C O L L O Q U I U M .... M A T H E M A T I C U M \noindentVOL periodCOLLOQUIUM 86 .... 2 0 .... NO period\ 1h f i l l MATHEMATICUM INFINITE FAMILIES OF NONCOTOTIENTS \noindentBY VOL . 86 \ h f i l l 2 0 \ h f i l l NO . 1 A period .. F L A M M E N K A M P AND F period .. L U C A .. open parenthesis BIELEFELD closing parenthesis\ centerline {INFINITECOLLOQUIUMMATHEMATICUM FAMILIES OF NONCOTOTIENTS } Abstract periodVOL For .any 86 positive integer n let phi open parenthesis 2 0 n closing parenthesis be the NOEuler . 1 function \ centerline {BY } of n period A positive INFINITE FAMILIES OF NONCOTOTIENTS integer n is called a noncototient if the equation x minus phi open parenthesis x closing parenthesis = n has \ centerline {A. \quad FLAMMENKAMPANDFBY . \quad LUCA \quad ( BIELEFELD ) } no solution x period .. InA.FLAMMENKAMP AND F . L U C A ( BIELEFELD ) this note comma weAbstract give a sufficient . For any condition positive integer on a positiven let φ integer(n) be kthe such Euler that function the geometrical of n. A positive \ hspace ∗{\ f i l l } Abstract . For any positive integer $ n $ let $ \phi ( n ) $ progression openinteger parenthesisn is called 2 to a the noncototient power of if m the k closing equation parenthesisx − φ(x) = subn has m greater no solution equalx. 1 consistsIn this entirely be the Euler function of $ n . $ A positive of noncototients periodnote , .. we We give then a sufficient use computations condition on to a positive integer k such that the geometrical detect seven suchprogression positive integers(2mk) k≥ period1 consists entirely of noncototients . We then use computations \noindent integer $ n $ ism called a noncototient if the equation $ x − \phi For any positiveto integer detectn seven let .. such phi positive open parenthesis integers k. n closing parenthesis .. b e the Euler phi function .. of ( x ) = n$ hasnosolution $x .$ \quad In n period .. A For any positive integer n let φ(n) b e the Euler φ function of n. A this note , we give a sufficient condition on a positive integer $ k $ such that the geometrical positive integerpositive n is called integer a noncototientn is called a ifnoncototient the equation xif minusthe equation phi openx − parenthesisφ(x) = n has x closing no parenthesis = n has no solution . For example , n = 10, 26, 34, 50, 52, 58, 86, 10 are all noncototients . \noindent progression $ ( 2 ˆ{ m } k ) { m }\geq 1 $ consists entirely of noncototients . \quad We then use computations to solution periodSierpi For examplen ´ ski and comma Erd hungarumlaut n = 10 comma− 26os comma ( see B 34 36 commain [ 3 50 ] ) comma conj ectured 52 comma that 58 there comma 86 detect seven such positive integers $ k . $ comma 1 0 are allare noncototients infinitely many period noncototients . Sierpi n-acute ski and Erd hungarumlaut-oAn affirmative answer s open t parenthesis o the above see conj B 36 ecture in open was square given bracketby Browkin 3 closing and square \ hspace ∗{\ f i l l }For any positive integer $ n $ let \quad $ \phi ( n ) $ \quad b e the Euler bracket closing parenthesisSchinzel conjin [ 1 ectured ] . In thatthe above there arementioned infinitely paper , they showed that 2n · 509203 is a $ \phi $ f u n c t i o n \quad o f $ n . $ \quad A many noncototientsnoncototient period for all positive integers n. Their proof used the number An affirmative509203 answer int o an the essential above conj way ecture . In this was work given , we by extend Browkin the and result of [ 1 ] by giving \noindent positive integer $ n $ is called a noncototient if the equation $ x Schinzel in opena general square bracketmethod 1 for closing finding square numbers bracketk such period that In 2nk theis abovea noncototient mentioned for paper comma they − \phi ( x ) = n$ hasno showed that 2 to theall powerpositive of integers n timesn. 509203 As a corollary of our method , we have the following : is a noncototient forTheorem all positive . integersLet m n≥ period1 be .. a Their positive proof integer used. the number Then each of the numbers \noindent509203 in ansolution.Forexample essential way period In this work $, comma n we = extend 10 the , result 26 of open , square 34 bracket , 50 1 closing , square52 bracket, 58 by giving , 86 , 1 0$ areall noncototients. a general method forn finding= 2mk, numbers wherek k∈ such {509203 that, 2554843 2 to the, 9203917 power of, 9545351 n k is a, noncototient1645867, 11942443 for , \noindentall positiveS integers i e r p i n $ period\acute ....{ Asn} a$ corollary ski andof our Erd method $ commahungarumlaut we have− theo following$ s ( :see B 36 in [ 3 ] ) conj ectured that there are infinitely manyTheorem noncototients period65484763 .. Let m .} greater, is a noncototient equal 1 .. be . a positive integer period .. Then each of the numbers n = 2 to the power of m k commaThe where following k in proposition open brace provides 509203 thecomma theoretical 2554843 background comma 920391 for our 7 comma 954535\ hspace 1 comma∗{\ f i l 1 lTheorem 645867}An affirmative comma . 1 1 942443 answer comma t o the above conj ecture was given by Browkin and 65484763 closing braceProposition comma is a .noncototientLet k periodbe a positive integer satisfying the following four \noindentThe followingSchinzel proposition in provides [ 1 ] .the In theoretical the above background mentioned for our paper , they showed that $ 2 ˆ{ n } \cdotTheorem509203 period $ isProposition a noncototient period .. Let for .. k all .. be positive a positive integer integersconditions satisfying $ n the: following . $ \quad four Their proof used the number conditions : line − parenleft i )k is an odd prime . \noindentline-parenleft509203 i closing in parenthesis an essential k is an way odd prime . In period this work , we extend the result of [ 1 ] by giving ( ii )k is not a Mersenne prime . open parenthesis ii closing parenthesis k is not a Mersenne prime period 2000 Mathematics Subject Classification : 1 1 A 2 5 , 1 1 L 20 , 1 1 L 26 . \noindent2000 Mathematicsa general Subject method Classification for finding : 1 1 A 2 numbers 5 comma 1 $ 1 L k 20 $ comma such 1 that 1 L 26 period $ 2 ˆ{ n } k $ Work by the second author was supported by the Alexander von Humboldt Founda - isWork a noncototient by the second author for was supported by the Alexander von Humboldt Founda hyphen tion . tion period \noindentopen squareall bracket positive 37 closing integers square bracket $ n . $ \ h f i l l As a corollary of our method , we have the following : [37] \ hspace ∗{\ f i l l }Theorem . \quad Let $ m \geq 1 $ \quad be a positive integer . \quad Then each of the numbers
\ [ n = 2 ˆ{ m } k , where k \ in \{ 509203 , 2554843 , 920391 7 , 954535 1 , 1 645867 , 1 1 942443 , \ ]
\noindent 65484763 \} , is a noncototient .
\ hspace ∗{\ f i l l }The following proposition provides the theoretical background for our
\noindent Theorem .
\ hspace ∗{\ f i l l } Proposition . \quad Let \quad $ k $ \quad be a positive integer satisfying the following four
\ begin { a l i g n ∗} conditions : \end{ a l i g n ∗}
\ centerline { $ l i n e −parenleft $ i $ ) k$ is anodd prime . }
\ centerline {( ii $ ) k $ is not a Mersenne prime . }
\ centerline {2000 Mathematics Subject Classification : 1 1 A 2 5 , 1 1 L 20 , 1 1 L 26 . }
\ hspace ∗{\ f i l l }Work by the second author was supported by the Alexander von Humboldt Founda −
\noindent t i o n .
\ [ [ 37 ] \ ] 38 .. A period F L AMMENKAMP AND F period L U CA \noindentopen parenthesis38 \quad iii closingA . F parenthesis L AMMENKAMP .. The AND number F . 2 L to U the CA power of t k minus 1 is composite for all integers t greater equal 1 period \ centerlineopen parenthesis{( i ivi i closing ) \quad parenthesisThe number .. The number $ 2 ˆ 2{ kt is a} noncototientk − period1 $ is composite for all integers $ tThen\ thegeq number1 2 to. $the} power of m k is a noncototient for all positive integers m period P r o o f period38 .. AssumeA . F L that AMMENKAMP 2 to the powerAND F .of L mU CA k is a cototient for some m greater equal 1 period Write \noindentEquation:( open i v parenthesis ) \quad( iii 1The )closing numberThe parenthesis number $ 2t ..k − x k minus1 is $ composite isphi aopen noncototient for parenthesis all integers x closing .t ≥ 1. parenthesis = 2 to theThen power the of m number k period( iv ) $The 2 ˆ number{ m } 2kk is $ a noncototientis a noncototient . Then the for number all2 positivemk is a noncototient integers $ m . $Clearly commafor x greater all positive 2 period integers .... Inm. particularP r o o f . comma Assume phi thatopen 2 parenthesismk is a cototient x closing for some parenthesism ≥ 1. i s even periodP r o.... o From f . equation\Writequad openAssume parenthesis that 1 $ closing 2 ˆ{ parenthesism } k $ comma is a it cototient follows for some $m \geq 1that . $x i s Write even period
Write x = 2 to the power of alpha y where y is odd period .. Ifm y = 1 comma we get \ begin2 to the{ a l power i g n ∗} of m k = x minus phi open parenthesisx − φ x(x closing) = 2 parenthesisk. = 2 to the power of alpha(1) minus 2\ tag to the∗{$ power ( of 1 alpha ) minus $} x 1 =− 2 to \thephi power( of alpha x minus ) = 1 comma 2 ˆ{ m } k . \end{ a l i g n ∗} Clearly , x > 2. In particular , φ(x) i s even . From equation ( 1 ) , it follows which is impossiblethat periodx i s even .. So . comma y greater 1 period .. We distinguish two cases : Case 1 : y is squarefree period ..Write If y =x p= i 2sα primey where commay is odd we get . If y = 1, we get \noindent2 to the powerC l e aof r lm y k = $ x , minus x phi> open2 parenthesis . $ x\ closingh f i l l parenthesisIn particular = 2 to the $ power , of\phi alpha p( minus x ) $ i s even . \ h f i l l From equation ( 1 ) , it follows 2 to the power of alpha minus 1 open parenthesis2mk = px minus− φ(x) 1 = closing 2α − 2 parenthesisα−1 = 2α−1, = 2 to the power of alpha minus 1 open parenthesis p plus 1 closing parenthesis period \noindentIt now followsthat thatwhich $x$ m is greater impossible i equal s even . alpha So . minus, y > 11. andWe p = distinguish 2 to the power two cases of t k : minusCase 1 comma1 : y whereis t = m minus alpha plussquarefree 1 period . .. ThisIf y = p i s prime , we get \ centerlinecontradicts condition{Write open $ x parenthesis = 2 iiiˆ{\ closingalpha parenthesis} y$ period where $y$ is odd . \quad I f $ y m α α−1 α−1 =Assume 1 , now $ that we y get = p} 1 period2 periodk = x − periodφ(x) = p 2s commap − 2 where(p − 1) p =1 less 2 period(p + 1) period. period less p s are odd primes and It now follows that m ≥ α − 1 and p = 2tk − 1, where t = m − α + 1. This contradicts \ [ 2 ˆ{ m } k = x − \phi ( x ) = 2 ˆ{\alpha } − 2 ˆ{\alpha s greater equalcondition 2 period .. ( iiiWe ) get . − 1 } = 2 ˆ{\alpha − 1 } , \ ] 2 to the power of m k = 2 to theAssume power nowof alpha that py 1= periodp1...ps, periodwhere periodp1 < ... p < s minus ps are 2 odd to the primes power and of alpha minus 1 open parenthesiss ≥ 2. pWe 1 minusget 1 closing parenthesis period period period open parenthesis p s minus 1 closing parenthesis period \noindent which is impossible . \quad So $ , y > 1 . $ \quad We distinguish two cases : It now follows that m greater equal alpha2mk = minus 2αp1 1...ps and− 2α−1(p1 − 1)...(ps − 1). CaseEquation: $1 open : parenthesis y$ is 2 closing squarefree parenthesis . \ ..quad 2 p 1If period $y period = period p$ p s isprime minus open parenthesis ,weget p It now follows that m ≥ α − 1 and 1 minus 1 closing parenthesis period period period open parenthesis p s minus 1 closing parenthesis = 2 to the power\ [ 2 of ˆ{ tm k comma} k = x − \phi ( x ) = 2 ˆ{\alpha } p − 2 ˆ{\alpha − 1 } ( p − 1 ) = 2 ˆ{\alpha − 1 } ( p + 1 ) . \ ] where t = m minus alpha plus 1 period ..2p Since1...ps s− greater(p1 − 1) equal...(ps 2− comma1) = 2t thek, product .. open parenthesis(2) p 1 minus 1 closing parenthesis period period period open parenthesis p s minus 1 closing parenthesis .. is a multiple of 4 periodwhere ..t Since= m − 2 pα 1+ period 1. Since periods period≥ 2, the p s product is a multiple (p1 of− 21) but...( notps − of1) 4 comma is a multiple from equation \noindentopen parenthesisIt nowof 2 4 closing . follows Since parenthesis 2 thatp1...ps is it$ afollows m multiple\geq that of t 2 =\ but 1alpha period not of .. 4− Now , from1 comma equation$ and if we ( $set2 ) pz it = follows = 2 p 1 2 period ˆ{ t } periodk − period1 p s ,$ commathat t where= equation 1. Now $t open , if parenthesiswe = set mz =− 2 2p1 closing...ps, \alphaequation parenthesis+ ( 2 )b b1 ecomes ecomes . $ \quad This contradictsz minus phi open condition parenthesis ( z iiiclosing ) parenthesis . = 2 k comma which contradicts condition open parenthesis iv closingz − φ parenthesis(z) = 2k, period \ hspace ∗{\ f i l l }Assumenowthat $y = p 1 . . . p s ,$ where $p Case 2 : y is notwhich squarefree contradicts period condition .. Let p ( to iv the ) . power of beta bar y for some odd prime p and some 1 < ... < p s $ are odd primes and beta greater 1 period ..Case Since2 p : toy theis powernot squarefree of beta minus. Let 1 barpβ k phiy for open some parenthesis odd prime xp closingand some parenthesis comma it follows thatβ > 1. Since pβ−1 | φ(x), it follows that \noindentp to the power$ s of beta\geq minus 12 bar open . $ parenthesis\quad We x minus get phi open parenthesis x closing parenthesis closing parenthesis = 2 to the power of m k period pβ−1 | (x − φ(x)) = 2mk. \ [Since 2 ˆ{ pm is odd} k and k = is prime 2 ˆ{\ commaalpha it follows} p that 1 p = k. and . beta . = 2 period p s .. Now− write2 ˆ{\alpha − Since p is odd and k is prime , it follows that p = k and β = 2. Now write y = k2z 1 }y =( k to p the power 1 − of 2 z1 for some ) odd . squarefree . . integer ( p z such s that− k nmid1 z period) . ..\ If] z = 1 comma we for some odd squarefree integer z such that k z. If z = 1, we get get - Equation: open parenthesis 3 closing parenthesis .. 2 to the power of m k = x minus phi open parenthesis x \noindent It now follows that $ m \geq \alpha − 1 $ and closing parenthesis = 2 to the power2mk of= alphax − φ k(x to) = the 2α powerk2 − 2α of−1 2k( minusk − 1) 2 = to 2α the−1k power(k + 1) of. alpha minus(3) 1 k open parenthesis k minus 1 closing parenthesis = 2 to the power of alpha minus 1 k open parenthesis k plus 1 closing parenthesis\ begin { a l period i g n ∗}Equation ( 3 ) implies that m ≥ α − 1 and k + 1 = 2t, where t = m − α + 1. This \ tagEquation∗{$ ( open 2contradicts parenthesis ) $} 2 ( 3 ii closing ) p . 1parenthesis . . implies . that p m greaters − equal( alpha p minus 1 1− and k1 plus ) 1 = 2. to. the . power ( of t p comma s where− t1 =Assume m ) minus now = alpha that 2 plusz ˆ{ > 1t 1. periodWrite} kz = ,p1...ps for some odd primes p1 < ... \endThis{ a contradictsl i g n ∗} open parenthesis ii closing parenthesis period Assume now that z greater 1 period Write z = p 1 period period period p s for some odd primes p 1 less period Weget ... < ps. period\noindent period where $ t = m − \alpha + 1 . $ \quad Since $ s \geq m α 2 α−1 2Equation: , $ the period product period period\quad2 lessk$ p= ( s 2 periodk pp1...ps .. 1 We− 2− get 2k(1 tok − the1)( ) powerp1 − .1) of... m( .ps k− = .1) 2, to ( the power p s of alpha− k1 to) the$ power\quad ofi 2 s p a 1 period period period p s minus 2 to the power of alpha minus 1 k open parenthesis k minus 1multiple closing parenthesis of 4 . open\quad parenthesisSince p 1 $2 minus p1 closing 1 parenthesis . . period . p period s$ period is open amultiple parenthesis of2butnotof4 p s ,fromequation minus(2)it 1 closing followsthat parenthesis comma $t = 1 .$ \quad Now,ifweset $z = 2 p 1 . . . p s ,$ equation(2)becomes
\ [ z − \phi ( z ) = 2 k , \ ]
\noindent which contradicts condition ( iv ) .
\ hspace ∗{\ f i l l }Case $2 : y$ is not squarefree . \quad Let $ p ˆ{\beta } \ parallel y $ for some odd prime $ p $ and some
\noindent $ \beta > 1 . $ \quad Since $ p ˆ{\beta − 1 }\mid \phi ( x ) ,$ it followsthat
\ [ p ˆ{\beta − 1 }\mid ( x − \phi ( x ) ) = 2 ˆ{ m } k . \ ]
\noindent Since $p$ is oddand $k$ is prime , it follows that $p = k$ and $ \beta = 2 . $ \quad Now wri te $ y = k ˆ{ 2 } z $ for some odd squarefree integer $ z $ such that $ k \nmid z . $ \quad If $z = 1 ,$ weget
\ begin { a l i g n ∗} \ tag ∗{$ ( 3 ) $} 2 ˆ{ m } k = x − \phi ( x ) = 2 ˆ{\alpha } k ˆ{ 2 } − 2 ˆ{\alpha − 1 } k ( k − 1 ) = 2 ˆ{\alpha − 1 } k ( k + 1 ) . \end{ a l i g n ∗}
\noindent Equation ( 3 ) implies that $ m \geq \alpha − 1 $ and $ k + 1 = 2 ˆ{ t } ,$ where $t = m − \alpha + 1 . $ This contradicts ( ii ) .
\ hspace ∗{\ f i l l }Assume now that $ z > 1 .$Write$z = p 1 . . . p s$ forsomeoddprimes $p 1 < . . . $
\ begin { a l i g n ∗} \ tag ∗{$ . . . < p s . $} We get \\ 2 ˆ{ m } k = 2 ˆ{\alpha } k ˆ{ 2 } p 1 . . . p s − 2 ˆ{\alpha − 1 } k ( k − 1 ) ( p 1 − 1 ) . . . ( p s − 1 ) , \end{ a l i g n ∗} INFINITE FAMILIES OF NONCOTOTIENTS .. 39 \ hspaceor ∗{\ f i l l }INFINITE FAMILIES OF NONCOTOTIENTS \quad 39 Equation: open parenthesis 4 closing parenthesis .. 2 kp 1 period period period p s minus open parenthesis k minus\noindent 1 closingor parenthesis open parenthesis p 1 minus 1 closing parenthesis period period period open parenthesis p s minus 1 closing parenthesis = 2 to the power of t period \ begin { a l i g n ∗} Since s greater 1 comma the product open parenthesisINFINITE k minus FAMILIES 1 closing OF parenthesis NONCOTOTIENTS open parenthesis39 p 1 minus\ tag ∗{ 1$ closing ( parenthesis 4or ) $} period2 kp period 1 period . open . parenthesis . p p s s minus− 1 closing( k parenthesis− 1 is a )multiple ( of p 41 period− ..1 Since ) . . . ( p s − 1 ) = 2 ˆ{ t } . \end2 kp{ a 1l i period g n ∗} period period p s i s even but i s not a multiple of 4 comma it follows that t = 1 period .. If we t write w = 2 kp 1 period period period2kp p1 s...ps comma− (k equation− 1)(p1 − open1)... parenthesis(ps − 1) = 2 4. closing parenthesis(4) b ecomes \noindentEquation:Since open parenthesis $ s > 5 closing1 parenthesis ,$ theproduct .. w minus phi $( open parenthesis k − 1 w closing ) parenthesis ( p 1 = 2 −period1 ) .Since .s > .1, the ( product p ( sk − 1)(− p1 −11)...( )$ps − 1) is is a a multiple multiple of 4 . of Since 4 . \ 2quadkp1...psSince $2The only kp solution 1i s even of . open but . parenthesis i s not . a multiple p 5 closing s$ of 4 parenthesis i , it sevenbuti follows is that w =t = 4 comma snotamultiple 1. If which we write i s impossiblew = 2 of4kp1...ps, since , it w is follows a that multiple$ t = 1equation . $ \quad ( 4 ) bI ecomes f we writeof k period $w = 2 kp 1 . . . p s ,$ equation(4)becomes The Proposition is therefore completely proved period w − φ(w) = 2. (5) \ beginProof{ ofa l thei g n Theorem∗} period .. It suffices to construct numbers k with properties \ tagopen∗{$ parenthesis ( 5 i closing ) $} w parenthesis− \ endashphi open( parenthesis w ) iv = closing 2 parenthesis . .. of the above Proposition \end{ a l i g n ∗} The only solution of ( 5 ) is w = 4, which i s impossible since w is a multiple of k. period .. We first look at .. open parenthesisThe Proposition iii closing is parenthesis therefore completely period .. The proved first . one t o prove the existence of infinitely many positive integers k fulfilling open parenthesis iii closing parenthesis was \noindent The onlyProof solution of the Theorem of ( . 5 )It is suffices $w to construct = 4 numbers , $ whichk with properties i s impossible ( i ) since H period Riesel– open ( iv ) square of the bracket above 4 Proposition closing square . bracket We first period look at .. Since ( iii ) then . comma The first .. one such t onumbers have$w$ been is called a multiple Riesel numbers period o f $ k . $prove the existence of infinitely many positive integers k fulfilling ( iii ) was H . Riesel [ The .. smallest4 .. ] Riesel . Since number then known , such .. numbers is .. k = have 509203 been .. and called it ..Riesel is .. numbersconj ectured. The smallest open parenthesis see .. open square bracket 6 closing square bracket closing parenthesis .. that .. this .. is .. \ centerline {TheRiesel Proposition number known is therefore is k = completely 509203 and proved it is .conj} ectured ( see [ 6 the .. smallest .. Riesel] ) .. that number this period is .. the Four .. smallest years .. later Riesel comma number .. W period. Four years later , Sierpi n-acute ski .. open parenthesis see .. open square bracket 5 closing square bracket closing parenthesis Proof of the TheoremW . Sierpi .n ´ \skiquad ( seeIt suffices [ 5 ] ) showed to construct that there numbers exist infinitely $ k many $ with positive properties .. showed that .. there exist infinitely manym positive inte hyphen ( i ) −− ( i vinte ) -\quad gers k ofsuch the that above 2 k + 1Proposition i s composite for . \ allquadm ≥We1 first. The look proofs at of\ bothquad ( i i i ) . \quad The first one t o gers k such thatRiesel 2 to the’ s result power and of m Sierpi k plusn ´ 1ski i s ’ composite s result rely for on all an m idea greater from equal a 1 1 950 period paper .. of The Erd proofs of bothprove the existence of infinitely many positive integers $ k $ fulfilling ( iii ) was H. Riesel [hungarumlaut 4 ] . \quad− oSince s ( see then[ 2 ] ) . , \quad such numbers have been called Riesel numbers . Riesel quoteright sIn result what and follows Sierpi , n-acute we shall ski explain quoteright this idea s result in the rely case on an of idea the existencefrom a 1 950 of Riesel paper of TheErd\ hungarumlaut-oquad s m a l l e s st open\quad parenthesisRiesel see number open square known bracket\quad 2 closingi s \quad square$ bracket k = closing 509203 parenthesis $ \quad and i t \quad i s \quad conj ectured ( see \quad [numbers 6 ] ) . \quad that \quad t h i s \quad i s \quad the \quad s m a l l e s t \quad R i e s e l \quad number . \quad Four \quad years \quad l a t e r , \quad W. period s Suppose that (ai, mi) is a covering system of congruences , that is S iIn e r what p i follows $ \acute comma{n} we$ shall s k iexplain\quad thisi=1( idea see in\ thequad case[ of 5 the ] existence ) \quad ofshowed that \quad there exist infinitely many positive inte − , every positive integer n satisfies at least one congruence n ≡ ai ( mod mi) for some gersRiesel $k$ numbers periodsuch that $2ˆ{ m } k + 1$ i s composite for all $m \geq 1 . $ \quadi =The 1, ..., proofs s. Moreover of both , assume that the moduli mi are chosen in such a way that Suppose that ..Q opens parenthesism a sub i comma m sub i closing parenthesis sub i = 1 to the power of s .. is a Riesel ’ s resulti=1(2 andi − 1) Sierpi is divisible $ \ byacute at least{n}s $distinct ski primes ’ s result . By Hall rely ’ s on theorem an idea , one from a 1 950 paper of .. covering system .. of congruences comma .. that is comma m Erd $ hungarumlautcan exhibit−o$ a set of s(see[2]).s different primes pi for i = 1, ..., s such that pi | (2 i − 1). Hence every positive integer, in order n satisfies t o assure at least that one a positive congruence integer n equivk fulfills a sub ( iii iopen ) , it parenthesissuffices to choose mod mk subsuch i closing parenthesis that Infor what some follows i = 1 comma , we period shall period explain period thiscomma idea s period in .. the Moreover case comma of the assume existence that the of moduli m subRiesel i are chosen numbers in . such a way that product sub i = 1 to the power2a ofi k s− open1 ≡ 0(mod parenthesispi). 2 to the power of m i minus(6) 1 closing parenthesisSuppose that is divisible\quad by at$ least ( s a distinct{ i primes} , period m ..{ Byi } ) ˆ{ s } { i = 1 }$ \quad i s a \quad covering system \quad of congruences , \quad that i s , everyHall quoteright positiveIndeed s theorem integer , if condition comma $ n one ( $ 6 can ) is satisfies exhibitsatisfied a , set then at of 2 sm least differentk−1 i ones primesalways congruence divisiblep i for i = by 1 some comma $ n pi, periodhence\equiv perioda { i } period( $ mod comma $ s m it{ cani never} ) b $ e prime . The fact that the system of congruences ( 6 ) is solvable is forsome$isuch that p i baran open immediate = parenthesis 1 consequence , 2 . to the .of power the . Chinese of ,m i Remainderminus s .$1 closing Lemma\quad parenthesis . Moreover period , .. assume Hence comma that the moduli in$m order{ ti o assure}$ areMoreover that chosen a positive , the in Chinese integer k Remainder Lemma guarantees that all solutions of the system suchfulfills a open way parenthesis thatof congruences $ \ iiiprod closingˆ ({ 6parenthesis )s } form{ i an comma arithmetical = it 1 suffices} progression( to choose 2 ˆ{ with km such} first thati − 1 ) $ is divisible by at least $ s $ distinct primes . \quadQBys Qs Equation: openterm parenthesisk0 and difference6 closing parenthesisi=1 pi. ..Clearly 2 to the, k power0 i s coprime of a sub t i o k minusi=1 pi. 1 equivIn partic 0 open - ular parenthesis moduloHall p ’ i s closing theorem, parenthesis by Dirichlet , one period ’ can s theorem exhibit , onea can set find of infinitely $ s many $ differentk ’ s fulfilling primes both conditions $ p ( i $ for $i=1,...Indeed commaiii if condition ) and ( i ) open . Fromparenthesis ,s$ density 6 closing arguments parenthesis , it follows is satisfied that most comma of the then primes 2 to fulfilling the power of m ksuch minus that1 i s always $p( i divisible ) and i ( iii by\mid ) some are not( Mersenne 2 ˆ{ primesm } .i The− only1 condition ) . that $ i\ squad left tHence o check i , in order t o assure that a positive integer $ kp i$ comma hences therefore it can never ( iv b ) e. prime period .. The fact that the system of congruences open parenthesis 6 closingfulfills parenthesis ( iii ) , it suffices to choose $ k $ such that is solvable is an immediate consequence of the Chinese Remainder Lemma period \ beginMoreover{ a l i gcomma n ∗} the Chinese Remainder Lemma guarantees that all solutions of \ tagthe∗{ system$ ( of 6 congruences ) $} 2 .. ˆ open{ a parenthesis{ i }} 6k closing− parenthesis1 \equiv .. form0 an arithmetical ( \bmod progressionp i with ) first. \endterm{ a lk i gsub n ∗} 0 and difference product sub i = 1 to the power of s p i period .. Clearly comma k sub 0 i s coprime t o product sub i = 1 to the power of s p i period .. In partic hyphen \noindentular commaIndeed by Dirichlet , if quoteright condition s theorem ( 6 ) comma is satisfied one can find , theninfinitely $ many 2 ˆ{ km quoteright} k s− fulfilling1 $ bothi s always divisible by some $pconditions i open , $ parenthesis hence itiii closing can never parenthesis be and prime open . parenthesis\quad The i closing fact parenthesis that the period system .. From of congruences ( 6 ) densityis solvable arguments is comma an immediate it follows that consequence most of the of the Chinese Remainder Lemma . primes fulfilling open parenthesis i closing parenthesis and open parenthesis iii closing parenthesis are not Mersenne\noindent primesMoreover period .. , The the only Chinese condition Remainder Lemma guarantees that all solutions of thethat system i s left t of o check congruences i s therefore\ openquad parenthesis( 6 ) \quad iv closingform parenthesis an arithmetical period progression with first \noindent term $ k { 0 }$ and difference $ \prod ˆ{ s } { i = 1 } p i . $ \quad C l e a r l y $ , k { 0 }$ i s coprime t o $ \prod ˆ{ s } { i = 1 } p i . $ \quad In p a r t i c − ular , by Dirichlet ’ s theorem , one can find infinitely many $ k $ ’ s fulfilling both conditions ( iii ) and ( i ) . \quad From density arguments , it follows that most of the primes fulfilling ( i ) and ( iii ) are not Mersenne primes . \quad The only condition that i s left t o check i s therefore ( iv ) . 40 .. A period F L AMMENKAMP AND F period L U CA \noindentA few words40 about\quad computationsA . F L AMMENKAMP period .. If one AND starts F . with L U the CA covering sys hyphen tem of congruences open parenthesis 0 comma 2 closing parenthesis comma open parenthesis 0 comma 3 Aclosing few parenthesis words about comma computations open parenthesis . 1\quad commaIf 4 closing one starts parenthesis with comma the open covering parenthesis sys 3− comma 8 closingtemof parenthesis congruences comma open (0 parenthesis ,2) , (0 7 comma ,3) 1 2closing , (1 parenthesis ,4) , (3comma ,8) open parenthesis , (7 ,12) 23 comma , (23 ,24) , thenonecan choose the six primes $p i $ t ob e \{ 3,7,5,17,13,241 \} , resulting in the following 24 closing parenthesis40 commaA . F L then AMMENKAMP one can AND F . L U CA systemchoose the of six congruences primesA few p i words t o for b e about open $ k computationsbrace : 3 $ comma . 7 comma If one starts 5 comma with 1the 7 comma covering 1 3 sys comma - tem 241 of closing brace comma resultingcongruences in the following ( 0 , 2 ) , ( 0 , 3 ) , ( 1 , 4 ) , ( 3 , 8 ) , ( 7 , 1 2 ) , ( 23 , 24 ) , then one can \ centerlinesystem of congruences{ $choose k the\ forequiv sixk : primes1pi t ($mod$3 o b e { 3 , 7 , 5 , 1 7 ), 1 3 , , 241 k} , resulting\equiv in the1 following ( $ mod $ 7k equiv ) 1 , opensystem 2 parenthesis k of congruences\equiv mod 3 closing for1k : (parenthesis $ mod 5comma ) , } k equiv 1 open parenthesis mod 7 closing parenthesis comma 2 k equiv 1 openk ≡ 1 parenthesis ( mod 3), mod k ≡ 5 closing1 ( mod parenthesis 7), 2k ≡ comma1 ( mod 5 ) , \ centerline8 k equiv 1 open{ $ parenthesis 8 k8k ≡\ mod1equiv ( 1 mod 7 closing 17)1, parenthesis2 ($mod$17k ≡ 1 ( comma mod 13) 2, to 7 the223k power )≡ 1 of , ( 7mod k 2ˆ equiv 241{ )17 . open} parenthesisk \equiv 1 ($mod$1 3 ) , 2ˆ{ 2 3 } k \equiv 1 ($ mod241) . } mod 1 3 closing parenthesisThis leads comma t o the 2 to arithmetical the power of progression 2 3 k equiv for 1 openk with parenthesis first t erm modk0 241= 509203 closing andparenthesis period difference 3 · 5 · 7 · 13 · 17 · 241. A quick computer check revealed that 509203 satisfies \noindentThis leadsThis t o theconditions leads arithmetical t ( i o ) ,the progression ( ii ) arithmetical and ( for ivk ) aswith well first progression . t ermCondition k sub for0( iv= 509203) needed $ k $ 2 sec with of CPU first - t t erm $ kand{ difference0 } = 3ime times 509203 . 5 The times next$ 7 times number 1 3 intimes the 1 above 7 times progression 24 1 period which .. A satisfies quick computer conditions check ( i ) revealed , ( ii that and509203 difference satisfies) conditions and $ ( 3iv ) open i\ scdot 65484763 parenthesis5 . i Condition\ closingcdot parenthesis (7 iv ) now\cdot comma needed open1 40 min parenthesis 3 of\ CPUcdot ii- t closing ime1 . parenthesis 7 \cdot and24 open 1 parenthesis . $ \quadThe iv closing otherA quick 5parenthesis values computer of k asclaimed well check period by the revealed .. Theorem Condition were that open found parenthesis choosing iv other closing systems parenthesis needed509203 2 satisfiesof covering conditions congruences ( .i ) , ( ii ) and ( iv ) as well . \quad Condition ( iv ) needed 2 secsec ofo f CPU CPU hyphen− tFinally ime t ime . period ,\ aquad few .. wordsThe The next nextabout number checkingnumber in the for in abovecondition the progressionabove ( iv ) progression . which Suppose satisfies that which 2k is satisfies a conditionsconditions open (cototient i parenthesis ) , ( . If iix ii closing) s anda solution parenthesis ( iv of ) i comma s 65484763 open parenthesis . \quad iiCondition closing parenthesis ( iv ) and now open needed 40 min parenthesiso f CPU − ivt closing ime .parenthesis i s 65484763 period .. Condition open parenthesis iv closing parenthesis now needed 40 min Theof other CPU hyphen 5 values t ime period of $ k $ claimed byx − theφ(x Theorem) = 2k, were found choosing (7) otherThe other systems 5 values of of covering k claimed by congruences the Theorem were . found choosing other systems ofthen covering , since congruencesφ(x) i s even period , it follows that x i s even as well . Moreover , it i s easy FinallyFinally comma, a fewt a o few seewords that words aboutx i about s squarefree checking checking . for Since for condition conditionφ(x) ≤ openx/2 whenever parenthesis ( iv )x .i iv s\quad even closing , anySuppose parenthesis solution that of period $ .. 2 Supposek $ that 2 k ( 7 ) satisfies x ≤ 4k. Since the largest value of k t ested satisfies isis a cototient cototient period . If If x i s $ a x solution $ i of s a solution1 of 0 Equation: open parenthesis 7 closing parenthesis .. x minus phi openY parenthesis x closing parenthesis = 2 k comma\ begin { a l i g n ∗} x ≤ 4k ≤ 4 · 65484763 < qi = Q, \ tag ∗{$ ( 7 ) $} x − \phi ( x ) = 2 k , then comma .. since phi open parenthesis x closing parenthesis i s eveni comma= 1 .. it follows that x i s even as well\end period{ a l i g .. n ∗} Moreover comma it i s where q1, ..., q10 are the first 10 primes , it follows that easy t o see that x i s squarefree period .. Since phi open parenthesis x closing parenthesis less or equal x slash 2\noindent whenever x ithen s even , comma\quad anys i n c e $ \phi ( x )$ iseven, \quad it follows that φ(x) > φ( Q) · x > 0.163588x $x$solution i of s open even parenthesis as well 7 . closing\quad parenthesisMoreover satisfiesQ , it x i less s or equal 4 k period .. Since the largest value easy t o see that $ x $ i s squarefree . \quad Since $ \phi ( x ) \ leq of k t ested satisfiesin the tested range . Hence , if x satisfies equation ( 7 ) , then x1 0 / 2$ whenever $x$ i seven ,any solution of ( 7 ) satisfies $ x \ leq 4 k . $ \quad Since the largest value of Line 1 x less or equal 4 k less or equal 4 times2k = 65484763x − φ(x) < less(1 product− 0.163588) q i =x, Q comma Line 2 i = 1 $ kwhere $ tq 1 ested comma satisfies period period period comma q 1 0 are the first 10 primes comma it follows that or x > 2.39k. To summarize , in order t o check ( iv ) , we wrote a computer phi open parenthesis x closing parenthesis greater phi open parenthesis Q Q closing parenthesis times x greater program which checked that equation ( 7 ) has no even squarefree solution 0\ centerline period 1 63588{1 x 0 } in the interval [2.39k, 4k] for every Riesel number k satisfying conditions ( i ) in the tested range period Hence comma if x satisfies equation open parenthesis 7 closing parenthesis comma and ( ii ) . then\ [ \ begin { a l i g n e d } x \ leq 4 k \ leq 4 \cdot 65484763 < \prod q The Theorem i s therefore proved . i2 k= = x Q minus , phi\\ open parenthesis x closing parenthesis less open parenthesis 1 minus 0 period 163588 closing It is interesting t o point out that all the known noncototients are parenthesisi = x 1comma\endRemark{ a l i g n e d . }\ ] even . To see why this is not coincidental , assume that y i s an odd or x greater 2 period 39 k period .... To summarize comma in order t o check open parenthesis iv closing noncototient . If y − 1 can b e written as a sum of two distinct primes p and q, then parenthesis comma we wrote a computer \noindentprogram whichwhere checked $q that 1equation , open . parenthesis . . 7 ,closing q parenthesis 1 0$ .... has arethe no even first10primes squarefree solution ,it followsthat pq − φ(pq) = y, in the interval open square bracket 2 period 39 k comma 4 k closing square bracket for every Riesel number k satisfying\ [ \phi conditions( x open ) parenthesis> \phi i closing( parenthesis{ Q } Q) \cdot x > 0 . 1 63588 x \and] open parenthesis ii closing parenthesis period The Theorem i s therefore proved period Remark period .. It is interesting t o point out that all the known noncototients \noindentare .. evenin period the .. tested To .. see range why this . .. Hence is .. not , .. if coincidental $ x $ comma satisfies .. assume equation that .. y i ( s 7.. an ) odd , then noncototient period .. If y minus 1 can b e written as a sum of two distinct primes p and \ [q 2 comma k then = x − \phi ( x ) < ( 1 − 0 . 163588 ) x , \pq] minus phi open parenthesis pq closing parenthesis = y comma
\noindent or $ x > 2 . 39 k . $ \ h f i l l To summarize , in order t o check ( iv ) , we wrote a computer
\noindent program which checked that equation ( 7 ) \ h f i l l has no even squarefree solution
\noindent inthe interval $[ 2 . 39 k , 4 k ]$ for every Rieselnumber $ k $ satisfying conditions ( i )
\noindent and ( i i ) .
\ centerline {The Theorem i s therefore proved . }
Remark . \quad It is interesting t o point out that all the known noncototients are \quad even . \quad To \quad see why t h i s \quad i s \quad not \quad coincidental , \quad assume that \quad $ y $ i s \quad an odd noncototient . \quad I f $ y − 1 $ can b e written as a sum of two distinct primes $ p $ and $ q , $ then
\ [ pq − \phi ( pq ) = y , \ ] INFINITE FAMILIES OF NONCOTOTIENTS .. 4 1 \ hspacewhich∗{\ contradictsf i l l }INFINITE the fact that FAMILIES y i s a noncototient OF NONCOTOTIENTS period .. Since\quad y minus4 1 1 i s even comma Goldbach quoteright s conj ecture asserts that y minus 1 .. can be written as a sum of two \noindentprimes commawhich .. at contradicts least when y greater the equal fact 5 thatopen parenthesis $ y $ notice i s a that noncototient .. 1 .. and 3 are . not\quad noncototientsSince closing$ y parenthesis− 1 $ period i s even , Goldbach ’ s conj ecture asserts that $ y − 1 $ \quad can be written as a sum of two Of course comma in Goldbach quoteright s conj ectureINFINITE one allows FAMILIES the two OF primes NONCOTOTIENTS t o b e equal period4 1 primesHowever , comma\quadwhich ..at a .. contradicts least conj ecture when the .. fact of $ Hardy thaty y ..\igeq and s a noncototient Littlewood5 ( $.. . asserts notice Since ..y that− that1 ..i s every even\quad large, Goldbach1 \quad and 3 are not noncototients ) . Ofenough course even , number in’ s conj Goldbach open ecture parenthesis asserts ’ s conj that probablyy ecture− 1 larger can one than be written allows 1 2 closing as the a parenthesissum two of two primes can primes be t written , o bat eleast as equal a sum of . Howevertwo primes , in\quad morewhena thany \quad one≥ 5 wayconj ( noticeperiod e c t that In u r particular e \ 1quad and commao f 3 Hardyare it not can\ noncototients bequad writtenand as Littlewood) a. sum Of course\ ,quad in a s s e r t s \quad that \quad every l a r g e enoughof two distinct even numberGoldbach primes period ( ’ s probably conj .. Thus ecture comma larger one allows it seems than the reasonabletwo 1 2 primes ) can t t o o conjb be e equal written ecture . However that as comma a , sum a in fact of conj comma twothere primes are no odd inecture morenoncototients ofthan Hardy one period and way .. Littlewood Notice . In also particular that asserts since that , it every can large be written enough even as number a sum ofp totwo the distinct power( of probably k minusprimes larger phi . open\ thanquad parenthesis 1 2Thus ) can be ,p written to it the seems power as a sum reasonable of k of closing two primes parenthesis t oin more conj = than p ecture to one the way power that of , k in fact , minusthere 1 comma are no. odd In particular noncototients , it can be . written\quad asNotice a sum of also two distinct that primes since . Thus , it seems for all k greaterreasonable equal 1 and t o all conj prime ecture numbers that , in p comma fact , there it follows are no that odd a noncototients noncototient . can Notice also \ [never p ˆ{ bek a} power −that of \since aphi prime period( p ˆ{ k } ) = p ˆ{ k − 1 } , \ ] We also point out that the authors of open square bracket 1 closing square bracket asked for the lower density of pk − φ(pk) = pk−1, \noindentthe set of noncototientsf o r a l l $ period k \ ..geq The above1 $ heuristic and all reasoning prime seems numbers t o indicate $ p , $ it follows that a noncototient can for all k ≥ 1 and all prime numbers p, it follows that a noncototient can never be neverthat the be upper a power density of of a this prime set is at . most 0 period 5 period a power of a prime . Acknowledgements period .. We thank Marek W o-acute j t owicz for pointing out t o We also point out that the authors of [ 1 ] asked for the lower density of the set Weus also reference point open out square that bracket the 1 authors closing square of [ bracket 1 ] asked period for the lower density of of noncototients . The above heuristic reasoning seems t o indicate that the upper theWe setalso thank of noncototients Professor Andreas . Dress\quad andThe his research above group heuristic in Biele reasoning hyphen seems t o indicate density of this set is at most 0 . 5 . thatfeld for the their upper hospitality density during of the this p eriod set when is this at paper most was 0 . written 5 . period Acknowledgements . We thank Marek Wo ´ j t owicz for pointing out t o us REFERENCES reference [ 1 ] . Acknowledgementsopen square bracket . 1 closing\quad squareWe thank bracket Marek .... J period W $ B\acute r ow k{ io n} and$ A j period t owicz S c h ifor n z e pointing l comma On out t o We also thank Professor Andreas Dress and his research group in Biele - feld for integersus reference not of the form [ 1 n ] minus . phi open parenthesis n closing parenthesis comma Colloq period Math period their hospitality during the p eriod when this paper was written . open square bracket 2 closing square bracket 68 P period open parenthesis E sub r d o-hungarumlaut s comma REFERENCES Weto the also power thank of 1 995 Professor closing parenthesis Andreas comma Dress 55 endash and his 5 On research sub integers group to the power in Biele of 8 period− of the form [ 1 ] J . B r ow k i n and A . S c h i n z e l , On integers not of the form n − φ(n), Colloq . 2feld to the forpower their of k plus hospitality p and related duringproblems the comma p Summaeriod whenBrasil period this paperMath period was written . Math . 2 open parenthesis 1 9501995) closing, parenthesis comma 1 1 3 endash 1 23 period [2] 68P.( 55−−5On8. of the form 2k+p and related problems \ centerlineopen square{ bracketREFERENCES 3 closingE } squarerdo−hungarumlauts, bracket .. R period K periodintegers G u y comma Unsolved Problems in Number , Summa Brasil . Math . Theory comma Springer comma 1 994 period 2 ( 1 950 ) , 1 1 3 – 1 23 . \noindentopen square[ 1bracket ] \ h 4 f i closing l l J . square Br bracketowk i .... n andAH period . R S i c e hs e i l n comma z e Nl oslash ,On integersgra stora primtal not of the form [3] R.K.Guy, Unsolved Problems in Number Theory , Springer , 1 994 . open$ n square− bracket \phi Some( large n primes ) closing ,$ squareColloq.Math. bracket comma Elementa 39 open parenthesis 1 956 closing [4] H.Riesel, N øgra stora primtal [ Some large primes ] , Elementa 39 ( 1 956 ) , 2 58 parenthesis comma 2 58 endash 260 – 260 \noindentopen parenthesis$ [ in 2Swedish ] closing 68{ P parenthesis} .( period{ E }ˆ{ 1 995 ) , } { r d o−hungarumlaut ( in Swedish ) . sopen , } square55 bracket−− 5 closing5 { On square}ˆ{ bracket8 . ....} W{ periodi n t e g Se r i s e r} p$ i n-acute of the s k form i comma $2ˆ Sur un{ problk } e-grave+ p $ [5] W.Sierpi n´ s k i , Sur un probl e` me concernant les nombres k · 2n + 1, Elem . meand concernant related les problems nombres k , times Summa 2 to Brasil the power . of Math n plus . 1 comma Elem period .... Math period Math . 1 5 open parenthesis 1 960 closing parenthesis comma 73 endash 74 semicolon Corrigendum comma ibid period 1 5 ( 1 960 ) , 73 – 74 ; Corrigendum , ibid . 1 7 ( 1 962 ) , 85 . 1\ centerline 7 open parenthesis{2(1950) 1 962 closing ,113 parenthesis−− comma1 23 85 . period} [ 6 ] The Riesel Problem , http : / / vamri . xray . ufl . edu / proths / rieselprob . html . open square bracket 6 closing square bracket .. The Riesel Problem comma http : slash slash vamri period FSP Mathematisierung Universit a¨ t Bielefeld xray\noindent period ufl[ period 3 ] \ eduquad slashR proths . K . slash G u rieselprob y , Unsolved period html Problems period in Number Theory , Springer , 1 994 . Postfach 1 0 1 31 FSP Mathematisierung 33 501 Bielefeld , Germany E - mail : achim @ uni - bielefeld . de \noindentUniversit dieresis-a[ 4 ] \ th Bielefeld f i l l H . R i e s e l , N \o gra stora primtal [ Some large primes ] , Elementa 39 ( 1 956 ) , 2 58 −− 260 fluca @ mathematik . uni - bielefeld . de Postfach 1 0 1 31 Received 2 July 1 999 ( 3786 ) \ centerline33 501 Bielefeld{( in comma Swedish Germany ) . } E hyphen mail : achim at uni hyphen bielefeld period de \noindentfluca at mathematik[ 5 ] \ h period f i l l W uni . hyphen S i e bielefeld r p i period $ \acute de {n} $ sk i , Surunprobl $ \grave{e} $ meReceived concernant 2 July les 1 999 nombres .. open parenthesis $ k \ 3786cdot closing2 parenthesis ˆ{ n } + 1 , $ Elem . \ h f i l l Math . \ centerline {15(1960 ) ,73 −− 74 ; Corrigendum , ibid . 1 7 ( 1 962 ) , 85 . }
\noindent [ 6 ] \quad The Riesel Problem , http : / / vamri . xray . ufl . edu / proths / rieselprob . html .
\noindent FSP Mathematisierung U n i v e r s i t $ \ddot{a} $ t Bielefeld
\noindent Postfach 1 0 1 31
\noindent 33 501 Bielefeld , Germany E − mail : achim $@$ uni − bielefeld . de
\ centerline { fluca $ @ $ mathematik . uni − bielefeld . de }
\ hspace ∗{\ f i l l } Received 2 July 1 999 \quad ( 3786 )