On elements with index of the form 2 a 3 b in a parametric family of biquadratic elds

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1 On elemens wh ndex of he form a 3 b n a paramerc famly of bquadrac elds Bora JadrevĆ Absrac In hs paper we gve some resuls abou prmve negral elemens p(c p n he famly of bcyclc bquadrac elds L c = Q ) c; (c + 4) c whch have ndex of he form () = a 3 b and coprme coordnaes n gven negral bases. recsely, we show ha f c and s an elemen wh ndex () = a 3 b c +, hen s an elemen wh mnmal ndex () = (L c) =. We also show ha for every neger C 0 3 we can nd e ecvely compuable consans M 0 (C 0) and N 0 (C 0) such ha f c C 0, han here are no elemens wh ndex of he form () = a 3 b ; where a > M (C 0) or b > N (C 0) : Inroducon Le be a prmve negral elemen of an algebrac number eld K of degree n wh rng of negers O K. Then ndex of s de ned as ndex of subgroup Z [] + n group O + K () = O + K : Z []+ ; where O + K and Z []+ denoe he addve groups of correspondng rngs. The mnmal ndex (K) of he eld K we de ne as he mnmum of he ndces of all prmve negers n he eld K: The eld ndex m (K) s he greaes common dvsor of ndces also aen for all prmve negers of K. Le f;! ; :::;! n g be an arbrary negral bass of K: Then dscrmnan of correspondng lnear form L (X) = X +! X + ::: +! n X n can be rewren as D K=Q (L (X)) = (I (X ; :::; X n )) D K ; where D K s dscrmnan of he eld K and I (X ; :::; X n ) s a homogeneous polynomal n n varables of degree n (n ) = wh raonal neger coe - cens. The polynomal I (X ; :::; X n ) s called he ndex form assocaed o he Mahemacs Subec Class caon. rmary: D57, A55, J86; Secondary: J68, Y50. Key words. ndex form equaons, mnmal ndex, oally real bcyclc bquadrac elds, smulaneous ellan equaons, p-adc case The auhor was suppored by Mnsry of Scence, Educaon and Spors, Republc of Croaa, gran

2 negral bass f;! ; :::;! n g. If he prmve negral elemen s represened n ha negral bass as = x +x! +:::+x n! n ; x ; x ; :::x n Z; hen he ndex of s us () = I (x ; :::; x n ) : Hence, he problem of deermnng elemens of gven ndex N can be reduced o he solvng ndex form equaons I (x ; :::; x n ) = n x ; :::; x n Z: Bcyclc bquadrac elds are quarc elds of he ype Q ( p m; p n), where m; n are dsnc square-free raonal negers. These elds were consdered several auhors. M. N. Gras, and F. Tanoe [9] have found necessary and su cen condons for bquadrac elds beng monogenc. I. Gaál, A. eh½o and M. ohs [7] gave an algorhm for deermnng he mnmal ndex and all elemens wh mnmal ndex n he oally real case usng he negral bass descrbed by K. S. Wllams [4]. G. Nyul [] class ed all monogene oally complex bquadrac elds and gave explcly all generaors of power negral bases n hem. In [0] and [] he auhor has deermned he mnmal ndex and all elemens wh mnmal ndex for hree n ne famles of oally real bcyclc bquadrac elds. Furher, I. Gaál and G. Nyul [8] provded an e cen algorhm for deermnng elemens of ndex dvsble by xed prmes n bquadrac number elds. In [] he auhor proved followng heorem. Theorem Le c 3 be an neger such ha c or 3 (mod 6) and c; c ; c + 4 are square-free negers. Then p p L c = Q c (c ); c (c + 4) () s a oally real bcyclc bquadrac eld and ) s eld ndex s m (L c ) = for all c; ) he mnmal ndex of L c s (L c ) = f c 7 and (L c ) = f c = 3; ) all negral elemens wh mnmal ndex are gven by p p p (c ) (c + 4) + (c ) c + p c (c + 4) x +x (c ) (c + 4)+x3 +x 4 ; () where x Z, (x ; x 3 ; x 4 ) = (0; ; ) ; (0; ; ) ; (; ; ) ; (; ; ) f c 7 and (x ; x 3 ; x 4 ) = ( ; ; 0) ; (0; ; 0) f c = 3: Snce he mnmal ndex of he eld () s of he form a 3 b, we wonder f here exs prmve negral elemens wh he ndex () of hs form excep hose wh he mnmal ndex. I su ces o observe elemens of he form () wh gcd (x ; x 3 ; x 4 ) = snce he ndex form I (x ; x 3 ; x 4 ) s a homogeneous polynomal of degree 6. For a (paral) answer on hs queson we need some addonal condons: an upper bound for he ndex () or an upper bound for he parameer c: The man resuls of he presen paper are gven n he followng heorems:

3 Theorem Le c 3 be an neger such ha c or 3 (mod 6) and c; c ; c + 4 are square-free negers. If s a prmve negral elemen of he eld () gven by (), where x ; x ; x 3 ; x 4 Z wh gcd (x ; x 3 ; x 4 ) = and ndex of s of he form () = a 3 b, where a 0; b 0 are negers, hen he followng holds: ) If c and () c +, hen () = : Furhermore, f c < and (), hen () = f c 6= 3 and () = or () = f c = 3. ) All elemens wh () = are gven by x Z, (x ; x 3 ; x 4 ) = (0; ; ) ; (0; ; ) ; (; ; ) ; (; ; ) excep when c = 3; n whch case we have furher soluons x Z, (x ; x 3 ; x 4 ) = (5; 9; ) ; (5; 9; ) ; (4; 9; ) ; (4; 9; ) : If c = 3, hen all elemens wh () = are gven by x Z, (x ; x 3 ; x 4 ) = ( ; ; 0) ; (0; ; 0). Theorem 3 For every neger C 0 3 we can nd e ecvely compuable consans M 0 (C 0 ) and N 0 (C 0 ) such ha f c C 0, hen here are no prmve negral elemens of he eld () gven by () wh gcd (x ; x 3 ; x 4 ) = and wh ndex of he form () = a 3 b where a > M 0 (C 0 ) or b > N 0 (C 0 ) : Drecly from Theorem and Theorem 3 we oban: Corollary For a gven parameer c; le := (c) denoe a correspondng prmve negral elemen of he eld () gven by () wh gcd (x ; x 3 ; x 4 ) =. Then he followng holds: ) Le a 0 and b 0 be arbrary bu xed negers such ha a 3 b > : If here exs a parameer c and an elemen (c) wh an ndex ( (c)) = a 3 b, hen c a 3 b : ) Le c > 3 be arbrary bu xed neger. If here exs an elemen (c) wh ndex of he form ( (c)) = a 3 b ; hen eher ( (c)) = or we can nd e ecvely compuable consans M 0 (c) and N 0 (c) such ha c + ( (c)) M0(c) 3 N0(c) : Remar For he parcular value of c here s an e cen algorhm for deermnng all elemens wh an ndex of he form a 3 b (see Secon 4) based on a more general algorhm gven by I. Gaál and G. Nyul []. relmnares Noe ha he eld () s oally real bcyclc bquadrac eld under he assumpon c, c, c + 4 are posve square-free, parwse relavely prme negers. Snce we use a mehod of I. Gaál, A. eh½o and M. ohs [7], we have o observe he congruence behavor of c; c ; c + 4 modulo 4. Hence, f c; c ; c + 4 are posve square-free negers, hen c 3 and c or 3 (mod 4) : Noe ha c; c ; c+4 are parwse relavely prme negers f and only f c or 3(mod 6): 3

4 Therefore, we observe cases when c 3, c ; 3; 7; 9 (mod ) and c; c ; c + 4 are square-free negers. Furhermore, n [, Secon 4] was shown, by usng he resul from [5], ha here are n nely many negers c wh he above properes whch agan mples ha here are n nely many oally real bcyclc bquadrac elds of he form (). Also, n [, Secon 4], by usng a mehod of I. Gaál, A. eh½o and M. ohs [7], we showed ha ndng all elemens wh gven ndex s equvalen o ndng all solvable sysems of he form (c ) U cv = F (3) (c ) Z (c + 4) V = F (4) cz (c + 4) U = 4F 3 ; (5) where F F F 3 = : Then all negral elemens wh ndex equal o are gven by () where U = x + x 3 ; V = x 4 ; Z = x 3 ; (6) and (U; V; Z) s passng hrough all soluons of all solvable sysems of he form (3), (4) and (5) wh F F F 3 = : Furhermore, snce he equaons (3), (4) and (5) are no ndependen, he relaon c (F ) (c + 4) (F ) = (c ) (4F 3 ) (7) holds. Therefore, f we wan o nd all negral elemens of he form () wh gcd (x ; x 3 ; x 4 ) = and wh ndex () = a 3 b ; hen we have o nd all solvable sysems of he form (3), (4) and (5) wh F F F 3 = a 3 b and all soluons (U; V; Z) of hese sysems whch are a form of (6), where gcd (x ; x 3 ; x 4 ) =. We noe here ha we have wo d eren approaches o hs problem, dependng on wheher we have gven an upper bound for he ndex () (as n Theorem ), or an upper bound for he parameer c (as n Theorem 3). If we have an upper bound for he ndex () ; hen we rs consder he sysem (3) and (5). Namely, we use he heory of connued fracons o deermne all possble small values of he rgh hand sde of (3) and (5) so ha he sysem of hese wo equaons has soluons. Afer ha, from equaon (7) by drec esng, we nd all possble rples (F ; F ; F 3 ) such ha F F F 3 s of he form a 3 b : If we have an upper bound on he parameer c, hen we rs consder he equaon (7). Snce, n our case F ; = ; ; 3 are of he form F = 3, we oban a S-un equaon over Z. Because we have an upper bound C 0 on he parameer c, we are able o nd upper bounds for he exponens a and b usng p adc esmaes and hose upper bounds depend only on C 0 : Snce our esmaes provdng large upper bounds for he exponens, we can dmnsh he upper bounds usng reducon procedure. Unforunaely, reducon procedure can be used only for parcular values of he parameer c; so reduced upper bounds can no be expressed as a funcon of C 0 : 4

5 3 Addonal condon: upper bound for he ndex We recall ha f we wan o nd all prmve negral elemens of he eld () wh ndex () = a 3 b, we have o nd all solvable sysems of he form (3), (4) and (5), where F F F 3 = a 3 b. Suppose ha (U; V; Z) s an neger soluon of he sysem (3), (4) and (5), where c 3, c ; 3; 7; 9 (mod ) and c; c ; c + 4 are square-free negers. If one of he negers U; V; Z s equal o zero, hen (3), (4) and (5) mply ha oher wo negers are no equal o zero. Furher, f U = 0, hen equaon (3) mples ha c dvdes F whch agan mples c = 3 snce F s of he form 3 and c 3 s an odd square-free posve neger. Smlarly, we nd f V = 0, hen equaon (4) mples c = 3: Furhermore, we oban ha here s no c whch sas es he equaon (4) f Z = 0: Therefore, f c > 3, hen s su cen o observe only soluons (U; V; Z) n posve negers. Le c 3 and le () K; where K s a posve neger. Furher, le (U; V; Z) be a soluon n posve negers of he sysem of ellan equaons (3) and (5). Snce () = F F F 3 K; hen F K and F 3 K: Now, from (3) and (5), we oban r c c U V = c c U V r c c + U V < F (c ) V r c c K p c (c )V (8) and r c + 4 c Z U = c + 4 c < 4F r 3 cu Z U r c Z c U c c + 4 4K p c (c + 4)U : (9) Noe ha for c > 3 s enough o assume K ; snce n ha case, by Theorem, mnmal ndex s equal o. 3. Case c > 3 Le c > 3: Addonally, suppose K c + : Noe, ha hs condon mples c ; and snce we have c ; 3; 7; 9 (mod ) ; hen c 3: Under hese condons, from (8) and (9), we oban ha all soluons (U; V; Z) n posve negers of he sysem of ellan equaons (3) and (5), sasfyng r c c U V < K p c + p < c (c )V c (c )V V (0) 5

6 and r c + 4 c Z U < 4K 4 (c + ) p p < 4 c (c + 4)U c (c + 4)U U : () Smlarly as n [, Secon 4.], we wll use heory of connued fracons o deermne all possble values of F and 4F 3 such ha equaons (3) and (5) have soluons n relavely prme negers. recsely, snce he nequales (0) and () are sas ed, we can apply Theorem (Worley [5], Duella [3]) and [4, Lemma ] (see also [, Theorem 3 and Lemma ]). We nd ha under above condons,.e. f F c + ; where c 3 and f equaon (3) has soluons n relavely prme negers U and V, hen F S (c) = f ; c; c g : Snce F s of he form F = 3 ; where 0; 0 are negers and snce c ; 3; 7; 9 (mod ), hen he only possbly s F =. Smlarly, f F 3 c + ; where c 3 and equaon (5) has soluons n relavely prme negers U and Z, hen 4F 3 S 3 (c) = f 4; ; 4c; 4c 9; c 9; c ; c 4; 3c 4; 3c 6g ; f c > 9: Addonally, we have where 4F 3 S 3 (c) [ S3 0 (c) f c = 9; 4F 3 S 3 (c) [ S3 0 (c) [ S3 00 (c) f c = 5; 4F 3 S 3 (c) [ S3 0 (c) [ S3 00 (c) [ S3 000 (c) f c = 3; S3 0 (c) = f6c 5; 5c 6g ; S3 00 (c) = fc ; 4c 69; 5c 96; 3c 44g ; S3 000 (c) = f6c 5; 8c 49; 0c 8; 7c 36; 9c 64; c 00g : Snce F 3 s of he form F 3 = 3, where 0; 0 are negers and snce c ; 3; 7; 9 (mod ) ; hen he only possbly s 4F 3 = 4 for all c 3: Now, suppose ha (U; V; Z) s a soluon of he sysem of ellan equaons (3) and (5) n posve negers. Le gcd (U; V ) = d and gcd (U; Z) = g: If F = 3 c + and F 3 = 3 c + ; where c 3; hen F = d and 4F 3 = 4g whch mples r c + d < c and g p c + < c : Le U = du = gu, V = dv and Z = gz. Then gcd (U ; V ) = ; gcd (U ; Z ) = and followng equaons are hold (c ) U cv = cz (c + 4) U = 4: 6

7 By [, Lemma 3], all such U are gven recurrenly n he followng way u 0 = ; u = c ; u m+ = (c ) u m+ u m ; m 0; () and all such U are gven recurrenly by v 0 = ; v = c + ; v n+ = (c + ) v n+ v n ; n 0: (3) Snce U = du = gu ; hen here exs nonnegave negers m and n such ha U = du m = gv n ; where u m and v n are de ned by () and (3), respecvely. By [, Lemma 4], for all m; n 0; we have u m ( ) m (m(m + )c ) (mod 4c ); v n n(n + ) c + (mod c ). Therefore, f du m = gv n ; hen du m gv n (mod c ) whch mples ( ) m d g(mod c). Snce 0 < d < c and 0 < g < c, we have d = g;.e. U = U : Thus, we oban a sysem of smulaneous ellan equaons (c ) U cv = ; cz (c + 4) U = 4: In [, Theorem 4] we nd ha for c 7 only soluons o hs sysem are (U ; V ; Z ) = (; ; ). Therefore, all soluons o he correspondng sysem of q ellan equaons (3) and (5) (wh F = d, 4F 3 = 4d and c+ d ) are of he form (U; V; Z) = (d; d; d) : If gcd (x ; x 3 ; x 4 ) =, hen (6) mples gcd (U; V; Z) = or. Therefore, we have d = or : ) If d = ; hen F = ; 4F 3 = 4, and from (7) we oban F = 6; whch mples () = F F F 3 = 6 = : Therefore, has he mnmal ndex,.e. () = (L c ) =. ) If d =, hen from (6), we oban x 4 = ; x + x 3 = ; x 3 = ; whch mples (x ; x 3 ; x 4 ) = (0; ; ) ; (0; ; ) ; (; ; ) ; (; ; ), a conradcon wh gcd (x ; x 3 ; x 4 ) = : When 3 < c < (whch mples c = 7); we ae K =. Snce, n hs case, he mnmal ndex of he eld L c s equal o ; hen () mples () = (L c ) = : For each c > 3; by Theorem, all elemens wh mnmal ndex () = (L c ) = are gven by (x ; x 3 ; x 4 ) = (0; ; ) ; (0; ; ) ; (; ; ) ; (; ; ) : To complee he proof of Theorem remans o consder he case c = 3 and K =. 7

8 3. Case c = 3 and K = Le c = 3 and () = F F F 3 K = : Le (U; V; Z) be a soluon n posve negers of he sysem (3) and (5) wh c = 3: Snce () = F F F 3 ; hen F and F 3 : Hence, from (8) and (9), we oban p 3 U V < F p p < 7 3V 3V V (4) and r 7 3 Z U < 4F 3 3U r p U < U ; (5) respecvely. If gcd (U; V ) = and F, hen from (4) by Theorem (Worley [5], Duella [3]) and [4, Lemma ], we oban F S 0 = f; ; 3; 6; g : Knowng ha, n our case, F s of he form F = 3 ; hen he only possbles are F = ; ; 3; 6: Smlarly, f gcd (U; Z) = and F 3 hen from (5) ; we oban 4F 3 S 3 = f ; 3; 4; 5; 7; ; 5; 7; 0; ; 5; 8; 35; 37; 4; 43; 47g : Snce we have F 3 = 3 ; hen he only possbles are 4F 3 = 4;,.e. F 3 = ; 3. Addonally, for c = 3; equaon (7) has form 3 (F ) 7 (F ) = 4F 3 ; (6) and snce F F F 3 ; we oban ha here are only wo possbles: ) (F ; F ; F 3 ) = (; ; ) whch mples () = : Therefore, () s equal o mnmal ndex (L 3 ) = : Now, by Theorem, all such are gven by (x ; x 3 ; x 4 ) = ( ; ; 0) ; (0; ; 0) : ) (F ; F ; F 3 ) = ( ; 6; ) whch mples () = : The correspondng sysem s U 3V = (7) 3Z 7U = 4: (8) Smlarly as n [, Secon 4.], we nd ha he only soluons o he sysem (7) and (8) are (U; V; Z) = (; ; ) and (U; V; Z) = (9; ; 9) : Snce negers U; V; Z are of he form gven n (6), where gcd (x ; x 3 ; x 4 ) =, all wh ndex () = are gven by (x ; x 3 ; x 4 ) = (0; ; ), (0; ; ), (; ; ), (; ; ), (5; 9; ), (5; 9; ), (4; 9; ), (4; 9; ) : 8

9 Noe ha he above resuls we oban by assumng (U; V; Z) s a soluon n posve negers o he sysem (3), (4) and (5) wh c = 3. I remans o observe he cases when (U; V; Z) s soluon n nonnegave negers wh U = 0 or V = 0: If c = 3 and V = 0; hen (3) and (4) mply U = F ; Z = F ; where U 6= 0 and Z 6= 0: Therefore, we have F F = U Z, whch mples (U; V; Z) = (; 0; ) ; (; 0; ) ; (; 0; 3) ; (; 0; ) ; (3; 0; ) : Snce F = U, F = Z and F F F 3 = a 3 b, from equaon (6), we oban ha he only possbly s (F ; F ; F 3 ) = (; ; ) and hs rple we have already obaned. If U = 0; hen (3) and (5) mply 3V = F ; 3Z = 4F 3 ; where V 6= 0 and Z 6= 0: Therefore, we have F F 3 = 9 4 V Z and Z s an even neger. Ths mples (U; V; Z) = (0; ; ) : Snce F = 3V = 3, 4F 3 = 3Z = ; from equaon (6), we nd F = 3: Therefore, we oban a rple (F ; F ; F 3 ) = ( 3; 3; 3) whch does no sasfy he condon F F F 3 : Therefore, we nshed he proof of Theorem. 4 Addonal condon: upper bound for he parameer c In hs secon we show ha f have an upper bound on he parameer c, hen we can nd an upper bound for he ndex: We wll follow he mehod of I. Gaal and G. Nyul gven n [8]. We bre y sech he man seps of our procedure. We sar wh equaon (7). Snce, n our case, unnowns F n (7); are of he form F = 3, = ; ; 3; we oban a S-un equaon over Z. In order o nd all elemens wh ndex () = F F F 3 = a 3 b ; we have o nd all prmve soluon of equaon (7) and all possbles for common facor A 3 B of F ; F ; F 3 (see Secon 4.). To nd all possbles for he exponens n he common facor A 3 B, we need o deermne how raonal prmes and 3 spl n hree dsnc quadrac sub elds of he quarc eld L c : We show ha he exponens A and B aan only very small values (see Secon 4.). If have an upper bound C 0 on he parameer c, hen we are able o nd an upper bound for he exponens n prmve soluons usng p adc lnear form esmaes and ha upper bound depends only on C 0 (see Secon 4.3): Snce our esmaes gvng large upper bounds for he exponens, we can dmnsh hose upper bounds usng reducon procedure (see Secon 4.4). Unforunaely, reduced upper bounds can no be expressed as a funcon of C 0 snce reducon procedure can be used only for parcular value of he parameer c: 4. S-un equaon Le be a prmve negral elemen of he form () wh gcd (x ; x 3 ; x 4 ) = and le ndex of be () = a 3 b where a 0 and b 0 are arbrary bu xed negers. We have shown ha () s of he form () = F F F 3, where 9

10 F ; F ; F 3 sasfy relaon (7). Snce () = a 3 b mples F = 3 ; 0 a; 0 b; = ; ; 3; hen (F ; F ; F 3 ) s soluon of he S-un equaon (7) over Z. We wll nd all prmve soluons (f ; f ; f 3 ) of (7) n posve negers (hose wh gcd (f ; f ; f 3 ) = ): Then all soluons of (7) are of he form F = f A 3 B ; = ; ; 3; where A 3 B = gcd (F ; F ; F 3 ) = : Se f = d 3 e ; = ; ; 3: Then equaon c (f ) (c + 4) (f ) = (c ) (4f 3 ) can be rewren n he form (c + 4) d 3 e c d 3 e = (c ) d3+ 3 e3 : (9) Noe ha, snce c 3, c ; 3; 7; 9 (mod ) and c; c ; c + 4 are squarefree negers, we have ord (c + 4) = ord (c) = ord (c ) = ord 3 (c + 4) = ord 3 (c ) = 0 and ord 3 (c) = where c = 3 c, 3 c and = 0 or. Now, f he equaon (9) we smplfy wh possble common facors and 3; we oban equaon (c + 4) d0 3 e 0 c d0 3 e 0 = (c ) d e 0 3 ; (0) where a mos one of he d 0 ; d 0 ; d 0 3 and a mos one of he e 0 ; e 0 ; e 0 s posve. Afer deermned d 0 ; d 0 ; d 0 3; e 0 ; e 0 ; e 0 3, values of f = d 3 e ; = ; ; 3; we can oban usng he followng: If (d 0 ; d 0 ; d 0 3) = (d 0 ; 0; 0), hen (d ; d ; d 3 ) = (d 0 + ; ; 0) ; If (d 0 ; d 0 ; d 0 3) = (0; d 0 ; 0), hen (d ; d ; d 3 ) = (; d 0 + ; 0) ; If (d 0 ; d 0 ; d 0 3) = (0; 0; ), hen (d ; d ; d 3 ) = (; ; 0) ; If (d 0 ; d 0 ; d 0 3) = (0; 0; d 0 3) ; d 0 3, hen (d ; d ; d 3 ) = (0; 0; d 0 3 ) ; If c ; 7 (mod ), hen (e ; e ; e 3 ) = (e 0 ; e 0 ; e 0 3) ; If c 3; 9 (mod ) and (e 0 ; e 0 ; e 0 3) = (e 0 ; 0; 0) ; hen (e ; e ; e 3 ) = (e 0 + ; 0; ) ; (e 0 ; e 0 ; e 0 3) = (0; 0; e 0 3), hen (e ; e ; e 3 ) = (; 0; e ) ; (e 0 ; e 0 ; e 0 3) = (0; e 0 ; 0) ; e 0, hen (0; e ; 0) = (0; e 0 ; 0) : I s easy o see ha exponens n (0) canno all be equal o zero. Furhermore, f c > 3; han we have: f here exs such ha d 0 6= 0; hen here exs such ha e 0 6= 0; and vce versa. Also, 6= mus hold: If c = 3; hen we have he followng excepons: (d 0 ; d 0 ; d 0 3; e 0 ; e 0 ; e 0 3) = (0; 0; 3; 0; 0; 0) ; (0; 3; 0; 0; 0; 0) ; (0; 0; ; 0; 0; ) ; (0; ; 0; 0; ; 0) : Therefore, from now on, we wll assume ha exacly one d 0 s posve and exacly one e0 s posve, where 6= : 0

11 4. gcd (F ; F ; F 3 ) calculaons In order o nd an upper bound for he exponens n = gcd (F ; F ; F 3 ) = A 3 B we wll follow of a mehod of I. Gaal and G. Nyul [8, Secon 6]. Frs, we need o deermne how raonal prmes and 3 spls n hree dsnc quadrac sub elds of he quarc eld L c = Q ( p m; p n) ; namely n he elds M = Q ( p n) ; M = Q ( p m), M 3 = Q p m n, where m = c + 4; n = c; m = (c + 4) (c ) ; n = c (c ) : Usng for example [, p. 45], we nd he facorzaon of he prncpal deals h and h3 no prme deals of rngs of negers O M, = ; ; 3 as follows: In he rng O M ; we have In he rng O M ; we oban h = ; + p n = h3 = 3; p n = ; f c 3; 9 (mod ) h3 = 3 ; f c ; 7 (mod ) : h = ; + p m = 4 ; h3 = 3; a + p m p 3; a m = 5 5 ; where a m(mod 3); snce x m(mod 3) s solvable. In he rng O M3 ; we have h = 6 ; h3 = h3; p m n = 7 ; f c 3; 9 (mod ) ; h3 = 8 ; f c ; 7 (mod ) : Usng prmve soluons f, = ; ; 3 of (7) we can rewre sysem (3), (4) and (5) as (cv ) nu = s () ((c + 4) V ) mz = s () (cz) m n U = s 3 (3) wh s = cf, s = (c + 4) f ; s 3 = 4cf 3 and = A 3 B. Le (U; V; Z) arbrary bu xed soluon of sysem (), () and (3). Then, followng [8, Secon 6], we se ' ' D D p p p c n cv nu cv + nu 0 3 p p p c + 4 m (c + 4) V mz (c + 4) V + mz 3 p 3 c m n cz p m n U cz+ p m n U 0 and by [8, Lemma 3] we oban he followng: Ideal h = 6 s prme n he rng of negers O M3 of he eld M 3. Snce c c + 4 (mod ), we have ord 6 (c) = and ord 6 p c (c + 4) =, and by [8, Lemma 3, ()], we oban A max nord 6 (c) ; ord 6 p o c (c + 4) +D 3 = max f; g+ = 4;

12 Le c ; 7 (mod ) : In hs case deal h3 = 3 s prme n he rng of negers O M of he eld M : Snce c (mod 3) and c (c ) (mod 3) we oban ord 3 (c) = 0 and ord 3 p c (c ) = 0, and by [8, Lemma 3, ()], we have B max nord 3 (c) ; ord 3 p o c (c ) + D = 0;.e. B = 0; Le c 3; 9 (mod ) : In hs case we have h3 = D 3; p c (c )E = n he rng of negers O M of he eld M : Snce we have 3 c and 3 - (c ), hen ord (c) = and ord p c (c ) =, and by [8, Lemma 3, ()], we oban B max nord (c) ; ord p o c (c ) + D = max f; g + 0 = : Hence, we oban: If c ; 7 (mod ) ; han gcd (F ; F ; F 3 ) = A ; where 0 A 4; If c 3; 9 (mod ) ; han gcd (F ; F ; F 3 ) = A 3 B ; where 0 A 4 and 0 B. 4.3 Upper bound for he exponens Le us denoe = c + 4; = c and 3 = c ; where c = c f c ; 7 (mod ) and c = c=3 f c 3; 9 (mod ) : Then ord ( l ) = ord 3 ( l ) = 0 for all l = ; ; 3: We assumed ha n (0) exacly one d 0 s posve, exacly one e0 s posve and 6=. Therefore, le d 0 6= 0 and e0 6= 0; where 6= : Then, from (0), for dsnc negers ; ; f; ; 3g ; we oban d 0 = ord 3 e0 3 e0 e 0 = ord 3 d0 d0 = ord 3 e0 = ord 3 d0 In all above cases we have expressons of he form ord p b ; (4) where b s posve neger and ; Q. We wll apply esmaes of Y. Bugeaud and M. Lauren [, Corollare ] on (4). Le us denoe by h () he absolue logarhmc hegh of he algebrac number ; gven by h (0) = 0 and by h () = a d log dq max f; l g l=

13 f a (x ) ::: (x d ) s he mnmal polynomal of 6= 0 over Z: We have h () = log ; h (3) = log 3; and h = log Q max ; = log (max f ; g), l= where = or : Therefore, f = or = ; hen h = log (c + 4) : If = c c or c c ; hen h = log (c ) ; f c 3; 9 (mod ) ; log c = log c; f c ; 7 (mod ) : Usng he noaons from [], we have K = Q ; = Q ; 3 f = and D = [K:Q] f =. Le A > and A > be real numbers such ha max h ( ) ; log p log A ; = ; : D In our case we have max h ( ) ; log p = log 3; D max h ( ) ; log p = max flog max f ; g ; log pg log (c + 4) ; D so we can ae log A = log 3 and log A = log (c + 4) : Now, we have 48 log 3 (log ) 4 b 0 = b D log A + b D log A = b log (c + 4) + log 3 : = Q; where b = d 0 f p = 3 and b = e 0 f p = : Hence, f c 3; hen [, Corolllare ] mples d 0 max log + log log + 0:4; 0 log (c + 4) and 48 log 3 (log ) 4 e 0 36 (log 3) 3 36 (log 3) 3 max log max log max log e 0 log (c + 4) + e 0 log 7 + log 3 d 0 log (c + 4) + d 0 log 7 + log 3 log 3 + log log + 0:4; 0 log (c + 4) (5) + log log 3 + 0:4; 0 log 3 log (c + 4) log 3 + log log 3 + 0:4; 0 log 3 log (c + 4) : (6) 3

14 Le 3 c C 0 and T (c) = max fd 0 ; d 0 ; d 0 3; e 0 ; e 0 ; e 0 3g = max d 0 ; e0 : If T (c) = d 0 ; hen from (5) we have ( e 0 d 0 8: 447 log 845 log (C 0 + 4) ; f e log log + 0:4 log (C0 + 4) ; f e : e 0 log 7 + log 3 If T (c) = e 0 ; hen from (6) we oban d 0 e 0 ( 7:6 log d 0 log 7 + log 3 (7) 376:87 log (C 0 + 4) ; f d log log 3 + 0:4 log (C0 + 4) f d : (8) Therefore, f 3 c C 0, hen T (c) T 0 (C 0 ) ; where T 0 (C 0 ) we can oban from nequales (7) and (8). Noe, for calculang T 0 (C 0 ) s enough o consder nequaly x 8: 447 log x log 7 + log 3 + log log + 0:4 log (C 0 + 4) : (9) Indeed, f nequaly (9) mples x K 0 ; hen T (c) K 0 ; so we ae T 0 (C 0 ) = K 0 : For example, f c C 0 = 00; hen from nequaly (9) we oban T (c) T 0 (00) = 35: Smlarly, we nd T = and T 0 (3) = 4534: Snce esmaes of Y. Bugeaud and M. Lauren [, Corollare ] gve a large upper bound for he exponens n (0), we can dmnsh ha upper bound usng [6, Lemma 4.]. Noe ha (4) s easy o conver no expressons wh p adc logarhms. Namely, by [3, Lemma II.9], we have d 0 = or and snce d 0 = ord 3 e0 = ord log + e 0 log 3 e 0 = ord d0 = ord 3 log 3 + d 0 log 3 3 e0 s adc un n and = ord = ord 3 e0 d0 (30) (3) d0 s 3 adc un n 3 for all dsnc negers ; ; f; ; 3g. By repeang he p adc reducon procedure gven n [6, Lemma 4.] for lnear forms n p adc logarhms from (30) and (3) as long as he reduced bounds are less han he orgnal one, for each c, 3 c C 0 ; we can oban d 0 := d 0 (c) M () R (c) and e0 := e 0 (c) N () R (c) ; 4

15 where M () () R (c) and N R (c) are he bes possble bounds for d0 and e0, respecvely. Denoe M R (C 0 ) = max M () R (c) and N R (C 0 ) = max N () R (c) ; f;;3g, cc 0 f;;3g, cc 0 (where four exceponal cases for c = 3 gven n Secon 4. are also ncluded). Then d 0 M R (C 0 ) T 0 (C 0 ) and e 0 N R (C 0 ) T 0 (C 0 ) ; for all 3 c C 0 and all ; f; ; 3g : The resuls from Secon 4. now mply ha he values of d ; d ; d 3 ; e ; e ; e are also bounded whch agan mples, ogeher wh he resuls from Secon 4., ha values of he F = d+a 3 e+b ; = ; ; 3 are bounded oo. recsely, we oban ord (F F F 3 ) = and 3X d +3A (M R (C 0 ) + 4)+34 = M R (C 0 )+6 = M 0 (C 0 ) ; = ord 3 (F F F 3 ) = 3X e + 3B N R (C 0 ) = N 0 (C 0 ) ; f c ; 7 (mod ) ; (N R (C 0 ) + ) + 3 = N R (C 0 ) + 8 = N 0 (C 0 ) ; f c 3; 9 (mod ) : Noe ha reducon procedure can be used only for parcular values of he parameer c, so, unforunaely, reduced upper bounds M R (C 0 ) and N R (C 0 ) are no e ecvely compuable consans f C 0 s oo large. Therefore, f we pu e ecvely compuable consan T 0 (C 0 ) nsead M R (C 0 ) and N R (C 0 ) n above formulas for M 0 (C 0 ) and N 0 (C 0 ), we have proved Theorem. 4.4 The reducon procedure Le 3 c C 0. Suppose = T (c) = max fd 0 ; d 0 ; d 0 3; e 0 ; e 0 ; e 0 3g = max d 0 ; e 0 T 0 (C 0 ) : where d 0 6= 0, e0 6= 0 and 6= : We consder lnear forms = log + e 0 log 3 and 3 = log 3 + d 0 log 3 ; (3) where ; ; are dsnc negers from he se f; ; 3g and = c+4; = c, 3 = c : We can dmnsh he upper bound T 0 (C 0 ) applyng [6, Lemma 4.] on lnear forms n (3). Usng he noaons from [6, Lemma 4.] we have n = ; p = or 3; X = max fx ; x g = max ; e 0 = e 0 T (c) T 0 (C 0 ) = X 0 f p = ; X = max fx ; x g = max ; d 0 = d 0 T (c) T 0 (C 0 ) = X 0 f p = 3; 5

16 Then we have # = log ; # = log 3 f p = ; # = log 3 ; # = log 3 f p = 3: ord ( ) = d 0 e 0 = 0 + e 0 f T (c) = d 0 ; ord 3 ( 3 ) = e 0 d 0 = 0 + d 0 f T (c) = e 0 : Therefore, consans c and c from he [6, Lemma 4.] are gven by (c ; c ) = (0; ) : Snce ord log ord (log 3) = ; ord 3 log 3 ord 3 (log 3 ) = ; for all dsnc negers ; ; from he se f; ; 3g (see below), hen, followng [6, Lemma 4.], we de ne 0 0 = log 3 log A + e 0 = (# ; ) + e 0 log 3, f e 0 and p = ; 0 3 = 3 log 3 = log 3 log 3 (33) A + d 0 = (# ; ) + d 0 ; f d 0 and p = 3: (34) For 0 < p Z and # ; p le # (p) ; be a unque raonal neger wh ord (# ; # (p) ; ) p and 0 # (p) ; p p ; where = f p = and = f p = 3. Denoe by p he lace spanned by he columns of he marx " # Le and 0 # (p) ; p p N () 0 (c) = + ord (log 3) 0 = + M () 0 (c) = 3 + ord 3 (log 3 ) 0 = 3 : Denoe by b (p) he rs vecor of he LLL reduced bass of p. If b (p) p > n T0 (C 0 ) = T 0 (C 0 ) ; hen, by [6, Lemma 4.] we have: : 6

17 - If T (c) = d 0 here s no e0 () wh N 0 (c) e 0 T 0 (C 0 ) : - If T (c) = e 0 here s no d0 wh M () 0 (c) d 0 T 0 (C 0 ) : Usng hese new bounds, he reducon can be repeaed, as long as he new bound for e 0 or d0 s less han he prevous one. Fnally, we oban: - If T (c) = d 0 ; hen e0 N () () R (c) ; where N R (c) s he bes possble bound for e 0 and all possble values for d0 ; e0 we oban from equaon d0 = 3 e 0 ; where e 0 N () R (c) and d0 e0 : - Smlarly, f T (c) = e 0, hen d0 M () () R (c) ; where M R (c) s he bes possble bound for d 0 and all possble values for d0 ; e0, we oban from equaon 3 e0 = d0 ; wh d 0 M () R (c) and e0 d0. In order o nd all possble elemens wh an ndex of he form a 3 b for gven c, 3 c C 0 ; we have o perform above reducon procedure for each of sx possble couples d 0 ; e0 ; where 6=. Havng deermned all possble sexuples (d 0 ; d 0 ; d 0 3; e 0 ; e 0 ; e 0 3), we have o nd all possble sexuples (d ; d ; d 3 ; e ; e ; e 3 ) (a connecon beween hem s gven n Secon 4.) whch gve us all possble prmve rples (f ; f ; f 3 ), where f = d 3 e ; = ; ; 3: Afer ha, all soluons of equaon (7) whch are of he form (f ; f ; f 3 ), we oban usng drec esng. Now all requred rples (F ; F ; F 3 ) are of he form F = f A 3 B, = ; ; 3; where 0 A 4 and B = 0 f c ; 7 (mod ) or 0 B f c 3; 9 (mod ) : For each explc value of he rple (F ; F ; F 3 ) we have o solve a correspondng sysem (3), (4) and (5). Every soluon (U; V; Z) of ha sysem whch s of he form (6), where gcd (x ; x 3 ; x 4 ) =, deermnes an negral elemen of he form () wh ndex () = F F F 3. For 3 c C 0 ; we can gve soluons n he form (c; () ; ) = c; a 3 b ; x ; x 3 ; x 4 : Compung ord p log p and # (p) ; From a de non of p adc logarhm follows ha, n our case, s enough o nd c log p = log p c + 4 ; log c c p and log c p c + 4 for p = and p = 3 snce log p = log p and log p Also, we have ord p log p = ord p log p : = log p : 7

18 Usng presenaon of he p adc logarhm as Taylor seres, rs we nd log 3 = log 3 = log 3 = log 3 ( ( 3)) = ( ) n 3n+ n + ; ( ) n 3 n+ (n + ) ; whch mples ord (log 3) = and ord 3 (log 3 ) =, respecvely. Smlarly, we nd all log p for p = and p = 3 from whch follows ha ord log ord (log 3) = ; ord 3 log 3 ord 3 (log 3 ) = : for all = c c+4, c c, c c+4 : Therefore, snce we follow [6, Lemma 4.], he p adc negers # ; and # ; n (33) and (34) have o be de ned as: # ; = log f p = and # ; = log 3 log 3 f p = 3: Addonally, usng he presenaon of he p each := # ; we can rewre n he form a n = = ; b n log 3 adc logarhm as Taylor seres, where ord p ( ) 0 and ord p ( ) = 0: Then, for every 0 < Z; we can nd su cenly large negers n and n such ha 0 = n a n and 0 = n b n sasfy () 0 (mod p ) ; where () s a unque raonal neger wh 0 ord p ( () ) and 0 () p : We oban he followng () log n log 3 where n = and () log 3 n log 3 (mod ) (mod 3 ) n n n a n ( ) n 3n n+ n a n ( ) n 3 n (n+) (mod ) ; (mod 3 ) ; 8

19 where n = gven n he followng wo ables: 3 f 3, and n = f = ;. The values of n a n are ) Case c ; 7 (mod ) : Noe, n hs case we have c = c: () (mod p ) 0 l n = n (l) 0 = n log c c log 3 (mod c ) ord l l+ n log ( c c+4 ) log 3 (mod c+ ) ord l l+ n log ( c c+4) log 3 (mod ) - n l log c 3 c log 3 (mod 3 l ; f l > 0 c ) ord 3 3 3; f l = 0; 3 n ; f l = 0, = ; log 3 ( c c+4 ) log 3 (mod 3 ) - log 3 ( c c+4) log 3 (mod 3 c+ ) ord 3 3 3; f 3 ; f = ; l l ; f l > 0 3; f l = 0; 3 ; f l = 0, = ; ) Case c 3; 9 (mod ) : Noe, n hs case we have c = c 3 : n n a n 3 ( c ) n+ (n+)c n+ ( c+ ) n+ (n+)(c+4) + (n+)(c+4) n+ + 3 n ( c 3 ) n+ (n+)c + n+ 3 n (n+)(c+4) n n ( c+ 3 ) n+ (n+)(c+4) + () (mod p ) 0 l n = n (l) 0 = n log c c log 3 (mod c 3 ) ord l l+ n log ( c c+4 ) log 3 (mod c+ ) ord l l+ log ( c c+4) log 3 (mod c+3 ) ord log c 3 c log 3 (mod 3 c 3 ) ord 3 3 log 3 ( c c+4 ) log 3 (mod 3 ) - log 3 ( c c+4) log 3 (mod 3 ) ord (c+3)(c+6) 3 9 If l > ; hen () (mod p ) 0: l 3 l+4 l l ; f l > 0 3; f l = 0; 3 ; f l = 0, = ; 3; f 3 ; f = ; l l ; f l 3; f l = ; 3 ; f l =, = ; n Tang he approprae values of 0 gven n he ables above, we can calculae he values of # (p) ; a each sep of he reducon procedure. 9 a n ( ) n ( c 3 ) n+ (n+)c n+ 3 3 n+ ( c+ ) n+ (n+)(c+4) + 4+ (c+6) n+ ( c+3 ) n+ 3 + (n+)(c+4) + n ( ) n n+ 3 n ( c 3 3 ) n+ (n+)c n+ n n+ 3 n (n+)(c+4) n+ n ( ) n 3+ ( (c+3)(c+6) 9 ) n+ 3(n+)(c+4) +

20 References [] S. Alaca, K. S. Wllams: Inroducory Algebrac Number Theory, Cambrdge Unv. ress, 004. [] Y. Bugeaud, M. Lauren, Mnoraon e ecve de la dsance p-adque enre pussances de nombres algébrques, J. Number Theory () 6 (996), [3] A. Duella, Connued fracons and RSA wh small secre exponen, Tara M. Mah. ubl. 9 (004), 0. [4] A. Duella and B. Jadrevć, A famly of quarc Thue nequales, Aca Arh. (00), [5]. Erdös, Arhmecal properes of polynomals. J. London Mah. Soc. 8 (953), [6] I. Gaál, I.Járás and F. Luca, A remar on prme dvsors of lenghs of sdes of Heron rangles, Expermen. Mah., (003), [7] I. Gaál, A. eh½o and M. ohs, On he resoluon of ndex form equaons n bquadrac number elds, III The bcyclc bquadrac case, J. Number Theory. 53 (995), [8] I. Gaál and G. Nyul, Index form equaons n bquadrac elds: he p adc case, ubl. Mah. Debrecen, 68 (006), 5-4. [9] M.N. Gras, and F. Tanoe, Corps bquadrac monogenes, Manuscrpa Mah., 86, (995), [0] B. Jadrevć, Esablshng he mnmal ndex n a paramerc famly of bcyclc bquadrac elds, erod. Mah. Hungar., 58 (009); [] B. Jadrevć, Solvng ndex form equaons n he wo paramerc famles of bquadrac elds, Mah. Commun. 4 (009), ; [] G. Nyul, ower negral bases n oally complex bquadrac number elds, Aca Acad. aed. Agrenss, Sec. Mah., 8 (00), [3] N.. Smar: The Algorhmc Resoluon of Dophanne Equaons, Cambrdge Unv. ress, 998. [4] K.S. Wllams, Inegers of bquadracs elds, Canad. Mah. Bull. 3 (970), [5] R. T. Worley, Esmang p=q, J. Ausral. Mah. Soc. 3 (98), Deparmen of Mahemacs, Faculy of Scence, Unversy of Spl, Teslna, 000 Spl, Croaa E-mal address: bora@pmfs.hr 0