Counting Solutions to Discrete Non-Algebraic Equations Modulo Prime Powers

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1 Rose-Hulman Insttute of Technology Rose-Hulman Scholar Mathematcal Scences Techncal Reorts (MSTR) Mathematcs Countng Solutons to Dscrete Non-Algebrac Equatons Modulo Prme Powers Abgal Mann Rose-Hulman Insttute of Technology, Advsors: Joshua Holden Follow ths and addtonal works at: htt://scholar.rose-hulman.edu/math_mstr Part of the Informaton Securty Commons, and the Number Theory Commons Recommended Ctaton Mann, Abgal, "Countng Solutons to Dscrete Non-Algebrac Equatons Modulo Prme Powers" (2016). Mathematcal Scences Techncal Reorts (MSTR) htt://scholar.rose-hulman.edu/math_mstr/153 MSTR Ths Dssertaton s brought to you for free and oen access by the Mathematcs at Rose-Hulman Scholar. It has been acceted for ncluson n Mathematcal Scences Techncal Reorts (MSTR) by an authorzed admnstrator of Rose-Hulman Scholar. For more nformaton, lease contact wer1@rose-hulman.edu.

2 Countng Solutons to Dscrete Non-Algebrac Equatons Modulo Prme Powers Abgal Mann May 20, 2016 Abstract As socety becomes more relant on comuters, crytograhc securty becomes ncreasngly mortant. Current encryton schemes nclude the ElGamal sgnature scheme, whch deends on the comlexty of the dscrete logarthm roblem. It s thought that the functons that such schemes use have nverses that are comutatonally ntractable. In relaton to ths, we are nterested n countng the solutons to a generalzaton of the dscrete logarthm roblem modulo a rme ower. Ths s acheved by nterolatng to -adc functons, and usng Hensels lemma, or other methods n the case of sngular lftng, and the Chnese Remander Theorem.

3 1 Introducton Socety has become ncreasngly relant on comuters for storng nformaton and communcatng securely. Peole exect that the crytograhc schemes currently n use wll kee ther nformaton confdental and wll allow them to verfy the authentcty of any ece of nformaton that they see. Publc key crytograhy schemes nvolve functons that are easy to comute one way usng a ublcly avalable key (to encryt or verfy sgnatures), but have nverses that are dffcult to comute wthout a rvate key, so that decryton or creatng a sgnature s only feasble for one user. Crytograhc schemes such as Dffe- Hellman key exchange and ElGamal encryton and sgnature schemes often use exonental modular mangs lke the dscrete exonentaton ma f : Z Z/Z, where x g x (mod ) and g Z, a rme. These are used snce they are generally beleved to be comutatonally nfeasble to nvert for large rme [10, Chater 7]. However, the securty of these schemes s stll beng analyzed, snce any nsght nto ther structure may reveal a vulnerablty. There has been revous analyss of the mas x g x (mod ) and x g x2 (mod ) usng functonal grahs n [8], [4], and [12]. Camensch and Stadler look at the double dscrete logarthm of y, g ax y (mod c) as well as the nth root of the dscrete logarthm of z, g xn z (mod c), where x, a, n, c Z, and g, y are n a cyclc grou G, for use n crytograhc sgnature schemes where there are multle keys that allow for the revelaton of artal nformaton [3]. We study the nth roots of dscrete logarthms n ths aer by countng nteger solutons to g xn x k (mod e ), where g, n, k, e Z, s a rme, and g. Ths may gve us some nsght nto the structure of nth roots of dscrete logarthms. Although t s not drectly used n any crytograhc schemes today, one may be bult off of ths equaton f ts structure acts suffcently random. The dea for countng solutons to these tyes of congruences was nsred by [6], whch uses -adc nterolaton, Hensel s lemma, and the Chnese remander theorem. Ths tye of analyss can also be found n [9] and [11], whch ales these methods to the Welch Equaton and the Dscrete Lambert ma. In ths aer we fnd that for x n a certan range, we can determne the exact number of solutons to g xn x k (mod e ) when k and when = k and n = Termnology and Background For ths aer, we count solutons to g xn x k (mod e ), where g, n, k, and e are fxed ntegers, s a rme, and g. In order to count solutons to our congruence modulo e for all ostve ntegers e, we wll fnd -adc ntegers helful, snce each -adc nteger descrbes our soluton modulo e for all e. Thus we wll be usng functons on the -adcs, or Q, whch are the comleton of Q under the -adc metrc. Frst, we note the defnton of the -adc valuaton of a ratonal number from [5, Secton 2.1]. 1

4 Defnton 1. Fx a rme number Z. The -adc valuaton on Z s the functon v : Z {0} R defned as follows: for each nteger n Z, n 0, let v (n) be the unque ostve nteger satsfyng n = v(n) n wth n. We extend v to the feld of ratonal numbers as follows: f x = a/b Q, then whch s well-defned. v (x) = v (a) v (b), We can now defne the -adc absolute value as follows: Defnton 2. For any x Q, we defne the -adc absolute value of x by f x 0, and we set 0 = 0. x = v(x) The comleton gves us all the ratonal -adc numbers, whle we need only to use a subset of Q. From [5, Secton 3.3], we fnd that the -adc ntegers Z are defned as Z = {x Q : x 1}. Now that we have defned Z, we let µ 1 be the set of all ( 1)-st roots of unty, where µ 1 Z by [5, Cor ]. As stated n [5, Cor ], we can wrte each element of Z unquely as an element of µ 1 (1 + Z ). So for each x Z we wrte x = ω(x) x for some ω(x) µ 1 and x 1 + Z. For odd rme, ths decomoston defnes a character of Z, whch s the surjectve homomorhsm ω : Z µ 1. Ths character ω s called the Techmüller character [5, Secton 4.5]. We wll use the factorzaton of x nto ω(x) x to ad n our analyss. Addtonally, we wll need the -adc exonental and logarthm functons. As n R, we can defne the -adc exonental and logarthm functons on certan subsets of the -adc numbers as formal ower seres: ex (x) = log (1 + x) = n=0 x n n!, ( 1) n+1 x n. n n=1 These functons have rad of convergence x < 1/( 1) and 1, resectvely. It s mortant to note that the denttes ex (log (1 + x)) = 1 + x and log (ex (x)) = x hold formally, and wll also hold functonally when we have convergence. For more on these functons, see [5, Secton 4.5]. 2

5 Lastly, we wll want to use a generalzaton of Hensel s lemma, whch allows the lftng of solutons to congruences modulo to solutons modulo e, that ales to the -adcs. Frst, we wll need to defne a restrcted ower seres. A formal ower seres s an object of the form =0 a x, where the a are unrestrcted coeffcents, and addton and multlcaton are erformed smlarly to olynomal oeratons. A restrcted ower seres s a formal ower seres where lm a = 0. Now we can take ths theorem from [6, Cor. 3.3]. Theorem 1. Let f(x) be a restrcted ower seres n Z [[x]] and a be n Z such that df dx (a) s n Z and f(a) 0 (mod ). then there exsts a unque x Z for whch x a (mod ) and f(x) = 0 n Z. Wth ths knowledge n mnd, we can now start our analyss. For ths aer, we let n, k, and e be ntegers, a rme, and g a unt modulo (.e. g, so g has an nverse modulo ). We wll be countng the nteger solutons of the congruence g xn x k (mod e ), or equvalently, the zeros of f : Z Z/ e Z, where f(x) = g xn x k (mod e ). We denote the multlcatve order of g modulo as m. 2 Perodcty The frst thng to note about our functon f s that t s erodc, snce t wll restrct the range of x to examne when countng solutons. The theorem n ths secton descrbes ts erodcty. Lemma 2. g m e 1 1 (mod e ). Ths lemma s obtaned from the roof of [9, Theorem 1], and allows us to conclude wth the followng theorem. Theorem 3. Fxng all varables excet x, we have that g (x+me ) n (x + m e ) k g xn x k (mod e ). In other words, f(x) = f(x + m e ). Proof. Frst, consder g (x+me ) n (mod e ). We know n ( ) n (x + m e ) n = x (m e ) n. =0 Snce m e 1 m e and m e dvdes all terms excet x n, by Lemma 2 we have g (x+me ) n g xn (mod e ). Now consder (x + m e ) k. We can also exand ths to (x + m e ) k = k =0 3 ( ) k x (m e ) k.

6 Snce e m e and m e dvdes all terms excet x k, we have (x + m e ) k x k (mod e ). Thus g (x+me ) n (x + m e ) k g xn x k (mod e ). 3 Interolaton Snce we would lke to analyze our equaton -adcally, our frst goal s to nterolate our functon f : Z Z/ e Z, f(x) = g xn x k (mod e ) to a functon from Z to Z. We fnd that although we cannot nterolate to a sngle contnuous -adc functon, we can nterolate to a fnte number of -adc functons that agree wth f(x) on certan values of x. Theorem 4. For 2, let g Z and x 0 Z/( 1)Z, and let I x0 = {x Z x x 0 (mod 1)} Z. Then F x0 (x) = ω(g) xn x 0 g n defnes a unformly contnuous functon on Z such that F x0 (x) = g xn x I x0. whenever Proof. By [5, Prooston 4.6.1], we need I x0 to be dense n Z and for each F x0 (x) be unformly contnuous and bounded. We know that f a functon f : Z Q s contnuous on Z, then t s also unformly contnuous and bounded [7, Theorem 4.1.4]. Thus, t suffces to show densty of I x0, contnuty of each F x0 as a functon on I x0, and that F x0 (x) = g xn wth the roer condtons on x. We frst need to rove densty of I x0 n Z. Ths s shown n the roof of [9, Theorem 16] when we let c = 1. Now we must show each F x0 (x) = ω(g) xn x 0 g n s unformly contnuous on I x0. Gven ɛ > 0, fnd N such that N < ɛ. Now f x, y I x0 such that x y N < (N 1) = δ, then x = y + N A for some A Z. Consder g xn g yn = g (y+n A) n g yn = g yn g (y+n A) n y n 1 = g (y+n A) n y n 1, and usng the bnomal theorem, we get g (y+n A) n y n n =1 = ( g n )y n ( N A). 4

7 If we factor out N from the exonent, we get g n =1 ( n )y n ( N A) = g N b, where b = n =1 ( n ) y n ( N ) 1 A, whch s an nteger. So we have g xn g yn = g (y+n A) n y n 1 = g N b 1 Usng the bnomal theorem agan, and the fact that g = 1 + M, we get (1 + M) N b = 1 + N bm + N b( N b 1) (M) (M) N b. 2 Because all terms excet for the frst are n N+1 Z, we see that g N A 1 (N+1) < N < ɛ. So the functon mang x g xn s unformly contnuous on I x0 and hence on Z by [7, Thm 4.15]. Snce each F x0 (x) = ω(g) xn x 0 g n for fxed x 0, and g, and ω(g) xn 0 s a constant, we have that Fx0 (x) s a constant tmes a unformly contnuous functon. Hence, each F x0 (x) s unformly contnuous on Z [7, Exercse 89]. Lastly, we show that F x0 (x) = g xn when x I x0. Snce x x 0 (mod 1), we have that g xn = ω(g) xn g xn = ω(g) xn 0 g x n = F x0 (x). We can extend ths theorem to multles of the order of g modulo : Theorem 5. For ths theorem only, we let m be any multle of the multlcatve order of g modulo, 2, so that m 1. Let g Z and x 0 Z/mZ, and let J x0 = {x Z x x 0 (mod m)} Z. Then F x0 (x) = ω(g) xn 0 g x n defnes a unformly contnuous functon on Z such that F x0 (x) = g xn whenever x J x0. Proof. Snce g m 1 (mod ), ω(g) mn = ω(g mn ) = ω(1) = 1. If x 0, x 0 Z/( 1)Z and x 0 x 0 (mod m), then the two functons F x0 and F x 0 gven by Theorem 4 are equal and are the same as g xn when x I x0 I x 0 J x0. 5

8 4 Countng Solutons Now that we have our -adc functons, we can use those to begn countng solutons. We begn by countng solutons to our modfed congruences modulo, and then roceed by lftng these solutons to -adc solutons modulo e. Lastly, we wll refer back to our theorems on nterolaton to fnd when the solutons to our modfed congruences wll gve us solutons to our orgnal congruence g xn x k (mod e ). The followng lemma analyzes solutons modulo. Lemma 6. Consder the equaton g xn 0 x k (mod ). q α 1 n 1 q α 2 n 2 q α Defne d = gcd(k, 1) gcd(k, 1 m ), and let qα1 1 qα2 2 qα be the rme factorzaton of d. m gcd(k, 1) Then there are N = soluton ars (x 0, x) to the above equaton, where x 0 {0, 1,..., m 1} and x {0, 1,..., 1}. Proof. Let h be a rmtve root modulo, so we can exress g h a (mod ) and x h b (mod ). So g xn 0 x k (mod ) becomes (h a ) xn 0 (h b ) k (mod ). Snce h s a rmtve root, we have that ax n 0 bk (mod 1). From [1, Theorem 5.1], we have that there are gcd(k, 1) mutually ncongruent solutons for b (whch corresond to a dstnct values of x) f gcd(k, 1) ax n 0, and no solutons otherwse. So we must now count x 0 where gcd(k, 1) ax n 0. We have that gcd(k, 1) ax n gcd(k, 1) a 0 f and only f gcd(k, 1,a) Note that gcd(k, 1, a) = gcd(gcd(k, 1), a) so gcd(k, 1) gcd(k, 1,a) gcd(k, 1,a) xn 0. s relatvely rme a to gcd(k, 1,a). Now we only need to count x 0 that satsfy gcd(k, 1) gcd(k, 1,a) xn 0. Because we defned h and a so that g h a (mod ), and g has order m, we know that gcd(a, 1) = 1 m. So gcd(k, 1, a) = gcd(k, gcd(a, 1)) = gcd(k, 1 m ). Now we are left wth countng x 0 that satsfy gcd(k, 1) gcd(k, 1 m ) xn 0, whch s the same as d x n 0. In order to count the number of solutons, we look at the rme factorzaton of d. We have that and thus we have q α1 1 qα2 2 qα x n 0 f and only f q α1 m q α 1 n 1 q α 2 n 2 q α 1 q α2 2 q α x 0, dstnct x 0 {0, 1,..., m} that satsfy our condtons. Snce there are gcd(k, 1) solutons x {0, 1,..., 1} for each x 0, m gcd(k, 1) we have a total of soluton ars (x 0, x) to g xn 0 x k (mod ). q α 1 n 1 q α 2 n 2 q α 4.1 Countng solutons when k When we lft the solutons we found modulo to solutons modulo e, we have to use dfferent methods for when k than when k. We wll be able to use 6

9 Hensel s lemma to lft to solutons modulo e when k. The followng lemma descrbes the result. Lemma 7. For 2, k, let g Z be fxed, and x 0 {0, 1,, m 1}. If a s a soluton n {0, 1,..., 1} to ω(g) xn 0 g x n 0 x k (mod ). Then there s a unque soluton n Z to the equaton where x a (mod ). Proof. Snce g s n 1 + Z, we get ω(g) xn x 0 g n = x k g xn = (ex (x n log ( g )) = (1 + x n log ( g ) + x 2n log ( g ) 2 /2! +hgher order terms n owers of log ( g )), where from [5, Prooston 4.5.9], we know that log ( g ) Z. Now that we have a convergent ower seres snce log ( g ) /! 0 as [2, Chater 2, Theorem 3.1], we examne f(x) = F x0 (x) x and ts dervatve to see f we can aly a generalzaton of Hensel s lemma. Consder f(x) = ω(g) xn 0 (1 + x n log ( g ) + x 2n log ( g ) 2 /2! + hgher order terms n owers of log ( g )) x k. Snce we know log ( g ) Z, so log ( g ) 0 (mod ), we have that f(a) ω(g) xn 0 (1 + a n (0) + a 2n (0) Addtonally, we have that so that + hgher order terms congruent to 0 (mod ) ) a k (mod ) ω(g) xn 0 a k 0 (mod ). f (x) = ω(g) xn 0 (nx n 1 log ( g ) + (2n)x 2n 1 log ( g ) 2 /2! +3nx 3n 1 log ( g ) 3 /3! +...) ka k 1 f (a) ω(g) xn 0 (na n 1 (0) + (2n)a 2n 1 (0) 2 /2! + 3na 3n 1 (0) 3 /3! +...) ka k 1 0 ka k 1 (mod ). We know a k ω(g) xn 0 (mod ) so then we know ak 0 (mod ) and thus a k 1 0 (mod ). Also, we have k. So then ka k 1 0 (mod ). Now we know we can aly Theorem 1, whch states that there s a unque x Z for whch x a (mod ) and f(x) = 0 n Z. 7

10 Now that we have found solutons to our modfed equatons, we need to be able to ece them together to gve us solutons to our orgnal equaton. The followng theorem uses the results from our lemmas to gve us the number of solutons to g xn x k (mod e ) when k. Theorem 8. For 2, let g Z and n, k Z be fxed and k. Then there m gcd(k, 1) are N = solutons x to the equaton q α 1 n 1 q α 2 n 2 q for x {1, 2,, e m}. α g xn x k (mod e ) Proof. We begn by consderng the number of solutons modulo to a slghtly m gcd(k, 1) dfferent equaton. By Lemma 6, we have soluton ars q α 1 n 1 q α 2 n 2 q (x 0, x 1 ) to g xn 0 x k 1 (mod ) where the x 0 are dstnct (mod m) and x 1 are dstnct (mod ). For each x 1 that aears n a soluton ar to g xn 0 x k 1 (mod ), then by Lemma 7 we have a unque soluton x n Z to ω(g) xn (x 0 g ) n = (x ) k where x x (mod ). By the Chnese Remander Theorem, we have that there s exactly one x Z/m e Z where x x 0 (mod m) and x x (mod e ). Thus by Theorem 5 we have exactly one soluton to g xn x k (mod e ) n Z/m e Z for every soluton ar to g xn 0 x k (mod ), and therefore there are m gcd(k, 1) solutons n Z/m e Z to the equaton g xn x k (mod e ). q α 1 n 1 q α 2 n 2 q α α We fnd that ths theorem s consstent wth our results. For examle, lookng at g xn x k (mod 7 e ) for 0 x m 7 e, we get the followng number of solutons for all n and e. Table 1: g xn x k (mod 7 e ) for 0 x < m 7 e g m # solns: k=1 # solns: k=2 # solns: k=3 # solns: k= and So f we look at the case when k = 4, we fnd that gcd(k, 1) d = gcd(k, 1 m gcd(4, 7 1) = ) gcd(2, 7 1 m ) = 2 gcd(4, 6 m ) = N = { m gcd(4,7 1) 1 = 2m f 2 m m gcd(4,7 1) 2 = m f 2 m, 8 { 1 f 2 m 2 f 2 m,

11 whch matches the fndngs n Table Countng solutons when = k and n = 1 Our fndngs for when = k dffers from our results when k. For examle, when = 11, we fnd the number of solutons detaled n Table 4.2. We see that our N solutons modulo lft to dfferent numbers of solutons modulo e than n the k case. Ths suggests that we must lft solutons modulo to solutons modulo e dfferently: we wll end u usng nducton on e. So, we wll count solutons modulo 2 and use that as the base case n our nducton. g m # solns: e=1 # solns: e=2 # solns: e=3 # solns: e= Table 2: g x x 11 (mod 11 e ) for 0 x < m 11 e As we lft, we fnd that the value of g 1 modulo 2 s mortant. By Fermat s Lttle Theorem, we have for rme and g, that g 1 1 (mod ). Lookng at ths equvalence modulo 2 gves the followng defnton. Defnton 3. An nteger g s called a Weferch base modulo f g 1 1 (mod 2 ). Now we are able to count solutons to g x x (mod 2 ), seeng that the result deends heavly on whether g s a Weferch base modulo. Lemma 9. Let 2, let a 0 be a soluton to g x x (mod ), and let x 0 a 0 (mod m). Then the followng are equvalent: 1. g = ω(g) g, where g 1 (mod 2 ). 2. g s a Weferch base modulo 3. a 0 lfts to at least one soluton a Z/ 2 Z to g x x (mod 2 ) where a a 0 (mod ) and a x 0 (mod m). Furthermore, n 3, we also have that f a 0 lfts to a soluton n Z/ 2 Z, t lfts to dstnct solutons n Z/ 2 Z. 9

12 Proof. For ths roof, we begn by fndng a congruence that holds exactly when we have a soluton a to g x x (mod 2 ) that satsfes the condtons that a x 0 (mod m) and a a 0 (mod ). We wll then use the equvalent statement to rove 3 = 2 and 1 = 3, and then fnsh by showng 2 = 1. Let a x 0 (mod m) and a a 0 (mod ), and consder when 0 g a a (mod 2 ). Recall from Theorem 5 that g a = ω(g) x0 g a when a x 0 (mod m). So snce a x 0 (mod m), we have 0 g a a ω(g) x0 g a a ( a (log ω(g) x0 g ) ) a (mod 2 ). (1)! =0 We have a a 0 (mod ), so we get a a 0 + a 1 (mod 2 ), and thus equaton (1) holds exactly when we get ω(g) x0 =0 ( (a0 + a 1 ) (log g ) Snce log g Z, ths reduces to! ) (a 0 + a 1 ) 0 (mod 2 ). ω(g) x0 (1 + a 0 log g ) (a 0 + a 1 ) 0 (mod 2 ). (2) When we exand the term (a 0 + a 1 ) modulo 2, we fnd that t s congruent to a 0, and we obtan Note that ω(g) x0 (1 + a 0 log g ) a 0 0 (mod 2 ). (3) log g = and snce g 1 Z, we have (( 1) =0 +1 ( g 1) log g g 1 (mod 2 ). So then we can relace log g n equaton (3) to obtan ω(g) x0 (1 + a 0 ( g 1)) a 0 ω(g) x0 a 0 ( g 1) + ω(g) x0 a 0 0 (mod 2 ). (4) We have that g 1 0 (mod ). We also know g a0 a 0 ω(g)a0 g a0 a 0 ω(g)x0 a 0 0 (mod ), (5) so we can wrte equaton (4) as ), ω(g) x0 a 0 ( g 1) + ω(g)x0 a 0 0 (mod ). (6) 10

13 Now consder the followng: by (4.2) and Fermat s Lttle Theorem we have ω(g) x0 a 0 ω(g)x0 a 0 0 (mod ). By defnton, ths s true exactly when Now ths s true f and only f ω(g) x0 a 0 = r, for some r Z. (ω(g) x0 ) 1 = (a 0 + r) 1. Snce ω(g) s a ( 1)th root of unty, we fnd that (ω(g) x0 ) 1 = 1. By usng ths fact and exandng (a 0 + r) 1, we fnd that the above s equvalent to 1 ( ( ) 1 1 = (r) (a 0 ) 1 ). =1 Examnng ths equaton modulo 2, we fnd that 1 a ( 1) 0 + ( 1)ra ( 2) 0 (mod 2 ), and snce the order of the grou Z/ 2 Z s ( 1), we get a ( 1) 0 0 (mod 2 ), so 1 a ( 1) 0 + ( 1)ra ( 2) 0 (mod 2 ), and thus 0 ( 1)ra ( 2) 0 (mod ). Snce 1, a 0 0 (mod ) (f a 0 0 (mod ), then 1 g (mod )), we must have r 0 (mod ). So then ω(g) x0 a 0 = (k) 0 (mod 2 ) for some k Z. ω(g) Thus we always have that x 0 a 0, and we can reduce equaton (6) further: ω(g) x0 a 0 ( g 1) + ω(g)x0 a 0 ω(g)x0 a 0 ( g 1) 0 (mod ). (7) Now we have that a solves g a a (mod 2 ) f and only f the above equvalence holds, and we contnue wth our roof. 1 = 3: Assumng 1, we have that g 1 (mod 2 ). So then 2 ( g 1), and so we get ω(g) x0 a 0 ( g 1) 0 (mod ). Thus all choces for a 1 {0, 1,..., 1} gve a soluton a a 0 + a 1 (mod 2 ) to g x x (mod 2 ) whenever a a 0 (mod ) and a x 0 (mod m). 11

14 By the Chnese Remander Theorem, we have exactly one a {0, 1,..., m 2 } where both a a 0 + a 1 (mod 2 ) and a x 0 (mod m) are satsfed. Snce there are dstnct choces for a 1, we have solutons to g x x (mod 2 ) n {0, 1,..., m 2 }. 3 = 2: Assumng 3, we know there s at least one a solvng ω(g)x 0 a 0( g 1) 0 (mod ). We know by equaton (4.2) and Fermat s Lttle Theorem that ω(g) x0 a 0 a 0 (mod ), and snce ω(g) 0 (mod ), then a 0. Thus we must have 2 ( g 1), so then g 1 ω(g) 1 g 1 ω(g) (mod 2 ), whch means g s a Weferch base modulo. 2 = 1: Assumng 2, we have that g 1 1 (mod 2 ). Wrte g = 1 + g 1. Then we have 1 g 1 ω(g) 1 g 1 g 1 (1 + g 1 ) 1 (mod 2 ). Exandng (1 + g 1 ) 1 we get 1 = 1 =0 ( 1 ) (g 1 ) 1 + ( 1)g 1 (mod 2 ). So then we have 0 ( 1)g 1 (mod 2 ), and dvdng through by, we obtan ( 1)g 1 0 (mod ). Snce 1 0 (mod ), we must have g 1 0 (mod ). Thus g 1 + g 1 1 (mod 2 ). The last theorem we gve uses our revous lemma as a base case to count solutons to g x x (mod e ) for e > 1. Theorem 10. For 2, let g Z be fxed. Let N be the same as n Theorem 8. Then there are N solutons x to the equaton g x x (mod ). Furthermore, for e > 1, the equaton g x x (mod e ) has N solutons x f g 1 1 (mod 2 ) (.e. g s a Weferch base modulo ), and no solutons otherwse. Proof. Frst consder when e = 1. We have by Lemma 6 that there are N soluton ars (x 0, x 1 ) Z/mZ Z/Z to g x0 x 1 (mod ). By the Chnese Remander Theorem, there s exactly one x Z/mZ where x x 0 (mod m) and x x 1 (mod ), so there s exactly one soluton x Z/mZ where g x g x0 x 1 x (mod ). Snce there are N soluton ars, then there are N solutons to g x x (mod ). 12

15 Now consder when e = 2. If g s a Weferch base modulo, then for each of the N solutons above, we let a 0 be a soluton and fnd that we have solutons to g x x (mod 2 ) by Lemma 9. Snce ths holds for all our N solutons modulo, we have N total solutons to g x x (mod 2 ). If g s not a Weferch base modulo, then by Lemma 9, none of the N solutons we found modulo lft to a soluton to g x x (mod 2 ), so there cannot be any solutons modulo 2. Furthermore, there cannot be any solutons to g x x (mod e ) for e 3. When g s a Weferch base modulo and e > 1, we use nducton. The base case (e = 2) s gven above, and note that the solutons modulo 2 take the form a 0 + a 1 (mod 2 ), where a 1 takes any value n Z/Z. Let ( f x0 (x) = F x0 (x) g = ω(g) x0 g x x x (log = ω(g) x0 g ) ) x.! For the nducton assumton, we assume that we have N solutons a Z/m e 1 Z s.t. g x x k 0 (mod e 1 ), wrtten a a 0 +a 1 + +a e 2 e 2 a + a e 2 e 2 (mod e 1 ), where a e 2 can take any value modulo. Note that we have g 1 (mod 2 ) by Lemma 9. We want to fnd a soluton x that solves g x x (mod e ), so t must also solve g x x (mod e 1 ). So we must have x a (mod e 1 ) for one of our solutons a. Thus we have that x a + a e 1 e 1 (mod e ) for some a. Set x 0 a (mod m). Note that f x0 (x) = g x x when x x 0 (mod m) by Theorem 5, so f x0 (a) 0 (mod e 1 ). So we have f x0 (x) f x0 (a + a e 1 e 1 ) ( (a + a e 1 e 1 ) (log ω(g) x0 g ) ) (a + a e 1 e 1 ) (mod e )! =0 (log g ) k Snce for all k Z +, k! (by [5, Lemma 4.5.4]), and because a a + a e 2 e 2 (mod e 1 ), we can smlfy to get ( (a + a e 2 e 2 ) (log f x0 (x) ω(g) x0 g ) ) (a + a e 2 e 2 + a e 1 e 1 )! =0 e 1 ω(g) (1 x0 + (a + a e 2 e 2 ) log g + =2 =0 (a ) (log g ) ) (a + a e 2 e 2 + a e 1 e 1 ) (mod e )! 13

16 Note that (a + a e 2 e 2 + a e 1 e 1 ) ( ) (a + a e 2 e 2 ) (a e 1 e 1 ) =0 (a + a e 2 e 2 ) ( ) (a ) (a e 2 e 2 ) =0 (a ) + (a ) 1 a e 2 e 1 (mod e ). So now we have that, gven a soluton x a (mod e 1 ), f t lfts to a soluton modulo e, then t lfts to any x a+a e 1 e 1. When we set f x0 (x) 0 (mod e ) to solve for x, we get e 1 0 ω(g) (1 x0 + (a + a e 2 e 2 ) log g + so collectng all the a e 2 terms, we fnd that a e 2 (ω(g) x0 e 2 log g (a ) 1 e 1 ) =2 (a ) (log g ) ) ((a ) + (a ) 1 a e 2 e 1 ) (mod e ), e 2 ω(g) (1 x0 + a log g +! =2 + ω(g)x0 (a ) e 1 (log g ) e 1 (e 1)! (a ) (log g )! (mod e ). Note that f x0 (x) f x0 (a) 0 (mod e 1 ) and so snce e 1 dvdes the left hand sde, e 1 must also dvde the rght hand sde, and we also know e 1 (log g ) e 1 ). So we can wrte a e 2 (ω(g) x0 e 2 log g (a ) 1 e 1 ) e 1 ω(g)x0 (1 + a log g + e 2 (a ) (log g ) =2! ) (a ) e 1 + ω(g)x0 (a ) e 1 (log g ) e 1 (e 1)! e 1 (mod ) ) (a ) It suffces to show that (ω(g)x 0 e 2 log g (a ) 1 e 1 ) 0 (mod ), snce e 1 dong so would mean t has an nverse modulo and we can solve unquely for a e 2. Then we would know that exactly one out of every solutons a a + a e 2 e 2 (mod e 1 ) lfts to a soluton modulo e. 14

17 If we assume t s congruent to 0 modulo, we have so (ω(g) x0 e 2 log g (a ) 1 e 1 ) e 1 0 (mod ) (ω(g) x0 e 2 log g ) e 1 (a ) 1 1 (mod ) by Fermat s Lttle Theorem. But (ω(g) x0 e 2 log g ) e 1 ω(g)x0 log g ω(g)x0 ( ( 1) +1 ( g 1) =1 ) (mod ) and g 1 (mod 2 ) so we get ω(g) x0 ( ( 1) +1 ( g 1) =1 ) 0 (mod ). Ths s a contradcton snce 1 0 (mod ). Thus we can solve unquely for a e 2, and for such an a e 2, any a e 1 Z/Z solves f x0 (a + a e 1 e 1 ) 0 (mod e ). By our nducton assumton, there are N solutons to f x0 (a) 0 (mod e 1 ), a a + a e 2 e 2 (mod e 1 ). Snce for each soluton a to f x0 (a ) 0 (mod e 2 ), a +a e 2 e 2 s a soluton modulo e for a unque a e 2 Z/Z, we have exactly N dstnct a that lft to solutons to f x0 (x) f x0 (a+a e 1 e 1 ) 0 (mod e ). Ths gves a total of N solutons modulo e. Agan, we see that ths s consstent wth our results. For an examle, we return to our results when = 11 detaled n Table 4.2. If we look at just g = 3 and g = 4, we have that N = 5 and so N = 55. We fnd that (mod 11 2 ) and (mod 11 2 ), so g = 3 s a Weferch base and has 55 solutons modulo 11 e for e > 1, whch g = 4 s not a Weferch base modulo 11 and thus has no solutons modulo e for e > 1. Both of these match our fndngs n Table Conclusons and Future Work In ths aer, we have aled the methods found n [6], [9], and [11] to count solutons to the equaton g xn x k (mod e ). When analyzng the equaton for x {1, 2,..., m e }, an odd rme, we have found an exact number of solutons for the case when k, secfcally N = m gcd(k, 1) α 2 n 2 q q α 1 n 1 q α solutons. In addton, we found that when odd, k = and n = 1, that there are N solutons when e = 1, and ether N or 0 solutons when e > 1, deendng on whether g s a Weferch base modulo. 15

18 It remans to be shown whether the same number of solutons s obtaned for general n n the k = case. We susect that smlarly to the k case, n wll only affect the value of N, and not the results of lftng solutons modulo to solutons modulo e, but ths has not been confrmed. Addtonally, the case where k but k has not been analyzed yet. Based on a few test results, we susect that k wll affect the value of N, but that the number of solutons modulo e for e > 1 wll be the same as n the = k case. Lastly, due to the fact that we need x 1 + 4Z 2 rather than x 1 + 2Z 2 for log 2 (ex 2 (x)) to converge, the case where = 2 must be analyzed dfferently. We have done some testng that confrms that = 2 yelds dfferent results than for odd rme, leavng another avenue of analyss. Besdes countng solutons, further analyss can be done by exlorng the dstrbuton of solutons for x among ntervals of length e rather than focusng on only the longer nterval of length m e, snce that range s generally more alcable to crytograhc schemes. There has been some analyss of the ma x g xn (mod c) when n = 2 by Wood [12], and some statstcal analyss of the ma when n = 1 from [8] and [4], but more work remans to be done. References [1] George E. Andrews, Number Theory, Dover Publcatons, Inc., [2] George Bachman, Introducton to -adc Numbers and Valuaton Theory, Academc Press Inc., [3] Jan Camensch and Markus Stadler, Effcent grou sgnature schemes for large grous, Lecture Notes n Comuter Scence 1294 (2006), [4] Danel R. Clouter, Mang the Dscrete Logarthm, Mathematcal Scences Techncal Reorts (MSTR) (2005). [5] Fernando Quadros Gouvea, -adc Numbers: An Introducton, 2nd ed., Srnger, [6] Joshua Holden and Margaret M. Robnson, Countng Fxed Ponts, Two Cycles, and Collsons of the Dscrete Exonental Functon Usng -adc Methods, Journal of the Australan Mathematcal Socety 92 (2012), no. 2, , DOI /S [7] Svetlana Katok, -adc Analyss Comared wth Real, Vol. 37, Amercan Mathematcal Socety, [8] Nathan Lndle, A Statstcal Look at Mas of the Dscrete Logarthm, Mathematcal Scences Techncal Reorts (MSTR) (2008). [9] Abgal Mann and Adelyn Yeoh, Deconstructng the Welch Equaton Usng -adc Methods, Rose-Hulman Undergraduate Mathematcs Journal 16 (2015), no. 1, [10] Wade Trae and Lawrence C. Washngton, Introducton to Crytograhy wth Codng Theory, 2nd ed., Pearson, [11] Anne Waldo and Cayun Zhu, The Dscrete Lambert Ma, Rose-Hulman Undergraduate Mathematcs Journal 16 (2015), no. 2, [12] A. Wood, The Square Dscrete Exonentaton Ma, Mathematcal Scences Techncal Reorts (MSTR) (2011). 16