilon}^m \sqrt{-F_{2m+1}}(\zeta^2 - \zeta^{-2})) \\
&= (1 + \epsilon^{2m} F_{2m+1}(\zeta - \zeta^{-1})^2)(1 + \bar{\epsilon}^{2m} F_{2m+1}(\zeta^2 - \zeta^{-2})^2) \\
&= \left(1 - \frac{5 + \sqrt{5}}{2}\epsilon^{2m} F_{2m+1}\right)\left(1 - \frac{5 - \sqrt{5}}{2}\bar{\epsilon}^{2m} F_{2m+1}\right) \\
&= 1 - \frac{5 + \sqrt{5}}{2}\epsilon^{2m} F_{2m+1} - \frac{5 - \sqrt{5}}{2}\bar{\epsilon}^{2m} F_{2m+1} \\
&\quad + 5N_{Q(\sqrt{5})}(\epsilon)^{2m} F_{2m+1}^2 \\
&= 1 - F_{2m+1} \left\{ \operatorname{Tr}_{Q(\sqrt{5})} \left( \frac{5 + \sqrt{5}}{2}\epsilon^{2m} \right) - 5F_{2m+1} \right\},
\end{align*}
$$

by using

$$
(\zeta - \zeta^{-1})^2 = \left(2i \sin \frac{2\pi}{5}\right)^2 = -\frac{5 + \sqrt{5}}{2}
$$

and

$$
(\zeta^2 - \zeta^{-2})^2 = \left(2i \sin \frac{4\pi}{5}\right)^2 = -\frac{5 - \sqrt{5}}{2}.
$$

Here, by (4.1) we have

$$
\begin{align*}
\operatorname{Tr}_{\mathbf{Q}(\sqrt{5})} \left( \frac{5 + \sqrt{5}}{2} \varepsilon^{2m} \right) &= \operatorname{Tr}_{\mathbf{Q}(\sqrt{5})} \left( \frac{5 + \sqrt{5}}{2} \cdot \frac{L_{2m} + F_{2m}\sqrt{5}}{2} \right) \\
&= \frac{5L_{2m} + 5F_{2m}}{2} \\
&= 5F_{2m+1}.
\end{align*}
$$

Therefore we get $N_M(\delta_m) = 1$. By similar calculations, we have

$$
\begin{align*}
NT(\delta_m) &= 4, \\
NT(\delta_m^{1+\tau}) &= 4 - 20F_{2m+1}^2, \\
NT(\delta_m^{2+\tau}) &= 4 - 40F_{2m+1}^2(1 + (-1)^n L_{2m+1}).
\end{align*}
$$
Substituting them into $f_{\delta_m}(X)$, we have

$$ 
\begin{aligned}
f_{\delta_m}(X) &= X^5 - 10X^3 - 20X^2 + 5(20F_{2m+1}^2 - 3)X \\
&\quad + 40F_{2m+1}^2((-1)^n L_{2m+1} + 1) - 4 \\
&= g_m(X).
\end{aligned}
$$

Assume that $\delta_m^{2+3\tau+\tau^2}$ is not a fifth power in $M_m$. Then we have

$$ \delta_m^{3+4\tau+2\tau^2+\tau^3} = N_M(\delta_m)\delta_m^{2+3\tau+\tau^2} = \delta_m^{2+3\tau+\tau^2} \notin k_m(\zeta)^5, $$

and hence $\delta_m^{3+4\tau+2\tau^2+\tau^3} \in \mathcal{M}(k_m)$. Then by Proposition 3.1, $\text{Spl}_Q(g_m) (= \text{Spl}_Q(f_{\delta_m}))$ is a $D_5$-extension of $Q$ containing $k_m$.

Assume in addition that $m \equiv 12 \pmod{5^2}$. We will show that the cyclic quintic extension $\text{Spl}_Q(g_m)/k_m$ is unramified. Let $\theta$ be a root of $g_m(X)$. Let $q$ be a prime number in general. A prime divisor of $q$ in $k_m$ is ramified in $\text{Spl}_Q(g_m)$ if and only if $q$ is totally ramified in $Q(\theta)$ because $[\text{Spl}_Q(g_m) : k_m]$ and $[k_m : Q]$ are relatively prime. Hence we have only to verify that no primes are totally ramified in $Q(\theta)$. This can be proved by using the following Sase's result. For a prime number $p$ and for an integer $m$, we denote the greatest exponent $\mu$ of $p$ such that $p^\mu \mid m$ by $v_p(m)$.

**Proposition 5.1.** [10, Proposition 2] Let $p \neq 2$ and $q$ be prime numbers. Suppose that the polynomial

$$ \varphi(X) = X^p + \sum_{j=0}^{p-2} a_j X^j, \quad a_j \in \mathbb{Z} $$

is irreducible over $Q$ and satisfies the condition

$$ v_q(a_j) < p-j \quad \text{for some } j, 0 \le j \le p-2. \qquad (5.1) $$

Let $\theta$ be a root of $\varphi(X)$.

(1) If $q$ is different from $p$, then $q$ is totally ramified in $Q(\theta)/Q$ if and only if

$$ 0 < \frac{v_q(a_0)}{p} \le \frac{v_q(a_j)}{p-j} \quad \text{for every } j, 1 \le j \le p-2. $$
ON $D_5$-POLYNOMIALS

(2) The prime $p$ is totally ramified in $\mathbb{Q}(\theta)/\mathbb{Q}$ if and only if one of the following conditions (S-i), (S-ii) holds:

(S-i) $0 < \frac{v_p(a_0)}{p} \le \frac{v_p(a_j)}{p-j}$ for every $j, 1 \le j \le p-2$;

(S-ii) (S-ii-1) $v_p(a_0) = 0$,
(S-ii-2) $v_p(a_j) > 0$ for every $j, 1 \le j \le p-2$,

(S-ii-3) $\frac{v_p(\varphi(-a_0))}{p} \le \frac{v_p(\varphi^{(j)}(-a_0))}{p-j}$ for every $j, 1 \le j \le p-2$,
and

(S-ii-4) $v_p(\varphi^{(j)}(-a_0)) < p-j$ for some $j, 0 \le j \le p-1$,

where $\varphi^{(j)}(X)$ is the $j$-th differential of $\varphi(X)$.

Now let us apply Proposition 4.1 to our polynomial $g_m(X)$. First, we easily verify that $g_m(X)$ satisfies (4.1) for each prime. Next, we see from Proposition 4.1 (1) that no primes except for 5 are totally ramified in $\mathbb{Q}(\theta)/\mathbb{Q}$ because the greatest common divisor of the coefficient of $X^3$ and that of $X$ is equal to 5. We will show, therefore, that 5 is not totally ramified. Denote the constant term of $g_m(X)$ by $c_0$;

$$c_0 := 40F_{2m+1}^2((-1)^n L_{2m+1} + 1) - 4.$$

Since $c_0$ is not divisible by 5, the condition (S-i) does not hold. By the assumption $m \equiv 12 \pmod{5^2}$, we have $2m + 1 \equiv 0 \pmod{5^2}$. Then by the property (C), in Section 3, we have $F_{2m+1} \equiv 0 \pmod{5^2}$, and hence $-c_0 \equiv 4 \pmod{5^5}$. Therefore we have

$$g_m(-c_0) \equiv 4^5 - 10 \cdot 4^3 - 20 \cdot 4^2 - 5(20d^2 - 3) \cdot 4 - 4 \equiv 0 \pmod{5^5},$$

$$g_m^{(1)}(-c_0) \equiv 5 \cdot 4^4 - 30 \cdot 4^2 - 40 \cdot 4 - 15 \equiv 0 \pmod{5^4},$$

$$g_m^{(2)}(-c_0) \equiv 20 \cdot 4^3 - 60 \cdot 4^2 - 40 \equiv 0 \pmod{5^3},$$

$$g_m^{(3)}(-c_0) \equiv 60 \cdot 4^2 - 60 \equiv 0 \pmod{5^2}.$$

Then the condition (S-ii-4) does not hold. (We can easily check that (S-ii-1), (S-ii-2) and (S-ii-3) hold.) Hence 5 is not totally ramified in $\mathbb{Q}(\theta)$. This completes the proof of the main theorem.

Next, we consider when $\delta_m^{2+3\tau+\tau^2}$ is not a fifth power in $M_m$.
Suppose that $\delta_m^{2+3\tau+\tau^2}$ is a fifth power in $M_m$; $\delta_m^{2+3\tau+\tau^2} = \alpha^5$, $\alpha \in M_m$. Since

$$ \delta_m^{3+4\tau+2\tau^2+\tau^3} = N_M(\delta_m)\delta_m^{2+3\tau+\tau^2} = \delta_m^{2+3\tau+\tau^2} = \alpha^5 $$

and $M_m$ is normal over $\mathbb{Q}$, we have

$$ \mathrm{Tr}_{E(\zeta)/E} (\sqrt[5]{\delta_m^{3+4\tau+2\tau^2+\tau^3}}) = \mathrm{Tr}_{E(\zeta)/E}(\alpha) \in M_m, $$

where $E = \mathrm{Spl}_{\mathbb{Q}}(g_m)$. By the last half of Proposition 3.1, therefore, we have

$$ \mathrm{Spl}_{\mathbb{Q}}(g_m) = k_m(\mathrm{Tr}_{E(\zeta)/E}(\alpha)) \subset M_m. $$

Hence the degree $[\mathrm{Spl}_{\mathbb{Q}}(g_m) : k_m]$ is less than 5. Then $g_m(X)$ must be reducible over $\mathbb{Q}$. Therefore, we have

**Proposition 5.2.** The element $\delta_m^{2+3\tau+\tau^2}$ is not a fifth power in $M_m$ if and only if $g_m(X)$ is irreducible over $\mathbb{Q}$.

Finally in this section, we prove the following.

**Proposition 5.3.** The set

$$ \{ \mathbb{Q}(\sqrt{-F_{50s+25}}) \mid s \ge 0 \} $$

is infinite.

*Proof.* On the contrary, we assume $\#\{\mathbb{Q}(\sqrt{-F_{50s+25}}) \mid s \ge 0\} < \infty$. For an integer $m$, we denote the square free part of $m$ by $\mathrm{sf}(m)$. By the assumption, the set

$$ \mathcal{P} := \bigcup_{s \ge 0} \{\text{prime factors of } \mathrm{sf}(F_{50s+25})\} $$

is finite. Then there exists a positive integer $t$ so that we have

$$ \mathcal{P} = \bigcup_{0 \le s \le t} \{\text{prime factors of } \mathrm{sf}(F_{25(2s+1)})\}. $$

Take a prime $q$ with $q > 2t + 1$. Then for each prime factor $r$ of $\mathrm{sf}(F_{25q})$, we have

$$ \mathrm{sf}(F_{25(2s+1)}) \equiv 0 \pmod{r} \quad \text{for some } s, 0 \le s \le t. \qquad (5.2) $$

On the other hand, because $q$ is prime, for each $s$, $0 \le s \le t$, we have

$$ \mathrm{gcd}(25(2s+1), 25q) = 25, $$
ON $D_5$-POLYNOMIALS

and hence by the property (D),

$$ \gcd(F_{25(2s+1)}, F_{25q}) = F_{25} = 3001 \cdot 5^2. $$

From this together with (5.2), we can express

$$ F_{25q} = 3001A_q^2 $$

for some $A_q \in \mathbb{Z}$. Then we have

$$ -1 = N_{\mathbb{Q}(\sqrt{5})}(\varepsilon^{25q}) = \frac{L_{25q}^2 - 5F_{25q}^2}{4} = \frac{L_{25q}^2 - 5 \cdot 3001^2 A_q^4}{4} $$

This implies that $(A_q, L_{25q})$ is an integer solution of the equation

$$ Y^2 = 5 \cdot 3001^2 X^4 - 4. \qquad (5.3) $$

The values of $L_{25q}$ ($q$ is prime), of course, are different from each other. However by Siegel's theorem, there are only finitely many integer solutions $(X, Y)$ of Eq. (5.3). This is a contradiction. $\square$

## 6. Numerical examples

*Example 6.1.* Let $m = 12$. By $F_{12} = 144$, $L_{12} = 322$ and $F_{25} = 3001 \cdot 5^2$, we have

$$ \delta_{12} = 1 + \frac{322 + 144\sqrt{5}}{2} \sqrt{-3001 \cdot 5^2 (\zeta - \zeta^{-1})} $$

and $g_{12}(X) = f_{\delta_{12}}(X)$ is given by

$$ X^5 - 10X^3 - 20X^2 + 562875062485X + 37771618494049996. $$

Since $g_{12}(X)$ is irreducible over $\mathbb{Q}$, it follows from Proposition 5.2 that $\delta_{12}$ is not a fifth power in $M_m$. Then by the main theorem, the splitting field of $g_{12}(X)$ is an unramified cyclic quintic extension of $\mathbb{Q}(\sqrt{-F_{25}})$.

In the following table, we list the prime decompositions of $-F_{2m+1}$ and the structure of the ideal class groups of $k_m = \mathbb{Q}(\sqrt{-F_{2m+1}})$ for $m \le 87$ with $m \equiv 12 \pmod{5^2}$.

*Remarks 6.2.* For this table we use GP/PARI (Version 2.1.5). By using the same calculator, the author verified that $g_m(X)$ is irreducible over $\mathbb{Q}$ for every $m$, $0 \le m \le 2000$. (The disit of the constant term of $g_{2000}(X)$ is 2510.)
Y. KISHI

<table><thead><tr><th>m</th><th>-F<sub>2m+1</sub></th><th>Structure of the ideal class group of k<sub>m</sub> = Q(√-F<sub>2m+1</sub>)</th></tr></thead><tbody><tr><td>12</td><td>-3001 · 5<sup>2</sup></td><td>C<sub>40</sub></td></tr><tr><td>37</td><td>-2 · 61 · 3001 · 230686501 · 5<sup>2</sup></td><td>C<sub>2461460</sub> × C<sub>2</sub> × C<sub>2</sub></td></tr><tr><td>62</td><td>-5 · 3001 · 158414167964045700001 · 5<sup>2</sup></td><td>C<sub>79285156360</sub> × C<sub>8</sub> × C<sub>2</sub></td></tr><tr><td>87</td><td>-13 · 701 · 3001 · 141961 × 17231203730201189308301 · 5<sup>2</sup></td><td>C<sub>1737032019043290</sub> × C<sub>6</sub> × C<sub>2</sub> × C<sub>2</sub> × C<sub>2</sub></td></tr></tbody></table>

## Acknowledgments

The author is grateful to Abdelmalek Azizi, Mohammed Ayadi, Moulay C. Ismaili and their colleagues for their kind hospitality during the conference *Congrès International Algèbre, Théorie des Nombres et leurs Applications*. Together, they made it possible to spend a wonderful week in Morocco.

## Added in proof

After this paper had been written, the author proved our conjecture (Conjecture 1.1) himself. For the details, see in “A new family of imaginary quadratic fields whose class number is divisible by five”, J. Number Theory **128** (2008), 2450–2458.

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YASUHIRO KISHI

Department of Mathematics

Fukuoka University of Education

1-1 Bunkyoumachi Akama, Munakata-shi

Fukuoka, 811-4192

Japan

ykishi@fukuoka-edu.ac.jp