An interesting result about subset sums. Nitu Kitchloo. Lior Pachter. November 27, Abstract

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1 A ieresig resul abou subse sums Niu Kichloo Lior Pacher November 27, 1993 Absrac We cosider he problem of deermiig he umber of subses B f1; 2; : : :; g such ha P b2b b k mod, where k is a residue class mod (0 < k ). If he umber of such subses is deoed N k he N k = 1 s 2 s s) s ) ( (k;s) s (k; s) ): Here ' deoes he Euler phi fucio ad is he Mobius fucio. This elaboraes o a resul by Erd}os ad Heilbro. We also derive a similar resul for ie abelia groups. 1 Iroducio Le A = f1; 2; : : : ; g. There have bee a umber of resuls i he pas abou how large a subse A A has o be so ha he sums of he elemes of A possess a cerai propery, [1], [2], [3]. I paricular, Erd}os ad Heilbro [2] proved he followig resul: Le be a posiive ieger, a 1 ; : : : ; a k disic residue classes mod, ad N a residue class mod. Le F (N; ; a 1 ; : : : ; a k ) deoe he umber of soluios of he cogruece where e 1 ; : : : ; e k ake he values of 0 or 1. e 1 a 1 + : : : + e k a k N mod Theorem 1 (Erd}os, Heilbro) Le a i be ozero for every i ad le p be a prime. The if k 3 p?2! 1 as p! 1. F (N; p; a 1 ; : : : ; a k ) = 2 k p?1 (1 + o(1)) 1

2 We cosider he relaed problem of explicily deermiig he umber of subses A A wih he propery ha he sum of he elemes of A is cogrue o k mod. Noe ha his is equivale o deermiig F (k; ; a 1 ; : : : ; a ) whe 0 < k. This follows if we accep he coveio ha he elemes of he empy se sum up o 0 mod. We will deoe F (k; ; a 1 ; : : : ; a ) by N. k Clearly N k 1. This is because for ay ; k, he subse fkg of A has he desired propery. Aoher subse of A wih his propery for 3; k = 0 is he subse B = fx 2 A : gcd(x; ) = 1g. This is a well kow resul. 2 Calculaio of N k Proposiio 2 Cosider he polyomial P (x) deed as follows: P (x) = =1 (1 + x ) = (+1) 2 r=0 a ;r x r : Le! = e 2i be a primiive h roo of uiy. The N k = 1 =1!?k P (! ): Proof: Noice ha each coecie of x r i P (x) is equal o he umber of subses of A ha sum o r. N k is he sum over he coecies of x r where divides r? k. Therefore, N k = :+k0 We will prove he proposiio usig (1) ad he followig Lemma: a ;+k : (1) Lemma 3 Le be a posiive ieger. The P?1 =0! = 0 whe 6 ad whe. Proof: Cosider he equaio x? 1 = 0. We facor his as (x? 1)(1 + x + x 2 + : : : + x?2 + x?1 ) = 0: Noe ha! is a roo of x? 1 for every. Hece i is a roo of he secod facor if ad oly if!? 1 6= 0. The resul follows. Now cosider (+1) 2 P (! ) = k=0 a ;k! k : 2

3 The =1!?k P (! ) =!?k =1 = = (+1) 2 r=0 (+1) 2 r=0 a ;r =1 :+k0 = (N k ): a ;r! r! (r?k) a ;+k Proposiio 4 P (! ) = 2 (;) if of ; (1 ). (;) i ad 0 oherwise. Here (; ) deoes he g.c.d. Proof: We shall rs prove wo echical lemmas ad he combie hem o obai he required resul. Lemma 5 Proof: Noe ha P (! ) = [P (;) (! (;) )](;) : P (! ) = = r=1 r=1 (1 + [! ] r (;) (;) ) (1 + [! (;) ] r (;) ): Now! (;) =! (;). Hece P(! ) = (;) r=1 (1 + [! (;) (;) Furhermore, ( ; ) = 1 so! is a primiive (;) (;) (;) rages from 1 o, he facors repea hemselves (; ) imes, i.e. Recallig ha ( (;) ; i P (;) (! (;) ). Hece, P P (! ) = [ (;) r=1 = [P (1 + [! (! (;) (;) (;) (;) ) = 1 we oice ha P (! (;) (;) (;) (;) (;) (! (;) (;) (;) ] r ): h roo of uiy. Therefore as r (;) )] (;) : ] r )] (;) ) = P (;) (! (;) ) 3 ) is us a permuaio of he facors

4 which gives he resul Lemma 6 P r (! r ) = 1? (?1) r : P (! ) = [P (;) (! (;) )](;) : Proof: Cosider he polyomial x r? 1. The 1;! r ;! r 2 ; : : : ;!r?1 r of his polyomial. Thus are he disic r roos Subsiuig x =?1 we ge i.e. 1? (?1) r = P k (! r ). Now This is equal o 2 (;) whe x r? 1 = (x? 1)(x?! r )(x?! 2 r ) (x?!r?1 r ): ((?1) r? 1) = (?1) r (1 +! r )(1 +! 2 r) (1 +! r r): (;) Proposiio 7 Suppose, =. The Proof: Firs oe ha! x Now rewrie! kx as! This is because reduces o ) (k;) ) k (;k) x (;k) P (! ) = [P (;) (! (;) )](;). Hece = [1? (?1) (;) ] (;) : i ad 0 oherwise.!?kx = ) (k;) ) (k;)! x : (k;) =! x. Also x ad?x are boh elemes of Z. Therefore!?kx =! kx :! kx = ) (k;) ) = (k;)!! k (;k) x (;k) k (;k) x (;k) : summads are ideical 8x 2 Z. Fially, sice ( k ; ) = 1 his (;k) (;k) ) (k;) ) which complees he proof of he proposiio. (k;) 4! x (k;)

5 Proposiio 8 2Z! = (): Proof: Le (x) deoe he h cycloomic polyomial. The P 2Z! egaive of he coecie of x )?1 i (x). is us he Claim 9 (x) = d (x m ) where = dm ad d is he produc of all he disic prime facors of. Proof: I is well kow ha (x) = Q r(x r? 1) (r). For a proof of his resul see [4], page 353. Now d (x m ) = Q sd(x s? 1) (s). If s ad s > d he s is divisible by he square of some prime ad so (s) = 0. Hece he claim. Claim 10 p (x) = (xp ) (x) if p is a prime ha does o divide. Proof: Oce agai we use he fac ha (x) = Q r(x r? 1) (r). I our case we have p (x) = rp = = (x p r rp:p=r? 1) (r) (x p r? 1) (r) rp:pr (x p r (x p? 1) () s(x s? 1) (sp) :? 1) (r) However is a muliplicaive fucio hece (sp) =?(s) so p (x) = ( (x p ))( (x))?1 : If p 2 for some prime p he by Claim 9 he coecie of x )?1 i (x) is 0. So assume ha = Q m p i=1 i, where he p i 's are disic. We ow use iducio o m ad Claim 10 o obai ha P 2Z! = (). Theorem 11 N k = 1 s 2 s s) s (k;s) )( s (k; s) ): Proof: Usig Proposiio 2 we obai ha N k = 1 =1!?k P (! ): 5

6 Now we use Proposiio 4 o obai Now le (; ) =. The N k = 1 N k = 1!?k : (;) odd 0 2 : odd 2 (;) :!?kx sice as x rages over Z, x rages over he elemes r such ha (r; ) =. Applyig Proposiio 10 we obai N k = 1 0 : odd 2 Fially, we use Proposiio 7 o coclude ha Subsiuig s = N k = 1 his reduces o N k = 1 : odd 2 s ) (k; )) 2 s ) (k; ) 1 C A! x (k; ) (k; ))( (k; )): s) s (k;s) )( s (k; s) ): For he case whe k = his formula ca easily be simplied o obai N = 1 s 2 s s): 3 A Theorem Abou Fiie Abelia Groups A aural geeralizaio of he problem discussed i he previous secio is a similar problem for ie abelia groups. Tha is, if G is a ie abelia group of order, we wa o calculae he umber of subses of G whose elemes sum up o he ideiy eleme (0) of G. For he purposes of his secio we will use he followig oaio: Le S deoe a k-uple of umbers, i.e. S = (s 1 ; s 2 ; : : : ; s k ). Give wo k-uples J ad N dee 0 < J N = 1 = 1 2 = 2 1 =1 2 =1 6 k = k k =1 : 1 C A :

7 Will will deoe he umber of subses of a ie abelia group G whose elemes sum up o 0 by N G. Theorem 12 Le G = Z 1 L Z2 L : : : L Zk be a ie abelia group of order = 1 2 k. Give a k-uple J dee T J =g.c.d.( 1 1 ; : : : ; k k ) ad le N = ( 1 ; 2 ; : : : ; k ). The N G = 1 [1? (?1) 0 < J N (;T J ) ](;T J) : We shall prove his heorem usig he same ideas as before. Proposiio 13 Cosider he polyomial F (x 1 ; x 2 ; : : : ; x k ) = 0 < S N (1 + x s 1 1 xs 2 2 x s k k ) = a x 1 1 x 2 2 x k k : The N G = 1 0 < J N F (! 1 1 ; : : : ;! k k ): Proof: The proof is ideical o ha of Proposiio 2. Proposiio 14 Proof: Noe ha! is i i =! F (! 1 1 ; : : : ;! k k ) = [1? (?1) i s i i F (! 1 1 ; : : : ;! k k ) =. Therefore = 0 < S N (;T J ) ](;T J) : (1 +! 1s 1 1! 2s 2 2! k s k k ) P i s i i i (1 +! ): 0 < S N Cosider he expoe i oe facor of he above produc for a xed S, i.e. i s i ( i i ) = T J ( i s i ( i i T J )): Claim 15 For every m (0 m ) here exiss a k-uple S such ha T J ( i s i ( i i T J )) T J m mod : 7

8 Proof: Noe ha g.c.d.( 1 ; : : : ; k 1 T J k T J ) = 1 ad herefore for ay ieger m here exiss s i 2 Z such ha s i i m = : i i T J Equivalely, T J m = T J ( i s i i i T J ): Now oe ha if ay s i is replaced by s i + i i he above equaio he we sill have equaliy (mod ). Thus every s i ca be chose o be less ha i. Therefore by he above claim we obai P i s i i i (1 +! ) = 0 < S N?1 m=0 (1 +! T J m = P (! T J ) = [1? (?1) ) (;T J ) ](;T J) ad so we have proved he proposiio. Proof (mai heorem): The heorem ow follows immediaely by combiig Proposiios 13 ad 14: 4 Furher Resuls N G = 1 0 < J N F (! 1 1 ; : : : ;! k k ) = 1 [1? (?1) 0 < J N (;T J ) ](;T J ) : Aoher problem relaed o he calculaio of N k is he calculaio of N ;m where 0 < m < (+1). N 2 ;m is deed o be he umber of subses B f1; 2; : : : ; g such ha P b2b b 0 mod m. We Remark ha N;m is easily obaied whe m. Proposiio 16 Le ; m be posiive iegers wih m. The N ;m = 1 m sm 2 s s): Proof: Usig Lemma 3 ad he same proof as give i Proposiio 2 we obai ha: N ;m = 1 m m =1 P (! m ): 8

9 Now 1 + (! m) m+i = 1 + (! m) i so he facors i P (! m) repea hemselves imes. Therefore m P (! ) = m [P m(! )] m m. Now we proceed as before o ge N ;m = 1 m sm 2 s s): Sevily, [5] has proposed he followig coecure: Coecure 17 The sequece fn;mg (+1) 2 m=1 is moooically decreasig. We also meio a ieresig coecio bewee our problem ad wo oher couig problems i combiaorics. Le C deoe he umber of circular sequeces of 0's ad 1's, where wo sequeces obaied by a roaio are cosidered he same. This problem is discussed i [6], page 75. The soluio is C = 1 )2 : This is ideical i form o our formula for N excep ha i our case we sum over all where i. Aoher relaed problem is he calculaio of he umber of moic irreducible polyomials of degree over a eld of q elemes where q is prime ([6], page 116). If he umber of such polyomials is deoed M q he M q = 1 d (d)q d : For q = 2 his has he exac same form as our formula for N k where (k; ) = 1. Oce agai, he oly dierece is ha our sum is over d such ha d i. Refereces [1] N. Alo ad G. Freima, O sums of a subse of a se of iegers, Combiaorica, 8(4) (1988), [2] P. Erd}os ad H. Heilbro, O he addiio of residue classes mod p, Aca Arihmeica, 9 (1964), [3] J. Olso, A addiive heorem modulo p, J. Combiaorial Theory, 5 (1968), [4] R. Dea, Classical Absrac Algebra, Harper ad Row, Publishers, New ork, [5] H. Sevily, persoal commuicaio. [6] J.H. va Li ad R.M. Wilso, A Course i Combiaorics, Cambridge Uiversiy Press

10 Niu Kichloo Deparme of Mahemaics MIT Cambridge, MA Lior Pacher Deparme of Mahemaics Calech Pasadea, CA 10