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Annals of Mathematics

On the Numerical Factors of the Arithmetic Forms n nAuthor(s): R. D. CarmichaelSource: The Annals of Mathematics, Second Series, Vol. 15, No. 1/4 (1913 - 1914), pp. 30-48Published by: Annals of MathematicsStable URL: http://www.jstor.org/stable/1967797 .

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ON THE NUMERICAL FACTORS OF THE ARITHMETIC FORMSn i34 *BY R. D. CARMICHAEL.Let a + ,3and ao3be any two relatively rime ntegers differentromzero). Thena and : arerootsofthe quadratic quation

Z2 - (a + 3)Z + Ca4= 0.It is obviousthat the numbers nand Sn,

Dn = ? on = a n-1 + a n-2 + ... + 13n-1 Sn = an + O3nare integers, since they are expressed as rational integral symmetricfunc-tions of the roots of an algebraic equation with integral coefficientswithleading coefficientunity. The principal object of the present paper isan investigationof the numericalfactors of the numbersDn and Sn. Thecasewhen a and : are rootsofunity s excludedfrom onsideration. (See ? 2.)The most valuable treatment of the questions connected with thesenumbers is that of Lucas. t The special case' in which a and i3are integershas been considered by Siebeck,4 Birkhoff nd Vandiver,? Dickson,11andCarmichael.?In Lucas's paper many results of interest and importance are obtained.The methods employed, however, are often ndirect and cumbersome. Inthe present paper a direct and powerfulmethodoftreatment** s employedthroughout; and in connection with the new results which are obtainedmany of Lucas's theorems are generalized and several errorstt in thestatementof his conclusions are pointed out.In ? 1 several fundamental algebraic formulae re obtained and a partialfactorization of Dn and Sn is effected. In ? 2 these algebraic formulaare employed to derive numerous elementary properties of the integers

*Presentedo theAmericanMathematical ociety, ecember, 912.tAmerican ournal fMathematics, (1878): 184-240, 89-321.1Crelle'sJournal,3 (1846):71-77.?Annals fMathematics,2) 5 (1904): 173-180.AmericanMathematical onthly, 2 (1905):86-89.?AmericanMathematicalMonthly,6 (1909): 153-159.** Compare hemethod mployedy Dickson nthe paper lready ited.ttCompare hereview fLucas's paper n theJahrbuchiber ie FortschritteerMathe-matik, 0 (1878): 134-136.

30


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ON THE NUMERICAL FACTORS OF THE ARITHMETIC FORMS. 31DA ndSnrelative o divisibility,nd thesepropertiesre stated nexplicittheorems.In ? 3 the important uestion of the appearanceof a given primefactor n the sequence D1, D2, D3, ... is investigated. The principalresults re contained n TheoremsXII and XIII. Attention s called tothe newnumber-theoreticunctionsntroducednconnection ithTheoremXIII and its corollary.In ? 4 a detailed study s made of the numerical actors f a set ofnumberswhich re the values of an algebraicform k(a, ,3)whichmaybedefineds that rreduciblelgebraicactorf -'k 1kwhichs not factorof any a" - p3" orwhichv < k (but see the definitionn ? 1). This in-vestigations fundamentaln the studyofthenumbers nand Sn, nd theresultswhich re here obtainedhave importantpplicationsn the theoryof numbers. Attention s called especially o Theorems XIV, XVI andXVIII.In ? 5 the theory f " characteristic actors ofF., Dn and Sn is de-veloped.In ? 6 very impleproofs regivenofcertain pecialcasesofDirichlet'scelebrated heoremoncerningheprime erms f narithmeticalrogressionof integers;n particular,t is shown thatthere s an infinitude f primenumbers feach oftheforms n + 1,4n - 1, 6n + 1, 6n - 1.In ? 7 are givena number f theoremswhich re useful n the identi-fication f argeprimenumbers. Among heresults btained hefollowingtwoalonewillbe mentioned ere:A necessary nd sufficientondition hata givenodd number is prime s thatan integer exists uchthat

F,-,(a, 1) 0 mod p;a necessarynd sufficientondition hat22" 1,n > 1, s prime s that

3221 + 1 _O mod229 1.1. Notation. FundamentalAlgebraic ormulae.Let Qn(X) = 0be thealgebraic quationwhose roots are the primitive th rootsof unitywithout epetition, he coefficientfthe highest owerof x in Qn(x)beingunity. The polynomialQn(x) has all its coefficientsntegers; nd it is ofdegree p(n),where (n) denotes henumber f ntegers otgreater han nand prime o n.From thetheory* f the primitive oots ofunitywe have twoformulae

*See Bachmann's reistheilung,speciallyhe thirdecture.


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32 R. D. CARMICHAEL.which re fundamental orourpurposes. Thus,(1) Zn - 1 = jjQd(X),dwhered ranges ver all thedivisors f n. Also,

() (Xn - 1) l I(XnIPiPi 1) ...(2) Qn(X) = l(XnP '- 1) _ (Xn/pipipk - 1) ...where hep's denote hedifferentrime actors f n andwhere heproductsdenoted y II extend verthe combinations, 4, 6, ** at a timeofPi, P2,p3 * * in thenumeratornd overthecombinations, 3, 5, * at a timein thedenominator.Let a + , and ao3be any tworelatively rime ntegersdifferentromzero); thena and 3are the rootsoftheequation

Z2 (a + O)z + ao3 = 0whose coefficients + , and a: are any two relativelyprime ntegersbothofwhichare differentrom ero. We shall exclude the trivialcaseCZ , = 1. It is then clear thata and ,3cannotbe equal.Now an + A representsn integer or veryvalue ofn, since the func-tionan + on is a symmetricolynomialn a and A and has integral oef-ficients. On the otherhand the functionn o ndoes not necessarilyhave an integralvalue. If, however, hisnumber s dividedby a - Btheresult s clearlyn integer,ince tmayobviously ewritten s a rationalintegral ymmetric unction f a and : with integralcoefficients.Ac-cordingly,etus definehe ntegers nand Sn, for veryvalue ofn, bytherelations

Dn = Cen- on On-1+ Cen-2 + ...+ 0n-1 Sin= Cen On.Then,obviously, Sn D2nDnso that a studyofthefactorizationftheform n, forvarying alues ofn,includesncidentallyhatoftheform n. We shallthereforee interestedprimarilyntheform n.We define k a, 3)bytherelation(3) Fk(ae, 3) = I3 (k)Qk(aII)We shallnow showthatFk(a, 3) is an integer oreveryvalue ofk exceptk = 1. The theorem s obviously true fork = 2; for,

F2(a, A) = a + he


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ON THE NUMERICAL FACTORS OF THE ARITHMETIC FORMS. 33Then supposethat k is greater han2. Let w be a primitive throotofunity. Then evidently,

4(k)(4) Fk(a, 0) - by (k)Qk(a/0) = II(a - wa8i)where or = 1, 2, ** , p(k), he s, are the wo(k)ositive ntegersess thank and prime o k. Hence

4(k)Fk(a, d) = I (a -since coc- = 1when s, + sk= kand thefactorsn the above equationobviously all ntopairs uchthatthesumofthes's in eachpairis k. Hence we see readily hat

16(k) (k)Fk (a, f3) = II (aWk" -( w8a),i=1 j=l

where n the last member j is written ork - si. By comparing hisequationwith 4) we find hatFk(a, 3) = Fk(1, a);

that s, k(a, 3) is symmetricith especto a and 3. But it is a poly-nomial n a and d with ntegral oefficients.Hencewe conclude hatThe number k(a, () is an integerorevery alueofk except = 1.Now from1) we have readily(5) DDn t = IT'Fd 3),a- =L dkawhered rangesover all the divisorsof n exceptunity. This importantformulagives (partial)factorizationfthe nteger n. Likewise, f v isanydivisor fn,(6) Dw.v= II' Fa(a (),awhere ranges verall the divisors fn/v xceptunity. If nowwedividethe first f these equationsby the second,member ormember,we have(7) n= a0(&-1)/&-+n (-2)/Yn/ + . . an"'(3n -2)1v4+ on(v-1)/v.=. JIFk(a, 3)wherek rangesover all thedivisors fn which re not at the same timedivisors fn/v.


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34 R. D. CARMICHAEL.From (2) we obtainreadily he equation

(8)(a, 0) - (aOn fn) . IIH(an/pipi -_ An/Pipi) ...(8) Fn~a, i) =I(an/Pi _ On/pi) . (an/PiPjPk - OPiPjPk) ...where he factors enotedby II extendover thecombinations, 4, 6, *.at a timeofpi, P2, * * * in the numeratornd over the combinations,3, 5,... at a time in the denominator.The total numberof factors n thenumerator f this equationis the same as that in the denominator; or,obviously, he first f these numberss the sumof thepositive erms ndthe secondis the sum of the negativeterms n the expansion f (1 - 1)?bythe binomial ormula, being he number f differentrime actors fn.Hence,dividing ach of thesefactorsn bothnumeratornd denominatorby a - f,we have(9) Fn(a, fi) = Dn HDn/ipip ..*H[Dn/i H[Dn/pjpC ..wheretheproductsdenotedby II have a meaning imilar o that above.Let p be anyprime actor fn and write

n = ppawhere heexponent is so chosen hatv is an integerwhich s notdivisibleby p. Considerthe factors n the secondmemberof (9) into whichpdoes notenter xplicitly; rom9) itselft it clearthat thesefactors lonehave thevalueF (aPa, fpa). In thesame-waywe see that thefactorsntowhichp enters xplicitly ave thevalue 1IF1 aPa-, fpPal1). Hence(10) Fn(a fi) - F,(pa, fpa) . F1 ,pa-l, pal)Since Fi(af) =a - a,equation 10) maybe usedas a recursionormula ordeterminingn(a, f)For n < 36, Sylvester's able*of cyclotomic unctionsmay convenientlybe employed orfindingn(ayf).In passingwe note withoutdemonstrationhat (10) may be proveddirectlynd thenbe employed or hederivation f 9).tIf,now, n equation (7) we replacen by 2n,giveto vthe value 2 andrememberhat D2n Sn,D=we have(11) Sn= an + fI Fk(a,f),A,*American ournalfMathematics, (1879):367-368.t Compare ickson, .c., p. 86.


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ON THE NUMERICAL FACTORS OF THE ARITHMETIC FORMS. 35wherek runsoverall thosedivisors f2nwhich ontain tothesamepoweras 2n itself. This importantormulagives (partial) factorizationf theinteger n.

Let vbe any odddivisor fn; then,writing /vforn in (11) we have(12) Sn/, flF(a, ( ),kwhere runs ver ll thosedivisors f2n/vwhich ontain tothe amepoweras 2n/vtself. Dividing 11) by (12), member ormember,we have(13) Sn = II Fk odd,Sn/Y kwherek runs verall thosedivisors f2nwhich ontain tothe samepoweras 2n itself nd whichdo notdivide2n/v.2. GeneralProperties f the IntegersDn and SnRelativeto Divisibility.

In view ofthefact that a rational ntegral ymmetricunction fa, 0jwith ntegral oefficientss an integerwe have readilythe two equations(a + fl)n = an + fOn af3Il = Sn + c43I1,

Dn = a 0 = a + 3n1 + a1312= Sn-i + a13I2,where , and I2 are integers. Since ad and a + ,3 are relatively rimeintegerst follows rom hefirst ftheseequationsthat Sn is prime o a13for veryvalue of n. Then from hesecondoftheequationswe concludethatDn is likewiseprimeto a: for veryvalue ofn. Hence we have thefollowing heorem:THEOREM I. The ntegersn andSn areboth rime oao(.

This theorem nables us to dispose of an exceptionalcase; namely,whenDm = 0 for ome valueof m. In thiscase alm = (3mandhenceSm= 2am.

But Sm is prime to aod nd henceto amolm".These tworesults gree onlywhen am. = Omso that n thiscase a and ( are bothrootsofunity. It is easy to see thatSk can assume no othervalue than - 2, - 1, 0, 1, 2; forISkl lka + (3kj = 2.Now (a _ ()2 = (a + ()2 - 4ao = integer;


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36 R. D. CARMICHAEL.and hence la -(3l 5 1,since a (3. Therefore

Dkj _ak - O3ki lakI + l(kI = 2,so thatDk can takeonly hevalues -2, - 1,0, 1,2. A correspondingis-cussion an be madewhen m= 0 for omevalueofm, ndwith ike results.The cases Dm= 0 for omem and Sm= 0 for omem are thereforeothtrivial. Theyarisewhen ndonlywhen andA arerootsofunity. Henceinwhatfollowsweshallexcluderom onsiderationhe asein which and (3are roots funity. ThenDm nd Sm realwaysdifferentrom ero.Now (aOn on)2 - (aOn - On)2 = 4 nn,andhence 52 - (a - 3)2Dn2 = 4an/n.It is clearthat (a - ()2 is an integer. Then fromthe above equation itfollows hatany common ivisor f n2 andDn2mustbe a divisor f4an~n;butbyTheorem sucha divisor s prime o ad. Hence it is a divisor f4.Therefore,ither n and Sn arerelatively rimeor theyhave thegreatestcommondivisor2. That bothof thesecases may arise is shownby thefollowingxamples:(1) a = 2, (3= 1. Dn andSnhavenot thecommon ivisor and henceare relatively rime;(2) a = 3, ( = 1. Dn and Sn have the common actor ifn is even.Hencewe have thefollowingheorem-:*THEOREMI. The integers n and Sn either re relativelyrimeorhavethegreatestommon ivisor .We shall nowdeterminehe character fDn and Sn relativeto divisi-bilityby 2. From Theorem it follows hatboth of themare odd whena(3 s even. Hence we have to treatfurthernlythe case whenad is odd.This will separatefurthernto twocases according s a + ( is odd or even.We startfrom he recurrenceormula

)D -+2-(a +f3)Dn+l + a(Dn = 0,Sn+2 - (a + (3)Sn+l + a(S,. = O0which are readilyverified y substitutingorDk and Sk, k = n, n + 1,n + 2, theirvalues in terms fa and (3. Sinceforthepresentdiscussiona(3 s odd,we have from14)

Dn+2 -Dn Sn+2 Sn mod 2or Dn+2 Dn+1 + Dn, Sn+2 Sn+1 + Sn mod2according s a + ( is evenorodd.*Lucas (1.c., p. 200) states naccuratelyhatD. andS. are relativelyrime.


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ON THE NUMERICAL FACTORS OF THE ARITHMETIC FORMS. 37Now D1 = 1 and D2 = a + g. Hence fromthe above congruenceswhich nvolveDn we see readily hat when + ,3 s evenDn is evenoroddaccording s n is even or odd; and that whena + f is odd,Dn is even or

odd according s n is or is not a multiple f 3.We treatthenumberSn n a similarmanner. We haveSi = at+t, S2 = a2+f2 = (a+f)2 - 2af.

Hence,ifa + ,3 s odd both S1 and S2 are odd; and ifa + fi s even bothSi and S2 are even. Therefore rom he above congruences,nvolving nwe concludereadilythat if a + g is even Sn is even for all values of n;andthat f + g is oddSn s even or odd according s n is or s nota multipleof3.Collecting heseresultswe have thefollowing heorem:THEOREM III. If aOs s evenboth n and Sn are odd. If ao3 is odd anda + f3s even, hen . is even or ll values fn while n s even rodd ccordingas n is even r odd. If both od nd a + FI re odd thenDn and S are botheven rboth dd according s n is or s not multiple f 3.From the properties f symmetricunctions f the rootsof an algebraicequationand thealgebraicdivisibilityfDn byD, when v is a divisor fn, it follows mmediately hat the integerD. is divisibleby the integerD. when v is a divisorof n. This is also an immediate onsequenceofequation 7); and the atter quation ngeneral tatesmore hanthis, hatis, it gives a partial factorizationf the nteger n/Dv. Thus wehave thefollowingheorem:THEOREM IV. If v is a divisor fn thenD, is a divisor fD,,and wehave

Dn = 11F g),

where ranges ver ll those ivisors fn which renot t the ame ime ivisorsof P.For v = 1 thistheorem ives a partialfactorizationfDn,sinceD1= 1.In the preceding ection we proved that the quantitiesFk(a, 43) haveinteger alues.By the aid ofequation 13) thefollowingheoremmaybe demonstrated:THEOREMV. If v is a divisorofn such thatn/v s odd then n is divisiblebyS. andwe haveS= Fk(a, 4),

where runsover ll those ivisorsf2n which ontain to the amepower s2n itself nd which o notdivide v.


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38 R. D. CARMICHAEL.Fromthe dentity

(am _- #m) an + 3n) - (an - An) (am + Am) = 2anfln(am-n - 3m-n), m > n,we havereadily(15) DmSn - DnSm = 2an/3nDm-n.Fromthisequationand the factthatDmandDn areprime o a13 t followsthat everycommon dd divisorofDmand Dn is also a divisorofDn;whencewe conclude eadily hateverycommon dd divisor fDm and Dnis a divisorof D, wherev is the greatest ommondivisorof m and n.But according o Theorem V D, is a divisorof Dm and Dn. Hence thegreatest ommondivisor fDmand Dn is D, providedthateitherDn/DyorDn/D, s odd. This latterfactwe shallnowproveby aid of TheoremsI and III.We have Dm am Om3mam/v am/vD~ a~-3~ -ifwe replacea", #Ivy a- 1. The last member f the above equationwedenoteby Dmv. We defineD-,,I n a similarmanner. It followsfromTheorem thatal03'and a" + 13vre relatively rime. They arebothdif-ferent rom ero. That is, aB and a + a arerelatively rimentegers othofwhichare differentrom ero. Hence wp may apply Theorem II toDmI,ndD,/I. Ifa-3 sevenbothofthesenumbersreodd. Ifa-fsoddanda + 1 isevenoneofthenumbers mIV ndDnV is odd; for itherm/v rn/vis odd,sincev is thegreatest ommondivisor f m and n. Likewise,fadand a + a arebothoddthenoneofthenumbers ).,, and DnyI s odd; foreitherm/vor n/v s not divisibleby 3, since v is the greatest ommondivisor fm and n. Hence Dai, and Dn/y ave not thecommon actor .RememberinghatD.1, = Din/D and Dn/,= Dn/Dy and making se oftheresults f the ast twoparagraphswehave thetheorem:*THEOREMVI. Thegreatestommonivisor fDA and Dn iS D, whereis thegreatestommon ivisor fm and n.SinceD1 = 1 we have at once thefollowingorollary:COROLLARY. The ntegersm ndDn arerelativelyrimewhenmandnarerelativelyrime.The example

S6(2, 1) = 26+ 1 = 5.13, S4 = 24+ 1 = 17, S2 = 22 + 1=5shows t oncethatthegreatest ommon ivisor fSm and Sn is notalways

*Thepartofthis heorem hich pplies o theodddivisors fDm ndD. is due to Lucas(1.c.,p. 206).


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ON THE NUMERICAL FACTORS OF THE ARITHMETIC FORMS. 39S, wherevis thegreatest ommondivisor fm and n. If, however,m/vand n/v are both odd thissimple aw obtains, s we now show. In thiscase it follows rom heoremV that S. is a common ivisor f Sm nd Sn.Now D22m SmDmand D2n = SnDn,whencewe concludebyaid ofTheoremVI thatthegreatest ommon ivisorofSmand Sn is a factor fD2,. Now

D2v = SvDvand hencewe have only to examinewhatfactors , has in commonwithSm and Sn. Now D, is a factor fD, and Dmand Smhave thegreatestcommon ivisor or 2. HenceD. haswithSmand Snthe greatest ommondivisor1 or2. Thereforemand Sn have the greatest ommon ivisor ,or 2S,; and in thenexttwoparagraphswe showthat thelattercase doesnot arise.To provethat the greatest ommondivisor nder onsiderations not2S, it is sufficient o show that either Sm/Svr Sn/S, s odd. This followsat oncefrom heorem II if a3 is even; forthen Smand Sn are odd. Ingeneral S' _ am + 3M anIv+ X1/'

Sv. all + TV1 -a + ~if a, = a and B = B. Denote the last numeratorabove by Sm./nd defineS,,, in a similarway. Then Theorem II is applicable to S,/v nd Snt,Now eitherm/v r n/v s primeto 3, and henceone of the numbers ./and Sn&vs odd ifa- and a + B are bothodd,that is, if a3 and a + 3 areboth odd. In thiscase, then, neat leastof thenumbers m/Svnd Sn/Svis odd.Let us nextconsider he case in whichaj3 is odd and a + ,3 s even;saythat a + $ is an oddmultiple f2k. Then,since

S, = a+3and S2 = a2 + /2 = (a + p)2 - 2a3,:it is easy to see that S, and S2 are odd multiplesof 2k and 2 respectively.By meansof thesecondrecursionormula14) one sees that n general nis an odd multiple f 2k or of2 according s n is odd or even. Hence inthis case Sm/Sand Sn/Svare both odd, since m and v and likewise n andv are bothodd or botheven.Thuswe have thefollowingheorem:


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40 R. D. CARMICHAEL.THEOREMVII. If v is the greatestommon ivisor fm and n, andm/v ndn/v reboth dd, hen he reatestommonivisor fSm ndSn s S.We turnnow to an interestingheoremf a differentharacter, amely:THEOREMVIII. Let ml, m2, *.., m8 and ni, n2, *.., n, be twosets ofpositiventegers hichhavethepropertyhat nypositiventeger , differentfromunity,which s a factor f (just) t integersfthe econd et s also afactor fat leastt integersfthe irstet;then henumber

D1m*Dfl2 . n,Dni -Dn2 *Ae Dn,is an integer.This theorems an immediate onsequence f the (partial)factorization

ofD, given n equation(5).COROLLARY . The product f any n consecutiveerms f the equenceD1, D2, D32 ... is divisible ytheproductfthe irst terms.*COROLLARY II. The numberD1D2 ... Dnl~n2+ +nk(D1D2 *.. Dn) (D1D2 *.* Dn2) ... (D1D2 *** Dn)

is an integer.This result s analogousto the theoremhatthepolynomial oefficient(n,+ n2 + ... + nk)!n i~n2! !.no!is an integer.Let mand n be anytworelatively rimepositive ntegersnd supposethat the positiventeger (d $ 1) is a divisor f s integers f the set 1, 2,*.. ,mand of t ntegersof the set 1, 2, * **,n. Then d is obviouslya divisorof at least s + t integers f theset 1, 2, ***, m+ n -1. Inview ofthisfactTheoremVIII yields hefurtherorollary:

COROLLARYII. If mand n are any tworelatively rimepositiventegers,then henumber D1D2 ... Dm+ni(D1D2 ... Dm)(DID2 ... Dn)is an integer.This theorems analogousto thatwhich sserts hat(m+n-1)!m! n!

is an integer,rovided hatmandn arerelatively rime.* The result ontainedn this orollarys dueto Lucas, whogave,however, very ifferentproof f t (Lucas,1. c., p. 203).


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ON THE NUMERICAL FACTORS OF THE ARITHMETIC FORMS. 41Similarly ne may provean extended nalogueof the theoremwhichstatesthat (km,) (kM2)! .. (kMk) k 2ml! M2! ..*mk! (ml + m2 + * +m)!' k>2

is an integer, amely:COROLLARY IV. The number(DlD2 ... Dk?nl) (DlD2 ... D;km2) .(DlD2 ..Dkm,)(D1D2 ... Dml)k1 ... (D1D2 ... Dm)k (DlD2 ... Dml+m,+...+mk)

is an integer.Justas equation (5) was used in the demonstrationf TheoremVIIIwe may employ quation 11) to provethe following heorem:THEOREM IX. Let ml, M2, *.., m8 and ni, n2, *... n, be two etsofpositiventegersuchthat very ositiventeger which s a factor f just) tofthenumbers i, n2, ** , n, withoddquotients also a factor fat least tofthenumbers l, i2 **, m8 with ddquotient. Then thenumber

Sm< Sm,2 * e e e Sm.Sn Snr * . Sn,is an integer.COROLLARY. The productof ny 2n - 1 consecutive erms f he equenceSi, S3, S5, ... is divisible ytheproductfthe irst terms.Ifmis any nteger nd q is any odd prime,t is obviousthat there xistintegers q- 1a1,a2, ...,a8, s- 2dependent n q alone,suchthat

-n _mq = (am _ am) q + alam om(Oem - om) q-2 + a2l2mfl2m(am _ Om)q4+ * + aga8mI3sm(am - r);whence

(16) Drnq = (a - j3) -lD + a,(& - f3) - + ...aaqm:mDmDLet us evaluatea8. Since t is independent fa, ,3 nd m, we maychooseany convenientvalues for these numbers. Then put m = 1, f = 1,a = r + 1, where is a positiventegero be chosen t convenience. Thenfrom16) we have (r + 1)q - 1(r~ - =a8(r+ r)amodr.If we supposer to be a primenumberdifferentrom we see that a. isnotdivisible yr. If we put r = q2it follows hata. is divisible y q butnot byq2. Hencea8 = q.


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42 R. D. CARMICHAEL.Supposenow that Dm s divisiblebyp', X * 0, and byno higher owerofp, p being prime umber; henfrom 16), since . = q,wehave

(17) Dmq qaJmI:mDmmod p3.Fromthiscongruencet follows hat pA+l is thehighestpowerof p con-tained nDmp, provided hatp is odd,and thatpA s the highest owerofp containednDmqwhenq is an odd primedifferentrom . We enquirefurther:What is the highestpower of p contained n D2m? We haveD2m= DmSm. In TheoremII we haveseenthatDm ndSmhave nocommonodd factordifferentrom nity). Hence, fp is an odd prime he highestpower f p containednD2mis pA. If p is even, o thatDm sdivisible y 2,itfollows rom heoremII thatSm s divisible y2. Then tfollows romTheoremI thatDm nd Smhave thehighest ommon actor . Hence inthiscase D2m contains A+1; and it containsno higherpowerof 2 unlessX = 1.These results ead to the followingheorem:

THEOREM X. If forX> 0, pA $ 2, pA is thehighest ower f a primecontainedn Dmthen hehighest ower f p containedn Dmpa is pa+A, .being ny number rimeto p. If pA = 2, thenDm,,2a contains hefactor2a+1 andDmns an oddmultiplef2.*Suppose thatSm is divisibleby pA,X> 0, but by no higherpoweroftheodd primep. Then D2m contains pA and no higher power of p, since

D2m = DmSmand Dmand Smhave no common dd primefactor. Therefore,ccordingto the preceding theorem,D2m.,,p, or Dmpa* .,.p-yAbeing primeto p,containspa+A and no higher power of p. Moreover Dm.apa nd Smupado not have a factor in common. Hence one ofthesenumbers ontainspa+k nd no higher owerofp whiletheother s pre to p. SinceD2mis a divisor fDmApf ji is even, we see that Dmnpa ontains pa+A when uis even. When u s odd Sm s a factor fSmpa nd hence in this case Smpacontains the factorpa+A.Thus we have thefollowingheorem:THEOREMXI. If pA, X > 0, is the highest power of an odd prime pcontainednSm nd u s a numberrime op; thenfA s even mpa is divisibleby a+A and byno higher ower f p andSmA.pas prime o p,whilefA s oddDMA, is prime op andSmupa s divisibley a+)A and bynohigherower f p.*The special aseof his heoremn whichA= 1 is given y Lucas (1.c., p. 210), butLucasfailed o notice he exceptionalharacterfthe casewhen A 2.


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ON THE NUMERICAL FACTORS OF THE ARITHMETIC FORMS. 433. On the Appearance f a Given Prime Factor n the SequenceD1, D2, D3,If it is known hat a primenumber is a factor fD, theoremsn theprecedingection nableus to say howp enters ntoDm#Lp-n thepresentsectionwe showthat any givenprimep, which s not a factor fao3, s afactor f a certain efinite umber f the sequenceD1,D2,D3, * .; we alsocarryout otherrelated nvestigations.We have need of twolemmas, sfollows:LEMMA I. If S(aP, ,3P) is any rational ntegralymmetricunction f

APT j3P withntegraloefficients,henS(aP, BP) S(a, A) mod p,p being primenumber.The proof s notdifficult.From Fermat'stheoremt follows hat

(18) aCOOP COmodp,since ca3 s an integer. Likewise

(a+fl)P- =a+ modp.But by the aid of thebinomialformulawe see that

(a +) )P acP +tP3modp,sincethebinomialcoefficientsor the primeexponentp are all multiplesof p and (a + ,3)P (acP iP) is thereforelearlyp timesa polynomialwhich s symmetricn a, ,3and has integral oefficients;hat is, (a + 3)- (aP + BP3)s p times n integer. Hence(19) aP + foP _a +f mod p.But, since ae andP are rootsof theequationX2 _ (aP + 3P)X + apfp = 0,it is a consequence fthetheory fsymmetricunctionsfthe rootsof analgebraic quationthatS(aP, Op)can be expressedn theform

S(aP, (P) = P(aP + (P", a


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44 R. D. CARMICHAEL.and therefore S(aog, 3P) -S(au, ) modp,as was to be proved.If m sany ntegerndq is an oddprime,wehavean identityftheform

(am - fm)Q = (am - 13qm) qalmlm(amn(q-2) - 13m(q-2)) +whence tfollowshat

(a - 3)-D q = Dmq + qI,where is an integer. Hence

Dmq -(a - 3)q-'Dmmod q.Hence,LEMMAI. If mis any ntegerndq is anyodd prime,wehaveDm (a- - 13) 'Dm mod q.In particular,

D =- (a - ) q-lDqa- (a* - 3)a( q-l)Dj mod q.Hence, sinceD1 = 1, it follows hatD a is divisibleby q whenand onlywhen (ca- 3)2 is divisible by q.TheoremIII gives exact informationoncerning he divisibility fDnand Snby2. We shallnowconsiderhequestionoftheentrance fanodd primefactor . If q is a factor f ao3 t follows rom heorem thatitdoesnotdivide ither AorS,. If t s a factor f a - 1)2 then tdividesD, as we readily ee from emmaII. In whatfollowswe shall considerthedivisibilityfDn and Sn by an odd primep which s not a divisor feither (a _ /3)2 or a13.If in equation 15) weputm = p andn = 1 wehave

D Sj - D1S, = 2aoD3pa4,or (a + j3)Dp -Sp = 2aceDp-.FromLemma I it follows hat

Dp =- (a - 13)Pl mod p,and from emma that Sp --a + 13 mod p.Hencefrom he astequationwehave

(a + 1)(a - 1)P-1 - (a + 1) 2a1Dp~1 mod p.Now (a _ 13)2 is an integer; nd thereforetfollows rom ermat'stheoremthat (a - 13)1 = 1 mod p.


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ON THE NUMERICAL FACTORS OF THE ARITHMETIC FORMS. 45Hence from heabove congruence e have the two casesD i_ O modp if (a- 3)P-l _ 1 modp,

a0D =_i - (a + A)mod p if (a- O)P-'=-1 mod p.Now it is easy to verifyhatDpsl -(ax + O3)Dp a0lDj = ;and hencewe see that

D,+ =_Omod p if (a- 3)P-1 -1 modp.Thereforewe have thefollowingheorem:*THEOREM XII. An odd prime p whichdoes not divide either a - 3)2or aof is a factor ofD 1 or ofD,+1 accordingas (a - (3) 1 is congruent o+ 1 or to - 1 modulop.Obviously, f a - 3 is an integer that is, if a and /3 re integers)wehave always thatD,1 is divisible y p.By meansofTheoremsX andXII weare nowtoprove result f funda-mental mportance. In order o be able to state this result uccinctlyweshall employa number-theoryunctionX,.(n)whichwe define elow. Itis convenient t the sametimeto define second functionop,(n)which sintimately elated oX,.(n).Let rs and r + s be any twointegers; hatis, let r and s be therootsofany quadratic equation ofthe form

.X2-ux + V = 0whereu and vareintegers. Whenp is an odd primewe define hesymbol(I ) by the congruence

(r s) P (1 ) modp,it being understood hat ( 8) is the residue of least absolute value;whence (rps) = 0, + 1, or-1 according s (r - 8)2 is divisibleby p,is a quadraticresidue f p, or is a quadraticnon-residuef p. The symbol(rY 8) is defined hus:( ') = 1, f s seven;( ,2s) = 0, ifrsis odd and r + s is even;( ) = -1, ifrsand + s areboth dd.

* Thistheorems due toLucas (1.c., pp. 290,296,297). Lucas's proof, owever,s differentfromhat bove.


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46 R. D. CARMICHAEL.Thenif n - paP al .a. pak,where i, P2, *--, pk are thedifferentrime actors fn,we define rs(n)

by the equation(P. (n) = fjpiaii [pi - (r's)]

This functions similar o one introduced y Lucas, 1. c., p. 300. It is,however, omewhatmore general. For r = 2 and s = 1 we have(P21(n) = spn),

where (n) is Euler's so-functionfn. The functionntroduced y Lucasdoes not have this interestingroperty f including he p-functions aspecialcase.The functional alue Xrs(n) is defined o be the east commonmultipleofthenumbers

-[P (pi)], i=1,2, ,k.It is obviousthatX,.(n) s a divisor fp,8(n).The functionsrs(n) and X,,(n)have several mportant roperties; utthis s not an appropriate lace to developthem n full.The fundamentalheoremo be provedmaynowbe statedas follows:THEOREMXIII. If thenumber ,

n = P laP2a2 . .. P kak,where 11 2, * P, arethedifferentrime actors fn, is prime o of ndif

X = B^(n),we have DA--0 mod n.To prove his heoremtis sufficiento show hatDA ontains hefactor

piai where is any number f the set 1, 2, *--,k. Thisfollows t oncefrom revious esults. For,X s a multiple fti,ti pi- [p- ( p d )1 =pi-1ki,

say. FromTheoremsXII and III and the remarkfollowing emma IIwe see thatDk, s in everycase divisibleby pi; and hencefromX thatDA is divisible y piai.COROLLARY.* If (p = (pa (n), then o e 0 mod n.

*Thiscorollarys essentiallyhesameas a certain undamentalesult ueto Lucas,1.c.,p. 300. It should enoted hatLucas'sstatementf his heoremsnotentirelyccurate.


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ON THE NUMERICAL FACTORS OF THE ARITHMETIC FORMS. 47In connectionwiththese simpletheorems oncerninghe divisorsofthenumbersn thesequenceD1,D2, ***it houldbe noticed hatno lawsofcorrespondingimplicitybtain n the case ofthesequenceSi, S2,

We have seen that an odd primep whichdoes not divide either a - p3)2or ad is a factor fD,, or of D,+i. But in the case of the sequenceSi, S2, ... it oftenhappensthat a given primenumber s not a factorof any term. Thus 7 is not a factorof Sn(2, 1), _ 2" + 1, foranyvalue of n. More generally, uppose that Dk, where k is odd,has anodd prime factorp while p is not a divisor f anyD, for v less thank.From TheoremVI it follows hatDm s divisible y p when nd onlywhenmis a multiple f k. Ifwe supposethatp is a divisor fS, forany givenvalue of n we shallbe led to a contradiction. For,sinceD2n DnSn,D2nis divisible by p; and therefore n is a multipleofk. But k is odd, and hencen is a multiple fk. Therefore n is divisibleby p; and Dn and Sn havethe commonodd prime factorp, which is impossible. Hence, an oddprimenumberp which dividesDk, wherek is odd, and does not divtide ny D, forv less thank, is not a factor of any Sn.4. On the Numerical Factors of the Forms Fk(cx,A).

We have already seen that the numbers Fk(a, ,3) are of fundamentalimportance in the factorization of Dn and Sno We turn thereforeto adetailed treatmentof these numbers.Let us suppose that F,(a, 3)Omodp, v >and that v is not a multiple of the primenumber p. Suppose that k is asubscriptfor which Fk(a,/) O mod p.Now* F. and Fk are divisors ofD. and Dk respectively,while the greatestcommon divisor of D, and Dk is Ds, where 8 is the greatest commondivisor of v and k. If we suppose that a is different rom v we shall beled to a contradiction; for,F, is then a factor ofD,/Ds, as we see from 5),whereas fromTheorem X it follows that Dl/Ds is not divisible by p sincep is a factorofDs and v/6 s primeto p. Hence 8 = v; and therefore is amultipleof v.We shall now show that FPa(a, A), > 0, is divisible by p but not by p2,exceptthatwhenp = 2, v = 3, F6may be divisibleby 22. [From TheoremIII it followsthat F6 is divisible by 2.] If we suppose that we do not havesimultaneously p = 2, v = 3, a = 1, we may proceed as follows: From

*Whennoconfusionanarisewe sometimes rite , forF, a, Al).


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48 R. D. CARMICHAEL.TheoremV wehave

Da= 7Fi(a, A),Dpa-i

where rangesover those divisorsof ypa whichcontainthe factor a.FromTheoremX it follows hatthefirstmemberfthis quation s divisibleby p but not byp2. Hence (only) oneofthe numbers i(a, d) of thesecondmembers divisibleby p and it is not divisibleby p2. Supposethatthisnumbers thatforwhich = k. Thenk is a multiple f a. But fromhediscussion n the preceding aragraphwe see that k is a multipleof P.Hencek = vpa, since his s the only ommonmultiple f v and pa occurringas a subscriptnthesecondnumber four equation.Fromthiswe conclude hateach of the numbers p, FVp * containsthe factor but that no one of themcontains 2, exceptthatwhenp = 2,P = 3, F6may contain 2.Now consider he number ,11Pa,whereA is greater hanunityand isprime o p. It is a divisorofDVXp-/DVPa;nd fromX it follows hatthelatternumber s not divisible y p. HenceFVEZPas prime o p.Let us supposethat F12= (a - )2, is divisibleby the odd primep.From the remarkfollowing emma II we see that each of the numbersFp, F,2, ... is divisible yp. Just s in the precedingrgumentwe mayshow thatno one of the numbers p2, p3, .. is divisibleby p2, and thatFla is not divisible y p if u s greater han 1 and is prime o p and a > 0.The example

a 1+ .6, A = 1- 16, (a - ,)2 = 24, F3 = a2+ a4+12 = 9shows hatF12may be divisible y p whileFp is at the sametimedivisibleby p2. If ji is greater han 1 and is prime op andiffurther., s divisibleby p, we see at once thatD,, andDp are bothdivisible yp-contrary tothe corollary o TheoremVI, which ssertsthatD,, and Dp are relativelyprime ince/i nd p are relatively rime. Hence FA is notdivisible y p.Now supposethatF12 s divisibleby 2. Then,since

F12 = (a - 1)2 = (a + 1)2 - 4aOit follows hat a + 13s divisible y 2. That is,F2 is divisibleby 2. Theexamplea = 2k + 1, 13 2- - 1 shows thatF2 may be divisibleby anypowerof2 whatever. By meansoftherelation

F2a = a2"2 + 12a1 = (a2a2 + 2a-2)2 - 2a2G212-itmaybe proved,however,hatF2a, > 1,is divisible y 2 butnotby22.To becontinued.

fibonacci factors

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