Rational Points of Some Elliptic Curves Related to the Tilings of the Equilateral Triangle

Abstract

Let n be a positive and squarefree integer. We show that the equilateral triangle can be dissected into $n\cdot k^2$ congruent triangles for some k if and only if $n\le 3$, or at least one of the curves $C_n :y^2 =x(x-n)(x+3n)$ and $C_{-n} : y^2 =x(x+n)(x-3n)$ has a rational point with $y\ne 0$. We prove that if p is a positive prime such that $p\equiv 7$ (mod 24), then $C_p$ and $C_{-p}$ do not have such points. Consequently, for these primes the equilateral triangle cannot be dissected into $p\cdot k^2$ congruent triangles for any k.

1 Introduction and Main Results

Let $C_n$ denote the elliptic curve $y^2 =x(x-n)(x+3n)$, where n is an integer. The group of rational points of $C_n$ will be denoted by $\Gamma _n$. We say that $(x,y)\in C_n$ is a nontrivial rational point of $C_n$ if x, y are nonzero rational numbers; that is, if the order of (x, y) as an element of the group $\Gamma _n$ is greater than two. Our first result shows that the existence of nontrivial rational points of $C_n$ is closely related to the number of pieces in certain tilings of the equilateral triangle.

Theorem 1.1

For every positive and squarefree integer n the following are equivalent.

• (i) There is a positive integer k such that the equilateral triangle can be dissected into $n\cdot k^2$ congruent triangles.

• (ii) Either $n\le 3$, or at least one of the curves $C_n$ and $C_{-n}$ has a nontrivial rational point.

The proof of Theorem 1.1 is based on the fact that the congruent copies of a triangle with sides a, b, c and corresponding angles $\alpha ,\beta ,\gamma$ tile an equilateral triangle if and only if either $\alpha , \beta , \gamma$ are multiples of $\pi /6$, or $\gamma \in \{ \pi /3 ,2\pi /3\}$ and a, b, c are pairwise commensurable (see [4, Thm. 3.3]). By the law of cosines, we have $\gamma =\pi /3$ or $2\pi /3$ if and only if $c^2 = a^2 +b^2 \pm ab$. Such triples are, e.g., $(a,b,c)=(7,8,13)$ or $(a,b,c)=(3,5,7)$.

Suppose that a, b, c are positive integers with $c^2 = a^2 +b^2 \pm ab$. Then the triangle with sides a, b, c tiles an equilateral triangle T. If the side length of T is m and the tiling has N pieces, then, comparing the areas we get $m^2 =N\cdot ab$, and thus the square free part of N is the same as that of ab. For example, if $(a,b,c)=(7,8,13)$, then the construction described in [3, Thm. 3.1] produces a tiling with $2,469,600=14\cdot 420^2$ pieces. For the triangle with sides 3, 5, 7, a tiling with $10,935 = 15\cdot 27^2$ pieces was found by Michael Beeson (see [2, Fig. 22, p. 28]).

As we shall see, a simple transformation maps these triples into nontrivial rational points of one of the corresponding curves $C_n$ or $C_{-n}$. Thus the triple (7, 8, 13) gives the point $(-6,48)$ of $C_{-14}$, and (3, 5, 7) gives the point $(-5,50)$ of $C_{-15}$.

In the other direction, every nontrivial rational point of $C_n$ or $C_{-n}$ determines a triple (a, b, c) as above. For example, from the point $(-1,8)$ of $C_{-5}$ we obtain the triple (5, 16, 19), and the from the point $(-1,30)$ of $C_{17}$ we get (17, 225, 217). The proof of Theorem 1.1 will be given in the next section.

Remarks 1.2

• 1. Since every triangle $\Delta$ can be dissected into $m^2$ congruent triangles similar to $\Delta$ for every m, it is clear that (i) of Theorem 1.1 is equivalent to the following statement.

  (i$'$) There are infinitely many positive integers k such that the equilateral triangle can be dissected into $n\cdot k^2$congruent triangles.

• 2. We shall prove in Lemma 3.1 that if p is a positive prime, then the only torsion points of $\Gamma _p$ and $\Gamma _{-p}$ are the points having zero y-coordinates. Therefore, if n is a positive prime, then (ii) of Theorem 1.1 is equivalent to the following statement.

  (ii$'$) Either $n\le 3$, or at least one of the groups $\Gamma _n$and $\Gamma _{-n}$has positive rank.

It is easy to see that if n, k are nonzero integers then $C_n$ has a nontrivial rational point if and only if $C_{nk^2}$ has one. Therefore, we have the following corollary of Theorem 1.1.

Corollary 1.3

If the equilateral triangle can be dissected into N congruent triangles, then either $N=k^2$, $N=2k^2$ or $N=3k^2$ for some k, or at least one of the curves $C_N$ and $C_{-N}$ has a nontrivial rational point.

We remark that the converse is not true. For example, $(-1,8)$ is a nontrivial rational point of $C_{-5}$, but the equilateral triangle cannot be dissected into 5 congruent triangles. This follows from a result of Beeson stating that the equilateral triangle cannot be dissected into p congruent triangles for any prime $p>3$ (see [1]). On the other hand, the equilateral triangle can be dissected into $5k^2$ congruent triangles for infinitely many positive integer k by Theorem 1.1.

In Sect. 3 we shall prove that if p is a positive prime and $p\equiv 7$ (mod 24), then the curves $C_p$ and $C_{-p}$ have no nontrivial rational points (see Corollary 3.6). Comparing with Theorem 1.1 we obtain the following.

Corollary 1.4

If p is a positive prime such that $p\equiv 7$ (mod 24), then the equilateral triangle cannot be dissected into $p\cdot k^2$ congruent triangles for any k.

2 Proof of Theorem 1.1

(i) $\Rightarrow$ (ii): Suppose that the equilateral triangle T can be tiled with $n\cdot k^2$ congruent triangles having angles $\alpha ,\beta ,\gamma$ and corresponding sides a, b, c. We may assume that the sides of T equal 1.

By [4, Thm. 3.3], one of the following cases holds: $\alpha =\beta =\pi /6$ and $\gamma =2\pi /3$; $\alpha =\pi /6 ,$$\beta =\pi /2,$$\gamma =\pi /3$; $\gamma \in \{ \pi /3 ,2\pi /3\}$ and a, b, c are pairwise commensurable.

Comparing the areas of T and the tiles we obtain $nk^2 \cdot ab\cdot \frac{\sqrt{3}}{4}=\frac{\sqrt{3}}{4}$; that is,

$$\begin{aligned} nk^2 \cdot ab=1. \end{aligned}$$

(1)

If $\alpha =\beta =\pi /6$, then $a=b$ and thus, by (1), $a=b=1/(k\cdot \sqrt{n} )$. By $c/a=\sqrt{3}$ we have $c=\sqrt{3} /(k\cdot \sqrt{n} )$. Since the side of the equilateral triangle is tiled with segments of length a and c, we obtain $1=ra+sc$ with suitable nonnegative integers r, s. Thus $r+s\sqrt{3} =k\cdot \sqrt{n}$. Since n is squarefree, this implies $n=1$ or $n=3$.

If $\alpha =\pi /6$, $\beta =\pi /2$ and $\gamma =\pi /3$, then $b=2a$ and thus, by (1), $a=1/(k\cdot \sqrt{2n})$. By $c/a=\sqrt{3}$ we have $c=\sqrt{3} /(k\cdot \sqrt{2n} )$. The side of the equilateral triangle is tiled with segments of length a, 2a and c, hence $1=ra+sc$ with suitable nonnegative integers r, s. Thus $r+s\sqrt{3} =k\cdot \sqrt{2n}$. Since n is squarefree, this implies $n=2$ or $n=6$. Now (9, 27) is a point of $C_6 :y^2 =x(x-6)(x+18)$, and thus the statement of (ii) is true in these cases.

In the remaining cases a, b, c are pairwise commensurable, and $\gamma =\pi /3$ or $\gamma =2\pi /3$. Then we have $c^2=a^2 +b^2 \pm ab$ by the law of cosines. Since $qa+rb+sc=1$ with nonnegative integers q, r, s, it follows that a, b, c are rational. Replacing a by $-a$ if necessary, we may assume $c^2=a^2 +b^2 +ab$. Under this change (1) becomes $\pm nk^2 \cdot ab=1$. We put $t=(c-b)/a$; then t is rational, and $b=c-ta$. We have

$$\begin{aligned} c^2= & {} a^2 +b^2 +ab=a^2 +(c-ta)^2 +ac-ta^2 \\= & {} a^2 (t^2 -t+1) -2act + ac+c^2 , \end{aligned}$$

$a^2 (t^2 -t+1) =ac(2t-1)$, and $a/c=(2t-1)/d$, where $d=t^2 -t+1$. Note that $d\ne 0$, as the polynomial $X^2 -X+1$ has no rational roots. Then we have $b/c=1-(ta/c)=(1-t^2)/d$. From (1) we get

$$\begin{aligned} 1=\pm nk^2 ab=\pm n \cdot (2t-1)(1-t^2 )\cdot (ck/d)^2 \end{aligned}$$

and $(2t-1)(t^2 -1)=\mp nv^2$, where $v=d/(nkc)$ is a nonzero rational number.

Putting $x=n(2t-1)$ we get $t=(x+n)/(2n)$, $t-1=(x-n)/(2n)$, $t+1=(x+3n)/(2n)$, and

$$\begin{aligned} x(x-n)(x+3n)= (2t-1)(t^2 -1) \cdot 4n^3 = \mp nv^2 \cdot 4n^3 = \mp y^2, \end{aligned}$$

where $y=2n^2 v$. Therefore, either (x, y) is a point of $C_{n}$ or $(-x,y)$ is a point of $C_{-n}$.

(ii) $\Rightarrow$ (i): It is clear that if $n\le 3$ then the equilateral triangle can be dissected into n congruent triangles.

Suppose that x, y are rational numbers, $y\ne 0$, and (x, y) is a rational point of either $C_n$ or $C_{-n}$. Then one of $t=x/n$ and $t=-x/n$ satisfies $t(t+1)(t-3)=\pm y^2 /n^3$. Fix such a t. Note that $t\ne 0, -1,3$. Putting $a=4t$, $b=t^2 -2t-3$ and $c=t^2 +3$ we have $ab\ne 0$ and $a^2 +b^2 +ab=c^2$. Then |a|, |b|, c are the sides of a rational triangle $\Delta$ such that $a^2 +b^2 \pm |a|\cdot |b|=c^2$, and thus, by the law of cosines, the angle between the sides of length |a| and |b| equals $\pi /3$ or $2\pi /3$. By [3, Thm. 3.1], there is an equilateral triangle T that can be dissected into triangles congruent to $\Delta$. Let m be the length of the side of T, and let N be the number of pieces of the decomposition. Then $N|ab|=m^2$, hence

$$\begin{aligned} m^2 /N= |ab|=4|t(t^2 -2t-3)|=4|t(t+1)(t-3)|=4y^2 /n^3 \end{aligned}$$

and $N=n^3 m^2 /(4y^2 ) =nk^2$, where $k=nm/(2y)$. Now k is rational and n is squarefree by assumption, so $N=nk^2$ implies that k must be an integer. We have found a dissection of T into $n\cdot k^2$ congruent triangles, proving (i). $\square$

3 Rational Points of $C_{\pm p}$

In this section we show that if p is a positive prime and $p\equiv 7$ (mod 24), then $C_p$ and $C_{-p}$ have no nontrivial rational points (see Corollary 3.6). Recall that the group of rational points of $C_n$ is denoted by $\Gamma _n$.

Lemma 3.1

Let p be a positive prime. Then the torsion points of the group $\Gamma _p$ are the points (0, 0), (p, 0), $(-3p ,0)$ and ${{\mathcal {O}}}$ (the point at infinity). The torsion points of $\Gamma _{-p}$ are the points (0, 0), $(-p,0)$, (3p, 0) and ${{\mathcal {O}}}$.

Proof

The points listed above, being of order two and one, are torsion points. Suppose there exists another torsion point (x, y). Since the discriminant of the curves equals $p^2 \cdot (3p)^2 \cdot (4p)^2 =3^2 \cdot 2^4 \cdot p^6$, it follows from the Nagell–Lutz theorem that $x,y\in {{\mathbb {Z}}}$, $y\ne 0$ and $y \mid 3 \cdot 2^2 \cdot p^3$. We distinguish between two cases.

Case I: $p\mid y$. Then $p\mid x$, $x=pz$, $p^2 \mid y$, $y=p^2 u$, $u\ne 0$, and

$$\begin{aligned} pu^2 =z(z\mp 1)(z\pm 3). \end{aligned}$$

(2)

Clearly, $z\ge -2$. It is easy to check that if $-2 \le z\le 13$ then $z(z\mp 1)(z\pm 3)$ is not of the form $qu^2$, where q is prime and $u\ne 0$, except when $z=4$ and $z(z+ 1)(z- 3)=5\cdot 2^2$. This gives the point $P_1 =(20,50)$ of $\Gamma _{-5}$. One can easily check that the x-coordinate of $2P_1$ is not an integer, hence $P_1$ is not a torsion point. (Thus $\Gamma _{-5}$ has positive rank.) Therefore, we may assume $z\ge 14$.

If $p=2$ or $p=3$ then $y=p^2 u\mid 3 \cdot 2^2 \cdot p^3$ implies that all prime factors of z and $z\pm 1$ are 2 and 3. Thus $z=2^\alpha$, $z\pm 1=3^\beta$ or the other way around. Then $z\le 10$ which is impossible.

Therefore, we may assume $p>3$. Then at most one of the terms z, $z\mp 1$, $z\pm 3$ is divisible by p. Since $u\mid 3\cdot 2^2 \cdot p^3$, it follows from (2) that the product of two of the terms z, $z\mp 1$$z\pm 3$ is a divisor of $3^2 \cdot 2^4 = 144$. By $z\ge 4$ this implies $z(z-3)\le 144$, hence $z\le 13$ which is impossible.

Case II: $p\not \mid y$. Then $y\mid 12$. Replacing x by $-x$ if necessary, we have $x(x+p)(x-3p)=\pm y^2$, and thus

$$\begin{aligned} |x(x+p)(x-3p)|=y^2 \mid 144. \end{aligned}$$

(3)

It is easy to see that if a is a positive integer and x is an integer different from 0 and a, then $|x(a-x)|\ge a-1$. Therefore, $|x(x+p)|\ge p-1$, $|x(x-3p)|\ge 3p-1$, $|(x+p)(x-3p)|\ge 4p -1$,

$$\begin{aligned} (p-1)(3p-1)(4p-1)\le |x(x+p)(x-3p)|^2 \le 144 ^2 , \end{aligned}$$

and thus $p\le 11$.

It follows from (3) that there are (positive or negative) divisors $d_1 ,d_2$ of 144 such that $d_2 -d_1 =4p$, $|x\cdot d_1 \cdot d_2 |$ is a square and is a divisor of 144, where $x=d_2 -p$. Checking the cases $p=2,3,5,7,11$, we find that the only possibility is $p=5$, $(d_1 , d_2 )=(-16,4)$ and $x=-1$. This gives the point $P_2 =(-1,8)$ of $\Gamma _{-5}$. One can easily check that $P_2 =P_1 +P_0$, where $P_0 =(-5,0)$ and $P_1 =(20,50)$. Since $P_0$ is a torsion point of $\Gamma _{-5}$ and $P_1$ is not, it follows that $P_2$ is not a torsion point either.

$\square$

Theorem 3.2

 

• (i) The rank of $\Gamma _p$ is at most two for every positive prime p.

• (ii) If $p\not \equiv 1$ (mod 24), then the rank of $\Gamma _p$ is at most one.

• (iii) If $p=2$, $p=3$ or $p\equiv 5$, 7 or 19 (mod 24), then the rank of $\Gamma _p$ is zero.

In the proof of Theorem 3.2 we apply the method described in [5, §5, Chap. III, pp. 92–94]. Consider the curves

$$\begin{aligned} C_p : y^2 =x^3 +2px^2 -3p^2 x \quad \text {and} \quad \overline{C} _p : y^2 =x^3 -4px^2 +16p^2 x \end{aligned}$$

with groups of rational points $\Gamma _p =C_p ({{\mathbb {Q}}})$ and $\overline{\Gamma }_p =\overline{C} _p ({{\mathbb {Q}}})$. We define $\alpha :\Gamma _p \rightarrow {{\mathbb {Q}}}^* /{{\mathbb {Q}}}^{* 2}$ by $\alpha ({{{\mathcal {O}}}}) =1$, $\alpha (0,0)=-3p^2 \equiv -3$ and, for $x\ne 0$, $\alpha (x,y)=x$ (mod ${{\mathbb {Q}}}^{* 2}$). Then $\alpha$ is a homomorphism from $\Gamma _p$ into ${{\mathbb {Q}}}^* /{{\mathbb {Q}}}^{* 2}$.

We also define $\overline{\alpha }:\overline{\Gamma }_p \rightarrow {{\mathbb {Q}}}^* /{{\mathbb {Q}}}^{* 2}$ by $\overline{\alpha }({{{\mathcal {O}}}}) =1$, $\overline{\alpha }(0,0)=16p^2 \equiv 1$ and, for $x\ne 0$, $\alpha (x,y)=x$ (mod ${{\mathbb {Q}}}^{* 2}$). Then $\overline{\alpha }$ is a homomorphism from $\overline{\Gamma }_p$ into ${{\mathbb {Q}}}^* /{{\mathbb {Q}}}^{* 2}$. The rank r of $\Gamma _p$ satisfies

$$\begin{aligned} 2^r = \frac{\#\, \alpha (\Gamma _p )\cdot \#\, \overline{\alpha }( \overline{\Gamma }_p )}{4} \end{aligned}$$

(4)

(see [5, p. 91]). Here $\alpha (\Gamma _p )$ equals the set of divisors $b_1$ of $b=-3p^2$ (mod ${{\mathbb {Q}}}^{* 2}$) such that the equation

$$\begin{aligned} N^2 =b_1 M^4 +2pM^2 e^2 +(-3p^2 /b_1 )\,e^4 \end{aligned}$$

(5)

is solvable in pairwise coprime integers N, M, e satisfying $M\ne 0$ and $\gcd (e,b_1 )=\gcd (M,-3p^2 /b_1 )=1$ (see [5, pp. 92–93]). Similarly, $\alpha (\overline{\Gamma }_p )$ equals the set of divisors $b_1$ of $\overline{b} =16p^2$ (mod ${{\mathbb {Q}}}^{* 2}$) such that the equation

$$\begin{aligned} N^2 =b_1 M^4 -4 pM^2 e^2 +(16p^2 /b_1 )\,e^4 \end{aligned}$$

(6)

is solvable in pairwise coprime integers N, M, e satisfying $M\ne 0$ and $\gcd (e,b_1 )=\gcd (M, 16p^2 /b_1 )=1$.

The statement of Theorem 3.2 is an immediate consequence of (4) and of the following lemma.

Lemma 3.3

 

• (i) $\#\, \alpha (\Gamma _p ) \le 8$ for every positive prime p.

• (ii) If $p=2$, $p=3$ or $p\equiv 5$, 7, 13 or 19 (mod 24), then $\#\, \alpha (\Gamma _p ) \le 4$.

• (iii) $\#\, \alpha (\overline{\Gamma }_p ) \le 2$ for every positive prime p.

• (iv) If $p\not \equiv 1$ (mod 12), then $\#\, \alpha (\overline{\Gamma }_p ) =1$.

Proof

 

• (i) is obvious from $b_1 \in \{ \pm 1, \pm 3, \pm p, \pm 3p\}$ (mod ${{\mathbb {Q}}}^{* 2}$).

• (ii) If $p=3$ then $b_1 \in \{ \pm 1, \pm 3\}$ (mod ${{\mathbb {Q}}}^{* 2}$), and $\#\, \alpha (\Gamma _p ) \le 4$. Therefore, we may assume $p\ne 3$. We have $(p,0), (-3p,0)\in \Gamma _p$ and $\alpha (0,0)=-3p^2 \equiv -3$, and thus $1,p, -3,-3p\in \alpha (\Gamma _p )$. Since $\alpha (\Gamma _p )$ is a subgroup of ${{\mathbb {Q}}}^* /{{\mathbb {Q}}}^{* 2}$, it follows that $\#\, \alpha (\Gamma _p )$ equals 4 or 8, and it equals 8 if and only if $-1\in \alpha (\Gamma _p )$.

Suppose that $\#\, \alpha (\Gamma _p )=8$. Then $-1\in \alpha (\Gamma _p )$ and thus, by $b_1 \mid 3p^2$, (5) is solvable for at least one of $b_1 =-1$ and $b_1 =-p^2$.

Suppose that $N^2 =-M^4 +2pM^2 e^2 +3p^2 e^4$ is solvable. If $p=2$, then M is odd by $\gcd (M, 3p^2 )=1$, and $N^2 \equiv -M^4$ (mod 4), which is impossible. If $p>3$, then $p\not \mid M$ by $\gcd (M, 3p^2 )=1$, and thus we have $\big (\frac{-1}{p}\big )=1$ and $p\equiv 1$ (mod 4).

We have $N^2 =(3pe^2 -M^2 )(pe^2 +M^2 )=A\cdot B$. Since $p\not \mid M$ and $\gcd (M,e)=1$, it follows that $\gcd (A,B) \mid 4$. If $\gcd (A,B)=1$ or 4, then A and B are squares. Thus $3pe^2 -M^2 =n^2$, hence $-M^2 \equiv n^2$ (mod 3), which is impossible, as $3\not \mid M$.

If $\gcd (A,B)=2$, then A / 2 and B / 2 are squares. Thus $3pe^2 -M^2 =2n^2$, hence $-M^2 \equiv 2n^2$ (mod p). Since $p\not \mid M$ and $p\equiv 1$ (mod 4), we get $\big (\frac{2}{p}\big )=1$ and $p\equiv 1$ (mod 8).

Next suppose that $N^2 =-p^2 M^4 +2pM^2 e^2 +3 e^4$ is solvable. Then we have $\gcd (M,3)=1$. If $p=2$, then e is odd (since otherwise both N and e would be even), and $N^2 \equiv 3e^4$ (mod 4), which is impossible. Suppose $p>3$. Then $p\not \mid e$ (since otherwise both e and N would be divisible by p), and thus $\big (\frac{3}{p}\big )=1$ and $p\equiv \pm 1$ (mod 12).

We have $N^2 =(3e^2 -pM^2 )(e^2 +pM^2 )=C\cdot D$. Since $p\not \mid e$ and $\gcd (M,e)=1$, it follows that $\gcd (C,D) \mid 4$. If $\gcd (C,D)=1$ or 4, then C and D are squares. Thus $3e^2 -pM^2 =n^2$, $-pM^2 \equiv n^2$ (mod 3), $p\equiv -1$ (mod 3) and $p\equiv - 1$ (mod 12).

If $\gcd (C,D)=2$, then C / 2 and D / 2 are squares. Thus $e^2 + pM^2 =2n^2$, hence $e^2 \equiv 2n^2$ (mod p), $\big (\frac{2}{p}\big )=1$, $p\equiv \pm 1$ (mod 8).

We proved that if $\#\, \alpha (\Gamma _p )=8$, then $p>3$ and either $p\equiv 1$ (mod 8), or $p\equiv - 1$ (mod 12). This proves (ii).

(iii) We have to estimate $\#\, \alpha (\overline{\Gamma }_p )$. It is clear that if $b_1 <0$ then (6) has no solutions, and thus, by $b_1 \mid 16p^2$, we have $b_1 \in \{ 2^\alpha p^\beta :0\le \alpha \le 4, 0\le \beta \le 2\}$. If $p=2$, then we obtain $\alpha (\overline{\Gamma }_p )\subset \{ 1,2\}$ (mod ${{\mathbb {Q}}}^{*2}$). Therefore, we may assume $p>2$.

Let $b_1 =2p^\beta$, and suppose that (6) is solvable. Then M is odd by $\gcd (M,16p^2 /b_1 )=1$, and thus the left hand side of (6) is divisible by 4, while the right hand side is not, which is impossible.

Next let $b_1 =8p^\beta$, and suppose that (6) is solvable. Then N is even and, consequently, e is odd. Thus the left hand side of (6) is divisible by 4, while the right hand side is not, which is impossible. We obtain that $b_1 \in \{ 1,p,p^2, 4, 4p, 4p^2 ,16, 16p, 16p^2\}$ and $b_1 \in \{ 1,p\}$ (mod ${{\mathbb {Q}}}^{*2}$). This proves (iii).

(iv) Suppose that $\#\, \alpha (\overline{\Gamma }_p ) =2$. Then $p\in \alpha (\overline{\Gamma }_p )$, and (6) is solvable for at least one of $b_1 =p$, $b_1 =4p$ and $b_1 =16p$.

Let $b_1 =p$, and suppose that $N^2 =p M^4 -4 pM^2 e^2 +16pe^4$ is solvable. Then M is odd by $\gcd (M,16p)=1$, and $N^2 \equiv pM^4$ (mod 4). Hence $p>2$ and $p\equiv 1$ (mod 4). We have $N=pN_1$ and

$$\begin{aligned} pN_1^2 =M^4 -4M^2 e^2 +16e^4 =(M^2 -2e^2 )^2 +12e^4 . \end{aligned}$$

Now $p\not \mid e$ by $\gcd (e,b_1 )=1$, and we get $\big (\frac{-12}{p}\big )=1$. Since $p\equiv 1$ (mod 4), we obtain $\big (\frac{3}{p}\big )=1$, $p\equiv \pm 1$ (mod 12) and $p\equiv 1$ (mod 12).

The case $b_1 =16p$ is similar with the roles of M and e exchanged. Therefore, if (6) is solvable for $b_1 =16p$, then $p\equiv 1$ (mod 12).

Finally, let $b_1 =4p$, and suppose that $N^2 =4p M^4 -4 pM^2 e^2 +4pe^4$ is solvable. Then $2\not \mid M$ by $\gcd (M,4p)=1$, and $2p\mid N$. Let $N=2pN_1$, then $pN_1 ^2 =M^4 -M^2 e^2 +e^4$. Since M is odd, we have $M^4 -M^2 e^2 +e^4 \equiv 1$ (mod 4), and thus $p\equiv 1$ (mod 4). We have

$$\begin{aligned} 4pN_1^2 =4M^4 -4M^2 e^2 +4e^4 =(2M^2 -e^2 )^2 +3e^4 . \end{aligned}$$

Now $p\not \mid e$ by $\gcd (e,b_1 )=1$, and we get $\big (\frac{-3}{p}\big )=1$. Since $p\equiv 1$ (mod 4), we obtain $\big (\frac{3}{p}\big )=1$, $p\equiv \pm 1$ (mod 12) and $p\equiv 1$ (mod 12).

We proved that if $\#\, \alpha (\Gamma _p )=2$, then $p\equiv 1$ (mod 12). This proves (iv). $\square$

Our next aim is to prove

Theorem 3.4

 

• (i) The rank of $\Gamma _{-p}$ is at most two for every positive prime p.

• (ii) If $p\not \equiv 1$ (mod 12), then the rank of $\Gamma _{-p}$ is at most one.

• (iii) If $p=2$, $p=3$ or $p\equiv 7$ (mod 24), then the rank of $\Gamma _{-p}$ is zero.

We consider the curves

$$\begin{aligned} C_{-p} : y^2 =x^3 -2px^2 -3p^2 x \quad \text {and} \quad \overline{C} _{-p} : y^2 =x^3 +4px^2 +16p^2 x. \end{aligned}$$

First we prove the following lemma.

Lemma 3.5

 

• (i) $\#\, \alpha (\Gamma _{-p} ) \le 8$ for every prime p.

• (ii) If $p=2$, $p=3$ or $p\equiv 7$ (mod 12), then $\#\, \alpha (\Gamma _{-p} ) \le 4$.

• (iii) $\#\, \alpha (\overline{\Gamma }_{-p} ) \le 2$ for every prime p.

• (iv) If $p\ne 3$ and $p\not \equiv 1$, 13 or 19 (mod 24), then $\#\, \alpha (\overline{\Gamma }_{-p} ) =1$.

Proof

The proof of the statement (i) is the same as in the case of Lemma 3.3.

(ii) Suppose $\#\, \alpha (\Gamma _{-p} ) =8$. As in the proof of (ii) of Lemma 3.3, this implies $p\ne 3$ and $-1\in \alpha (\Gamma _{-p} )$. Therefore, by $b_1 \mid -3p^2$, $N^2 =b_1 M^4 -2pM^2 e^2 +(-3p^2 /b_1 )e^4$ is solvable for at least one of $b_1 =-1$ and $b_1 =-p^2$.

Suppose that $N^2 =-M^4 -2pM^2 e^2 +3p^2 e^4$ is solvable. If $p=2$, then M is odd by $\gcd (M, 3p^2 )=1$, and $N^2 \equiv -M^4$ (mod 4), which is impossible. If $p>3$, then $p\not \mid M$ by $\gcd (M, 3p^2 )=1$, and thus we have $\big (\frac{-1}{p}\big )=1$ and $p\equiv 1$ (mod 4).

Next suppose that $N^2 =-p^2 M^4 -2pM^2 e^2 +3 e^4$ is solvable; then $\gcd (M,3)=1$. If $p=2$, then e is odd (since otherwise both N and e would be even), and $N^2 \equiv 3e^4$ (mod 4), which is impossible. Suppose $p>3$. Then $p\not \mid e$ by $\gcd (e, b_1 )=1$, and thus $\big (\frac{3}{p}\big )=1$ and $p\equiv \pm 1$ (mod 12).

We have $N^2 =(3e^2 +pM^2 )(e^2 -pM^2 )=C\cdot D$. Since $p\not \mid e$ and $\gcd (M,e)=1$, it follows that $\gcd (C,D) \mid 4$. If $\gcd (C,D)=1$ or 4, then C and D are squares. Thus $3e^2 +pM^2 =n^2$, $pM^2 \equiv n^2$ (mod 3), $p\equiv 1$ (mod 3) and $p\equiv 1$ (mod 12).

If $\gcd (C,D)=2$, then C / 2 and D / 2 are squares. Thus $3e^2 +pM^2 =2n^2$, hence $p\equiv pM^2 \equiv 2n^2 \equiv 2$ (mod 3). Since $p\equiv \pm 1$ (mod 12), we get $p\equiv -1$ (mod 12).

We proved that if $\#\, \alpha (\Gamma _{-p} )=8$, then $p\equiv 1$ (mod 4) or $p\equiv -1$ (mod 12). This proves (ii).

(iii) The argument proving (iii) of Lemma 3.3 shows that $\alpha (\overline{\Gamma }_{-p} ) \subset \{ 1,p\}$ (mod ${{\mathbb {Q}}}^{*2}$).

(iv) Suppose $\#\, \alpha (\overline{\Gamma }_{-p} ) =2$. Then $p\in \alpha (\overline{\Gamma }_{-p} )$, and

$$\begin{aligned} N^2 =b_1 M^4 +4pM^2 e^2 +(16p^2 /b_1 )\,e^4 \end{aligned}$$

is solvable for at least one of $b_1 =p$, $b_1 =4p$ and $b_1 =16p$.

Let $b_1 =p$, and suppose that $N^2 =p M^4 +4 pM^2 e^2 +16pe^4$ is solvable. Then M is odd by $\gcd (M,16p)=1$, and $N^2 \equiv pM^4$ (mod 4). Hence $p>2$ and $p\equiv 1$ (mod 4). We have $N=pN_1$ and

$$\begin{aligned} pN_1^2 =M^4 +4M^2 e^2 +16e^4 =(M^2 +2e^2 )^2 +12e^4 . \end{aligned}$$

Now $p\not \mid e$ by $\gcd (e,b_1 )=1$, and we get $\big (\frac{-12}{p}\big )=1$. Since $p\equiv 1$ (mod 4), we obtain $\big (\frac{3}{p}\big )=1$, $p\equiv \pm 1$ (mod 12) and $p\equiv 1$ (mod 12).

The case $b_1 =16p$ is similar with the roles of M and e exchanged. Therefore, if (6) is solvable for $b_1 =16p$, then $p\equiv 1$ (mod 12).

Finally, let $b_1 =4p$, and suppose that $N^2 =4p M^4 +4 pM^2 e^2 +4pe^4$ is solvable. Then M is odd by $\gcd (M,4p)=1$. Also, $2p\mid N$, and thus e is odd. Let $N=2pN_1$, then $pN_1 ^2 =M^4 +M^2 e^2 +e^4$. Thus $pN_1^2 \equiv 3$ (mod 8), hence $p\equiv 3$ (mod 8). We have

$$\begin{aligned} 4pN_1^2 =4M^4 +4M^2 e^2 +4e^4 =(2M^2 +e^2 )^2 +3e^4 . \end{aligned}$$

Now $p\not \mid e$ by $\gcd (e,b_1 )=1$, and we get $p=3$ or $\big (\frac{-3}{p}\big )=1$. Suppose $p\ne 3$. Since $p\equiv 3$ (mod 4), we obtain $\big (\frac{3}{p}\big )=-1$, $p\equiv 5$ or 7 (mod 12). Since $p\equiv 3$ (mod 8), we get $p\equiv 19$ (mod 24).

We proved that if $\#\, \alpha ({\overline{\Gamma }}_{-p} )=2$, then $p=3$ or $p\equiv 1$ (mod 12) or $p\equiv 19$ (mod 24), This proves (iv). $\square$

Proof of Theorem 3.4

Statements (i) and (ii) of the theorem follow from Lemma 3.5 and from (4). If $p=2$ or $p\equiv 7$ (mod 24), then the rank of $\Gamma _{-p}$ is zero by Lemma 3.5 and (4).

What remains to prove is that the rank of $\Gamma _{-3}$ is zero. Since $\#\, \alpha (\overline{\Gamma }_{-3} ) \le 2$ by Lemma 3.5, it is enough to show that $\#\, \alpha (\Gamma _{-3} ) \le 2$.

Consider the curve $C_{-3} :y^2 =x^3 -6x^2 -27x$. Then $b_1 \in \{ \pm 1, \pm 3, \pm 9, \pm 27\}$, and thus $\alpha (\Gamma _{-3} ) \subset \{ \pm 1, \pm 3\}$ (mod ${{\mathbb {Q}}}^{* 2}$). We show that $3 \notin \alpha (\Gamma _{-3} )$. Suppose $3 \in \alpha (\Gamma _{-3} )$. Then the equation $N^2 =b_1 M^4 -6M^2 e^2 -(27/b_1 )e^4$ is solvable for at least one of $b_1 =3$ and $b_1 =27$.

Suppose that $N^2 =3M^4 -6M^2 e^2 -9e^4$ is solvable. Then $3\not \mid M$ by $\gcd (M,9) =1$, and $3\not \mid e$ since $3\mid N$. Let $N=3N_1$. Then $3N_1^2 =M^4 -2M^2 e^2 -3e^4$, hence $M^4 \equiv 2M^2 e^2$ (mod 3), which is impossible.

Finally, suppose that $N^2 =27M^4 -6M^2 e^2 -e^4$ is solvable. Then $3\not \mid e$ by $\gcd (e,b_1 )=1$. Thus $N^2 \equiv - e^2$ (mod 3), which is impossible. $\square$

Corollary 3.6

If $p=2$, $p=3$ or $p\equiv 7$ (mod 24), then the curves $C_p$ and $C_{-p}$ have no nontrivial rational points. $\square$

4 Numerical Examples

As the following table shows, for all primes $3<p<100$, if $p\not \equiv 7$ (mod 24), then at least one of the curves $C_p$ and $C_{-p}$ has nontrivial rational points and, consequently, $\Gamma _p$ or $\Gamma _{-p}$ has positive rank. Note that the point (75, 210) belongs to both $C_{-23}$ and $C_{73}$.

The points below were found by searching for integer solutions of $N^2 =b_1 M^4 \pm 2p M^2 e^2 +b_2 e^4$ with $b_1 b_2 =-3p^2$, and putting $x=b_1 M^2 /e^2$, $y=b_1 MN/e^3$. The solutions for $p \ne 83$ were found by using GNU Octave (https://www.gnu.org/software/octave/). I am grateful to Peter Salvi for finding a solution for $p=83$; he used Julia 1.0 (https://julialang.org/blog/2018/08/one-point-zero).

$$\begin{aligned} p= & {} 5: (-1, 8)\in \Gamma _{-5}, \\ p= & {} 11: (75, 720)\in \Gamma _{11},\\ p= & {} 13: (-12, 90)\in \Gamma _{13},\\ p= & {} 17: (-1,30)\in \Gamma _{17},\\ p= & {} 19: \Big ( \displaystyle {\frac{17689}{225}, \frac{1374688}{3375}}\Big )\in \Gamma _{-19},\\ p= & {} 23: (75, 210)\in \Gamma _{-23},\\ p= & {} 29: \Big ( \displaystyle {-\frac{529}{25}, \frac{16744}{125}}\Big ) \in \Gamma _{-29}, \\ p= & {} 37: \Big ( \displaystyle {\frac{231361}{324}, \frac{116481365}{5832}} \Big )\in \Gamma _{37}, \\ p= & {} 41: ( -121, 198)\in \Gamma _{41}, \\ p= & {} 43: \Big ( \displaystyle {\frac{4165798849}{21538881}, \frac{171543655606240}{99961946721}} \Big )\in \Gamma _{-43}, \\ p= & {} 47: (1875, 79050)\in \Gamma _{-47}, \\ p= & {} 53: \Big ( -\displaystyle {\frac{167281}{4225}, \frac{89165272}{274625}} \Big ) \in \Gamma _{-53}, \\ p= & {} 59: \Big ( -\displaystyle {\frac{930433009}{6076225}, \frac{13189530387264}{14977894625}} \Big )\in \Gamma _{59}, \\ p= & {} 61: ( -108, 1170)\in \Gamma _{61}, \\ p= & {} 67: \Big ( \displaystyle {\frac{909373939321}{51279921}, \frac{863887766632341760}{367215514281} } \Big )\in \Gamma _{-67}, \\ p= & {} 71: (507, 9282)\in \Gamma _{-71}, \\ p= & {} 73: (75, 210)\in \Gamma _{73}, \\ p= & {} 83: \Big ( -\displaystyle {\frac{2140232721200}{59682001401}, \frac{13897116923228469980}{14580253260262899}} \Big )\in \Gamma _{83}, \\ p= & {} 89: \Big ( -\displaystyle {\frac{121}{289}, \frac{489280}{4913}} \Big )\in \Gamma _{-89}, \\ p= & {} 97: \Big ( -\displaystyle {\frac{121}{25}, \frac{45408}{125}} \Big )\in \Gamma _{-97}. \\ \end{aligned}$$