Quartic Polynomials with a Given Discriminant

(Mordell, 1920). For any given 0 ≠ k ∈ ℤ, the equation

Equation (1.3) is often called Mordell’s equation, in honour of the contribution Louis Joel Mordell [17] has made to this subject. An extension to Theorem 1.1 was later made by Carl Ludwig Siegel [18]. In its simplest form, Siegel’s result can be formulated as follows:

(Siegel, 1929). Let α, β ∈ ℤ be such that 4α³ + 27β² ≠ 0. Then the equation

Mordell’s equation has had a long history. First discoveries concerning (1.3) were given in Dickson [2: pp. 533–539] going back to the work of Bachet from 1621. Many interesting historical notes to (1.3) can be found in [1,5,7,16]. Perhaps the most extensive historical comments related to Mordell’s contribution to (1.3) can be found in the recent paper [6].

Let n ∈ ℕ, n ≥ 2, 0 ≠ D ∈ ℤ and let

Proof. First, by induction on k, it can be proved that k!|f^(k)(w) for any k ∈ {0, 1, 2,… }. Hence, f_w(x) ∈ ℤ[x]. Further, Taylor’s theorem yields

Let α₁,…,α_n be the roots of f(x) in the set of complex numbers ℂ. Then

Next, by (2.8), for any α ∈ {α₁,…, α_n}, we have

Combining (2.7) with (2.9), we get f_w(α – w) = 0, and thus,

are the roots of f_w(x) in ℂ. Using (2.10), we now get

as desired. □

First we show that (i) is equivalent to (ii). Using Taylor’s theorem, we obtain

for any w ∈ ℤ. Therefore,

Combining (2.12) with (2.11), we get f(x + w) = f_w(x). Hence, (i) and (ii) are equivalent.

Further we prove that (i) is equivalent to (iii). Assume that g(x) = f(x + w) for some w ∈ ℤ. Then (2.12) yields

Hence, r_g(x) = g(x – (4w + a)/4) = g(x – w – a/4) = f(x – w – a/4 + w) = f(x – a/4) = r_f(x).

Finally, let r_f(x) = r_g(x). Then f(x – a/4) = g(x – a̅/4). Hence, f(x – a/4 + a̅/4) = g(x – a̅/4 + a̅/4) = g(x) and g(x) = f(x – (a̅ – a)/4) follows. Put w = (a̅ – a)/4. Clearly, if a ≡ a̅ (mod 4), then w ∈ ℤ. Suppose that a ≢ a̅ (mod 4). Using (2.5) we obtain R = 8b – 3a² = 8b̅ – 3a̅², S = 8c – 4ab + a³ = 8c̅ – 4a̅b̅ + a̅³, which implies a² ≡ a̅² (mod 8) and a³ ≡ a̅³ (mod 4). Therefore, a² ≡ a̅² (mod 4), which yields, without loss of generality, that either a ≡ 0 (mod 4), a̅ ≡ 2 (mod 4) or a ≡ 1 (mod 4), a̅ ≡ 3 (mod 4). If a ≡ 0 (mod 4), a̅ ≡ 2 (mod 4), then a² ≡ 0 (mod 8), a̅² ≡ 4 (mod 8), which is in contradiction to a² ≡ a̅² (mod 8). Similarly, if a ≡ 1 (mod 4), a̅ ≡ 3 (mod 4), then a³ ≡ 1 (mod 4), a̅³ ≡ 3 (mod 4), which is in contradiction to a³ ≡ a̅³ (mod 4). □

Let 0 ≠ D ∈ ℤ and let Q_D ≠ ∅. For f(x), g(x) ∈ Q_D put

It is evident that ∼ is an equivalence relation on the set Q_D. Moreover, Q_D/∼ has only finitely many equivalence classes. In Section 4, this fact will be proved using the results of Mordell and Siegel presented in Theorem 1.1 and Theorem 1.2. On the other hand, this claim also follows as a consequence of a more general theorem that has been proved by Kálmán Györy [8: p. 419]. See also [9: p. 475] or consult [3: p. 109].

Let 0 ≠ D ∈ ℤ. If Mordell’s equation

Proof. Let f(x) = x⁴ + ax³ + bx² + cx + d ∈ Q_D and let r_f(x) = x⁴ + Ax² + Bx + C ∈ ℚ[x]. Direct calculation will verify that (2.3) can be written in the form

Substituting (2.4) into (3.2), after short calculation, we obtain

Put

Then X, Y ∈ ℤ and (3.3) yields

Since D_rf = D_f = D, the proof is complete.

If x⁴ + ax³ + bx² + cx + d ∈ Q_D, then (2.2) yields A, B, C ∈ ℤ ⇔ 4|a. In this case, we can write (3.2) in the form V² = 4U³ – 27D_rf, where U = A² + 12C and V = 2A³ –72AC+27B².

Hence, we have

Since D_rf = D, the substitutions X = 4U, Y = 4V reduce (3.5) to

It is interesting that Mordell’s equation (3.6) plays a fundamental role also in the theory of cubic polynomials with the same discriminant D. Consult [10: p. 313].

The following notation will be useful. For an arbitrary 0 ≠ D ∈ ℤ, let M_D denote the set of all [X₀, Y₀], where X₀, Y₀ ∈ ℤ and Y02=X03−21633D.

Let 0 ≠ D ∈ ℤ and let [X₀, Y₀] ∈ M_D. Then (i), (ii), (iii) and (iv) hold:

1. If 2|X₀, then 4|X₀, 8|Y₀.

2. If 2|Y₀, then 4|X₀, 8|Y₀.

3. If 3|X₀, then 9|Y₀.

4. If 3|Y₀, then 3|X₀, 9|Y₀.

Proof. The conclusions (i)–(iv) immediately follow from Y02=X03−21633D. □

They will be used in Section 4 and Section 5. □

Letξ₀, η₀, e ∈ ℤ be such that

1. ξ₀ ≡ 0 (mod 12) andη₀ ≡ 0 (mod 216).

2. There exists exactly one e ∈ {0, 1, 2, 3} satisfying (4.1).

Proof. (i) Since the validity of the congruence ξ₀ ≡ 0 (mod 12) is evident, we only prove that η₀ ≡ 0 (mod 216). First, observe that η₀ ≡ 9eξ₀ – 108e³ (mod 216). Further, ξ₀ ≡ 36e² (mod 96) is equivalent to 9ξ₀ ≡ 324e² (mod 864). Hence, 9ξ₀ ≡ 108e² (mod 216). This, together with η₀ ≡ 9eξ₀ – 108e³ (mod 216), yields η₀ ≡ 0 (mod 216).

(ii) Let ξ₀, η₀ ∈ ℤ satisfy (4.1) for some e ∈ {0, 1, 2, 3}. Suppose that e is not unique. Then it follows from ξ₀ ≡ 36e² (mod 96) that e ∈ {1, 3} and that ξ₀ ≡ 36 (mod 96). On the other hand, using η₀ ≡ 9eξ₀ – 108e³ (mod 1728), we obtain 9ξ₀ – 108 ≡ 27ξ₀ – 2916 (mod 1728), which yields ξ₀ ≡ 60 (mod 96), a contradiction. □

Let 0 ≠ D ∈ ℤ and let M_D ≠ ∅. Then Q_D ≠ ∅ if and only if there exists an [X₀, Y₀] ∈ M_D such that the elliptic equation

Proof. First, assume that Q_D ≠ ∅. Then there exists an f(x) = x⁴ + ax³ + bx² + cx + d ∈ Q_D such that r_f(x) = x⁴ + (R/8)x² + (S/8)x + T/256 ∈ ℚ[x] where R, S, T are integers satisfying (2.5). Further, from Theorem 3.1 it follows that there exists a [X₀, Y₀] ∈ M_D such that R² + 3T = X₀ and R³ – 9RT + 108S² = Y₀. Substituting 3T = X₀ – R² into R³ – 9RT + 108S² = Y₀, we obtain

and multiplying (4.6) by 432, we get

Put ξ₀ = –12R and η₀ = 216S. Now, (4.7) implies immediately that [ξ₀, η₀] is an integer solution of (4.2).

Finally, we have to prove that [ξ₀, η₀] satisfies (4.3)–(4.5) for some e ∈ {0, 1, 2, 3}. Since a ∈ ℤ, there exist uniquely determined w ∈ ℤ and e ∈ {0, 1, 2, 3} such that a = 4w + e. Substituting a = 4w + e into the first equation of (2.5), we obtain R ≡ –3e² (mod 8) and –12R ≡ 36e² (mod 96) follows. This together with ξ₀ = –12R yields ξ₀ ≡ 36e² (mod 96). Hence, (4.3).

Further, from the second equation of (2.5), it follows

Putting a = 4w + e, 8b = R + 3a², ξ₀ = –12R and η₀ = 216S into (4.8), we obtain

Reducing (4.9) by modulus 1728 and using ξ₀ ≡ 36e² (mod 96), we get (4.4).

Finally, the third equation of (2.5) implies

For the left-hand side of (4.10), we have 432T=144(X0−R2)=144X0−ξ02 and the right-hand side of (4.10) can be rewritten, substituting a = 4w+e, 8b = R+3a², 8c = S+4ab – a³, ξ₀ = –12R and η₀ = 216S into

Since η₀ ≡ 9e_o – 108e³ (mod 1728) and ξ₀ ≡ 36e² (mod 96), we get

Hence, (4.5).

Conversely, assume that there exists a [X₀, Y₀] ∈ M_D such that equation (4.2) has an integer solution [ξ₀, η₀] satisfying (4.3)–(4.5) for some e ∈ {0, 1, 2, 3}. Put

Then, by part (i) of Lemma 4.1, we have R, S ∈ ℤ. We now prove that T ∈ ℤ. From the first and third equation in (4.11) we obtain T = (X₀ – R²)/3. First we show that

Let 3|X₀. Then, by part (iii) of Lemma 3.3, we have 9|Y₀. Further, by (4.11), we have 3|ξ₀ and 3³|η₀. Since η02=ξ02−108X0ξ0+432Y0, we also have 0≡η02≡ξ03 (mod 3⁵) and ξ₀ ≡ 0 (mod 3²) follows. This together with ξ₀ = – 12R yields 3|R.

Let 3|R. Since ξ₀ = –12R, we have 3²|ξ₀ and 36|ξ03 follows. Next, by (4.11), 36|η03. Since η02≡ξ03−108X0ξ0+432Y0, we have 432Y₀ ≡ 0 (mod 3⁵), and Y₀ ≡ 0 (mod 3²) follows. By part (iv) of Lemma 3.3, we get 3|X₀. This proves (4.12).

Further, suppose that X₀ ≡ 2 (mod 3). Then from [X₀, Y₀] ∈ M_D it follows that Y02≡2 (mod 3), which is a contradiction. Combining this fact with (4.12), we get

Clearly, in both cases (4.12) and (4.13), we have X₀ – R² ≡ 0 (mod 3). Hence, T ∈ ℤ.

Consider now the polynomial

We prove that the discriminant D_r of r(x) is equal to D. First, direct calculation verifies that

On the other hand, substituting ξ₀ = -12R, η₀ = 216S into η02=ξ03−108X0ξ0+432Y0 we obtain

Hence, we get R³ –3R(X₀ –R²)+108S² = Y₀. Since, X₀ –R² = 3T, we have R³–9RT +108S² = Y₀. This, together with Y02≡X03−21633D yields

Hence, D_r = D.

Finally, let e ∈ {0, 1, 2, 3} satisfy (4.3)–(4.5). Then, by part (ii) of Lemma 4.1, e is uniquely determined. Put g(x) = r(x + e/4). Then we obtain after some calculation that

and

Since D_g = D_rg = D_r = D, we have g(x) ∈ Q_D, as desired. The proof is complete. □

Before proceeding, the following notations will be adopted. For any [X₀, Y₀] ∈ M_D, let E_D(X₀, Y₀) denote the set of all [ξ₀, η₀] where ξ₀, η₀ ∈ ℤ and Y02≡X03−108X0ξ0+432Y0. Next, let E_D denote the set of all [X₀, Y₀, ξ₀, η₀, e] where [X₀, Y₀] ∈ M_D, [ξ₀, η₀] ∈ E_D(X₀, Y₀) and e ∈ {0, 1, 2, 3} satisfy (4.3)–(4.5).

Let 0 ≠ D ∈ ℤ and let f(x) = x⁴ + ax³ + bx² + cx + d ∈ ℤ[x]. Then f(x) ∈ Q_D if and only if there exists [X₀, Y₀, ξ₀, η₀, e] ∈ E_D and w ∈ ℤ such that

Let 0 ≠ D ∈ ℤ and let M_D ≠ ∅. Then (i), (ii) and (iii) hold:

1. E_D (X₀, Y₀) is a finite set for any [X₀, Y₀] ∈ M_D.

2. E_D is a finite set.

3. Q_D/∼ has only finitely many equivalence classes for any Q_D ≠ ∅.

Proof. (i) Put α = –108X₀ and β = 432Y₀. Then 4α3+27β2=2839(−X03+Y02)=−224312D≠0.

Conclusion (i) now follows from Theorem 1.2.

1. Conclusion (ii) is a direct consequence of Theorem 1.1 and part (i) of Proposition 4.4.

2. Let φ : E_D → Q_D/∼ be the mapping defined by φ(X₀, Y₀, ξ₀, η₀, e) = {f_w(x); w∈ℤ}, where

   f0(x)=x4+ex3+36e2−ξ096x2+108e3−9eξ0+η01728x+432e4−ξ02−72e2ξ0+16eη0+144X0110592.

Then φ is bijective. Injectivity of φ is evident and surjectivity of φ immediately follows from Corollary 4.3. Hence, #Q_D/∼= #E_D. This proves (iii). □

Let [X₀, Y₀], [X0*,Y0*] ∈ M_D and let [X0*,Y0*]. By an example we will prove that the set ED(X0,Y0)∩ED(X0*,Y0*) can be nonempty. For D = –23, we have [64, 6400], [–320, –2816] ∈ M_–23 and [96, ±1728] ∈ E_–23(64, 6400) ∩ E_–23(–320, –2816).

Now we are ready to formulate the method for determining the set Q_D. It can be formally divided into five steps as follows:

1. Let 0 ≠ D ∈ ℤ. First we find the set M_D of all integer solutions [X₀, Y₀] of Mordell’s equation Y² = X³ – 2¹⁶3³D. By Theorem 1.1, M_D is a finite set and Theorem 3.1 states that, if M_D = ∅, then Q_D = ∅.

2. Let M_D ≠ ∅. Next we find, for any [X₀, Y₀] ∈ M_D, the set E_D(X₀, Y₀) of all integer solutions [ξ₀, η₀] of the elliptic equation η² = ξ³ – 108X₀+432Y₀. By part (i) of Proposition 4.4, E_D (X₀, Y₀) is a finite set for any [X₀, Y₀] ∈ M_D and Theorem 4.2 says that, if E_D(X₀, Y₀) = ∅ for any [X₀, Y₀] ∈ M_D, then Q_D = ∅.

3. In step (iii), we establish the set E_D. By part (ii) of Proposition 4.4, E_D is a finite set and Corollary 4.3 states that Q_D ≠ ∅ if and only if E_D ≠ ∅.

4. Let E_D ≠ ∅ and let #E_D = n. In this step, we assign to each [X₀, Y₀, ξ₀, η₀, e] ∈ E_D the polynomial

   g(x)=x4+ex3+36e2−ξ096x2+108e3−9eξ0+η01728x+432e4−ξ02−72e2ξ0+16eη0+144X0110592.

   In this way, we obtain the full system of representatives G_D = {g₁(x),…, g_n(x)} of Q_D/∼. By part (iii) of Proposition 4.4, Q_D/∼ is a finite set.

5. Finally, applying Corollary 2.3 to each g_i(x) ∈ G_D, i ∈ {1,…,n}, we obtain the n sets {f_i,w(x); w ∈ ℤ} where

   fi,w(x)=x4+gi′′′(w)3!x3+gi′′(w)2!x2+gi′(w)1!x+gi(w).

   Hence, we get

   QD=∪i=1n{fi,w(x);w∈ℤ}.

The below example illustrates our method.

Let D = –87. Then we have

Hence,

Further, we have

Hence, it follows that #E_–87 = #Q_–87//∼= 4 and that G_–87 = {g₁(x), g₂(x), g₃(x), g₄(x)} where

Finally,

and

Applying the method, the validity of Theorem 4.7 can be verified.

Let 0 ≠ D ∈ ℤ and let 1 ≤ |D| ≤ 1000. Then we have:

1. If 1 ≤ D ≤ 1000, then Q_D ≠ ∅ if and only if

   D∈{5,12,20,21,32,40,45,48,49,60,77,81,85,96,104,112,117,125,140,144,148,165,169,189,192,216,221,224,229,252,256,257,260,272,285,288,320,321,333,357,361,392,400,404,432,437,468,469,473,480,488,500,512,525,528,533,544,549,564,572,580,592,605,621,629,656,672,697,725,729,733,761,768,785,788,792,816,832,837,864,892,896,900,916,957,981,985}

2. If –1 ≥ D ≥ –1000, then Q_D ≠ ∅ if and only if

   D∈{−3,−16,−23,−27,−31,−44,−59,−76,−83,−87,−107,−108,−112,−135,−139,−140,−175,−176,−199,−211,−231,−236,−240,−247,−255,−256,−268,−275,−279,−283,−288,−304,−331,−332,−335,−351,−367,−400,−416,−428,−432,−448,−464,−475,−491,−500,−507,−527,−556,−560,−563,−575,−588,−608,−643,−671,−684,−688,−695,−731,−751,−783,−800,−816,−844,−848,−863,−864,−891,−931,−944,−959,−972,−976,−983}.

Let 0 ≠ D ∈ ℤ and let e ∈ {1, 3}. Then there exists a one-to-one correspondence between the sets Q_D(1) and Q_D (3) given by the relation

Proof. First observe that [ξ₀, η₀] ∈ E_D (X₀, Y₀) if and only if [ξ₀, –η₀] ∈ E_D (X₀, Y₀).

Let 0 ≠ D ∈ ℤ and let e ∈ {0, 2}. Then we have:

1. [X₀, Y₀, ξ₀, η₀, 0] ∈ E_D ⇔ [X₀, Y₀, ξ₀, –η₀, 0] ∈ E_D.

2. [X₀, Y₀, ξ₀, η₀, 2] ∈ E_D ⇔ [X₀, Y₀, ξ₀, –η₀, 2] ∈ E_D.

Proof. Since part (i) of Proposition 5.2 immediately follows from (4.3)–(4.5), we prove (ii). Let [X₀, Y₀, ξ₀, η₀, 2] ∈ E_D. Then (4.3)–(4.5) yields that ξ₀ ≡ 48 (mod 96), η₀ ≡ 18ξ₀ – 864 (mod 1728), and

Since ξ₀ ≡ 48 (mod 96) is equivalent to 18ξ₀ ≡ 864 (mod 1728), we have –η₀ ≡ – 18ξ₀ + 864 ≡ 1710ξ₀ – 864 = 18ξ₀ – 864 + 94 · 18ξ₀ ≡ 18ξ₀ – 864 + 47 · 1728 ≡ 18ξ₀ – 864 (mod 1728). Further, ξ₀ ≡ 48 (mod 96) is equivalent to 1152ξ₀ ≡ 55296 (mod 110592) and η₀ ≡ 18ξ₀ – 864 (mod 1728) is equivalent to 64η₀ ≡ 1152ξ₀ – 55296 (mod 110592). Hence, we obtain 64η₀ ≡ 0 (mod 110592) and 32η₀ ≡ – 32η₀ (mod 110592) follows. Now we see that (5.6) is equivalent to 6912 – ξ02 – 288ξ₀ – 32η₀ + 144X₀ ≡ 0 (mod 110592). This proves (ii). □

If Q_D ≠ ∅, then any of the below seven cases can occur:

1. Q_D(0)≠ ∅, Q_D(1) ∪ Q_D(3) ≠ ∅, Q_D(2) ≠ ∅, D = –23,

2. Q_D(0) ≠ ∅, Q_D(1) ∪ Q_D(3) ≠ ∅, Q_D(2) ≠ ∅, D = 32,

3. Q_D(0) ≠ ∅, Q_D(1) ∪ Q_D(3) ≠ ∅, Q_D(2) ≠ ∅, D = 5,

4. Q_D(0) = ∅, Q_D(1) ∪ Q_D(3) ≠ ∅, Q_D(2) ≠ ∅, D = –87,

5. Q_D(0) ≠ ∅, Q_D(1) ∪ Q_D(3) = ∅, Q_D(2) ≠ ∅, D = –27,

6. Q_D(0) = ∅, Q_D(1) ∪ Q_D(3) ≠ ∅, Q_D(2) = ∅, D = 12,

7. Q_D(0) = ∅, Q_D(1) ∪ Q_D(3) = ∅, Q_D(2) ≠ ∅, D = –3.

The above values of D are the least, in absolute value, for which the case occurs.

Let a, b, m ∈ ℤ, m > 1 and let g = gcd(a, m). Then the congruence ax ≡ b (mod m) is soluble if and only if g|b.

Using Proposition 6.1, we now prove Lemma 6.2.

If X₀, Y₀ ∈ ℤ, then the congruence 3αX₀ + Y₀ – 4α³ ≡ 0 (mod 27) holds for at most one α ∈ {0,1, 2}.

Proof. The proof consists of three steps. (i) First, suppose that Y₀ ≡ 0 (mod 27) and 3X₀ + Y₀ – 4 ≡ 0 (mod 27). Then Y₀ ≡ 4 – 3X₀ ≡ 0 (mod 27), which yields 3X₀ ≡ 4 (mod 27). Hence, gcd(3, 27) = 3 ∤ 4, which is a contradiction.

(ii) Further, suppose that Y₀ ≡ 0 (mod 27) and 6X₀ + Y₀ – 32 ≡ 0 (mod 27). Then Y₀ ≡ 5 – 6X₀ ≡ 0 (mod 27), which yields 6X₀ ≡ 5 (mod 27). Hence, gcd(6, 27) = 3 ∤ 5, which is a contradiction.

(iii) Finally, suppose that 3X₀ + Y₀ – 4 ≡ 0 (mod 27) and 6X₀ + Y₀ – 32 ≡ 0 (mod 27). Then Y₀ ≡ 4 – 3X₀ ≡ 5 – 6X₀ (mod 27), which yields 3X₀ ≡ 1 (mod 27). Hence, gcd(3, 27) = 3 ∤ 1, which is a contradiction. Combining (i)–(iii) proves the lemma. □

We are now ready to prove the main result of this section.

Let 0 ≠ D ∈ ℤ and let [X₀, Y₀] ∈ ℰD.

1. If 3αX₀ + Y₀ – 4α³ ≡ 0 (mod 27) does not hold for any α ∈ {0, 1, 2}, then the system

   (6.1)R2+3T=X0andR3−9RT+108S2=Y0

   is not solvable in integers.

2. If 3αX₀ + Y₀ – 4α³ ≡ 0 (mod 27) holds for some α ∈ {0, 1, 2}, thenα is uniquely determined, X₀ – 4α² ≡ 0 (mod 12), 3αX₀ + Y₀ – 4α³ ≡ 0 (mod 108) and the set K of all integer solutions of (6.1) can be obtained from the set L of all integer solutions of the elliptic equation

   (6.2)η2=ξ2−αξ2−X0−4α212ξ+3αX0+Y0−4α3108.

   Moreover, between K and L, there exists a one-to-one correspondence.

Proof. (i) Let [R₀, S₀, T₀] be an arbitrary integer solution of (6.1). Since R₀ ∈ ℤ, there exits a uniquely determined r ∈ ℤ and α ∈ {0, 1, 2} such that R₀ = 3r + α.

First, we prove

Let 2|X₀. Then, by part (i) of Lemma 3.3, 4|X₀, and thus 4|X₀ – 4α². Next, the first equation in (6.1) yields X0−R02≡X0−α2≡0 (mod 3), and 3|X₀ – 4α² follows. Hence, 12|X₀ – 4α², which means (X₀ – 4α²)/12 ∈ ℤ. Because the validity of the converse implication is evident, we get (6.3).

Further, substituting 3T0=X0−R02 into R03−9R0T0+108S02−Y0, we obtain

Since R₀ = 3r + α, (6.4) can be written in the equivalent form

Reducing (6.5) by the modulus 108, using (6.3), we get 3αX₀ + Y₀ – 4α³ ≡ 0 (mod 108). This proves (i).

(ii) Assume that there exists an α ∈ {0,1, 2} such that 3αX₀ + Y₀ – 4α³ ≡ 0 (mod 27). Then Lemma 6.2 states that α is uniquely determined. Since [X₀, Y₀] ∈ ℰD, by part (i) of Lemma 3.3, 4|X₀ and 8|Y₀, which yields 4|3αX₀ + Y₀ – 4α³ for any α ∈ {0, 1, 2}. Hence,

We now prove that

First, by part (i) of Lemma 3.3, X₀ – 4α² ≡ 0 (mod 4). Next, from the assumption 3αX₀ + Y₀ – 4α³ ≡ 0 (mod 27), we get Y₀ ≡ α³ ≡ α (mod 3). Hence, Y02≡α2 (mod 3). Furthermore, it is clear from [X₀, Y₀] ∈ ℰD that Y02≡X03 (mod 3), which, together with X03≡X0 (mod 3), yields X₀ ≡ α² (mod 3). Hence, X₀ – 4α² ≡ 0 (mod 3), and (6.6) follows.

Let [R₀, S₀, T₀] ∈ K. Then there exist a uniquely determined r ∈ ℤ and an α ∈ {0,1, 2} such that R₀ = 3r + α. In much the same way as in the proof of part (i), we get (6.5). Hence,

where (X₀ – 4α²)/12 and (3αX₀ + Y₀ – 4α³)/108 are integers. Put [ξ₀, η₀] = [–r, S₀]. Then it follows from (6.7) that [ξ₀, η₀] ∈ L.

Conversely, let [ξ₀, η₀] ∈ L where α ∈ {0,1, 2} satisfies 3αX₀ + Y₀ – 4α³ ≡ 0 (mod 27). Put R₀ = –3ξ₀ + α and S₀ = η₀. Then R₀, S₀ ∈ ℤ, and substituting ξ₀ = (α – R₀)/3, η₀ = S₀ into (6.2) some calculation will yield