On the difference of two fourth powers

Abstract

We investigate the equation $D=x^4-y^4$ in field extensions. As an application, for a prime number p, we find solutions to $p=x^4-y^4$ if $p\equiv 11$ (mod 16) and $p^3=x^4-y^4$ if $p\equiv 3$ (mod 16) in all cubic extensions of $\mathbb{Q}(i)$.

1. Introduction

We are interested in the following problem.

Problem 1.

Let k be a perfect field of characteristic not equal to 2. Let K be a finite extension of k. Let $D\in k^*$. Find all solutions to the equation

(1)\begin{equation}D=x^4-y^4.\\\end{equation}

By a solution (x, y) to Equation (1), we always mean $(x,y)\in \mathbb{A}^2(K)$ satisfying Equation (1) and xy ≠ 0.

When $k=K=\mathbb{Q}$, using a variety of methods, many authors have shown that Equation (1) has no solutions if $D=nz^p$ for integers n and prime numbers p; see [1, 2, 4, 6, 7, 11].

It is natural to ask for solutions of Equation (1) when k and K are not the rational field. When k and K are number fields, since Equation (1) defines a curve of genus 3, by Faltings’ theorem [8], Equation (1) only has a finite number of solutions, but to find all solutions to Equation (1) is in general a difficult task. We adopt here the method of Cassels’ [5] and Bremner’s [3], which is effective in finding solutions to Equation (1) in all cubic extensions of the base field in many situations. It is also worth mentioning that the work of Silverman [13] on the equation $x^4+y^4=D$ (and $x^6+y^6=D$) over number fields. But Silverman’s method does not apply when finding solutions in cubic extensions of the base field. The main result of this paper is as follows:

Theorem 1. Let k be a perfect field of characteristic not equal to 2. Let $D\in k$ such that $D\not\in \pm k^2$. Assume that

1. (i) every solution $(X,y,z)\in \mathbb{A}^3(k)$ to $X^2-y^4=Dz^4$ satisfies z = 0,

2. (ii) every solution $(x,Y,z)\in \mathbb{A}^3(k)$ to $x^4-Y^2=Dz^4$ satisfies z = 0.

If (x, y) is a solution to $D=x^4-y^4$ in a cubic extension K of k, then

1. (1) (if $-1\not \in k^2$)

   \begin{equation*}K=k(\theta),\quad x=\pm \left(\dfrac{D}{s\theta}-\dfrac{s}{4}\right),\quad y= \pm \left(\dfrac{D}{s\theta}+\dfrac{s}{4}\right), \end{equation*}
   where $\theta^3-s^2\theta^2/8-2D^2/s^2=0$ and $s\in k^{*}$, and

2. (2) (if $-1\in k^2$)

   \begin{equation*}K=k(\theta),\quad x=\pm \left(\dfrac{\theta}{s}-\dfrac{s}{4}\right),\quad y=\pm i\left(\dfrac{\theta}{s}+\dfrac{s}{4}\right), \end{equation*}
   where $i=\sqrt{-1}$, $\theta^3+s^4\theta/16+Ds^2/2=0$, and $s\in k^{*}$.

A nice corollary of Theorem 1 is

Corollary 1. Let p be a prime number. Let D = p if $p\equiv 11$ (mod 16), and let $D=p^3$ if $p\equiv 3$ (mod 16). Then solutions to $D=x^4-y^4$ in all cubic extensions of $\mathbb{Q}(i)$ are

1. (1)

   \begin{equation*}x=\pm \left(\dfrac{D}{s\theta}-\dfrac{s}{4}\right),\qquad y= \pm \left(\dfrac{D}{s\theta}+\dfrac{s}{4}\right),\end{equation*}

   where $\theta^3-s^2\theta^2/8-2D^2/s^2=0$ for some $s\in \mathbb{Q}(i)^{*}$; and

2. (2)

   \begin{equation*} x=\pm \left(\dfrac{\theta}{s}-\dfrac{s}{4}\right),\qquad y=\pm i\left(\dfrac{\theta}{s}+\dfrac{s}{4}\right), \end{equation*}

   where $\theta^3+s^4\theta/16+Ds^2/2=0$ for some $s\in \mathbb{Q}(i)^{*}$.

Remark 1. Theorem 1 finds all possible cubic extensions K of k and solutions to $D=x^4-y^4$ in K. The defining polynomial of K, $F_s(x)=x^3-s^2x^2/8-2D^2/s^2$ or $F_s(x)=x^3+s^4x/16+Ds^2/2$, must be irreducible in $k[x]$, which in general is difficult to check since the irreducibility of $F_s(x)$ depends on s. However, if k is a number field, by Hilbert’s irreducibility theorem [12, Theorem 3.4.1], there exist infinitely many $s\in k$ such that $F_s(x)$ is irreducible in $k[x]$ and Theorem 1 finds all solutions to $D=x^4-y^4$ in these cases.

2. Proof of Theorem 1

We follow Cassels [5]. Equation (1) can be written in the homogeneous form

(2)\begin{equation} x^4-y^4=Dz^4. \end{equation}

Let $\mathcal{C}$ be the projective curve over k defined by Equation (2). Suppose that $P=[x_1:y_1:z_1]$is a point on (2) whose coordinates generate a cubic extension K of k. If $z_1=0$, then $[x_1:y_1:z_1]=[\pm 1:1:0]$; therefore K = k, which is impossible. Therefore, $z_1\neq 0$. Since $(x_1/z_1)^4-(y_1/z_1)^4=D$ and $|K:k|=3$, we have $x_1/z_1,y_1/z_1\not\in k$. Thus,

(3)\begin{equation} k\left(\dfrac{x_1}{z_1}\right)=k\left(\dfrac{y_1}{z_1}\right)= k\left(\dfrac{x_1^2}{z_1^2}\right)=k\left(\dfrac{y_1^2}{z_1^2}\right)=K. \end{equation}

Fix an algebraic closure $\overline{k}$ of k. Let $P_i=[x_i:y_i:z_i] \in \mathbb{P}^{2}(\overline{k})$, $i=1,2,3$, be the Galois conjugates of P. The equation

(4)\begin{equation} X^2-Y^2=DZ^2 \end{equation}

has a parametrization

\begin{equation*}[X:Y:Z]=[l^2+Dm^2:l^2-Dm^2:2lm].\end{equation*}

Since $[x_1^2:y_1^2:z_1^2]$ satisfies Equation (4), there exist $\lambda,\mu\in k$ such that

(5)\begin{equation}{ [x_1^2:y_1^2:z_1^2]=[\lambda^2+D\mu^2:\lambda^2-D\mu^2:2\lambda\mu]}.\end{equation}

Since $z_1\neq 0$, it follows from Equation (5) that µ ≠ 0, λ ≠ 0, and

(6)\begin{equation} {[\lambda:\mu]=[x_1^2+y_1^2:z_1^2]=[Dz_1^2:x_1^2-y_1^2]}. \end{equation}

Let $\theta=\lambda/\mu$. Then Equations (3) and (6) show that $\theta \not \in k$. Hence, $k(\theta)=K$. Therefore, there exists an irreducible cubic polynomial $P(x)=ax^3+bx^2+cx+d \in k[x]$ such that $P(\theta)=0$. In particular, ad ≠ 0. From Equation (5), we have

(7)\begin{equation} \left(\dfrac{x_1}{z_1}\right)^2:\left(\dfrac{y_1}{z_1}\right)^2=\dfrac{\theta^2+D}{2\theta}:\dfrac{\theta^2-D}{2D}. \end{equation}

Step 1: Consider the weighted projective curve

(8)\begin{equation} \mathcal{C}_1\colon X^2-y^4=Dz^4. \end{equation}

By points on $\mathcal{C}_1$, we mean the equivalence classes of points $[X:y:z]$ in $\mathbb{P}^2_{2,1,1}(\overline{k})$ satisfying Equation (8). Since $x_i^2,y_i^2,y_iz_i,z_i^2$ are linearly dependent over k and $x_i\neq 0$, there exist $r,s,t\in \mathbb{Q}$ such that

\begin{equation*}x_i^2=ry_i^2+sy_iz_i+tz_i^2,\end{equation*}

for $i=1,2,3$. Consider the weighted projective curve

(9)\begin{equation} \mathcal{D}_1\colon X=ry^2+syz+tz^2. \end{equation}

By the weighted Bézout theorem [Theorem VIII.2][10], the two curves $\mathcal{C}_1$ and $\mathcal{D}_1$ intersect at 4 points in $\mathbb{P}_{2,1,1}^2(\overline{k})$. We know that three of these four points are $[x_i^2:y_i:z_i]$ for $i=1,2,3$. Let $v_1(T)$ be the fourth point of intersection. Since the set $\{[x_i^2:y_i:z_i]:i=1,2,3\}$ is stable under the action of $\operatorname{Gal}(\overline{k}/k)$, $v_1(T)$ is fixed by $\operatorname{Gal}(\overline{k}/k)$. Therefore, $v_1(T)$ is a k-rational point. By the assumption in Theorem 1, we have $v_1(T)=[\pm 1:1:0]$.

$\bullet$ $v_1(T)=[1:1:0]$. Then Equation (9) gives r = 1. Since

\begin{equation*}(X-y^2-tz^2)^2-s^2y^2z^2=0,\end{equation*}

the homogeneous quartic in $l,m$,

\begin{equation*}(l^2+Dm^2-(l^2-Dm^2)-2tlm)^2-s^2(l^2-Dm^2)(2lm),\end{equation*}

has factors m and $P(l,m)$. Therefore, there exists $q\in k$ such that

(10)\begin{align} (l^2+Dm^2-(l^2-Dm^2)-2tlm)^2-2lms^2(l^2-Dm^2)=& 2qm(al^3+bl^2m\\& +clm^2+dm^3). \end{align}

Thus,

\begin{equation*}2m(Dm-tl)^2-ls^2(l^2-Dm^2)=q(al^3+bl^2m+clm^2+dm^3).\end{equation*}

Therefore,

(11)\begin{equation}\begin{cases}qa&=-s^2,\\ qb&=2t^2,\\ qc&=Ds^2-4Dt,\\ qd&=2D^2. \end{cases}\end{equation}

Hence,

\begin{equation*} q(c+Da)=-4Dt,\quad q^2bd=4D^2t^2,\ q\neq 0.\end{equation*}

Therefore,

(12)\begin{equation} (c+Da)^2=4bd. \end{equation}

Since $a,d\neq 0$, system (11) gives

(13)\begin{equation} \dfrac{a}{d}\equiv -2\pmod{k^2}. \end{equation}

$\bullet$ $v_1(T)=[-1:1:0]$. Then Equation (9) gives $r=-1$. Since

\begin{equation*}(X+y^2-tz^2)^2-s^2y^2z^2=0,\end{equation*}

the homogeneous quartic in $l,m,$

\begin{equation*}(l^2+Dm^2+l^2-Dm^2-2tlm)^2-s^2(l^2-Dm^2)(2lm),\end{equation*}

has factors l and $P(l,m)$. Therefore, there exists $q\in k$ such that

(14)\begin{equation} (2l^2-2tlm)^2-s^2(l^2-Dm^2)(2lm)=2ql(al^3+bl^2m+clm^2+dm^3).\end{equation}

Hence,

\begin{equation*}2l(l-tm)^2-ms^2(l^2-Dm^2)=q(al^3+bl^2m+clm^2+dm^3).\end{equation*}

Therefore,

(15)\begin{equation} \begin{cases} qa&=2,\\ qb&=-4t-s^2,\\ qc&=2t^2,\\ qd&=Ds^2. \end{cases}\end{equation}

Hence,

\begin{equation*} q^2ac=4t^2,\quad q\left(b+\dfrac{d}{D}\right)=-4t,\ q\neq 0.\end{equation*}

Therefore,

(16)\begin{equation} \left(b+\dfrac{d}{D}\right)^2=4ac.\end{equation}

Since $a,d\neq 0$, system (15) also gives

(17)\begin{equation} \dfrac{a}{d}\equiv 2D\pmod{k^2}. \end{equation}

Step 2: Consider the weighted projective curve

(18)\begin{equation} \mathcal{C}_2\colon x^4-Y^2={D}z^4.\end{equation}

By points on $\mathcal{C}_2$, we mean the equivalence classes of points $[x:Y:z]$ in $\mathbb{P}^2_{1,2,1}(\overline{k})$ satisfying Equation (18). Since $y_i^2,x_i^2,x_iz_i,z_i^2$ are linearly dependent over k and $y_i\neq 0$, there exist $r,s,t\in \mathbb{Q}$ such that

\begin{equation*}y_i^2=rx_i^2+sx_iz_i+tz_i^2,\end{equation*}

for $i=1,2,3$. Consider the weighted projective curve

(19)\begin{equation} \mathcal{D}_2\colon Y=rx^2+sxz+tz^2. \end{equation}

By the weighted Bézout theorem [Theorem VIII.2][10], the two curves $\mathcal{C}_2$ and $\mathcal{D}_2$ intersect at 4 points in $\mathbb{P}_{1,2,1}^2(\overline{k})$. We know that three of these four points are $[x_i:y_i^2:z_i]$ for $i=1,2,3$. Let $v_2(T)$ be the fourth point of intersection. Since the set $\{[x_i:y_i^2:z_i]:i=1,2,3\}$ is stable under the action of $\operatorname{Gal}(\overline{k}/k)$, $v_2(T)$ is fixed by $\operatorname{Gal}(\overline{k}/k)$. Therefore, $v_2(T)$ is a k-rational point. By the assumption in Theorem 1, we have $v_2(T)=[1:\pm 1:0]$.

$\bullet$ $v_2(T)=[1:1:0]$. Then Equation (19) gives r = 1. We have

\begin{equation*}(Y-x^2-tz^2)^2-s^2{x^2z^2}=0,\end{equation*}

so that the homogeneous quartic in $l,m,$

\begin{equation*} (l^2-Dm^2-(l^2+Dm^2)-2tlm)^2-s^2(l^2+Dm^2)(2lm), \end{equation*}

has factors m and $P(l,m)$. Therefore, there exists $q\in k$ such that

(20)\begin{equation} (l^2-Dm^2-(l^2+Dm^2))^2-2tlm)^2-s^2(l^2+Dm^2)(2lm)=2qmP(l,m). \end{equation}

Thus,

\begin{equation*}2m(Dm+rl)^2-ls^2(l^2+Dm^2)=q(al^3+bl^2m+clm^2+dm^3).\end{equation*}

Hence,

(21)\begin{equation}\begin{cases} qa&=-s^2,\\ qb&=2r^2,\\ qc&=4Dr-Ds^2,\\ qd&=2D^2. \end{cases} \end{equation}

Therefore,

\begin{equation*} q(c-Da)=4Dr,\quad q^2bd=4{D^2}r^2,\ q\neq 0.\end{equation*}

Hence,

(22)\begin{equation} (c-Da)^2=4bd. \end{equation}

Since $a,d\neq 0$, system (21) gives

(23)\begin{equation} \dfrac{a}{d}\equiv -2\pmod{k^2}. \end{equation}

$\bullet$ $v_2(T)=[1:-1:0]$. Then Equation (19) gives $r=-1.$ We have

\begin{equation*}(Y+x^2-tz^2)^2-s^2{x^2z^2}=0,\end{equation*}

so that the homogeneous quartic in $l,m,$

\begin{equation*}(l^2-Dm^2+(l^2+Dm^2)-2rlm)^2-s^2(l^2+Dm^2)(2lm),\end{equation*}

has factors l and $P(l,m)$. Thus, there exists $q\in k$ such that

(24)\begin{align} (l^2-Dm^2+(l^2+Dm^2)-2tlm)^2-s^2(l^2+Dm^2)(2lm)=&2lq(al^3+bl^2m\\&+clm^2+dm^3). \end{align}

Hence,

\begin{equation*}2l(l-tm)^2-ms^2(l^2+Dm^2)=q(al^3+bl^2m+clm^2+dm^3).\end{equation*}

Therefore,

(25)\begin{equation}\begin{cases} qa&=2,\\ qb&=-4t-s^2,\\ qc&=2t^2,\\ qd&=-Ds^2. \end{cases} \end{equation}

Hence,

\begin{equation*} q^2ac=4t^2,\quad q\left(b-\dfrac{d}{D}\right)=-4r,\ q\neq 0.\end{equation*}

Thus,

(26)\begin{equation} \left(b-\dfrac{d}{D}\right)^2=4ac. \end{equation}

System (25) also gives

(27)\begin{equation} \dfrac{a}{d}\equiv -2D\pmod{k^2}. \end{equation}

It follows from (23), (27), (13) and (17) and the assumption that $D\not\in \pm k^2$ that there are only two compatible cases for $v_1(T)$ and $v_2(T)$.

Case 1: $v_1(T)=[1:1:0]$ and $v_2(T)=[1:1:0]$. From (12) and (22), we have

\begin{equation*}(c-Da)^2=(c+Da)^2.\end{equation*}

Since aD ≠ 0, we have c = 0. From (21), we have $r=s^2/4$. Thus,

\begin{equation*}{qP(x)=-s^2x^3+\dfrac{s^4}{8}x^2+2D^2.}\end{equation*}

Therefore θ satisfies

(28)\begin{equation} \theta^3-\dfrac{s^2}{8}\theta^2-2\dfrac{D^2}{s^2}=0. \end{equation}

Then (10) and (20) give

(29)\begin{equation} \dfrac{\theta^2+D}{2\theta}=\left(\dfrac{s}{4}+\dfrac{D}{2\theta}\right)^2,\qquad \dfrac{\theta^2-D}{2\theta}=\left(\dfrac{s}{4}-\dfrac{D}{2\theta}\right)^2. \end{equation}

From (7) and (29), we have

(30)\begin{equation} \dfrac{x_1}{z_1}=\pm \left(\dfrac{D}{s\theta}-\dfrac{s}{4}\right),\qquad\dfrac{y_1}{z_1}= \pm \left(\dfrac{D}{s\theta}+\dfrac{s}{4}\right). \end{equation}

Case 2: $v_1(T)=[-1:1:0]$ and $v_2(T)=[1:-1:0]$. This case also implies that $-1\in k^2$. From (17) and (25), we have

\begin{equation*}\left(b+\dfrac{d}{D}\right)^2=\left(b-\dfrac{d}{D}\right)^2.\end{equation*}

Since d ≠ 0, we have b = 0. Hence, from (15), we have $t=-s/4$. Therefore,

\begin{equation*}qP(x)=2x^3+\dfrac{s^4}{8}x+Ds^2.\end{equation*}

Therefore, θ satisfies

(31)\begin{equation} \theta^3+\dfrac{s^4}{16}\theta+\dfrac{Ds^2}{2}=0. \end{equation}

It follows from (14) and (24) that

(32)\begin{equation} \dfrac{\theta^2+D}{2\theta}=\left(\dfrac{\theta}{s}-\dfrac{s}{4}\right)^2,\qquad \dfrac{\theta^2-D}{2\theta }=\left(i\left(\dfrac{\theta}{s}+\dfrac{s}{4}\right)\right)^2,\end{equation}

where $i\in k$ such that $i^2=-1\in k$. From (7) and (32), we have

(33)\begin{equation} \dfrac{x_1}{z_1}=\pm \left(\dfrac{\theta}{s}-\dfrac{s}{4}\right),\qquad \dfrac{y_1}{z_1}=\pm i\left(\dfrac{\theta}{s}+\dfrac{s}{4}\right). \end{equation}

3. Proof of Corollary 1

Corollary 1 is a consequence of Theorem 1 and the following lemma due to Izadi et al. [9].

Lemma 1.

• (1) For prime numbers, $p\equiv 3$ (mod 16), then the equations $x^2{-}y^4={\pm} p^3z^4$ only have solutions $X=\pm y^2$ and z = 0 in $\mathbb{Q}(i)$.

• (2) For prime numbers, $p\equiv 11$ (mod 16), then the equations $X^2{-}y^4={\pm} pz^4$ only have solutions $X=\pm y^2$ and z = 0 in $\mathbb{Q}(i)$.

Proof. See Izadi [9, Theorems 3.2 and 3.4].