Regular Rational Diophantine Sextuples


 Margery Ward
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1 Regulr Rtionl Diophntine Sextuples Philip E Gis A polynomil eqution in six vriles is given tht generlises the definition of regulr rtionl Diophntine triples, qudruples nd quintuples to regulr rtionl Diophntine sextuples. The definition n e used to extend rtionl Diophntine quintuple to wek rtionl Diophntine sextuple. In some ses regulr sextuple is full rtionl Diophntine sextuple. Ten exmples of this re provided. Introdution A rtionl Diophntine mtuple is set of m positive rtionl numers { 1,, m } suh tht the produt of ny two is one less thn rtionl numer squred. i j + 1 = x ij 2, i j, i, x ij Q The prolem of finding suh mtuples ws originlly introdued in the third entury AD y Diophntus of Alexndri who ws le to find triples nd qudruples of suh numers [1]. Diophntus ws interested in solving vriety of lgeri prolems in rtionl numers. In this se it is not ler why he hoose suh n esoteri prolem without ny nturl motivtion ut it hs turned out to e rih sujet onneting Fioni numers [2] ellipti urves [3] nd lgeri invrints [4] while providing mny onjetures, generlistions, nd of ourse some results. During the renissne Pierre de Fermt reinvented Diophntine numer theory s the serh for solutions in integers rther thn rtionls nd provided the sequene 1,3,8,120 s the first Diophntine qudruple in positive integers [5]. It ws not until the twentieth entury tht Bker nd Dvenport showed tht no fifth integer n e dded to Fermt s sequene to mke Diophntine quintuple [6]. However there re mny suh Diophntine qudruples nd it is n outstnding prolem to determine whether ny suh Diophntine quintuple exists. It is now known tht no Diophntine sextuple exists in integers nd there is ound on the numer of possile quintuples [7,8]. After so muh progress on the prolem in integers, fous is returning to the prolem in rtionls. Euler disovered tht fifth rtionl n e dded to Fermt s sequene to give the following rtionl Diophntine quintuple (the FermtEuler sequene) [9] , 3, 8, 120, No sixth rtionl tht extends this sequene further hs een found, nor hs ny lterntive vlue for the fifth rtionl. However, rtionl Diophntine quintuples re lso very undnt nd there re now known exmples of rtionl Diophntine sextuples suh s [10,11,12]
2 96 847, , , , , Very reently some infinite fmilies of sextuples hve een found [13,14,15] ut no sustntil progress hs een mde towrds finding rtionl Diophntine septuple with seven frtions or proving their nonexistene. Regulr Diophntine mtuples If the existene of Diophntine mtuples were pseudorndom proess where the proility of positive integer N eing squre is N 1 2, how mny of them would e expeted? For n mtuple of height H (height eing the lrgest numertor or denomintor) the proility of it eing rtionl Diophntine mtuple would e of order H m(m 1) nd the numer of mtuples of this size is of order H 2m 1. The expeted numer of rtionl Diophntine mtuples would therefore e given y n integrl of order H 3m m2 1 dh. This integrl diverges logrithmilly for m = 3 nd onverges rpidly for m > 3. This mens infinitely mny rtionl Diophntine triples would e expeted ut they should e rre. Rtionl Diophntine qudruples nd lrger mtuples would only e finite in numer, if they existed t ll. Only Diophntine pirs (m = 2) should exist in lrge numers. In relity mtuples re not pseudorndom in this wy nd it is only the pirs tht follow this predition. There is lso n undne of mtuples up to t lest m = 6. This mens tht there must e some priniple t work tht mkes them more ommon thn the pseudorndom rgument suggests. This unexpeted plenitude of rtionl Diophntine mtuples n in prt e explined y the existene of symmetri polynomil equtions whih n e solved to extend rtionl Diophntine mtuples to rtionl Diophntine (m + 1)tuples for = 2,3,4. Given two distint positive rtionl numers, Q suh tht + 1 = x 2, x Q (lled rtionl Diophntine pir), third rtionl numer n e defined in two wys to mke rtionl Diophntine triple using the formul This is equivlent to the polynomil formul = + ± 2x P(,, ) = ( + ) 2 4( + 1) = 0 When expnded, this expression is found to e symmetri under permuttions of,, nd whih mens it n lso e written s ( + ) 2 = 4( + 1) ( + ) 2 = 4( + 1)
3 Therefore, given the rtionl Diophntine pir {, }, n e found s solution to P(,, ) = 0 nd then + 1 nd + 1 will e squres giving the rtionl Diophntine triple {,, } When using the minus sign to give = + 2x the triple n fil to e vlid euse my e zero or negtive or repetition of or, ut when using the plus sign = + + 2x, is lwys positive nd distint from nd, so vlid rtionl Diophntine triple is lwys formed. A rtionl Diophntine triple tht stisfies the eqution P(,, ) = 0 is sid to e regulr nd one tht does not is irregulr [10]. There re mny exmples of either in oth rtionls nd positive integers. Similr polynomils exist for regulr qudruples nd quintuples. For qudruples the polynomil is defined y [16] P(,,, d) = ( + d) 2 4( + 1)(d + 1) Agin this is symmetri under permuttions of the four vriles. It is qudrti in eh vrile individully ut is qurti overll due to the inlusion of the term 4d The eqution P(,,, d) = 0 n e solved for d y ompleting the squre nd finding tht the disriminnt ftorizes giving, P(,,, d) = ( d) 2 4( + 1)( + 1)( + 1) This shows tht if {,, } is Diophntine triple, then the eqution n e solved for d giving two solutions t lest one of whih is positive nd not equl to, or. {,,, d} will then e rtionl Diophntine qudruple [17] (e.g. d + 1 is squre when + 1 is squre euse of the defining eqution) It stisfies P(,,, d) = 0 so we ll it regulr. For quintuples the orresponding polynomil is defined y P(,,, d, e) = (de d e) 2 4( + 1)( + 1)( + 1)(de + 1) One gin this n e solved for e to extend rtionl Diophntine qudruple to regulr rtionl Diophntine quintuple [18]. This time the expression hs ftor (d 1) 2 in the denomintor nd it n fil in exeptionl irumstnes inluding when {,,, d} is regulr nd d = 1 (It is n interesting exerise to work out the generl solution to this se.) The polynomils re relted y P(,, ) = P(,,, 0) nd P(,,, d) = P(,,, d, 0). For ompleteness P(, ) = P(,, 0) = ( ) 2 4 nd P() = P(, 0) = 2 4 The FermtEuler sequene is then the positive solution to = 1, P(, ) = P(,, ) = P(,,, d) = P(,,, d, e) = 0 Is there similr polynomil for extending rtionl Diophntine quintuples to sextuples nd eyond? This would require polynomil P(,,, d, e, f) whih is symmetri under
4 permuttions of ll its rguments nd whose disriminnt s qudrti in f ftorises to four times the produt of ll squres formed in the remining quintuple. To ontinue the sequene we lso expet tht P(,,, d, e) = P(,,, d, e, 0). Until now it hs een ssumed tht no solution to this exists ut in ft it does nd is given s follows, P(,,, d, e, f) = (de + df + ef def def def +2 2def d e f) 2 4( + 1)( + 1)( + 1)(de + 1)(df + 1)(ef + 1) A rtionl Diophntine sextuple will e lled regulr if it stisfies P(,,, d, e, f) = 0. Given ny rtionl Diophntine quintuple {,,, d, e}, this eqution n e solved for f with two roots exept in speil ses. The wekness of this extension method ompred to those for smller mtuples is tht {,,, d, e, f} is often not rtionl Diophntine sextuple. From the definition we only get tht the produt (df + 1)(ef + 1) is squre nd similrly when {d, e} is repled with other pirs of elements from the originl quintuple. In other words the five produts (f + 1),, (ef + 1) re squres multiplied y single ommon ftor. For exmple this eqution n e used to dd sixth element f to the FermtEuler sequene , 3, 8, 120, , This does not mke it Diophntine sextuple. The produt of f with ny of the previous five numers is one less thn squre divided y the denomintor of f. Nevertheless, f is the nturl next element in the sequene euse it solves the eqution P(,,, d, e, f) = 0 Despite its filings s n eqution for extending quintuples to sextuples, it does hve some vlue in the theory of sextuples euse some of the known exmples of sextuples re in ft regulr. Here re ten exmples: 33/ / / /152 19/2 1920/19 249/ / / /128 38/3 920/3 2261/ /78 989/ / / /104 6/ / / / / / / / / / / / / /11 143/ / / / / / / / / / / / / / / / / / / / / / / / / / / /143
5 For referene the polynomil P(,,, d, e, f) whih hs 105 terms when fully expnded n e onveniently written in symmetri form s: P(,,, d, e, f) = (de) 2 + (df) 2 + (ef) 2 + (def) 2 + (def) 2 + (def) 2 4(def) 2 d + e + f + de + df + ef + de 2(def + 2) ( +df + ef + def + de + df + ef + def + def ) + + d + e + f + + d + e 2(2def + 1) ( +f + d + e + f + de + df + ef ) 8def 2( d + e + f)(de + df + ef + def + def + def) Wek Diophntine mtuples d 2 + e 2 + f 2 4 The vlue of the eqution for regulr Diophntine sextuples n e understood little etter in the ontext of wek Diophntine mtuples defined s follows. A set of positive rtionl numers { 1,, m } is wek Diophntine mtuple if ( i j + 1)( i k + 1)( j k + 1) = x 2, x Q, i < j < k A wek Diophntine mtuple up to m = 6 will e lled regulr when it stisfies the sme polynomil equtions tht define rtionl Diophntine mtuples s regulr. Here re some properties of wek Diophntine mtuples: A rtionl Diophntine mtuple is lso wek Diophntine mtuple If { 1,, m } is wek Diophntine mtuple then so is the set of its reiprols { 1,, }. 1 m This is euse ( 1 i 1 j + 1) ( 1 i 1 k + 1) ( 1 j 1 k + 1) = ( i j +1)( i k +1)( j k +1) ( i j k ) 2 A wek Diophntine triple {,, } n e extended to wek Diophntine qudruple {,,, d} y solving P(,,, d) = 0. A wek Diophntine triple {,, } n lso e extended to wek Diophntine qudruple {,,, d} y solving d = 1. A wek Diophntine mtuple in positive integers is lwys Diophntine mtuple. Proof: ny wek Diophntine triple in positive integers n e extended to wek Diophntine qudruple in integers y solving P(,,, d) = 0. However, it is known tht ny solution of this eqution in positive integers is Diophntine qudruple (proof is y infinite deent.) This implies tht the wek Diophntine triple is Diophntine triple. Sine this pplies to ny triple in the wek Diophntine mtuple it mens tht it is Diophntine mtuple. 1
6 A regulr wek Diophntine quintuple is regulr rtionl Diophntine mtuple. This follows from the defining eqution for regulr quintuples. In generl wek Diophntine qudruple nnot e extended to wek Diophntine quintuple using the eqution for regulr quintuples. In wek Diophntine quintuple, the produt of 10 ftors D = i<j ( i j + 1) is squre. This is euse the produt i<j<k( i j + 1)( i k + 1)( j k + 1) = D 3 nd sine eh triple ftor is squre this mkes D 3 squre. Therefore D is squre. Sine D is squre, the eqution for regulr Diophntine sextuple n usully e solved in rtionls to extend wek Diophntine quintuple to sextuple. The defining eqution for the polynomil eqution then mkes this wek Diophntine sextuple. If wek Diophntine sextuple is regulr then its reiprol is lso regulr. This follows from the identity P ( 1, 1, 1, 1 d, 1 e, 1 f ) (def)2 = P(,,, d, e, f) whih n e verified from the definition. The unusul se when P(,,, d, e, f) nnot e solved for f given wek Diophntine quintuple {,,, d, e} is when the polynomil oeffiient of f 2 is zero nd the polynomil terms independent if f give zero. In other words when oth the quintuple nd its reiprol re regulr, ut then they re oth rtionl Diophntine quintuples. It is not known if there re ny exmples of regulr rtionl Diophntine quintuples whose reiprols re lso regulr rtionl Diophntine quintuples.
7 The Squre Identities A numer of identities for the polynomils P(,, ) hve lredy een given in the form B 2 P = A 2 4Π Where A nd B re polynomils nd Π is produt of ftors like ( + 1) et. In order to understnd why the polynomil for regulr sextuples does not lwys give full rtionl Diophntine sextuples nd why there is no generlistion to septuples it is helpful to reord the full list of these identities. For qudruples upwrds these identities do not exist for ll possile produts Π ut the ses in whih they do exist n e hrterised s follows: Prtition the mtuple into two susets of vriles X nd Y. The produts re formed in of two wys, y tking ll ftors ( + 1) where either nd re in the sme suset X or Y, or nd re in different susets X nd Y. The produt in the first se will e written s Π{X; Y} nd in the seond se s Π[X; Y]. For exmple with four vriles Π{, ; p, q} = ( + 1)(pq + 1) Π{,, ; p} = ( + 1)( + 1)( + 1) Π{,,, d} = ( + 1)( + 1)(d + 1)( + 1)(d + 1)(d + 1) Π[, ; p, q] = (p + 1)(p + 1)(q + 1)(q + 1) Π[,, ; p] = (p + 1)(p + 1)(p + 1) Π[,,, d] = 1 For eh of these produts there is n identity written B{X; Y} 2 P(X, Y) = A{X; Y} 2 4Π{X; Y} B[X; Y] 2 P(X, Y) = A[X; Y] 2 4Π[X; Y]
8 The identities n e summrised in two tles. For the lrgest ses YES/NO re used to indite where they exist or not. A{X; Y} p p q p q r 0 p p q 2pqr p q r p p q 2pqr p q r + + p + p q pqr 2pqr p d d e d e f p pq p q d( d) d + 2d + 2d d YES NO d( d) d + 2d + 2d d p(1 d) 2 NO YES q r (pq + pr + qr) pqr( + + ) +2 2pqr p q r The polynomils B{X; Y} hve not een shown. For the first three rows they re 1 nd for the fourth row they re (1 d). How n we e sure tht no solution exists for the sextuple se where indited with NO? If solution did exist in either of these ses then it ould e used to show tht the extension formul would lwys give full rtionl Diophntine sextuple ut the FermtEuler sequene is lredy ounterexmple to tht possiility. The entries in the tle hve een written so tht it is possile to move upwrds or leftwrds y setting one of the vriles to zero. This mens tht no entries for septuple ses re possile euse if they were it would e possile to move either up or left nd provide solution for one of the foridden sextuple ses.
9 A[X; Y] p p q 2 2 p 2 2 (p q) p 2 + p 2 (p l) 2 + p + q 2 2 p 2 + p + p 2 (p q) 2 + p + p + q + q + 2pq 2 2 p 2 + p + p + p 2 (p q) 2 + pq(p + q) +( + + )(p + q) +2( + + )pq d d e d e f 2 2 dp 2 p 2 +p + p + p + dp 2 2 NO 2 (p q) 2 + d(2p 2 q 2 (p q) 2 ) +( + d + d + d)pq(p + q) +( d)(p + q) +2( d + d + d)pq The polynomil B[X, Y] is zero for the first olumn (trivil se), p for the seond olumn nd (p q) for the third olumn. One gin the NO se n t exist euse it would imply tht extension gives full sextuples nd no se for septuples fit. No fourth olumn n e dded preserving the rule for moving up nd left y setting vriles to zero. In summry, the identities for sextuples only exist with produts Π{X; Y} or Π[X; Y] when the numer of times eh vrile ppers in the produt is even. The extension formul therefore works to extend wek Diophntine mtuples ut does not normlly sueed in extending rtionl Diophntine quintuples to full rtionl Diophntine sextuples. Nevertheless it only requires one of the new produts to e one less thn squre nd they will ll e. There re multiple instnes where this hppens nd extension does then produe full rtionl Diophntine sextuple.
10 Further Polynomil Generlistions The polynomils tht define regulr mtuples n in prt e explined from the theory of ellipti urves [19,20], yet the full level of symmetry remins mysterious. Some further explntion rises from the oservtion tht the polynomil eqution for regulr qudruples is speil se of Cyley s hyperdeterminnt whih generlises the 2 x 2 determinnt to n expression for 2 y 2 y 2 rry. This is done in suh wy s to extend its properties s polynomil invrint nd s disriminnt [4]. A polynomil generlising P(,,, d) n e defined s H(,,, d, k, l, m, n) = (k + l m dn) 2 4( + nm)(d + kl) Then Cyley s hyperdeterminnt for three dimensionl rry of numers ijk is given y Det( ijk ) = H( 000, 110, 101, 011, 111, 001, 010, 100 ) The eqution for regulr Diophntine qudruples n e reovered from P(,,, d) = H(,,, d, 1,1,1,1) = H(1,1,1,1,,,, d) The following identity n lso e verified n 2 H(,,, d, k, l, m, n) = (2 + nk + nl + nm dn 2 ) 2 4( + nm)( + nl)( + nk) This shows tht the qudrti disriminnt for Cyley s hyperdeterminnt when treted s qudrti in ny one of its vriles ftorises into three ftors whih re 2 y 2 determinnts. This hd never een noted efore the omprison with the formul for regulr Diophntine qudruples hd een mde. When redued to expressions for regulr Diophntine qudruples this identity for the hyperdeterminnt yields two ses one from eh of the two tles ove for A[, ;, d] nd A{, ;, d} Given tht P(,,, d) generlises to P(,,, d, e) nd then to P(,,, d, e, f), it is nturl to investigte whether H(,,, d, k, l, m, n) lso generlises to expressions in more vriles whih redue to the expressions for regulr quintuples nd sextuples. Cyley s hyperdeterminnt n e generlises to invrints for multidimensionl rrys of ny size ut it does not pper tht ny of these n e redued s required. Nevertheless, the generlistions do exist, ut they re not invrints, disriminnts or ny other kind of previously reognised polynomils. Their origins nd signifine therefore remins mysterious nd nothing more n e done other thn to desrie wht they re.
11 The generlistion for P(,,, d, e) is polynomil of degree ten in fifteen vriles defined in terms of simple lok design. Fifteen vriles n e grouped into six loks of five suh tht eh vrile ppers in two loks. The polynomil is formed from just the produts of eh lok nd its ompliment. I.e. T 1 = de, T 2 = sghk, T 3 = slmn, T 4 = glpq, T 5 = dhmpr, T 6 = eknqr (Notie tht this n lso e regrded s the prmetri solution to the prolem of finding six squre free integers whose produt is squre numer) T 1 T 1 = T 2 T 2 = T 3 T 3 = T 4 T 4 = T 5 T 5 = T 6 T 6 = deghklmnpqrs = T 1 T 2 T 3 T 4 T 5 T 6 Then the polynomil is given y H(,,, d, e, g, h, k, l, m, n, p, q, r, s) = T 2 i 2 T i T j 4 T i Identities stisfied y this polynomil whih redue to the known identities for P(,,, d, e) inlude the following two, i i<j r 2 H(,,, d, e, g, h, k, l, m, n, p, q, r, s) = (der + 2sgl + sghkr + slmnr + glpqr dhmpr 2 eknqr 2 ) 2 4(s + pqr)(g + mnr)(l + hkr)(der + slg) i (hdmp kenq) 2 H(,,, d, e, g, h, k, l, m, n, p, q, r, s) = (2lqngpksmh + 2delmnpq + 2deghkpq + 2dehkmns + d 2 ehmp + de 2 knq + dgh 2 kmps + eghk 2 nqs + dhlm 2 nps + eklmn 2 qs + dghlmp 2 q + egklnpq 2 r(eknq dhmp) 2 ) 2 4(dh + lqn)(dm + gqk)(dp + snk)(ek + pml)(en + gph)(eq + smh)
12 The mster generlistion of the eqution for regulr sextuples is polynomil of degree 32 in 32 vriles whih n e defined y the following identity p 2 P 2 H(,, Z) = (defstuvwxyzrp 2 P + defstuvwxyzrp 2 P + DefstUvwXyzRp 2 P + BCDEFstuvwxYZRp 2 P + ACDEFstuVWXyzrp 2 P + ABDEFSTUvwxyzrp 2 P BdefSTUvwxYZRpP 2 AdefSTUVWXyzrpP 2 CdefstuVWXYZRpP 2 ABCdEFSTuVWxYZrpP 2 ABCDeFStUVwXYzRpP 2 ABCDEfsTUvWXyZRpP^2 + 2DEFstuvwxyzrp 3 2ABCdefSTUVWXYZRP 3 ) 2 4 (stup + ABSTUP) (vwxp + ACVWXP) (yzrp + BCYZRP) (desvyp + DEsvyp) (dftwzp + DFtwzp) (efuxrp + EFuxrp) Referenes [1] T. L. Heth, Diophntus of Alexndri. A Study in the History of Greek Alger. (Cmridge, Englnd, 1910), Mrtino Pulishing, 2003, pp , , [2] F. Lu, A. O. Mungi, Diophntine triples with vlues in the sequenes of Fioni nd Lus numers, Gls. Mt. Ser. III [3] A. Dujell, On MordellWeil groups of ellipti urves indued y Diophntine triples, Gls. Mt. Ser. III 42 (2007), [4] P. Gis, Diophntine qudruples nd Cyley's hyperdeterminnt, rxiv mth.nt/ [5] P. Fermt, Oservtions sur Diophnte, Oeuvres de Fermt, Vol. 1 (P. Tnnery, C. Henry, eds.), 1891, p. 303 [6] A. Bker nd H. Dvenport, The equtions 3x22 = y2 nd 8x27 = z2, Qurt. J. Mth. Oxford Ser. (2) 20 (1969), [7] A. Dujell, There re only finitely mny Diophntine quintuples, J. Reine Angew. Mth. 566 (2004), [8] M. Cipu, T. Trudgin, Serhing for Diophntine quintuples, At Arith. 173 (2016), [9] L. Euler, Opusul Anlyti I, 1783, pp [10] P. Gis, Some rtionl Diophntine sextuples, Gls. Mt. Ser. III 41 (2006),
13 [11] P. Gis, A generlised SternBroot tree from regulr Diophntine qudruples, rxiv mth.nt/ [12] A. Dujell, Rtionl Diophntine sextuples with mixed signs, Pro. Jpn Ad. Ser. A Mth. Si. 85 (2009), [13] A. Dujell, M. Kzliki, M. Miki, M. Szikszi, There re infinitely mny rtionl Diophntine sextuples, Int. Mth. Res. Not. IMRN [14] T. Piezs, Extending rtionl Diophntine triples to sextuples, [15] A. Dujell nd M. Kzliki, More on Diophntine sextuples, in Numer Theory  Diophntine prolems, uniform distriution nd pplitions, Festshrift in honour of Roert F. Tihy's 60th irthdy (C. Elsholtz, P. Grner, Eds.), SpringerVerlg, Berlin [16] P. E. Gis, Computer Bulletin 17 (1978), 16 [17] J. Arkin, V. E. Hoggtt nd E. G. Struss, On Euler's solution of prolem of Diophntus, Fioni Qurt. 17 (1979), [18] A. Dujell, On Diophntine quintuples, At Arith. 81 (1997), [19] A. Dujell, Diophntine mtuples nd ellipti urves, J. Théor. Nomres Bordeux 13 (2001), [20] A. Dujell, Irregulr Diophntine mtuples nd highrnk ellipti urves, Pro. Jpn Ad. Ser. A Mth. Si. 76 (2000),
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