and decompose in cycles of length two

Transcription

1 Permutaton of Proceedng of the Natona Conference On Undergraduate Reearch (NCUR) 006 Domncan Unverty of Caforna San Rafae, Caforna Apr - 4, 007 that are gven by bnoma and decompoe n cyce of ength two Yeena Cruz Roado Department of Mathematc The Unverty of Puerto Rco at Humacao Humacao, Puerto Rco Facuty Advor: Dr Ivee Rubo Abtract Th paper tude bnoma that gve permutaton of In partcuar, tude the neceary and uffcent condton to obtan permutaton that decompoe n cyce of ength two Thee type of permutaton are uefu for appcaton to codng theory and cryptography The paper preent ome bnoma that are never permutaton and permutaton bnoma that never decompoe n cyce of ength two Furthermore, t gve the neceary condton for certan permutaton bnoma to decompoe n cyce of ength two Keyword: Permutaton, Cycc decompoton, Bnoma Introducton Let be the fnte fed wth = p r eement, where p a prme We tudy permutaton of gven by bnoma We tarted by tudyng bnoma wth term that are permutaton monoma of We proved that certan type of bnoma never gve permutaton of We ao tudy bnoma n [ x] of the form π (x) = x + ax We found neceary condton on thee bnoma o that the permutaton gven by them decompoe n cyce of ength two Ao we found the neceary condton on the bnoma o that the permutaton gven by them doe not decompoe n cyce of ength two Furthermore, we preent a conjecture that tate the uffcent condton for a permutaton bnoma to decompoe n cyce of ength two We tart by preentng ome background matera that needed for the ret of the paper fnte fed We are ntereted on permutaton of fnte fed A fed an agebrac tructure n whch the operaton of addton, ubtracton, mutpcaton and dvon (except dvon by cero) may be performed, and the ame rue whch are famar from the arthmetc of rea number hod A fnte fed a fed wth fntey many eement It can be proved that any fnte fed ha p r eement, where p a prme number and there a fnte fed wth p r for any pprme, rpotve nteger It eaer to tudy fnte fed f we wrte t eement a power of one eement: a prmtve root A prmtve root n an eement that generate a *:= {0} Th, α a prmtve root

2 n f *:= {α o,α,α,,α } It can be proved that every fnte fed ha prmtve eementα and the maet potve nteger uch thatα = Exampe The eement a prmtve root n Z becaue t generate a eement n Z *: 0 =, =, = 4, 3 = 8, 4 = 5, 5 = 0, 6 = 9, 7 = 7, 8 = 3, 9 = 6, 0 = Lemma Let a *, then a = Proof: Let a * andα be a prmtve root of Then a = α, for ome {0,, - } Th mpe that a = (α ) = (α ) = () = Lemma Let be an odd prme and et a Proof: Snce odd, even and Therefore, (a ) = 0 Factorng, we obtan (a * Then a {,} an nteger Let a * Then, by Lemma, a = )(a +) = 0 and therefore a = or a = Lemma 3 Let α be a prmtve root n * and be an odd prme Thenα = Proof: By Lemma, α ε{,} But nce the maet potve nteger uch thatα =, we mut haveα = permutaton We want to fnd bnoma that produce permutaton of A permutaton of a et a reorderng of the eement of the et In other word, a permutaton of a et X a functon π : X X that one to one and onto A bnoma that produce a permutaton caed a permutaton bnoma We want to tudy bnoma that produce permutaton of that decompoe n cyce of ength two The foowng propoton tate that to get permutaton of a fnte et t enough to check that the functon one to one Propoton Let A be a fnte et and conder f : A A Then f one to one f and ony f f onto Proof: We know that A a fnte fed Suppoe that f one to one, th mpe that each eement of A par up wth ony one eement of A Therefore a eement of A are covered Hence f onto Suppoe that f unto, th mpe that each eement of A covered Therefore a the eement are par up wth ony one eement of A becaue that way a the eement w be covered Hence f one to one A we w ee n the foowng propoton t enough to tudy bnoma of the form x +αx j, to fnd thoe that produce permutaton that decompoe n cyce of ength two Propoton The bnoma π (x) = ax + bx j a permutaton of f and ony f π (x) = x + α x j a permutaton of, whereα = b a Proof: Suppoe that π (x) = ax + bx j gve a permutaton of Then π (x) = ax + bx j one to one Th mpe that for every, where, we have that π () π () Th, a() + b() j a() + b() j Now dvdng both de by a, we have

3 π () = () + b a ()j () + b a ()j = π () Hence, mpe that π () π ()and π (x) one to one The other mpcaton can be proved mary Exampe The permutaton gven by the bnoma π ( x) = x + ax n , and the cycc decompoton of th permutaton : ( ) ( ) Oberve that we repreented the permutaton ung the foowng notaton, the frt row are the eement of the fed fnte Z and the econd row are ther mage under π Note that the ength of the cyce 5 We are ntereted n cyce of ength two We frt tudy how thee cyce are contructed On th exampe the cyce tart wth (we can chooe any eement for th) and contnue wth 5 Th ( ) becaueπ ( ) = 5 The thrd eement 3 becaueπ ( 5 ) = 3 In genera, we contruct the cyce by takng α and evauatng t nπ(x) whch gve u another eement or the ame that we evauated If π ( α ) = α then α a fxed pont and we do not wrte thee cyce of ength On the other hand, f the bnoma k π (x) produce another eement π ( α ) = α the cycc decompoton of th ( α α k ) Snce we have a dfferent eement we evauate t n the functon and repeat unt we get frt eement A cyce w ook ke( α π (α ) π (α ) π n (α ) = α ), where π (α ), Ζ, th mean that π compoe wth tef tme Note that f the permutaton gven by a bnoma n decompoe n cyce of ength two, then π (α ) = α for a 0 3 uadratc redue One of the bnoma n whch we focu π (x) = x + ax It know that π (x) a permutaton bnoma n F f and ony f a a uadratc redue moduo We have that an nteger a a uadratc redue moduo h f there ext another nteger b uch that a = b (mod h) We denote that a a uadratc redue byη (a) = Exampe 3 Note that a uadratc redue moduo 3, becaue 5 = (mod 3) Z Exampe 4 Note that that b = (mod 5) not a uadratc redue moduo 5, becaue there not another nteger b, uch Bnoma wth permutaton monoma a term The frt bnoma that we tudy are gven by permutaton monoma that decompoe n cyce of ength two Lou Cruz found the neceary and uffcent condton on the coeffcent uch that the a 3 monoma π (x) = ax and π (x) = ax are permutaton of that decompoe n cyce of ength We contructed bnoma wth thee monoma a term to ee f they ao produce permutaton of that decompoe n cyce of ength two By Propoton there are eentay three bnoma that are gven by 3 3 thee monoma and x : π (x) = x + ax, π (x) = x ax and x ax The bnoma 3 π (x) = x + ax 3 wa tuded by Cácere and Coón They preented the neceary and uffcent

4 3 condton for π (x) = x + ax over a fnte fed of odd charactertc to be a permutaton poynoma They determned when π ef nvertbe and hence decompoe n cyce of ength two If π not ef nvertbe they found the nvere Here we prove that the bnoma π (x) = x ax not a permutaton for a 0 For the thrd cae, n a of our exampe, the bnoma We preent th a a conjecture that reman to be proved 3 x ax never a permutaton for a 0 Lemma 4 Let be an odd prme, α a prmtve root n F and k Ζ If uch that ( ) / ( 4 + k) and = k Then α α n F Proof: Suppoe thatα = α n F Then (mod ) Th k (mod ), and th mpe that 4 ( ) k (mod ) Th ( ) ( 4 + k ( )) Hence ( ) ( 4 + k) and th contradct the hypothe, thereforeα α n Propoton 3 The bnoma π ( x) = x ax never gve a permutaton n F, for a 0 k Proof: Let k Ζ, α a prmtve root and a = α Ao et be uch that be ( ) / ( 4 + k) and uch that = k By Lemma 4, α α n F We w how that π ( α ) = π ( α ) and th w mpy that π not one to one and hence not a permutaton Now K = k mpe that + + α = α = and therefore k + = 0 Th mpe that: 0 = [( α ) α ][ + α α α ] + + k α ( ) = k k k k ( α ) α ( α )( α ) ( α ) + α ( α )( α ) = ( α ) ( α ) α ( α )( α ) ( α )( α ) + α ( α )( α ) = mutpyng by ( α ) ( α ) we have: 0 ( ) =, now 0 k 0 k 0 ( α ) ( α ) α ( α )( α ) α ( α ) + α ( α )( α ) π ( α ) π ( α ) Th mpe that 0 = π ( α ) π ( α ) and π ( α ) = π ( α ) Hence we found α α uch thatα α and π ( α ) = π ( α ) and therefore π not one on one n Conjecture Let be odd and > 5 The bnoma π( x) = x ax never a permutaton bnoma of, for a 0 4 Permutaton gven by π ( x) = x + ax that decompoe n cyce of ength The bnoma that we tred n Secton II where not permutaton bnoma or have been tuded before In th ecton we w work wth bnoma that are know to be bnoma permutaton and tudy the neceary and uffcent condton for them to decompoe n cyce of ength two Thee permutaton bnoma are gven by π ( x) = x + ax The foowng theorem tate when the bnoma π (x) are permutaton bnoma Theorem Gven an odd number = and ony f η(a ) = r p 3, the bnoma π ( x) = x + ax, a permutaton bnoma of f

5 We preent dfferent reut reated to thee permutaton bnoma, that were derved from the tudy of the neceary and uffcent condton on a uch that the permutaton decompoe n cyce of ength two Ao we etabh neceary and uffcent condton for the bnoma π(x) to have fxed pont Propoton 4 Let π ( x) = x + ax be a permutaton of F and et α be a prmtve root n Then α a fxed pont f and ony f even and a = 0 or odd and a = Proof: Letα be a prmtve root n Suppoe thatα a fxed pont of F Then π ( α ) = α Th ( α ) + aα = α Now we have that (( α ) α ) + aα = ( α ) + aα = α Th mpe that, f even, we have α + aα = α, whch mean that a = 0 Now f odd, factorzng we haveα (a ) = 0 whch mean that a = Let a = and be odd Then π ( α ) = ( α ) + α = (( α ) α ) + α = ( α ) + α = α + α = α Hence π ( α ) = α and α a fxed pont Let a = 0 and be even Then andα a fxed pont ( α ) = ( α ) + 0 α = (( α ) α ) = ( α ) α Hence π ( α ) = π = Propoton 5 Suppoe that π ( x) = x + x a permutaton of that decompoe n cyce of ength two Then 4 / ( ) Proof: We w prove by the counterpotve Let α be a prmtve root n Snce the coeffcent of x n π (x) a =, then from Propoton 4 we have that for a 0, where odd π ( α ) = α Snce there are pobe vaue for, th mean that there are vaue of that are odd and therefore +, fxed pont countng the fxed pont 0 Suppoe that π(x) decompoe n cyce of ength two, th mean that the eement that are not fxed pont are contan n cyce of ength two Therefore we have, cyce of ength two But th a contradcton becaue 4 / ( ) Hence π (x) 4 doe not decompoe n cyce of ength two n α Propoton 6 Let π (x) = x a permutaton of F Then π(x) decompoe n cyce of ength two f and ony f 4 ( ) Proof: ( ) We w prove the counterpotve Th, we w prove that f 4 / ( ), then π (x) doe not decompoe n cyce of ength two Suppoe that 4 / ( ) and etα be a prmtve root n Snce the coeffcent of x n π (x) a = 0, then from Propoton 4 we have that for a 0, where even π ( α ) = α Snce there are pobe vaue for, th mean that there are vaue of that are even and therefore +, fxed pont countng the fxed pont 0 Suppoe that π(x) decompoe n cyce of ength two, th mean that the eement that are not fxed pont are contaned n cyce of ength two Therefore we have, cyce of ength two But th a contradcton becaue 4 / ( ) 4 Hence π (x) doe not decompoe n cyce of ength two n

6 ( ) Let 4 ( ) and α be a prmtve root n F Suppoe that π (x) decompoe n cyce of ength two Th ( ) ( ) π ( π ( α)) = α, and (( α ) ) = (( α) α) = ( α ) = ( α ) ( α ) = ( ) ( α) ( α ) = ( α) Now, nce 4 ( ), th mean that 4k =, k Ζ and = k Th mean that = ( ) k = and π ( π ( α)) = α Hence π (x) decompoe n cyce of ength two Propoton 7 Let π ( x) = x + ax, a 0 be a permutaton of that decompoe n cyce of ength two, then a = Proof: Letα be a prmtve root n F Snce π(x) a permutaton that decompoe n cyce of ength two Then for a 0, that not a fxed pont we have that π ( α ) = α Th α α ) + aα ) + a(( α α ) + aα ) = (( α uch that (( α ) + aα ) + a(( α ) + aα ) = α Th ha to be true for a even and odd Therefore, when even or odd, we have that α ( a )( a ) = α or α ( a + )( a + ) = α whch mpe that a = 0 or a =, but thee coeffcent generate fxed pont Ao when odd or even we have that α ( a + )( a ) = α whch mpe that a = Hence f π(x) gve a permutaton of F that decompoe n cyce of ength two, then a = Studyng the neceary and uffcent condton for π(x) to decompoe n cyce of ength two we have reach an mportant conjecture, that etabh beow Conjecture Letα be uch that a a uadratc redue mod and uppoe that a = mod Then the bnoma π ( x) = x + ax decompoe n cyce of ength two n F or π ( x) = x ax decompoe n cyce of ength two n Propoton 8 There are fed for whch no bnoma of the form π ( x) = x + ax where a a uadratc redue, produce a permutaton of that decompoe n cyce of ength two Exampe 5 Z a fed for whch no bnoma of the form π ( x) = x + ax produce a permutaton that decompoe n cyce of ength two 4 Concuon and Work n Progre In th ecton, we preent neceary condton for the bnoma permutaton π ( x) = x + x to decompoe n cyce of ength two Ao we preented that certan type of bnoma never gve permutaton of Furthermore we preented a varety of conjecture The frt one tate that a certan type of bnoma never gve a permutaton and the other one tate the uffcent condton for a permutaton bnoma to decompoe n cyce of ength two We pan for our future work to fnd and characterze the fed for whch no bnoma of the form π ( x) = x + ax a permutaton bnoma of, that decompoe n cyce of ength two Ao we want to tudy other bnoma Lat but not eat we want to prove the conjecture or fnd a counterexampe

7 5 Acknowedgement Th work ha been upported n part by the Natona Securty Agency, Grant Number H C-0486; and the Natona Scence Foundaton CSEMS Program at the UPRH, Grant Number Reference Journa R Ld, H Nederreter, Encycopeda of Mathematc and t appcaton: Fnte Fed Cambrdge Unverty Pre, 997, 0 L Cruz, "Permutaton that Decompoe n Cyce of Length and are Gven by Monoma", Proceedng of the NCUR (Apr, 005) 3 A Cácere, & O Coón, Some Crtera for Permutaton Bnoma, (September, 997), Unverty of Puerto Rco at Humacao