An Account of Congruences Mod p k Using Halley s Method

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1 World Applied Scieces Joural 16 (11): , 01 ISSN IDOSI Pulicatios, 01 A Accout of Cogrueces Mod p Usig Halley s Method M. Khalid Mahmood ad M. Aslam Mali Departmet of Mathematics, Uiversity of the Puja, Quaid-e-Azam Campus, Lahore-490, Paista Astract: I this paper we fid solutios of cogrueces of the form ax (mod p ), 1 where a, ad >0 are itegers which are ot divisile y a prime p usig Halley's iterative algorithm. A algorithm is also proposed to reduce the degree of a oliear cogruece modulo p. AMS Mathematics Suject Classificatio (000): 0C 11E04 0G1 Key words: Halley's method Cogrueces mod p Euler's theorem Degree reductio INTRODUCTION I recet years, the use of iterative methods for solvig oliear equatios has ecome a foremost hu for umerical aalysts ad umer theorists. These ideas were used i [1-] earlier, where iterative methods from umerical aalysis were discussed to calculate the iverses of umers modulo prime powers. To fid solutio of cogrueces give i equatio (1) through Numerical Aalysis thus ecomes a iterestig prolem. ax (mod p ), 1 (1) where a, ad >0 are itegers ot divisile y p. Hesel s Lemma [4, ], is a well-ow result i umer theory showig that if a equatio has a osigular solutio modulo a sufficietly large power of a prime p, the it has solutios modulo all powers of p. I Hesel s Lemma the idea is to try to get a solutio modulo p, the move o to modulo p, the to p ad y iteratio to p. The proof of Hesel s Lemma is ased o Newto's method, whereas we use Halley's method to calculate solutios of cogrueces, of the form ax (mod p ), 1. Precisely we have proved that if x, 1 is the solutio of the cogruece ax (mod p ) the x +1 is the solutio of the cogruece ax (mod p ), 1 satisfyig the equatio ( + 1) ( 1)ax x + ( 1) + ( + 1)ax where a, ad >0 are itegers ot divisile y a prime p. Also we propose a algorithm to reduce the degree of a oliear cogruece mo dulo p. We first ecoutered this idea i [6], where a method ased o reductio of a oliear cogruece to either liear or quadratic is itroduced for solvig some oliear cogrueces. The idea was to cosider a cogruece of the form x a(mod p) with a 0 (mod p) ad p is a prime satisfyig the coditio p =.l+ (or p = l+) for some,l N. This cogruece was reduced to either a liear cogruece ax (mod p), a, Z, or to the quadratic oe x c(mod p), c Z. It is importat to ote that i [6], author used Fermat's little theorem to reduce the degree of a oliear cogruece modulo a sigle prime p. Thus it ecomes iterestig to reduce the degree of a oliear cogruece modulo some power of primes. I this wor, we use Euler's theorem to reduce the degree of a oliear cogruece modulo p, 1. Notatios used i this paper are stadard. We eed the followig result of [4, ], for use i the sequel. Theorem 1.1: Let m e a positive iteger havig a primitive root ad suppose (a,m) = 1. The the cogruece x a(mod m) has a solutio if ad oly if (, a 1 (mod m) If the cogruece x a(mod m) is solvale, the it has exactly d = (, ) icogruet solutios. MATERIAL AND METHODS The itetio of the preset paper is two-fold. First is to reduce the degree of a give cogruece to a lower degree cogruece ad the to fid its solutios y Correspodig Author: M. Khalid Mahmood, Departmet of Mathematics, Uiversity of the Puja, Quaid-e-Azam Campus, Lahore-490, Paista 166

2 employig Numerical Aalysis. I the first part we use Euler s Theorem to reduce the degree of a give cogruece ad the we use Halley s method to fid the solutio of reduced degree cogruece. RESULTS AND DISCUSSION Degree reductio algorithm: Here we use Euler's theorem i order to reduce the degree of a oliear cogruece modulo some powers of a prime p. Let r m= p α i i i= 1 where α i 1 ad p 1, p,, p r e distict primes. Let a e ay iteger relatively prime to m. The Euler's theorem states that a 1 (mod m) () I the followig theorem, we illustrate that how oe ca use Euler's theorem to reduce the degree of a oliear cogruece. For a odd prime p, we cosider the followig oliear cogruece Now it remais oly to show that a solutio of cogruece (4) also satisfies (). For this, we show that a l d is a solutio of (). To prove the assertio, we have l l φ (p ) + d φ(p ) d (a ) = a d = a d = a d a 1 φ(p ) d = (a ) a a (mod p ) l Hece, x d 0 a (mod p ) is a solutio of the cogruece x a(mod p ). Corollary.: If is eve the d must e eve. That is, a eve degree cogruece will e reduced to a eve degree cogruece. Proof: Let e a eve iteger. The gcd (, φ(p )) must e a eve iteger sice φ( p ) is always a eve iteger. But the gcd (, φ(p )) = gcd (, l+d) = gcd (, d). As the left had side is a eve iteger so does the right had side ad hece d must e a eve iteger. Corollary.: If is odd the d is odd (or eve) accordig as l is odd (or eve). x a (mod p ) () Theorem.1 Let p e a odd prime such that 1 φ (p ) = p (p 1) = l + d, d < where,l,d N. The the cogruece () is equivalet to the cogruece a l x d 1 (mod p ). Proof: To prove this, let x RRS(mod p ) e ay iteger. The it is clear that (x,p) = 1. So y Euler's theorem, we have, 1 x (mod p ) 1 p (p 1) x (mod p ) l+ d x (mod p ) l d (x ) x (mod p ) l d ( a ) x (mod p ). usig() This implies that the cogruece () is equivalet to cogruece of lower degree l d a x 1 (mod p ) (4) Proof: Let e a odd iteger. Sice φ(p ) is always a eve iteger so we let φ(p ) = l+d = t for some iteger t. The d = t-l. This clearly shows that d is odd (or eve) accordig as l is odd (or eve). Note: It is importat to ote that if (4) is ot solvale the ecessarily () is ot solvale. For istace, let s reduce the degree of x 1 (mod 81) y meas of degree reductio algorithm. Sice φ (81)) = 4 =.1+1, so the reduced degree cogruece is the liear cogruece 1x 1 (mod 81). Clearly, this is ot solvale. Hece the cogruece x 1 (mod 81) is ot solvale. To discuss the umer of solutios we see that (, φ(p )) is 1 or eve accordig as is odd or eve respectively. Sice (, φ(p )) = (, l+d) = (, d). So if is odd the the cogruece a l x d 1(mod p ) has oly oe solutio for (). If is eve the either it has o solutio or have at most d solutios for (). The followig three examples illustrate Theorem.1, Corollaries. ad.. Example(s).4 (1) Sice (6, φ( )) = 4 ad (mod ), so y Theorem 1.1, the cogruece x 6 6(mod ) is 167 φ()

3 solvale ad has four icogruet solutios. To fid these solutios, we apply Theorem.1 successively. For this, we see that φ( ) = 00 = , so l = 44 ad d = 6. Thus from (4), we otai, 6 44 x 6 1 (mod ) which ca e simplified as x 6 01(mod ). Now φ( ) = 00 = , so l = 69 ad d = 16. The agai usig (4), we have, 01 x 1 (mod ) which ca e 16 simplified as x 16 (mod ). Agai, φ( ) = 00 = , so l = 16 ad d = 4. The agai 16 4 usig (4), we have, 16 x 1 (mod ) which 4 ca e simplified as x 196 (mod ). This implies that x 14(mod ) ad x (mod ). By solvig these quadratic cogrueces we fid that x 4, 8, 71, 94 (mod ) are the desired four icogruet solutios of the 6 cogruece x 6 (mod ). () Sice (9, φ (7 )) = 1 ad 1 1 (mod ). So y 9 Theorem 1.1, the cogruece x (mod 7 ) is solvale ad has a uique solutio. To fid the desired solutio, we apply Theorem. twice. For this, we see that, φ (7 ) = 94 = , so l = 10 ad d = 4. Thus y (4), we have, 10 x 4 1 (mod7 ) 4 which ca e simplified as x 17 (mod 7 ). Now, φ ( ) = 94 = 4 7 +, so l = 7 ad d =. The y 7 (4), 17 x 1 (mod 7 ) which ca e simplified as x 68 (mod 7 ). By solvig the quadratic cogrueces we fid that x 18,18 (mod 7 ) are solutios of the reduced degree cogruece. But oly oe of them amely x 18 (mod 7 ) satisfies the origial cogruece. φ() () Sice (49, φ ( )) = 1 ad 9 1 (mod ). So y φ(7 ) Theorem 1.1, the cogruece 9 1 (mod ) is solvale ad has a uique solutio. To fid out these solutios, we apply Theorem.1. For this, we see that, φ ( ) = 00 = , so l = 1 ad d = Thus from (4), we have, 9 x 1 (mod ) which implies that x 764 (mod ) is the desired root of 49 the cogruece x 9 (mod ). φ() Halley's method: Halley's method [9], is a well-ow iterative algorithm for solvig a oliear equatio f(x) = 0. It is a good tool i may ways for the solutio of a oliear prolem. Its simplicity ad high speed always attract i attemptig a 168 oliear prolem. Assume that a iitial estimate x 0 is ow for the desired root α of ƒ(x) = 0. The to perform iteratios the formula for Halley's method is: f(x ) f(x )f(x ) 1 x = x [1 ], = 0,1,, + 1 f(x ) (f ( x )) I the followig theorem, we use Halley's method i order to fid solutios of cogrueces of the form ax (mod p ), 1, from the solutios of the cogruece ax (mod p ). Theorem.4: Let a, ad e positive itegers which are ot divisile y a prime p. If x, 1, is the solutio of the cogruece ax (mod p ) the x +1 is the solutio of the cogruece ax (mod p ) satisfyig the equatio ( + 1) ( 1)ax x + ( 1) + ( + 1)ax Proof: To prove this, we see that, if the a 0 (mod p ) x a (mod p ) x ad hece ax (mod p ) ecause a (mod p ) wheever a (mod p ) for ay iteger. Thus it is reasoale to tae f(x) = a ad solve ƒ(x) = 0 usig (), we get, x x = x ( a)( )[ x x x ( + 1) 1 ( a)( )] + x x ( ax )x + = x + ( 1)( ax ) ( ax ) = x [1 + ] ( + 1)( ax ) + + ax ax = x [ ] ( + 1)( ax ) ( + 1) ( 1)ax = ( 1) + ( +1)ax Next we show that x +1 is a root of the cogruece ax (mod p ). Now if x is the solutio of ax (mod p ) the for some iteger t, we have, ax = + tp. Puttig i (6), we otai, (6)

4 This implies that ( + 1) ( 1)( + tp ) x + ( 1) + (+ 1)( + tp ) tp = [1 ]x + ( + 1)tp tp ax + [1 ] ( + tp ) + ( + 1)tp Usig Biomial Series expasio, we otai tp ax + 1 = [1 ]( + tp ) + ( + 1)tp tp ( + 1)tp 1 = [1 (1 + ) ]( + tp ) tp ( + 1)tp = [1 (1 )]( + tp ) tp + ( + 1)t p = [ ]( + tp ) This further ca e writte as ax + 1 = [ tp + ( + 1)t p ]( + tp ) ( tp )( + tp ) (mod p ) ( t p ) (mod p ) (mod p ) Fially, p is a odd prime ad, are ot divisile y p. Thus, (, p) = 1 ad hece (, p ) = 1. By Cacellatio Law, (7) yields that x +1 is the solutio of the cogruece ax (mod p ). The followig example illustrates Theorem.4. Example.: Let us solve the cogruece x (mod 7 ). I order to solve a polyomial cogruece of the form ax (mod p ), 1. We first see that it is sufficiet to solve ax (mod p), sice every solutio of ax (mod p ) is a solutio of ax (mod p). Oce we do this, the we ca apply Theorem.4, for fidig the solutios of ax (mod p ) from the solutios of the cogruece ax (mod p ). We see that x (mod 7) is the solutio of the cogruece x (mod 7). Thus we choose x 1 = as our iitial guess. The y (6), we otai So 488x = (mod 7 ) x 4 (mod 7 ) (7) 169 We repeat aove process util we get the solutio of the cogruece x (mod 7 ). The ecessary computatios are summarized i the followig Tale 1. Tale 1: Shows solutio of the cogruece x (mod 7 ) x x + 1(mod P ) 1 4 (mod 7 ) (mod 7 ) (mod 7 ) (mod 7 ) (mod 7 ) CONCLUSION From the umerical computatios of Tale 1, it ca e see that algorithm developed usig Halley s method eeds oly five iteratios to fid a solutio of the cogruece x (mod 7 ) whereas Hessel s lemma requires iteratios to compute the same solutio. Thus i cotrast, the techique suggested i this paper performs much faster for values of m i powers of for solvig polyomial cogrueces. For the cogrueces of prescried algorithm performs faster ad costs less i terms of time. As the solutio yielded y the give method is a iteger therefore the level of precisio is asolute ad desired covergece is volutarily achieved. Hece the solutio is a asolute solutio rather tha a approximate oe. Both the methods would geerate exactly same solutio hece elimiatig ay chace of error. Moreover Theorem.1 provides a method for reducig the degree of the cogruece, this provides eve a greater divided for larger powers. Reductio of degree assists to coverge to the solutio i eve lesser time for prolems which are otherwise irrepressile. REFERENCES 1. Krishamurthy, E.V. ad V.K. Murthy, 198. Fast Iterative Divisio of P-adic Numers. IEEE Trasactios o Computers, : Krishamurthy, E.V., O optimal iterative schemes for high-speed divisio. IEEE Tras. Computers, C-0: Kapp, M.P. ad C. Xeophotos, 010. Numerical Aalysis Meets Numer Theory: Usig Rootfidig Methods to Calculate Iverses Mod p. Appl. Aal. Discrete Math., 4: Adler, A ad J.E. Coury, 199. The Theory of Numers, Joes ad Bartlett Pulishers, Bosto.

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