Quartic polynomials with a given discriminant

a o DOI: 10.1515/ms-2022-0003 Math. Slovaca 72 (2022), No. 1, 35–50 QUARTIC POLYNOMIALS WITH A GIVEN DISCRIMINANT Jiˇ r ´ ı Klaˇ ska Dedicated to the eminent Czechoslovak mathematician Ladislav Skula (Communicated by Milan Paˇst´eka ) ABSTRACT. Let 0 6=D∈Zand let QDbe the set of all monic quartic polynomials x4+ax3+bx2+ cx +d∈Z[x] with the discriminant equal to D. In this paper we will devise a method for determining the set QD. Our method is strongly related to the theory of integral points on elliptic curves. The well-known Mordell’s equation plays an important role as well in our considerations. Finally, some new conjectures will be included inspired by extensive calculations on a computer. c 2022 Mathematical Institute Slovak Academy of Sciences 1. Introduction Let 0 6=D∈Zand let QD={f(x) = x4+ax3+bx2+cx +d∈Z[x]; Df=D}(1.1) where Df=a2b2c2−4a2b3d−4a3c3+ 18a3bcd −27a4d2−4b3c2 + 16b4d+ 18abc3−80ab2cd −6a2c2d+ 144a2bd2 −27c4+ 144bc2d−128b2d2−192acd2+ 256d3 (1.2) is the discriminant of f(x). In this paper, the set QDwill be studied in detail. Most of the focus will be given to the problem of determining all polynomials in QD. Clearly, this is equivalent to finding all integer solutions of the Diophantine equation Df=D. In proving the main results, the following two known theorems will be needed. Theorem 1.1 (Mordell, 1920) . For any given 06=k∈Z, the equation Y2=X3+k(1.3) has at most finitely many integer solutions. 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: Theorem 1.2 (Siegel, 1929) . Let α, β ∈Zbe such that 4α3+ 27β26= 0. Then the equation η2=ξ3+αξ +β(1.4) has at most finitely many integer solutions. 2020 Mathematics Subject Classification: Primary 11D25, 11D45, 11Y50. Keywords: Quartic polynomial, discriminant, Mordell’s equation, elliptic curve. 35 JIˇ R´ I KLAˇ SKA There is a standard method for computing all integer solutions of (1.3) and (1.4) using David’s bounds and lattice reduction. This method can be found, for example, in [19]. At present, this method is implemented in several computer algebra packages, including Magma and Pari (Sage). Remark 1.3 . 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]. Throughout this paper, the following notation will be adopted. If Ais a finite set, #Adenotes the number of elements of A. 2. Equivalence on the set QD Let f(x) = x4+ax3+bx2+cx +d∈Z[x] and let Dfbe the discriminant of f(x). Next, let rf(x) = f(x−a/4). Then rf(x) = x4+Ax2+Bx +C∈Q[x] (2.1) where A=b−3a2 8, B =c−ab 2+a3 8, C =d−ac 4+a2b 16 −3a4 256.(2.2) Moreover, we have Drf=Df= 16A4C−4A3B2−27B4−128A2C2+ 144AB2C+ 256C3.(2.3) From (2.2), it follows that there exist R, S, T ∈Zsuch that A=R 8, B =S 8, C =T 256,(2.4) where R= 8b−3a2, S = 8c−4ab +a3, T = 256d−64ac + 16a2b−3a4.(2.5) Hence, we can write (2.1) in the form rf(x) = x4+R 8x2+S 8x+T 256 ∈Q[x] with R, S, T ∈Z.(2.6) We start with a more general theorem. Theorem 2.1 . Let n∈N,n≥2,06=D∈Zand let f(x) = xn+an−1xn−1+· · · +a1x+a0∈Z[x] be an arbitrary polynomial with the discriminant equal to D. Further, for any w∈Z, let fw(x) = n X k=0 f(k)(w) k!xk,(2.7) where f(k)(w)denotes the k-th derivative of f(x)at w. Then fw(x)∈Z[x]and all polynomials in {fw(x); w∈Z}have the same discriminant equal to D. 36 QUARTIC POLYNOMIALS WITH A GIVEN DISCRIMINANT P r o o f. First, by induction on k, it can be proved that k!|f(k)(w) for any k∈ {0,1,2, . . . }. Hence, fw(x)∈Z[x]. Further, Taylor’s theorem yields f(x) = n X k=0 f(k)(w) k!(x−w)kfor any w∈Z.(2.8) Let α1, . . . , αnbe the roots of f(x) in the set of complex numbers C. Then D= n−1 Y i=1 n Y j=i+1 (αj−αi)2. Next, by (2.8), for any α∈ {α1, . . . , αn}, we have f(α) = n X k=0 f(k)(w) k!(α−w)k= 0.(2.9) Combining (2.7) with (2.9), we get fw(α−w) = 0, and thus, β1=α1−w, . . . , βn=αn−w(2.10) are the roots of fw(x) in C. Using (2.10), we now get Dfw= n−1 Y i=1 n Y j=i+1 (βj−βi)2= n−1 Y i=1 n Y j=i+1 (αj−w−(αi−w))2= n−1 Y i=1 n Y j=i+1 (αj−αi)2=D, as desired.  Remark 2.2 . Observe that, in Theorem 2.1, f(x) = f0(x). Hence, f(x)∈ {fw(x); w∈Z}. Corollary 2.3 . Let 06=D∈Zand let f(x) = x4+ax3+bx2+cx +d∈QD. Further, for any w∈Z, let fw(x) = x4+f000(w) 3! x3+f00(w) 2! x2+f0(w) 1! x+f(w).(2.11) Then (i) and (ii) hold: (i) QDis an infinite set and {fw(x); w∈Z} ⊆ QD. (ii) For any w∈Z, we have rfw(x) = rf(x) = x4+Ax2+Bx +C∈Q[x], where A, B, C satisfy (2.2). P r o o f. Part (i) of Corollary 2.3 is a direct consequence of Theorem 2.1 for n= 4. Part (ii) can be verified by direct calculation.  Lemma 2.4 . Let 06=D∈Zand let f(x), g(x)∈QD. Then (i),(ii) and (iii) are equivalent: (i) There exists w∈Zsatisfying g(x) = f(x+w). (ii) There exists w∈Zsatisfying g(x) = fw(x). (iii) rf(x) = rg(x). P r o o f. Let f(x) = x4+ax3+bx2+cx +d,g(x) = x4+ax3+bx2+cx +d∈QD. First we show that (i) is equivalent to (ii). Using Taylor’s theorem, we obtain f(x) = (x−w)4+f000(w) 3! (x−w)3+f00(w) 2! (x−w)2+f0(w) 1! (x−w) + f(w) for any w∈Z. Therefore, f(x+w) = x4+f000(w) 3! x3+f00(w) 2! x2+f0(w) 1! x+f(w).(2.12) 37 JIˇ R´ I KLAˇ SKA Combining (2.12) with (2.11), we get f(x+w) = fw(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 g(x) = x4+ (4w+a)x3+ (6w2+ 3aw +b)x2+ (4w3+ 3aw2+ 2bw +c)x+w4+aw3+bw2+cw +d. Hence, rg(x) = g(x−(4w+a)/4) = g(x−w−a/4) = f(x−w−a/4 + w) = f(x−a/4) = rf(x). Finally, let rf(x) = rg(x). Then f(x−a/4) = g(x−a/4). Hence, f(x−a/4 + a/4) = g(x− a/4 + a/4) = g(x) and g(x) = f(x−(a−a)/4) follows. Put w= (a−a)/4. Clearly, if a≡a (mod 4), then w∈Z. Suppose that a6≡ a(mod 4). Using (2.5) we obtain R= 8b−3a2= 8b−3a2, S= 8c−4ab+a3= 8c−4ab+a3, which implies a2≡a2(mod 8) and a3≡a3(mod 4). Therefore, a2≡a2(mod 4), which yields, without loss of generality, that either a≡0 (mod 4), a≡2 (mod 4) or a≡1 (mod 4), a≡3 (mod 4). If a≡0 (mod 4), a≡2 (mod 4), then a2≡0 (mod 8), a2≡4 (mod 8), which is in contradiction to a2≡a2(mod 8). Similarly, if a≡1 (mod 4), a≡3 (mod 4), then a3≡1 (mod 4), a3≡3 (mod 4), which is in contradiction to a3≡a3(mod 4).  Let 0 6=D∈Zand let QD6=∅. For f(x), g(x)∈QDput f(x)∼g(x)⇐⇒ ∃ w∈Z:g(x) = f(x+w) = fw(x)⇐⇒ rf(x) = rg(x). It is evident that ∼is an equivalence relation on the set QD. Moreover, QD/∼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 K´alm´an Gy¨ory [8: p. 419]. See also [9: p. 475] or consult [3: p. 109]. 3. Connection between Mordell’s equation Y2=X3−21633D and the set QD Theorem 3.1 . Let 06=D∈Z. If Mordell’s equation Y2=X3+kwith k=−1769472D=−21633D(3.1) has no integer solution, then QD=∅. P r o o f. Let f(x) = x4+ax3+bx2+cx +d∈QDand let rf(x) = x4+Ax2+Bx +C∈Q[x]. Direct calculation will verify that (2.3) can be written in the form Drf=4 27(A2+ 12C)3−1 27(2A3−72AC + 27B2)2.(3.2) Substituting (2.4) into (3.2), after short calculation, we obtain Drf=1 1769472 (R2+ 3T)3−(R3−9RT + 108S2)2.(3.3) Put X=R2+ 3Tand Y=R3−9RT + 108S2.(3.4) Then X, Y ∈Zand (3.3) yields Y2=X3+kwhere k=−1769472Drf=−21633Drf. Since Drf=Df=D, the proof is complete.  38 QUARTIC POLYNOMIALS WITH A GIVEN DISCRIMINANT Remark 3.2 . If x4+ax3+bx2+cx+d∈QD, then (2.2) yields A, B, C ∈Z⇐⇒ 4|a. In this case, we can write (3.2) in the form V2= 4U3−27Drf, where U=A2+12Cand V= 2A3−72AC+27B2. Hence, we have (4V)2= (4U)3−432Drf.(3.5) Since Drf=D, the substitutions X= 4U,Y= 4Vreduce (3.5) to Y2=X3−432D. (3.6) 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 6=D∈Z, let MDdenote the set of all [X0, Y0], where X0, Y0∈Zand Y2 0=X3 0−21633D. Lemma 3.3 . Let 06=D∈Zand let [X0, Y0]∈MD. Then (i),(ii),(iii) and (iv) hold: (i) If 2|X0, then 4|X0,8|Y0. (ii) If 2|Y0, then 4|X0,8|Y0. (iii) If 3|X0, then 9|Y0. (iv) If 3|Y0, then 3|X0,9|Y0. P r o o f. The conclusions (i)–(iv) immediately follow from Y2 0=X3 0−21633D. They will be used in Section 4 and Section 5. 4. Method for determining the set QD The next lemma will be needed in the proof of Theorem 4.2. Lemma 4.1 . Let ξ0, η0, e ∈Zbe such that ξ0≡36e2(mod 96) and η0≡9eξ0−108e3(mod 1728).(4.1) Then we have: (i) ξ0≡0 (mod 12) and η0≡0 (mod 216). (ii) There exists exactly one e∈ {0,1,2,3}satisfying (4.1). P r o o f. (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∈Zsatisfy (4.1) for some e∈ {0,1,2,3}. Suppose that eis 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.  The following Theorem 4.2 provides the necessary and sufficient condition for QD6=∅. In addition, Theorem 4.2 makes it possible to determine a particular polynomial in QD. 39 JIˇ R´ I KLAˇ SKA Theorem 4.2 . Let 06=D∈Zand let MD6=∅. Then QD6=∅if and only if there exists an [X0, Y0]∈MDsuch that the elliptic equation η2=ξ3−108X0ξ+ 432Y0(4.2) has at least one integer solution [ξ0, η0]satisfying conditions (4.3)–(4.5) 36e2−ξ0≡0 (mod 96),(4.3) 108e3−9eξ0+η0≡0 (mod 1728),(4.4) 432e4−ξ2 0−72e2ξ0+ 16eη0+ 144X0≡0 (mod 110592).(4.5) for some e∈ {0,1,2,3}. In this case, g(x) = x4+ex3+36e2−ξ0 96 x2+108e3−9eξ0+η0 1728 x+432e4−ξ2 0−72e2ξ0+ 16eη0+ 144X0 110592 ∈QD and rg(x) = x4−ξ0 96x2+η0 1728x+144X0−ξ2 0 110592 . P r o o f. First, assume that QD6=∅. Then there exists an f(x) = x4+ax3+bx2+cx +d∈QD such that rf(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]∈MDsuch that R2+ 3T=X0 and R3−9RT + 108S2=Y0. Substituting 3T=X0−R2into R3−9RT + 108S2=Y0, we obtain 4R3−3X0R+ 108S2=Y0,(4.6) and multiplying (4.6) by 432, we get (216S)2= (−12R)3−108X0(−12R) + 432Y0.(4.7) Put ξ0=−12Rand 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∈Zand e∈ {0,1,2,3}such that a= 4w+e. Substituting a= 4w+einto the first equation of (2.5), we obtain R≡ −3e2(mod 8) and −12R≡36e2 (mod 96) follows. This together with ξ0=−12Ryields ξ0≡36e2(mod 96). Hence, (4.3). Further, from the second equation of (2.5), it follows 216S= 1728c−864ab + 216a3.(4.8) Putting a= 4w+e, 8b=R+ 3a2,ξ0=−12Rand η0= 216Sinto (4.8), we obtain η0= 1728(c−4w3−3ew2) + 36w(ξ0−36e2)+9eξ0−108e3.(4.9) Reducing (4.9) by modulus 1728 and using ξ0≡36e2(mod 96), we get (4.4). Finally, the third equation of (2.5) implies 432T= 110592d−27648ac + 6912a2b−1296a4.(4.10) For the left-hand side of (4.10), we have 432T= 144(X0−R2) = 144X0−ξ2 0and 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= 216Sinto 110592(d−w4−ew3)−64w(η0−9eξ0+ 108e3)−1152w2(36e2−ξ0)−16eη0+ 72e2ξ0−432e4. Since η0≡9eξ0−108e3(mod 1728) and ξ0≡36e2(mod 96), we get 144X0−ξ2 0≡ −16eη0+ 72e2ξ0−432e4(mod 110592). Hence, (4.5). 40 QUARTIC POLYNOMIALS WITH A GIVEN DISCRIMINANT Conversely, assume that there exists a [X0, Y0]∈MDsuch that equation (4.2) has an integer solution [ξ0, η0] satisfying (4.3)–(4.5) for some e∈ {0,1,2,3}. Put R=−ξ0 12 , S =η0 216, T =144X0−ξ2 0 432 .(4.11) 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 X0≡0 (mod 3) ⇐⇒ R≡0 (mod 3).(4.12) Let 3|X0. Then, by part (iii) of Lemma 3.3, we have 9|Y0. Further, by (4.11), we have 3|ξ0and 33|η0. Since η2 0=ξ3 0−108X0ξ0+ 432Y0, we also have 0 ≡η2 0≡ξ3 0(mod 35) and ξ0≡0 (mod 32) follows. This together with ξ0=−12Ryields 3|R. Let 3|R. Since ξ0=−12R, we have 32|ξ0and 36|ξ3 0follows. Next, by (4.11), 36|η3 0. Since η2 0=ξ3 0−108X0ξ0+ 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]∈MDit follows that Y2 0≡2 (mod 3), which is a contradiction. Combining this fact with (4.12), we get X0≡1 (mod 3) ⇐⇒ R≡1 (mod 3) or R≡2 (mod 3) ⇐⇒ R2≡1 (mod 3).(4.13) Clearly, in both cases (4.12) and (4.13), we have X0−R2≡0 (mod 3). Hence, T∈Z. Consider now the polynomial r(x) = x4+R 8x2+S 8x+T 256 ∈Q[x]. We prove that the discriminant Drof r(x) is equal to D. First, direct calculation verifies that Dr=(R2+ 3T)3−(R3−9RT + 108S2)2 21633. On the other hand, substituting ξ0=−12R,η0= 216Sinto η2 0=ξ3 0−108X0ξ0+ 432Y0, we obtain (216S)2= (−12R)3−108X0(−12R) + 432Y0. Hence, we get R3−3R(X0−R2)+108S2=Y0. Since, X0−R2= 3T, we have R3−9RT+108S2=Y0. This, together with Y2 0=X3 0−21633Dyields D=(R2+ 3T)3−(R3−9RT + 108S2)2 21633. Hence, Dr=D. Finally, let e∈ {0,1,2,3}satisfy (4.3)–(4.5). Then, by part (ii) of Lemma 4.1, eis uniquely determined. Put g(x) = r(x+e/4). Then we obtain after some calculation that g(x) = x4+ex3+R+ 3e2 8x2+eR + 2S+e3 16 x+2e2R+ 8eS +T+e4 256 =x4+ex3+36e2−ξ0 96 x2+108e3−9eξ0+η0 1728 x+432e4−ξ2 0−72e2ξ0+ 16eη0+ 144X0 110592 and rg(x) = r(x) = x4−ξ0 96x2+η0 1728x+144X0−ξ2 0 110592 . Since Dg=Drg=Dr=D, we have g(x)∈QD, as desired. The proof is complete.  41 JIˇ R´ I KLAˇ SKA Before proceeding, the following notations will be adopted. For any [X0, Y0]∈MD, let ED(X0, Y0) denote the set of all [ξ0, η0] where ξ0, η0∈Zand η2 0=ξ3 0−108X0ξ0+ 432Y0. Next, let EDdenote the set of all [X0, Y0, ξ0, η0, e] where [X0, Y0]∈MD, [ξ0, η0]∈ED(X0, Y0) and e∈ {0,1,2,3}satisfy (4.3)–(4.5). Corollary 4.3 . Let 06=D∈Zand let f(x) = x4+ax3+bx2+cx +d∈Z[x]. Then f(x)∈QD if and only if there exists [X0, Y0, ξ0, η0, e]∈EDand w∈Zsuch that a= 4w+e, b= 6w2+ 3ew +36e2−ξ0 96 , c= 4w3+ 3ew2+36e2−ξ0 48 w+108e3−9eξ0+η0 1728 , d=w4+ew3+36e2−ξ0 96 w2+108e3−9eξ0+η0 1728 w+432e4−ξ2 0−72e2ξ0+ 16eη0+ 144X0 110592 . Proposition 4.4 . Let 06=D∈Zand let MD6=∅. Then (i),(ii) and (iii) hold: (i) ED(X0, Y0)is a finite set for any [X0, Y0]∈MD. (ii) EDis a finite set. (iii) QD/∼has only finitely many equivalence classes for any QD6=∅. P r o o f. (i) Put α=−108X0and β=432Y0. Then 4α3+27β2= 2839(−X3 0+Y2 0) = −224312D6=0. Conclusion (i) now follows from Theorem 1.2. (ii) Conclusion (ii) is a direct consequence of Theorem 1.1 and part (i) of Proposition 4.4. (iii) Let ϕ:ED→QD/∼be the mapping defined by ϕ(X0, Y0, ξ0, η0, e)={fw(x); w∈Z}, where f0(x) = x4+ex3+36e2−ξ0 96 x2+108e3−9eξ0+η0 1728 x+432e4−ξ2 0−72e2ξ0+ 16eη0+ 144X0 110592 . Then ϕis bijective. Injectivity of ϕis evident and surjectivity of ϕimmediately follows from Corollary 4.3. Hence, #QD/∼= #ED. This proves (iii).  Remark 4.5 . Let [X0, Y0],[X∗ 0, Y ∗ 0]∈MDand let [X0, Y0]6= [X∗ 0, Y ∗ 0]. By an example we will prove that the set ED(X0, Y0)∩ED(X∗ 0, Y ∗ 0) 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 QD. It can be formally divided into five steps as follows: (i) Let 0 6=D∈Z. First we find the set MDof all integer solutions [X0, Y0] of Mordell’s equation Y2=X3−21633D. By Theorem 1.1, MDis a finite set and Theorem 3.1 states that, if MD=∅, then QD=∅. (ii) Let MD6=∅. Next we find, for any [X0, Y0]∈MD, the set ED(X0, Y0) of all integer solutions [ξ0, η0] of the elliptic equation η2=ξ3−108X0ξ+432Y0. By part (i) of Proposition 4.4, ED(X0, Y0) is a finite set for any [X0, Y0]∈MDand Theorem 4.2 says that, if ED(X0, Y0) = ∅for any [X0, Y0]∈MD, then QD=∅. (iii) In step (iii), we establish the set ED. By part (ii) of Proposition 4.4, EDis a finite set and Corollary 4.3 states that QD6=∅if and only if ED6=∅. (iv) Let ED6=∅and let #ED=n. In this step, we assign to each [X0, Y0, ξ0, η0, e]∈EDthe polynomial g(x) = x4+ex3+36e2−ξ0 96 x2+108e3−9eξ0+η0 1728 x+432e4−ξ2 0−72e2ξ0+ 16eη0+ 144X0 110592 . 42 QUARTIC POLYNOMIALS WITH A GIVEN DISCRIMINANT In this way, we obtain the full system of representatives GD={g1(x), . . . , gn(x)}of QD/∼. By part (iii) of Proposition 4.4, QD/∼is a finite set. (v) Finally, applying Corollary 2.3 to each gi(x)∈GD,i∈ {1, .. . , n}, we obtain the nsets {fi,w(x); w∈Z}where fi,w(x) = x4+g000 i(w) 3! x3+g00 i(w) 2! x2+g0 i(w) 1! x+gi(w). Hence, we get QD= n [ i=1 {fi,w(x); w∈Z}. The below example illustrates our method. Example 4.6 . Let D=−87. Then we have M−87 ={[−320,±11008],[−92,±12376],[448,±15616]}. Hence, E−87(−320,11008) = {[−80,±1216],[−48,±1728],[240,±5184],[384,±8640],[8592,±796608]}, E−87(−320,−11008) = ∅, E−87(−92,12376) = {[−156,0]}, E−87(−92,−12376) = {[156,0]}, E−87(448,15616) = {[−156,±3240],[96,±1728]}, E−87(448,−15616) = ∅. Further, we have E−87 =[−320,11008,240,5184,2],[−320,11008,240,−5184,2], [448,15616,−156,−3240,1],[448,15616,−156,3240,3]. Hence, it follows that #E−87 = #Q−87/∼= 4 and that G−87 ={g1(x), g2(x), g3(x), g4(x)}where g1(x) = x4+ 2x3−x2+x, g2(x) = x4+ 2x3−x2−5x−3, g3(x) = x4+x3+ 2x2−x, g4(x) = x4+ 3x3+ 5x2+ 6x+ 3. Finally, f1,w(x) = x4+ (4w+ 2)x3+ (6w2+ 6w−1)x2+ (4w3+ 6w2−2w+ 1)x+w4+ 2w3−w2+w, f2,w(x) = x4+ (4w+ 2)x3+ (6w2+ 6w−1)x2+(4w3+ 6w2−2w−5)x+w4+2w3−w2−5w−3, f3,w(x) = x4+ (4w+ 1)x3+ (6w2+ 3w+ 2)x2+ (4w3+ 3w2+ 4w−1)x+w4+w3+ 2w2−w, f4,w(x) = x4+ (4w+ 3)x3+(6w2+ 9w+ 5)x2+(4w3+9w2+10w+6)x+w4+3w3+5w2+6w+3, and Q−87 = 4 [ i=1 {fi,w(x); w∈Z}. Applying the method, the validity of Theorem 4.7 can be verified. 43 JIˇ R´ I KLAˇ SKA [13] KLAˇ SKA, J.—SKULA, L.: Law of inertia for the factorization of cubic polynomials – the case of discriminants divisible by three, Math. Slovaca 66(4) (2016), 1019–1027. [14] KLAˇ SKA, J.—SKULA, L.: Law of inertia for the factorization of cubic polynomials – the case of primes 2 and 3, Math. Slovaca 67(1) (2017), 71–82. [15] KLAˇ SKA, J.—SKULA, L.: On the factorizations of cubic polynomials with the same discriminant modulo a prime, Math. Slovaca 68(5) (2018), 987–1000. [16] LONDON, J.—FINKELSTEIN, M.: On Mordell’s Equation y2−k=x3, Bowling Green, Ohio Bowling Green State University, 1973. [17] MORDELL, L. J.: A statement by Fermat, Proc. Lond. Math. Soc. (2) 18 (1920), pp. v–vi. [18] SIEGEL, C. L.: ¨ Uber einige Anwendungen diophantischer Approximationen, Abh. Preuss Akad. Wiss., 1929, pp. 1–41. [19] SMART, N. P.: The Algorithmic Resolution of Diophantine Equations, Cambridge University Press, Cambridge, 1998. Received 16. 11. 2020 Accepted 25. 1. 2021 Institute of Mathematics Faculty of Mechanical Engineering Brno University of Technology Technick´a 2 616 69 Brno CZECH REPUBLIC E-mail: klask[email protected] 50