Quartic polynomials with a given discriminant

a o DOI: 10.1515/ms-2022-0003 Ma h. Slo aca 72 (2022), No. 1, 35–50 QUARTIC POLYNOMIALS WITH A GIVEN DISCRIMINANT Jiˇ ´ ı Klaˇ ska Dedica ed o he eminen Czechoslo ak ma hema ician Ladisla Skula (Communica ed by Milan Paˇs ´eka ) ABSTRACT. Le 0 6=D∈Zand le QDbe he se o all monic qua ic polynomials x4+ax3+bx2+ cx +d∈Z[x] wi h he disc iminan equal o D. In his pape we will de ise a me hod o de e mining he se QD. Ou me hod is s ongly ela ed o he heo y o in eg al poin s on ellip ic cu es. The well-known Mo dell’s equa ion plays an impo an ole as well in ou conside a ions. Finally, some new conjec u es will be included inspi ed by ex ensi e calcula ions on a compu e . c 2022 Ma hema ical Ins i u e Slo ak Academy o Sciences 1. In oduc ion Le 0 6=D∈Zand le QD={ (x) = x4+ax3+bx2+cx +d∈Z[x]; D =D}(1.1) whe e D =a2b2c2−4a2b3d−4a3c3+ 18a3bcd −27a4d2−4b3c2 + 16b4d+ 18abc3−80ab2cd −6a2c2d+ 144a2bd2 −27c4+ 144bc2d−128b2d2−192acd2+ 256d3 (1.2) is he disc iminan o (x). In his pape , he se QDwill be s udied in de ail. Mos o he ocus will be gi en o he p oblem o de e mining all polynomials in QD. Clea ly, his is equi alen o inding all in ege solu ions o he Diophan ine equa ion D =D. In p o ing he main esul s, he ollowing wo known heo ems will be needed. Theo em 1.1 (Mo dell, 1920) . Fo any gi en 06=k∈Z, he equa ion Y2=X3+k(1.3) has a mos ini ely many in ege solu ions. Equa ion (1.3) is o en called Mo dell’s equa ion, in honou o he con ibu ion Louis Joel Mo dell [17] has made o his subjec . An ex ension o Theo em 1.1 was la e made by Ca l Ludwig Siegel [18]. In i s simples o m, Siegel’s esul can be o mula ed as ollows: Theo em 1.2 (Siegel, 1929) . Le α, β ∈Zbe such ha 4α3+ 27β26= 0. Then he equa ion η2=ξ3+αξ +β(1.4) has a mos ini ely many in ege solu ions. 2020 Ma hema ics Subjec Classi ica ion: P ima y 11D25, 11D45, 11Y50. Keywo ds: Qua ic polynomial, disc iminan , Mo dell’s equa ion, ellip ic cu e. 35 JIˇ R´ I KLAˇ SKA The e is a s anda d me hod o compu ing all in ege solu ions o (1.3) and (1.4) using Da id’s bounds and la ice educ ion. This me hod can be ound, o example, in [19]. A p esen , his me hod is implemen ed in se e al compu e algeb a packages, including Magma and Pa i (Sage). Rema k 1.3 . Mo dell’s equa ion has had a long his o y. Fi s disco e ies conce ning (1.3) we e gi en in Dickson [2: pp. 533–539] going back o he wo k o Bache om 1621. Many in e es ing his o ical no es o (1.3) can be ound in [1,5,7,16]. Pe haps he mos ex ensi e his o ical commen s ela ed o Mo dell’s con ibu ion o (1.3) can be ound in he ecen pape [6]. Th oughou his pape , he ollowing no a ion will be adop ed. I Ais a ini e se , #Adeno es he numbe o elemen s o A. 2. Equi alence on he se QD Le (x) = x4+ax3+bx2+cx +d∈Z[x] and le D be he disc iminan o (x). Nex , le (x) = (x−a/4). Then (x) = x4+Ax2+Bx +C∈Q[x] (2.1) whe e A=b−3a2 8, B =c−ab 2+a3 8, C =d−ac 4+a2b 16 −3a4 256.(2.2) Mo eo e , we ha e D =D = 16A4C−4A3B2−27B4−128A2C2+ 144AB2C+ 256C3.(2.3) F om (2.2), i ollows ha he e exis R, S, T ∈Zsuch ha A=R 8, B =S 8, C =T 256,(2.4) whe e R= 8b−3a2, S = 8c−4ab +a3, T = 256d−64ac + 16a2b−3a4.(2.5) Hence, we can w i e (2.1) in he o m (x) = x4+R 8x2+S 8x+T 256 ∈Q[x] wi h R, S, T ∈Z.(2.6) We s a wi h a mo e gene al heo em. Theo em 2.1 . Le n∈N,n≥2,06=D∈Zand le (x) = xn+an−1xn−1+· · · +a1x+a0∈Z[x] be an a bi a y polynomial wi h he disc iminan equal o D. Fu he , o any w∈Z, le w(x) = n X k=0 (k)(w) k!xk,(2.7) whe e (k)(w)deno es he k- h de i a i e o (x)a w. Then w(x)∈Z[x]and all polynomials in { w(x); w∈Z}ha e he same disc iminan equal o D. 36 QUARTIC POLYNOMIALS WITH A GIVEN DISCRIMINANT P o o . Fi s , by induc ion on k, i can be p o ed ha k!| (k)(w) o any k∈ {0,1,2, . . . }. Hence, w(x)∈Z[x]. Fu he , Taylo ’s heo em yields (x) = n X k=0 (k)(w) k!(x−w)k o any w∈Z.(2.8) Le α1, . . . , αnbe he oo s o (x) in he se o complex numbe s C. Then D= n−1 Y i=1 n Y j=i+1 (αj−αi)2. Nex , by (2.8), o any α∈ {α1, . . . , αn}, we ha e (α) = n X k=0 (k)(w) k!(α−w)k= 0.(2.9) Combining (2.7) wi h (2.9), we ge w(α−w) = 0, and hus, β1=α1−w, . . . , βn=αn−w(2.10) a e he oo s o w(x) in C. Using (2.10), we now ge D w= 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 desi ed.  Rema k 2.2 . Obse e ha , in Theo em 2.1, (x) = 0(x). Hence, (x)∈ { w(x); w∈Z}. Co olla y 2.3 . Le 06=D∈Zand le (x) = x4+ax3+bx2+cx +d∈QD. Fu he , o any w∈Z, le w(x) = x4+ 000(w) 3! x3+ 00(w) 2! x2+ 0(w) 1! x+ (w).(2.11) Then (i) and (ii) hold: (i) QDis an in ini e se and { w(x); w∈Z} ⊆ QD. (ii) Fo any w∈Z, we ha e w(x) = (x) = x4+Ax2+Bx +C∈Q[x], whe e A, B, C sa is y (2.2). P o o . Pa (i) o Co olla y 2.3 is a di ec consequence o Theo em 2.1 o n= 4. Pa (ii) can be e i ied by di ec calcula ion.  Lemma 2.4 . Le 06=D∈Zand le (x), g(x)∈QD. Then (i),(ii) and (iii) a e equi alen : (i) The e exis s w∈Zsa is ying g(x) = (x+w). (ii) The e exis s w∈Zsa is ying g(x) = w(x). (iii) (x) = g(x). P o o . Le (x) = x4+ax3+bx2+cx +d,g(x) = x4+ax3+bx2+cx +d∈QD. Fi s we show ha (i) is equi alen o (ii). Using Taylo ’s heo em, we ob ain (x) = (x−w)4+ 000(w) 3! (x−w)3+ 00(w) 2! (x−w)2+ 0(w) 1! (x−w) + (w) o any w∈Z. The e o e, (x+w) = x4+ 000(w) 3! x3+ 00(w) 2! x2+ 0(w) 1! x+ (w).(2.12) 37 JIˇ R´ I KLAˇ SKA Combining (2.12) wi h (2.11), we ge (x+w) = w(x). Hence, (i) and (ii) a e equi alen . Fu he we p o e ha (i) is equi alen o (iii). Assume ha g(x) = (x+w) o 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, g(x) = g(x−(4w+a)/4) = g(x−w−a/4) = (x−w−a/4 + w) = (x−a/4) = (x). Finally, le (x) = g(x). Then (x−a/4) = g(x−a/4). Hence, (x−a/4 + a/4) = g(x− a/4 + a/4) = g(x) and g(x) = (x−(a−a)/4) ollows. Pu w= (a−a)/4. Clea ly, i a≡a (mod 4), hen w∈Z. Suppose ha a6≡ a(mod 4). Using (2.5) we ob ain R= 8b−3a2= 8b−3a2, S= 8c−4ab+a3= 8c−4ab+a3, which implies a2≡a2(mod 8) and a3≡a3(mod 4). The e o e, a2≡a2(mod 4), which yields, wi hou loss o gene ali y, ha ei he a≡0 (mod 4), a≡2 (mod 4) o a≡1 (mod 4), a≡3 (mod 4). I a≡0 (mod 4), a≡2 (mod 4), hen a2≡0 (mod 8), a2≡4 (mod 8), which is in con adic ion o a2≡a2(mod 8). Simila ly, i a≡1 (mod 4), a≡3 (mod 4), hen a3≡1 (mod 4), a3≡3 (mod 4), which is in con adic ion o a3≡a3(mod 4).  Le 0 6=D∈Zand le QD6=∅. Fo (x), g(x)∈QDpu (x)∼g(x)⇐⇒ ∃ w∈Z:g(x) = (x+w) = w(x)⇐⇒ (x) = g(x). I is e iden ha ∼is an equi alence ela ion on he se QD. Mo eo e , QD/∼has only ini ely many equi alence classes. In Sec ion 4, his ac will be p o ed using he esul s o Mo dell and Siegel p esen ed in Theo em 1.1 and Theo em 1.2. On he o he hand, his claim also ollows as a consequence o a mo e gene al heo em ha has been p o ed by K´alm´an Gy¨o y [8: p. 419]. See also [9: p. 475] o consul [3: p. 109]. 3. Connec ion be ween Mo dell’s equa ion Y2=X3−21633D and he se QD Theo em 3.1 . Le 06=D∈Z. I Mo dell’s equa ion Y2=X3+kwi h k=−1769472D=−21633D(3.1) has no in ege solu ion, hen QD=∅. P o o . Le (x) = x4+ax3+bx2+cx +d∈QDand le (x) = x4+Ax2+Bx +C∈Q[x]. Di ec calcula ion will e i y ha (2.3) can be w i en in he o m D =4 27(A2+ 12C)3−1 27(2A3−72AC + 27B2)2.(3.2) Subs i u ing (2.4) in o (3.2), a e sho calcula ion, we ob ain D =1 1769472 (R2+ 3T)3−(R3−9RT + 108S2)2.(3.3) Pu X=R2+ 3Tand Y=R3−9RT + 108S2.(3.4) Then X, Y ∈Zand (3.3) yields Y2=X3+kwhe e k=−1769472D =−21633D . Since D =D =D, he p oo is comple e.  38 QUARTIC POLYNOMIALS WITH A GIVEN DISCRIMINANT Rema k 3.2 . I x4+ax3+bx2+cx+d∈QD, hen (2.2) yields A, B, C ∈Z⇐⇒ 4|a. In his case, we can w i e (3.2) in he o m V2= 4U3−27D , whe e U=A2+12Cand V= 2A3−72AC+27B2. Hence, we ha e (4V)2= (4U)3−432D .(3.5) Since D =D, he subs i u ions X= 4U,Y= 4V educe (3.5) o Y2=X3−432D. (3.6) I is in e es ing ha Mo dell’s equa ion (3.6) plays a undamen al ole also in he heo y o cubic polynomials wi h he same disc iminan D. Consul [10: p. 313]. The ollowing no a ion will be use ul. Fo an a bi a y 0 6=D∈Z, le MDdeno e he se o all [X0, Y0], whe e X0, Y0∈Zand Y2 0=X3 0−21633D. Lemma 3.3 . Le 06=D∈Zand le [X0, Y0]∈MD. Then (i),(ii),(iii) and (i ) hold: (i) I 2|X0, hen 4|X0,8|Y0. (ii) I 2|Y0, hen 4|X0,8|Y0. (iii) I 3|X0, hen 9|Y0. (i ) I 3|Y0, hen 3|X0,9|Y0. P o o . The conclusions (i)–(i ) immedia ely ollow om Y2 0=X3 0−21633D. They will be used in Sec ion 4 and Sec ion 5. 4. Me hod o de e mining he se QD The nex lemma will be needed in he p oo o Theo em 4.2. Lemma 4.1 . Le ξ0, η0, e ∈Zbe such ha ξ0≡36e2(mod 96) and η0≡9eξ0−108e3(mod 1728).(4.1) Then we ha e: (i) ξ0≡0 (mod 12) and η0≡0 (mod 216). (ii) The e exis s exac ly one e∈ {0,1,2,3}sa is ying (4.1). P o o . (i) Since he alidi y o he cong uence ξ0≡0 (mod 12) is e iden , we only p o e ha η0≡0 (mod 216). Fi s , obse e ha η0≡9eξ0−108e3(mod 216). Fu he , ξ0≡36e2(mod 96) is equi alen o 9ξ0≡324e2(mod 864). Hence, 9ξ0≡108e2(mod 216). This, oge he wi h η0≡9eξ0−108e3(mod 216), yields η0≡0 (mod 216). (ii) Le ξ0, η0∈Zsa is y (4.1) o some e∈ {0,1,2,3}. Suppose ha eis no unique. Then i ollows om ξ0≡36e2(mod 96) ha e∈ {1,3}and ha ξ0≡36 (mod 96). On he o he hand, using η0≡9eξ0−108e3(mod 1728), we ob ain 9ξ0−108 ≡27ξ0−2916 (mod 1728), which yields ξ0≡60 (mod 96), a con adic ion.  The ollowing Theo em 4.2 p o ides he necessa y and su icien condi ion o QD6=∅. In addi ion, Theo em 4.2 makes i possible o de e mine a pa icula polynomial in QD. 39 JIˇ R´ I KLAˇ SKA Theo em 4.2 . Le 06=D∈Zand le MD6=∅. Then QD6=∅i and only i he e exis s an [X0, Y0]∈MDsuch ha he ellip ic equa ion η2=ξ3−108X0ξ+ 432Y0(4.2) has a leas one in ege solu ion [ξ0, η0]sa is ying condi ions (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) o some e∈ {0,1,2,3}. In his case, g(x) = x4+ex3+36e2−ξ0 96 x2+108e3−9eξ0+η0 1728 x+432e4−ξ2 0−72e2ξ0+ 16eη0+ 144X0 110592 ∈QD and g(x) = x4−ξ0 96x2+η0 1728x+144X0−ξ2 0 110592 . P o o . Fi s , assume ha QD6=∅. Then he e exis s an (x) = x4+ax3+bx2+cx +d∈QD such ha (x) = x4+(R/8)x2+(S/8)x+T/256 ∈Q[x] whe e R, S, T a e in ege s sa is ying (2.5). Fu he , om Theo em 3.1 i ollows ha he e exis s a [X0, Y0]∈MDsuch ha R2+ 3T=X0 and R3−9RT + 108S2=Y0. Subs i u ing 3T=X0−R2in o R3−9RT + 108S2=Y0, we ob ain 4R3−3X0R+ 108S2=Y0,(4.6) and mul iplying (4.6) by 432, we ge (216S)2= (−12R)3−108X0(−12R) + 432Y0.(4.7) Pu ξ0=−12Rand η0= 216S. Now, (4.7) implies immedia ely ha [ξ0, η0] is an in ege solu ion o (4.2). Finally, we ha e o p o e ha [ξ0, η0] sa is ies (4.3)–(4.5) o some e∈ {0,1,2,3}. Since a∈Z, he e exis uniquely de e mined w∈Zand e∈ {0,1,2,3}such ha a= 4w+e. Subs i u ing a= 4w+ein o he i s equa ion o (2.5), we ob ain R≡ −3e2(mod 8) and −12R≡36e2 (mod 96) ollows. This oge he wi h ξ0=−12Ryields ξ0≡36e2(mod 96). Hence, (4.3). Fu he , om he second equa ion o (2.5), i ollows 216S= 1728c−864ab + 216a3.(4.8) Pu ing a= 4w+e, 8b=R+ 3a2,ξ0=−12Rand η0= 216Sin o (4.8), we ob ain 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 ge (4.4). Finally, he hi d equa ion o (2.5) implies 432T= 110592d−27648ac + 6912a2b−1296a4.(4.10) Fo he le -hand side o (4.10), we ha e 432T= 144(X0−R2) = 144X0−ξ2 0and he igh -hand side o (4.10) can be ew i en, subs i u ing a= 4w+e, 8b=R+3a2, 8c=S+4ab−a3,ξ0=−12R and η0= 216Sin o 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 ge 144X0−ξ2 0≡ −16eη0+ 72e2ξ0−432e4(mod 110592). Hence, (4.5). 40 QUARTIC POLYNOMIALS WITH A GIVEN DISCRIMINANT Con e sely, assume ha he e exis s a [X0, Y0]∈MDsuch ha equa ion (4.2) has an in ege solu ion [ξ0, η0] sa is ying (4.3)–(4.5) o some e∈ {0,1,2,3}. Pu R=−ξ0 12 , S =η0 216, T =144X0−ξ2 0 432 .(4.11) Then, by pa (i) o Lemma 4.1, we ha e R, S ∈Z. We now p o e ha T∈Z. F om he i s and hi d equa ion in (4.11) we ob ain T= (X0−R2)/3. Fi s we show ha X0≡0 (mod 3) ⇐⇒ R≡0 (mod 3).(4.12) Le 3|X0. Then, by pa (iii) o Lemma 3.3, we ha e 9|Y0. Fu he , by (4.11), we ha e 3|ξ0and 33|η0. Since η2 0=ξ3 0−108X0ξ0+ 432Y0, we also ha e 0 ≡η2 0≡ξ3 0(mod 35) and ξ0≡0 (mod 32) ollows. This oge he wi h ξ0=−12Ryields 3|R. Le 3|R. Since ξ0=−12R, we ha e 32|ξ0and 36|ξ3 0 ollows. Nex , by (4.11), 36|η3 0. Since η2 0=ξ3 0−108X0ξ0+ 432Y0, we ha e 432Y0≡0 (mod 35), and Y0≡0 (mod 32) ollows. By pa (i ) o Lemma 3.3, we ge 3|X0. This p o es (4.12). Fu he , suppose ha X0≡2 (mod 3). Then om [X0, Y0]∈MDi ollows ha Y2 0≡2 (mod 3), which is a con adic ion. Combining his ac wi h (4.12), we ge X0≡1 (mod 3) ⇐⇒ R≡1 (mod 3) o R≡2 (mod 3) ⇐⇒ R2≡1 (mod 3).(4.13) Clea ly, in bo h cases (4.12) and (4.13), we ha e X0−R2≡0 (mod 3). Hence, T∈Z. Conside now he polynomial (x) = x4+R 8x2+S 8x+T 256 ∈Q[x]. We p o e ha he disc iminan D o (x) is equal o D. Fi s , di ec calcula ion e i ies ha D =(R2+ 3T)3−(R3−9RT + 108S2)2 21633. On he o he hand, subs i u ing ξ0=−12R,η0= 216Sin o η2 0=ξ3 0−108X0ξ0+ 432Y0, we ob ain (216S)2= (−12R)3−108X0(−12R) + 432Y0. Hence, we ge R3−3R(X0−R2)+108S2=Y0. Since, X0−R2= 3T, we ha e R3−9RT+108S2=Y0. This, oge he wi h Y2 0=X3 0−21633Dyields D=(R2+ 3T)3−(R3−9RT + 108S2)2 21633. Hence, D =D. Finally, le e∈ {0,1,2,3}sa is y (4.3)–(4.5). Then, by pa (ii) o Lemma 4.1, eis uniquely de e mined. Pu g(x) = (x+e/4). Then we ob ain a e some calcula ion ha 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 g(x) = (x) = x4−ξ0 96x2+η0 1728x+144X0−ξ2 0 110592 . Since Dg=D g=D =D, we ha e g(x)∈QD, as desi ed. The p oo is comple e.  41 JIˇ R´ I KLAˇ SKA Be o e p oceeding, he ollowing no a ions will be adop ed. Fo any [X0, Y0]∈MD, le ED(X0, Y0) deno e he se o all [ξ0, η0] whe e ξ0, η0∈Zand η2 0=ξ3 0−108X0ξ0+ 432Y0. Nex , le EDdeno e he se o all [X0, Y0, ξ0, η0, e] whe e [X0, Y0]∈MD, [ξ0, η0]∈ED(X0, Y0) and e∈ {0,1,2,3}sa is y (4.3)–(4.5). Co olla y 4.3 . Le 06=D∈Zand le (x) = x4+ax3+bx2+cx +d∈Z[x]. Then (x)∈QD i and only i he e exis s [X0, Y0, ξ0, η0, e]∈EDand w∈Zsuch ha 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 . P oposi ion 4.4 . Le 06=D∈Zand le MD6=∅. Then (i),(ii) and (iii) hold: (i) ED(X0, Y0)is a ini e se o any [X0, Y0]∈MD. (ii) EDis a ini e se . (iii) QD/∼has only ini ely many equi alence classes o any QD6=∅. P o o . (i) Pu α=−108X0and β=432Y0. Then 4α3+27β2= 2839(−X3 0+Y2 0) = −224312D6=0. Conclusion (i) now ollows om Theo em 1.2. (ii) Conclusion (ii) is a di ec consequence o Theo em 1.1 and pa (i) o P oposi ion 4.4. (iii) Le ϕ:ED→QD/∼be he mapping de ined by ϕ(X0, Y0, ξ0, η0, e)={ w(x); w∈Z}, whe e 0(x) = x4+ex3+36e2−ξ0 96 x2+108e3−9eξ0+η0 1728 x+432e4−ξ2 0−72e2ξ0+ 16eη0+ 144X0 110592 . Then ϕis bijec i e. Injec i i y o ϕis e iden and su jec i i y o ϕimmedia ely ollows om Co olla y 4.3. Hence, #QD/∼= #ED. This p o es (iii).  Rema k 4.5 . Le [X0, Y0],[X∗ 0, Y ∗ 0]∈MDand le [X0, Y0]6= [X∗ 0, Y ∗ 0]. By an example we will p o e ha he se ED(X0, Y0)∩ED(X∗ 0, Y ∗ 0) can be nonemp y. Fo D=−23, we ha e [64,6400],[−320,−2816] ∈M−23 and [96,±1728] ∈E−23(64,6400) ∩E−23(−320,−2816). Now we a e eady o o mula e he me hod o de e mining he se QD. I can be o mally di ided in o i e s eps as ollows: (i) Le 0 6=D∈Z. Fi s we ind he se MDo all in ege solu ions [X0, Y0] o Mo dell’s equa ion Y2=X3−21633D. By Theo em 1.1, MDis a ini e se and Theo em 3.1 s a es ha , i MD=∅, hen QD=∅. (ii) Le MD6=∅. Nex we ind, o any [X0, Y0]∈MD, he se ED(X0, Y0) o all in ege solu ions [ξ0, η0] o he ellip ic equa ion η2=ξ3−108X0ξ+432Y0. By pa (i) o P oposi ion 4.4, ED(X0, Y0) is a ini e se o any [X0, Y0]∈MDand Theo em 4.2 says ha , i ED(X0, Y0) = ∅ o any [X0, Y0]∈MD, hen QD=∅. (iii) In s ep (iii), we es ablish he se ED. By pa (ii) o P oposi ion 4.4, EDis a ini e se and Co olla y 4.3 s a es ha QD6=∅i and only i ED6=∅. (i ) Le ED6=∅and le #ED=n. In his s ep, we assign o each [X0, Y0, ξ0, η0, e]∈ED he 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 his way, we ob ain he ull sys em o ep esen a i es GD={g1(x), . . . , gn(x)}o QD/∼. By pa (iii) o P oposi ion 4.4, QD/∼is a ini e se . ( ) Finally, applying Co olla y 2.3 o each gi(x)∈GD,i∈ {1, .. . , n}, we ob ain he nse s { i,w(x); w∈Z}whe e i,w(x) = x4+g000 i(w) 3! x3+g00 i(w) 2! x2+g0 i(w) 1! x+gi(w). Hence, we ge QD= n [ i=1 { i,w(x); w∈Z}. The below example illus a es ou me hod. Example 4.6 . Le D=−87. Then we ha e 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) = ∅. Fu he , we ha e E−87 =[−320,11008,240,5184,2],[−320,11008,240,−5184,2], [448,15616,−156,−3240,1],[448,15616,−156,3240,3]. Hence, i ollows ha #E−87 = #Q−87/∼= 4 and ha G−87 ={g1(x), g2(x), g3(x), g4(x)}whe e 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, 1,w(x) = x4+ (4w+ 2)x3+ (6w2+ 6w−1)x2+ (4w3+ 6w2−2w+ 1)x+w4+ 2w3−w2+w, 2,w(x) = x4+ (4w+ 2)x3+ (6w2+ 6w−1)x2+(4w3+ 6w2−2w−5)x+w4+2w3−w2−5w−3, 3,w(x) = x4+ (4w+ 1)x3+ (6w2+ 3w+ 2)x2+ (4w3+ 3w2+ 4w−1)x+w4+w3+ 2w2−w, 4,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 { i,w(x); w∈Z}. Applying he me hod, he alidi y o Theo em 4.7 can be e i ied. 43 JIˇ R´ I KLAˇ SKA [13] KLAˇ SKA, J.—SKULA, L.: Law o ine ia o he ac o iza ion o cubic polynomials – he case o disc iminan s di isible by h ee, Ma h. Slo aca 66(4) (2016), 1019–1027. [14] KLAˇ SKA, J.—SKULA, L.: Law o ine ia o he ac o iza ion o cubic polynomials – he case o p imes 2 and 3, Ma h. Slo aca 67(1) (2017), 71–82. [15] KLAˇ SKA, J.—SKULA, L.: On he ac o iza ions o cubic polynomials wi h he same disc iminan modulo a p ime, Ma h. Slo aca 68(5) (2018), 987–1000. [16] LONDON, J.—FINKELSTEIN, M.: On Mo dell’s Equa ion y2−k=x3, Bowling G een, Ohio Bowling G een S a e Uni e si y, 1973. [17] MORDELL, L. J.: A s a emen by Fe ma , P oc. Lond. Ma h. Soc. (2) 18 (1920), pp. – i. [18] SIEGEL, C. L.: ¨ Ube einige Anwendungen diophan ische App oxima ionen, Abh. P euss Akad. Wiss., 1929, pp. 1–41. [19] SMART, N. P.: The Algo i hmic Resolu ion o Diophan ine Equa ions, Camb idge Uni e si y P ess, Cam- b idge, 1998. Recei ed 16. 11. 2020 Accep ed 25. 1. 2021 Ins i u e o Ma hema ics Facul y o Mechanical Enginee ing B no Uni e si y o Technology Technick´a 2 616 69 B no CZECH REPUBLIC E-mail: klask[email p o ec ed] 50