Degree project CUBIC CONGRUENCE EQUATIONS

Degree project CUBIC CONGRUENCE EQUATIONS Author: Qadeer Ahmad Supervisor: Per-Anders Svensson Date: 2012-05-09 Subject: Mathematics and Modeling Level: Master Course code:5ma11e

Abstract Let N m(f(x)) denote the number of solutions of the congruence equation f(x) 0 (mod m), where m 2 is any composite integer and f(x) is a cubic polynomial. In this thesis, we use different theorems and corollaries to find a number of solutions of the congruence equations without solving then we also construct the general expression of corresponding congruence equations to demonstrate the solutions of the equations. In this thesis, we use Mathematica software as a tool.

Contents 1 Introduction 4 2 Theory 6 3 Cubic Congruence (mod p) 8 3.1 Cubic Congruence with linear term (mod p)........... 8 4 Cubic Congruence (mod p k ) 12 4.1 Special case.............................. 13 4.2 Cubic Congruence with linear term (mod p k )........... 18 4.2.1 Quadratic Equation (mod p 3 )............... 20 4.2.2 General Expression...................... 21 4.3 Cubic Congruence with linear and quadratic term (mod p k ).. 23 4.4 Root of multiplicity three of cubic congurence (mod p k )..... 24 5 Cubic congruence (mod 2 k ) 29 5.1 Case I when j k.......................... 30 5.2 Case II when j < k.......................... 31 6 Cubic Congruence (mod m) 35 6.1 Cubic Congruence with linear term (mod m)........... 36 7 Bibliography 38 3

Chapter 1 1 Introduction A congruence is like a statement about divisibility, but the difference is that this is more convenient notation. It often makes easier to discover proofs. The theory of congruence was introduced by Carl Friedrich Gauss (1777 1855) one of the best mathematicians. Even though Pierre de Fermat (1601 1665) earlier studied number theory in systematic way, Gauss was first to introduce the congruence subject as a branch of mathematics and developed the language of congruence in beginning of 19th century. Definition 1.1. Let m be any positive integer and a, b be integers with a 0. Then we say that a is congruent to b modulo m if m divide the difference of a b and write as a b (mod m). If a b is not divisible by m then we say that a is not congruent to b write as a b (mod m). Since a b is divisible by m if and only if a b is also divisible by m, we shall suppose throughout the modulus m is a positive integer. Example 1.1. Let 6, 3, 9 Z, where Z is the set of integers and 6 congruent to 3 modulo 9. We denote as 6 3 (mod 9). Linear Congruence Equations A linear congruence equation is of the form ax b (mod m), where a, b, m Z. Theorem 1.1. Let a, b, and m > 0 be given integers, and d =gcd(a, m). The linear congruence equation is ax b (mod m), has no solution at all, if d b. If d b then there are d solutions For proof see [1] Example 1.2. Solve the congruence equation 8x 4 (mod 12). (mod m). Solution. Since (8, 12) = 4 and 4 4, there are exactly four solutions. We solve the congruence equation and x 2, 5, 8, 11 (mod 12) are the solutions. Quadratic Equation A quadratic equation is of the form ax 2 b (mod m), where a, b, m Z. 4

Lemma 1.1. Let p be an odd prime. Let a Z, (a, p) = 1 then x 2 a (mod p) either has no solution or, exactly two solutions. More detail of lemma can find in [5] Example 1.3. Solve x 2 3 (mod 11). Solution. We see that f(3) = 22 0 (mod 11) and f(6) = 33 0 (mod 11). Hence x 3, 6 are the solutions of above congruence. Example 1.4. Solve x 2 5 (mod 3). Solution. We note that f(t) 0 (mod 3) where t = 0, 1, 2, from which we deduce that the quadratic congruence has no solution. Cubic Congruences Equations Let the cubic congruence equation be f(x) = x 3 a (mod m) where a Z and m be any composite integer. Chapter s Overview Chapter 2 is about some basic definitions that we will use throughout in our thesis. In Chapter 3 we will solve the different kind of cubic congruence equations modulo p, where p be a prime number, to find all possible solutions of the congruence equations. In the literature the discussion of polynomial congruence equations is mostly focused on how to solve such an equations, and not on computing the number of solutions. By using Mathematica as a tool, we will derive a general expression for how the solutions look like for the different kind of cubic congruence equations, and from these general expressions can also find the number of possible solutions. This is described in Chapters 4 and 5. In Chapter 4 we will solve different kinds of cubic congruence equation s mod p k where k Z +. We shall use Hensel s lifting method. In Chapter 5 we will solve cubic congruence equations mod 2 k. In Chapter 6 we will solve different kinds of cubic congruence equations modulo m where m is any composite positive integer. 5

Chapter 2 2 Theory The theory of the number is related with the properties of the natural numbers 1, 2, 3, also called the positive integers, these number s combine with negative integers and zero from set of integers. Thousands of people work on communal number problems over the internet. The solution of the main problem in number theory is reported on the PBS television series NOVA. People study the number theory to understand the system for making secret messages. Definition 2.1. An integer a 0 is divisor of an integer b if there exist an integer y such that b = ay and we denote this by a b. In case a does not divide b then we write a b. If p k a and p k+1 a where p is a prime, that is p k is the highest power dividing a, then we denote it by p k a. Example 2.1. Suppose that 6, 12 Z, where Z is the set of integers. The 6 is divisor of 12 becuase 12 = 6 2. This is denoted by 6 12. Definition 2.2. A prime is an integer greater than 1 that is only divisible by 1 and itself. In other words, a prime is an integer with two positive integer divisors. Example 2.2. Let 11 Z, where Z is the set of integers. The divisors of 11 are 1 and itself only so 11 is a prime number. Definition 2.3. An element a Z is said to have a multiplictaive inverse if there exist an element b Z such that ab 1 (mod m). Theorem 2.1. An element a Z has a multiplicative inverse modulo m, if and only if greatest common divisor of a and m are relative prime and we denote it by a 1. Further, if a multiplicative inverse exist, then it can be determined uniquely (mod m). proof. By definition 2.3 multiplicative inverse of a is solution of ax 1 (mod m) define in theorem 1.1, if and only if (a, m) = 1. Example 2.3. Since gcd(3, 7) = 1, the element 3 Z has a multiplicative inverse modulo 7. We see that 3 1 = 5, since 3 5 15 1 (mod 7). Definition 2.4. Let m be any integer, and (a, m) = 1. If the congruence x 2 a (mod m) has solution then a is called a quadratic residue modulo m. If the congruence has no solution, then a is called quadratic non residue mod m. Example 2.4. Let x 2 4 (mod 7). This congruence has two solutions x = 2 and 5, therefore 4 is a quadratic residue. 6

Definition ( ) 2.5. If p is an odd prime, a Z and (a, p) = 1 then Legendre symbol is defined as (i) (ii) ( ( a p a p a p ) = 1 if a is quadratic residue (mod p). ) = 1 if a is quadratic non-residue (mod p). (iii) 0 if a 0 (mod p). Example 2.5. Let x 3 5 (mod 11). quadratic residue therefore ( 5 11) = 1. Then x = 3 is a solution, hence 5 is Definition 2.6. The discriminant of a cubic equation x 3 + ax + b = 0 is defined as = 4a 3 27b 2. Further information on discriminant can find in [2]. Example 2.6. Suppose that x 3 + 4x + 7 0 (mod 13). The discriminant of this congruence is = 4 4 3 27 7 2 = 1579. 7

Chapter 3 3 Cubic Congruence (mod p) A cubic congruence equation be x 3 a (mod p), where p is a prime positive integer and a Z. Theorem 3.1. Let p be a prime and (a, p) = 1. Then x n a (mod p) has (n, p 1) solutions if a (p 1)/(n,p 1) 1 (mod p) and no solution if a (p 1)/(n,p 1) 1 (mod p). For proof of this theorem see [1]. Example 3.1. Determine the number of solutions of the congruence x 3 6 (mod 7). Solution. We note that (6, 7) = 1. As (3, 7 1) = (3, 6) = 3 and 6 (6/3) = 6 2 1 (mod 7), we deduce that the congruence has x 3, 5, 6 (mod 7) three solutions. 3.1 Cubic Congruence with linear term (mod p) Suppose that we have cubic congruence equation with the linear term x 3 + bx a (mod p), where p is prime and a, b Z. The following lemma is stated in [3]. Lemma 3.1. Let p > 3 be a prime, let a, b Z, If (p, b) = 1 and suppose the congruence is x 3 + ax + b 0 (mod p). Let t Z be such that tb 2 a 3 (mod p), and Ψ p (T ) = T k 1 (p 5)(p 7) 3 2 2 T k 2 + + 3! ( 1) s 1 [p (2s + 1)][p (2s + 3)] [p (4s 1)] (2s 1) T k s + + 2 2(s 1) (2s 1)! ( 1) k 1 [p (2k + 1)][p (2k + 3)] [p (4k 1)] (2k 1). (3.1) 2 2(k 1) (2k 1)! (i) If (4t + 27) is a quadratic non residue modulo p, then cubic congruence has a unique solution x, 0 < x < p. 8

(ii) If (4t + 27) 0 (mod p), then the congruence has one repeated root and one another means two distinct solutions x 1, x 2, 0 < x 1, x 2 < p. (iii) If (4t+27) is a quadratic residue modulo p and if Ψ p (t) 0 (mod p) then cubic congruence has three distinct solution x 1, x 2, x 3, 0 < x 1, x 2, x 3 < p. (iv) If (4t + 27) is a quadratic residue modulo p and if Ψ p (t) 0 (mod p) then cubic congruence has no solution. (v) The values of t are k such that congruence has many solutions and 2k values of for which there is no solution for congruence. Example 3.2. Solve the congruence x 3 6x + 15 0 (mod 13). Solution. We get the value of t = 11, after solving the linear congruence equation t15 2 ( 6) 3 (mod 13). Now we see that (4t + 27) = (4 11 + 27) = 5 8 (mod 13). The 8 is a quadratic non-residue so this congruence equation has unique solution. The above example satisfy case (i) and now we take another example to check case (ii) lemma 3.1. Example 3.3. Solve the congruence x 3 13x + 23 0 (mod 7). Solution. We solve t23 2 ( 13) 3 (mod 7) to get t = 2. We obtain (4t+27) = (8+27) 0 (mod 7) after verifying the equation. So this equation has x = 4, 4 and 6 two distinct solutions. This satisfy the case (ii) of lemma 3.1. Now we take another example to verify case (iii). Example 3.4. Solve the congruence x 3 14x + 15 0 (mod 23). Solution. Start to solve t15 2 ( 14) 3 (mod 23) for t, to get t = 6. Further we check that (4t + 27) = (4 6 + 27) 18 (mod 23). The integer 18 is a quadratic residue so this equation has three solutions or no solution. We check it as Ψ 23 (T ) = T 3 3 (23 5)(23 7) 2 2 T 2 + ( 1) 3 1 (23 7)(23 9)(23 11) (5) 3! 2 2 2 T (5)! + ( 1) 4 1 (23 9)(23 11)(23 13)(23 15) (7) 2 2 3. (3.2) (7)! We obtain that Ψ 23 (T ) = T 3 36T 2 +7T 7 24 1 use multiplicative inverse then Ψ 23 (T ) = T 3 36T 2 +7T 7. We replace T by t so Ψ 23 (t) = t 3 36t 2 +7t 7 0 (mod 23). We have t = 6 therefore we get Ψ 23 (6) = 6 3 36 6 2 +7 6 7 13 0 (mod 23). The congruence equation has no solution according to case (iv) of lemma 3.1. However we have checked also on Mathematica that this equation has three solutions. The result we obtained from Mathematica tool and above example, deviate from the result of case (iii) of lemma 3.1. The deviation the result might 9

be due to the misunderstanding or mistakes in the formulation of lemma 3.1. We do not have proof of lemma 3.1 and its assumptions in [3]. Another result, which agrees with Mathematica calculations is given by theorem 3.2. Let cubic congruence equation with linear term is x 3 + ax + b 0 (mod p) where p > 3 is prime, k( ) Z + and a, b Z. The discriminant of this equation is = 4a 3 27b 2 and is Legendre symbol. (i) p ( ) p = 1 if is quadratic non-residue (mod p) then the congruence has unique solution. (ii) 0 (mod p) then the congruence has three solutions with one repeated solution. ( ) (iii) p = 1 if is quadratic residue (mod p) then the congruence has 0 or 3 solutions. The first two cases have taken from lemma 3.1 and case (iii) has stated in theorem 3.2. Example 3.5. Determine the number of solutions of congruence x 3 +5x+6 0 (mod 7). Solution. The discriminant of congruence is = 4 5 3 27 6 2 = 1472 and Legendre ( ) 1472 7 = 1 because 1472 is quadratic non-residue mod 7. Hence the congruence has unique solution that is x = 6. Example 3.6. Determine the number of solutions of congruence x 3 +2x+3 0 (mod 5). Solution. The = 4 2 3 27 3 2 = 275 is discriminant of cubic congruence equation and 275 0 (mod 5) therefore congruence equation has two distinct solutions. The solutions are x = 2, 4, 4. Theorem 3.2. Let ( A, B Z, ) and let p = 6k ± 1 be a prime number such that (p, AB) = 1 and = 1. If v n is defined as v 0 = 2, v 1 = B and 4a 3 27b 2 p v n+1 = Bv n + ( ) A 3 3 vn 1, then { N p (x 3 3, if v + Ax B) = (p ( p 3 ))/3 2( p 3 )( A 3 )(1 ( p 3 ))/2 (mod p) 0, if v (p ( p 3 ))/3 ( p 3 )( A 3 )(1 ( p 3 ))/2 (mod p). Here ( p 3 ) is Legendre while ( A 3 ) is not. For proof of this theorem see [4]. 10

Example 3.7. Find the number of solutions of x 3 14x + 15 0 (mod 23). Solution. Let A = 14, B = 15 and A B = 210, we have that ( 210, 23) = 1. Now = 4 ( 14) 3 27 15 2 = 4901 is discriminant of the cubic congruence and ( ) 4901 23 = 1 is Legendre with discriminant. This means that equation has three solutions or no solution. We also verify it by theorem 3.2 such as ( ) 23 23 v (23 ( 23 3 ))/3 (1 ( 2 ( 14) 3 ))/2 (3 1 23 (1 ( ) 3 ))/2 (mod 23), 3 we get v 8 17 (mod 23). By theorem 3.2 we have that v 0 = 2, A = 14 and v 1 = B = 15. We calculate v 2 3 (mod 23), v 3 10 (mod 23), v 4 0 (mod 23), v 5 17 (mod 23), v 6 21 (mod 23), v 7 6 (mod 23) and v 8 17 (mod 11). The result is verified, hence prove that above congruence equation has three solutions. Example 3.8. Find number of solutions of the congruence x 3 + 2x + 15 0 (mod 13). Solution. Starting with A = 2, B = 15 so A B = 30, we see that (30, 13) = 1. The discriminant is = 4 2 3 27 15 2 = 6107 and Legendre ( ) 6107 13 = 1 because 6107 is a quadratic residue. This means that equation has three solutions or no solution, we check it by theorem 3.2 according as ( ) 13 13 v (13 ( 13 3 ))/3 (1 ( (2) 3 ))/2 (3 1 13 (1 ( ) 3 ))/2 (mod 13), 3 and we get v 4 = 1 12 (mod 13). By theorem 3.2 we have that v 0 = 2, A = 2 and v 1 = B = 15. We calculate v 2 = 15 15 + 2 3 (3 1 ) 3 2 7 (mod 13), v 3 = 15 7 15 2 3 9 3 9 (mod 13) and v 4 = 15 9 + 2 3 9 3 7 12 (mod 13). Hence it is verified that above congruence equation has no solution. 11

Chapter 4 4 Cubic Congruence (mod p k ) Let the cubic congruence be where p be a prime, k Z + and a Z. x 3 a (mod p k ), Theorem 4.1. Hensel s Lemma Let f(x) be a polynomial with integer coefficients and k an integer with k 2. Let r be a solution of f(x) 0 (mod p k 1 ). Then (i) if f (r) 0 (mod p), then we get a unique integer t, 0 t p, such that f(r + tp k 1 ) 0 (mod p k 1 ), given as t f (r)(f(r)/p k 1 ) (mod p), where f (r) is the multiplicative inverse of f (r) modulo p. (ii) if f (r) 0 (mod p) and f(r) 0 (mod p k ), then f(r + tp k 1 ) 0 (mod p k ) for all integers t where 0 t p 1. (iii) if f (r) 0 (mod p) and f(r) 0 (mod p k ), then f(x) 0 (mod p k ) has no solution with x r (mod p k 1 ). In case (i), we note that f(x) 0 (mod p k 1 ) lifts a unique solution of f(x) 0 (mod p k ), and in case (ii), t is arbitrary so solution lifts to p incongruent solutions modulo p k or to none at all. For proof of lemma see [5]. Example 4.1. Solve the congruence f(x) = 2x 3 + 7x 4 0 (mod 5 2 ). Solution. Starting with f(x) = 2x 3 + 7x 4 0 (mod 5), we note that x = 1 is the only solution. Here f (1) = 13 3 0 (mod 5), so root is non-singular. Taking f (1) 2 (mod 5) and f(1) = 5. We use case (i) that is t = 2 5/5 = 2 3 (mod 5) to get t = 3. We conclude that x = 1 + 3 5 16 (mod 5 2 ) is the only solution of f(x) = 2x 3 + 7x 4 0 (mod 5 2 ). Corollary 4.1. Let r be a solution of the polynomial congruence equation f(x) 0 (mod p), where p is prime. If f (r) 0 (mod p), then we have a unique solution r k modulo p k, k = 2, 3, such that r k = r k 1 f(r k 1 )f (r). Here f (r) is the multiplicative inverse of f (r) modulo p. For proof of this corollary see [5]. 12

Example 4.2. Solve x 3 + 2x + 3 0 (mod 7 3 ). Solution. First we note that x 6 (mod 7) is solution of x 3 + 2x + 3 0 (mod 7). Since f (x) = 3x 2 + 2, we see that f (6) 5 0 (mod 7). Taking f (6) 3 (mod 7), we see by corollary 4.1 that the root r 1 = 6 (mod 7) lifts to r 2 = 6 231 3 48 (mod 7 2 ). Further r 2 lifts r 3 = 48 110691 3 342 (mod 7 3 ) is only solution of x 3 + 2x + 3 0 (mod 7 3 ). Theorem 4.2. Let m = 1, 2, 4, p α, or 2p α, where is p is an odd prime. If (a, m) = 1 then the congruence equation x n a (mod m) has (n, φ(m)) solutions if a φ(m)/(n,φ(m)) 1 (mod m) but no solutions if a φ(m)/(n,φ(m)) 1 (mod m). For proof of the theorem see [1]. Example 4.3. Find the number of solutions of x 3 7 (mod 19 2 ). Solution. Since (7, 19 2 ) = 1 the congruence has (3, φ(19 2 )) = (3, 342) = 3 solutions or no solutions according as 7 φ(192 )/(3,φ(19 2 )) = 7 342/3 1 (mod 19 2 ). The congruence x 3 7 (mod 19 2 ) has three solutions. 4.1 Special case Among the refrences there is no generalization of theorem 4.2 that describes the case when (a, p k ) 1. Now we are going to solve the special case which is not recovered by theorem 4.2. Let p > 3 be a prime number, if (a, p k ) 1 then the congruence equation x 3 a np α (mod p k ) (4.1) has two cases where n an integer and k, α Z +. Suppose that x β + tp k 1 (mod p k ) are solutions of congruence equation (4.1) where 0 t p 1 and k 2. Put this value of x in equation (4.1) with k replaced by k + 1 we see that f(β + tp k 1 ) = (β + tp k 1 ) 3 + np α = β 3 + t 3 p 3k 3 + 3β 2 tp k 1 + 3βt 2 p 2k 2 + np α. Since 3k 3 = k + (2k 3) k + 1 we have that f(β + tp k 1 ) (β 3 + np α ) + 3β 2 tp k 1 f(β) + 3tf (β)p k 1 cp k + 3dtp k 0 (mod p k+1 ) 13

Since throughout the congruence is divisible by p k, on dividing p k we get that c + dt 0 (mod p), (4.2) which is a linear equation, where c = f(β)/p k and d = f (β)/p by second case of Hensel s lemma. This linear congruence has s = gcd(d, p) solutions if s c otherwise no solution by theorem 1.1. This shows that number of solutions of congruence equation (4.1) are dependent of f (x) = 3x 2, p and independent of n. Case I When α k When α k then cubic congruence is x 3 p α 0 (mod p k ), (4.3) where k, α Z +. The number of solutions will depend on k. There exist two sub cases for this congruence equation. (i) When k 0, 1 (mod 3) the number of solutions will be p 2q where q = k/3. General Expression (a) When k 0 (mod 3), to find all possible solutions we will generate general expression as follows. Suppose that x mp q (mod p k ), (4.4) where q = k/3 and m = 0, 1,, p 2q 1. Put this value in (4.3) as (mp q ) 3 0 (mod p k ). Take left hand side and put value of k/3 = k 0 3 by division algorithm in above congruence equation such as (mp ( k 0 3 ) ) 3 m 3 p k 0 (mod p k ), shows the right side. This show that the congruence equation x 3 0 (mod p k ) has the solutions of form x mp q (mod p k ). (b) When k 1 (mod 3), to find all possible solutions we will general generate expression. Let us x mp q+1 (mod p k ), (4.5) where q = k/3 and m = 0, 1,, p 2q 1. Put this value in (4.3) we find that (mp q+1 ) 3 0 (mod p k ). Take left hand side and put value of k/3 = k 1 3 by division algorithm in above congruence equation such as (mp ( k 1 3 +1) ) 3 m 3 p k+2 0 (mod p k ), the above result is equal to right hand side. This shows that the congruence equation x 3 0 (mod p k ) has the solutions of form x mp q+1 (mod p k ). 14

(ii) When k 2 (mod 3) the number of solutions will be p 2q+1 where q = k/3. General Expression In order to find all possible solutions we will derive a general expression as follows. Suppose that x mp q+1 (mod p k ), (4.6) where q = k/3 and m = 0, 1,, p 2q+1 1. Put this value in (4.3) (mp q+1 ) 3 0 (mod p k ). Take left hand side and put value of k/3 = k 2 3 by division algorithm in above congruence equation such as (mp ( k 2 3 +1) ) 3 m 3 p k+1 0 (mod p k ), gives right hand side. This shows that the congruence equation x 3 0 (mod p k ) has the solutions of form x mp q+1 (mod p k ). Example 4.4. Determined the solutions of congruence x 3 0 (mod 5 3 ). Solution. We see that k 0 (mod 3), the congruence equation has 5 2 1 = 25 solutions because q = 3/3 = 1. To find all possible solutions we use equation (4.4), that is x m5 (mod 5 3 ) where m = 0, 1,, 5 2 1. The roots are x 0, 5, 10, 15,, 120 (mod 5 3 ). Example 4.5. Calculate the solutions of the congruence equation x 3 0 (mod 7 4 ). Solution. We note that k 1 (mod 3), the congruence equation has 7 2 1 = 49 solutions because q = 4/3 = 1. To determine all possible solutions by equation (4.5) that is x m7 2 (mod 7 4 ), where m = 0, 1,, 7 2 1. The possible solutions are x 0, 49, 98, 147,, 2352 (mod 7 4 ). Example 4.6. Find the solutions of the congruence equation x 3 0 (mod 11 5 ). Solution. Since k 2 (mod 3), the congruence equation has 11 2 1+1 = 1331 solutions because q = 5/3 = 1. We use equation (4.6) to find the all possible solutions of above congruence that is x m(11) 1+1 (mod 11 5 ), where m = 0, 1,, 11 4 1. Hence solutions are Case II When α < k x 0, 121, 242, 363,, 160930 (mod 11 5 ). When α < k then cubic congruence is x 3 p α (mod p k ), (4.7) where k, α Z +. The congruence equation (4.7) has solutions if 3 α otherwise not. 15

(a) when k 0 (mod 3) Let α = k then the cubic congruence is x 3 p k (mod p k ). (4.8) This cubic congruence equation (4.8) has solutions of form x mp q (mod p k ) where q = k/3 by equation (4.4) as we have discussed in above general expression. Put this value of x in equation (4.8) and replace k by k + 1 we see that f(mp q ) = (mp q ) 3 p k = (mp k 3 ) 3 p k m 3 p k p k 0 (mod p k+1 ). Divide throughout the congruence by p k we obtain that m 3 1 0 (mod p). (4.9) The congruence equation (4.9) is solveable and has (3, Φ(p)) solutions by theorem 4.2. The congruence equation is cubic therefore n = 3 is fix has always 1 or 3 solutions. (b) when k 1 (mod 3) Let α = k then the cubic congruence is x 3 p k (mod p k ), (4.10) has the solutions of form x mp q+1 (mod p k ) where q = k/3 by equation (4.5). Put this value of x in equation (4.8) and replace k by k + 1 we see that f(mp q+1 ) = (mp q+1 ) 3 p k = (mp k 1 3 +1 ) 3 p k m 3 p k+2 p k 0 (mod p k+1 ). Divide throughout the congruence by p k we obtain that m 3 p 2 1 1 0 (mod p). This congruence is not solveable. (c) when k 2 (mod 3) Let α = k then the cubic congruence is x 3 p k (mod p k ). (4.11) The congruence equation (4.8) has solutions of form x mp q+1 (mod p k ) where q = k/3 by equation (4.4). Put this value of x in equation (4.8) and replace k by k + 1 we see that f(mp q+1 ) = (mp q+1 ) 3 p k = (mp k 2 3 +1 ) 3 p k m 3 p k+1 p k 0 (mod p k+1 ). 16

Divide throughout the congruence by p k we obtain that m 3 p 1 1 0 (mod p). This congruence equation cannot be solved. This shows that the congruence equation x 3 p α (mod p k ) has solutions if 3 α. The congruence equation has p 2q or 3p 2q solutions where q = α/3 depend on equation (4.9). General Expression We can construct the general expression for cubic congruence x 3 p α (mod p k ) (4.12) where α, k Z + to find all the solutions when α < k. congruence equation (4.12) is Let the solutions of x mp q + np b (mod p k ), (4.13) where q = α/3, b = k 2q and n = 0, 1,, p 2q 1. Here m is a solution of the cubic congruence m 3 1 (mod p). Put this value of x in cubic congruence equation (4.12) such as (mp q + np b ) 3 p α (mod p k ). Take Left hand side and by division algorithm α/3 = (α 0)/3 we have that (mp q + np b ) 3 = m 3 p 3q + n 3 p 3b + 3m 2 p 2q p b + 3mp q n 2 p 2b = m 3 p α + n 3 p 3k 2α + 3m 2 p 2α/3 p k 2α/3 + 3mp α/3 n 2 p (6k 4α)/3 = m 3 p α + n 3 p 3k 2α + 3m 2 p k + 3mp α/3 n 2 p (2k α)/3 m 3 p α (mod p k ). Since m 3 1 (mod p) therefore the congruence is (mp q + np b ) 3 p α (mod p k ). The above result is equal right hand side of equation (4.12). This shows that the solutions of x 3 p α (mod p k ) always in the form x mp q + np b (mod p k ). Example 4.7. Determine the solutions of x 3 7 3 (mod 7 4 ). Solution. We start to solve the congruence equation m 3 1 (mod 7) by equation (4.9) has m 1, 2, 4 (mod 7) three solutions. The original congruence equation has 3 7 2 3/3 = 147 solutions. To find all possible solutions we use the general equation (4.13) that is x 7m + 7 2 n (mod 7 4 ) where n = 0, 1, 2,, 7 2 1 and m 1, 2, 4 (mod 7). For m = 1 the solutions are x 7, 56, 105,, 2359 (mod 7 4 ). 17

For m = 2 we have the following solutions The possible solutions when m = 4 are x 14, 63, 112,, 2366 (mod 7 4 ). x 28, 77, 126,, 2380 (mod 7 4 ). These are all solutions of the congruence equation x 3 7 3 (mod 7 4 ). Example 4.8. Find the solutions of x 3 11 3 (mod 11 5 ). Solution. The congruence m 3 1 (mod 11) has a unique solution m 1 (mod 11) by equation (4.9). The original cubic congruence equation has 11 2 3/3 = 121 solutions. To demonstrate the solutions use equation (4.13) that is x 11m + 11 3 n (mod 11 5 ) where n = 0, 1, 2,, 11 2 1 and m = 1. The solutions of congruence equation are x 11, 1342, 2673,, 159731 (mod 11 5 ). 4.2 Cubic Congruence with linear term (mod p k ) Let cubic congruence equation with linear term is where p is prime, k Z + and a, b Z. The discriminant of (4.14) is ( p ) is Legendre symbol. (i) ( p x 3 + ax + b 0 (mod p k ), (4.14) = 4a 3 27b 2. ) = 1 if is quadratic non-residue (mod p) and root x r (mod p) staisfy first case of Hensel s lemma then the congruence has unique solution (mod p k ). (ii) 0 (mod p) then the congruence has different solutions (mod p k ). ( ) (iii) = 1 if is quadratic residue (mod p) and root x r (mod p) p staisfy first case of Hensel s lemma then the congruence has 0 or 3 solutions (mod p k ). The first two cases have taken from lemma 3.1 and case (iii) has stated in theorem 3.2. 18

Example 4.9. Show that the congruence equation x 3 + 2x + 3 0 (mod 13 8 ) has unique solution. Solution. The discriminant is = 4.2 3 27.3 2 = 275 11 (mod 13). The Legendre ( 11 13) = 1 because 11 is quadratic non-residue. We note that x = 12 is the only solution mod 13. Here f (12) 0 staisfy case (i) of Hensel s lemma therefore cubic congruence has a unique solution (mod 13 8 ). Example 4.10. Find the number of solutions of congruence equation x 3 +12x+ 6 0 (mod 5 6 ). Solution. Starting with = 4 12 3 27 6 2 = 9099 1 (mod 5). Now we see that ( 1 5) = 1 because 1 is quadratic residue so cubic congruence has 0 or 3 solutions (mod 5 6 ) verify by theorem 3.2. Here A = 12, B = 6 so A B = 72 and (72, 5) = 1 then we have ( ) 5 v (5 ( 5 3 ))/3 (12) (1 ( 5 3 ))/2 (3 1 ) (1 ( 5 3 ))/2 (mod 5). 3 We obtain v 2 = 24 4 (mod 5). By theorem 3.2 since v 0 = 2, A = 12 and v 1 = B = 6, we find v 2 = 6 6 + 12 3 (3 1 ) 3 2 4 (mod 5). Result is verified and above equation has no solution (mod 5). It is clear, that by Hensel s lemma (mod 5 6 ) also has no solution. Example 4.11. Find the number of solutions of congruence equation x 3 +4x+ 6 0 (mod 11 5 ). Solution. ( We have = 4 4 3 27 6 2 = 1228 4 (mod 11). The Legendre 4 11) = 1 because 4 is quadratic residue. It is clear that cubic congruence has 0 or 3 solutions (mod 11 5 ). it is verified by theorem 3.2. First we solve this congruence (mod 11), here A = 4, B = 6 so A B = 24 then we see that v (11 ( 11 3 ))/3 2 ( 11 3 ) 11 (1 ( (4) 3 ))/2 (3 1 11 (1 ( ) 3 ))/2 (mod 11), we get v 4 1 (mod 11). Using theorem 3.2 v 0 = 2, A = 4 and v 1 = B = 6. We determine that v 2 = 6 6 + 4 3 (3 1 ) 3 2 0 (mod 11), v 3 = 6 0 4 3 (3 1 ) 3 6 9 (mod 11) and v 4 = 6 9 4 3 (3 1 ) 3 0 1 (mod 11). The above result that congruence has three solutions (mod 11). By uniqueness of Hensel s lemma case (i) mod 11 5 also has three solutions. 19

4.2.1 Quadratic Equation (mod p 3 ) Let us the cubic congruence equation is f(x) = x 3 + ax + b 0 (mod p k ), (4.15) where p is prime, k Z + and a 0 such that a, b Z. When the discriminant of the equation will be = 4a 3 27b 2 = 0 then this congruence has always two distinct solutions such as x (x α) 2 (x β) (mod p), where α is repeated root. If α staisfy (ii) case of Hensel s lemma that is f (α) 0 (mod p) and f(α) 0 (mod p 2 ) then α + tp are solutions (mod p 2 ) where 0 t p 1. If α satisfied the third case of Hensel s lemma then the repeated root will not lift and congruence equation has a unique solution (mod p k ). To find which solutions and how many will lift for next level mod p 3 we can generate a general quadratic equation. Put α + tp in (4.15) and take k = 3 we get f(α + tp) = (α + tp) 3 + a(α + tp) + b = α 3 + t 3 p 3 + 3α 2 tp + 3αt 2 p 2 + aα + atp + b = (α 3 + aα + b) + (3α 2 + a)tp + 3αt 2 p 2 = f(α) + f (α)tp + 3αt 2 p 2 3αt 2 p 2 + kp 2 t + p 2 m (mod p 3 ), finally we get the quadratic equation that is 3αt 2 + kt + m 0 (mod p), (4.16) where k = f (α)/p (mod p) and m = f(α)/p 2 (mod p). This quadratic has 0 or 2 solutions (mod p). ( ) 0, if N p (3αt 2 p = 1 (mod p) + kt + m) = ( ) 2, if = 1 (mod p). p (4.17) When two solutions will survive, and second case of Hensel s lemma will satisfy then these two solutions will yield 2p solutions on the next level, and congruence equation will have 2p + 1 solutions. Since We have a unique solution on each level by first case of Hensel s lemma. Theorem 4.3. Suppose that f(x) is a polynomial with integral coefficients. Let f(a) 0 (mod p j ), that p τ f (a), and that j 2τ + 1. If b a (mod p j τ ) then f(b) f(a) (mod p j ) and p τ f (b). Moreover there is unique t (mod p) such that f(a + tp j τ ) 0 (mod p j+1 ). In this condition, a collection of p τ solutions (mod p j ) give rise to p τ solutions (mod p j+1 ), while power of p dividing f remain constant. Since hypothesis of theorem apply with a replaced by a + tp j τ and (mod p j ) replaced by 20

(mod p j+1 ) but with τ unchanged, the lifting may be repeated and continuous indefinitely. For proof this theorem see [1] 4.2.2 General Expression Let cubic congruence equation is f(x) = x 3 + ax + b 0 (mod p j+1 ), (4.18) where p is prime, j Z + and a, b Z. Suppose that x α + tp j τ (mod p j ) are solutions of equation (4.18) and p τ f (α), where 0 t p 1 and j 2τ + 1. Put value of x in cubic congruence equation we have f(α + tp j τ ) = (α + tp j τ ) 3 + a(α + tp j τ ) + b = α 3 + t 3 p 3(j τ) + 3α 2 tp j τ + 3αt 2 p 2(j τ) + aα + atp j τ + b, since 3j 3τ = j + (2j 3τ) j + 1 therefore (α 3 + aα + b) + (3α 2 + a)tp j τ f(α) + f (α)tp j τ kp j + mtp j 0 (mod p j+1 ). Since both term on left side are divisible by p j and the right side is also we find k + mt 0 (mod p), (4.19) where (m, p) = 1, k = f(α)/p j (mod p) and m = f (α)/p τ (mod p). So use this linear equation we can find the value of unique t. Example 4.12. Find the number of solutions of the congruence x 3 +12x+4 0 (mod 17 4 ). Solution. The congruence has discriminant = 4 12 3 27 4 2 0 (mod 17) so this congruence has three solutions given as x 1, 8, 8 (mod 17), with one repeated root. The polynomial is f(x) = x 3 + 12x + 4. We note that x = 1 is one solution. Here f (1) = 3.1 2 + 12 0 (mod 17) satisfy case (i) of Hensel s lemma. Now f (8) = 204 0 (mod 17) and f(8) = 612 34 0 (mod 17 2 ), this hold third case of Hensel s so 8 will not lift to the next solutions and only one solution x = 1 will survive uniquely (mod 17 4 ), from which we deduce that the congruence has a unique solution (mod 17 4 ). 21

Example 4.13. Find the number of solutions of the congruence equation x 3 + 5x + 17 0 (mod 19 5 ). Solution. We have = 4 5 3 27 17 2 0 (mod 19) so, three solutions of congruence are, x 12, 12, 14 (mod 19), with one repeated root. Here f (14) = 593 4 0 (mod 19) hold case (i) of Hensel s lemma. We note that x = 12 is repeated root. We have that f (12) = 437 0 (mod 19) and f(12) = 1805 0 (mod 19 2 ) satisfy case (ii), so that x 12, 31, 50, 69,, 354 (mod 19 2 ), are the solutions of x 3 + 5x + 17 0 (mod 19 2 ). Now we will use equation (4.16) to find the solutions (mod 19 3 ). First we find value of k and m such than k = f (12)/19 = 437/19 4 (mod 19), m = f(12)/19 2 = 1805/19 2 4 (mod 19). By quadratic equation (4.16) we see that 36t 2 + 4t + 5 0 (mod 19). The discriminant is = 4 2 4 36 5 = 704. Legendre symbol of discriminant is ( ) 704 19 = 1 because 704 is quadratic non-residue therefore by equation (4.17) no solution will survive (mod p 3 ). We conclude that congruence has a unique solution (mod 19 4 ) that will lift by solution x = 14 due to first case of Hensel s lemma. Example 4.14. Find the number of solutions of congruence equation x 3 + x + 10 0 (mod 13 4 ). Solution. Starting with = 4 1 3 27 10 2 = 2704 0 (mod 13). So this equation has three solutions that are x 11, 11, 4 (mod 13), with one repeated root. We see that f (4) = 49 10 0 (mod 13) hold case (i) of Hansel s. We note that x = 11 is repeated root. We have that f (11) = 364 0 (mod 13) and f(11) = 1352 0 (mod 13 2 ) satisfy case (ii) of Hensel s lemma. So that x 11, 24, 37, 50,, 167 (mod 13 2 ), are solutions of x 3 + x + 10 0 (mod 13 2 ). Now we will use equation (4.16) to find solutions (mod 13 3 ). First we find value of k and m such that k = f (11)/13 = 364/13 2 (mod 13), m = f(11)/13 2 = 1352/13 2 8 (mod 13). By quadratic equation (4.16) we have 33t 2 + 2t + 8 0 (mod 13). We have ) = 2 2 4 33 8 = 1052. Legendre symbol with discriminant is = 1 because 1052 is quadratic residue therefore by equation (4.17) ( 1052 13 22

two solution will survive (mod 13 2 ). We solve this quadratic we get t = 10, 12 which gives that the 11 + 10 13 = 141 and 11 + 12 13 = 167 these two roots give 2 13 + 1 solutions of the form 141 + 13 2 t, 167 + 13 2 t (mod 13 3 ), where 0 t 12. By theorem 4.3 we can see that 13 f (141), f(141) 0 (mod 13 3 ) and 3 2 + 1. Similarly 13 f (167), so that f(167) 0 (mod 13 3 ). Hence there will be unique t that will lift for every next level and number of solutions will remain same after modulo 13 3. We have found for each k 3 there are precisely 2 13+1 solutions. We find this unique t by general expression 4.2.2. We have that k = f(141)/13 3 2 (mod 13) and m = f (141)/13 12 (mod 13). Hence (12, 13) = 1 from which 12t + 2 0 (mod 13) give that t = 2. Similarly k = f(167)/13 3 2 (mod 13) and m = f (167)/13 1 (mod 13). We see (1, 13) = 1 from which t + 2 0 (mod 13) give t = 12. This shows that t = 2, 12 will lift for (mod 13 4 ), from which we deduce that and x = 479 + 13 2 t 479, 2676, 4873,, 26843 (mod 13 4 ), x = 2195 + 13 2 t 2195, 4392, 6589,, 28559 (mod 13 4 ), where 0 t p 1 are solutions of x 3 + x + 10 0 (mod 13 4 ). We conclude that (mod 13 4 ) has 27 solutions. 4.3 Cubic Congruence with linear and quadratic term (mod p k ) Suppose the congruence equation is x 3 + bx 2 + cx + d 0 (mod p k ), (4.20) where p is prime and b, c, d Z. We substitute x = y + α in (4.20), to get (y + α) 3 + b(y + α) 2 + c(y + α) + d 0 (mod p k ), y 3 + (3α + b)y 2 + (3α 2 + 2bα + c)y + (α 3 + bα 2 + cα + d) 0 (mod p k ). If we see at coefficient of quadratic 3α + b 0 (mod p k ) is a linear congruence. There could be an α, but this is not uniquely determined at p = 3. The prime p = 3 is an evil case because multiple of α is 3 so 3α + b b 0 (mod 3) cannot find value of α. When p > 3, then all the time we can get rid from quadratic term such as find multiplicative inverse of 3 modulo p that is β and put in linear congruence 3βα + βb 0 (mod p k ) to find value of α = γ. Finally we put value of α in linear congruence equation and it will be incongruent zero modulo p. The modification form of equation (4.20) is written as x 3 +cx+d 0 (mod p k ). We can use lemma 3.1 and theorem 3.2 to find the solutions of this cubic congruence equation. 23

Example 4.15. Simplify the congruence equation x 3 + 4x 2 + 5x + 8 0 (mod 5 3 ) into x 3 +cx+d 0 (mod 5 3 ) and find number of solutons of simplified congruence equation. Solution. starting with substitute x = y + α in congruence equation,so that f(y + α) = (y + α) 3 + 4(y + α) 2 + 5(y + α) + 8 We obtain that = (y 3 + α 3 + 3y 2 α + 3yα 2 ) + 4(y 2 + α 2 + 2yα) + 5(y + α) + 8. y 3 + (3α + 4)y 2 + (3α 2 + 8α + 5)y + (α 3 + 4α 2 + 5α + 8) 0 (mod 5 3 ). (4.21) We take cofficient of quadratic term to find value of α that is 3α + 4 0 (mod 5 3 ) gives α = 82 by Hensel s lemma. Put this value α = 82 in congruence equation (4.21) we obtain y 3 + 250y 2 + 20833y + 578682 0 (mod 5 3 ). Further, we simplify this equation we have that y 3 + 83y + 57 0 (mod 5 3 ). The discriminant = 4 83 3 27 57 2 4 (mod 5). The Legendre is ( 4 5) = 1 because 4 is quadratic residue therefore congruence has 0 or 3 solutions (mod 5 3 ) verify by theorem 3.2 such as v (5 ( 5 3 ))/3 ( 5 3 ) (83) (1 ( 5 3 ))/2 (3 1 ) (1 ( 5 3 ))/2 (mod 5), and we get v 2 1 (mod 5). By theorem 3.2 we have that v 0 = 2, A = 83 and v 1 = B = 57. We calculate v 2 = 57 1 + 83 3 (3 1 ) 3 2 1 (mod 5). Hence it is verified that above congruence y 3 + 83y + 57 0 (mod 5 3 ) has no solution. 4.4 Root of multiplicity three of cubic congurence (mod p k ) Let the cubic polynomial is f(x) = (x p)(x ap)(x bp), where a, b Z and p is a prime number. Then f(x) = x 3 (p + ap + bp)x 2 + (ap 2 + bp 2 + abp 2 )x abp 3. 24

The coefficient of x 2, x and the constant term can be divided by at least p, p 2 and p 3 respectively. To get rid of the quadratic term we divide coefficient of x 2 by 3 and then put in original polynomial such as f ( x α ) 3 = ( x α 3 ) 3 α ( x α 3 ) 2 + (ap 2 + bp 2 + abp 2 ) where α = p + ap + bp and after simplifying we will obtain f(x) = x 3 cp2 x 3 dp3 27, ( x α ) mnp 3, 3 where c, d Z. Now for more simplicity we take this polynomial (mod p 3 ). f(x) = x 3 cp2 x dp3 3 27 x 3 np 2 x 0 (mod p 3 ), where n Z. In this thesis we just focus on the cubic polynomial of form f(x) = x 3 np 2 x 0 (mod p k ), (4.22) has always solution x 0 (mod p) of multiplicity three. The solution zero always satisfy case (ii) of Hensel s lemma therefore x = pt will be the solutions (mod p 2 ) where 0 t p 1. Let k = 3 and put x = pt in equation (4.22) as (pt) 3 np 2 (pt) 0 (mod p 3 ). (4.23) This shows that always all solutions will survive from mod p 2 to mod p 3 and that the number of solutions will be p 2. To find the number of solutions (mod p 4 ) we take Legender symbol. (i) If ( n p ) = 1 then three solutions will survive and modulo p4 has 3p 2 solutions. (ii) If ( n p ) = 1 then one solutions will survive and modulo p4 has p 2 solutions. To find which solutions will survive for (mod p 4 ) we can construct a general expression. Suppose that the cubic congruence is f(x) = x 3 np 2 x 0 (mod p 4 ). (4.24) Let x α + p 2 t be the solutions of the above congruence. By putting this value in (4.23) we obtain f(α + p 2 t) = (α + p 2 t) 3 np 2 (α + p 2 t) = α 3 + t 6 p 6 + 3α 2 tp 2 + 3αp 4 t 2 np 2 α np 4 t f(α) + f (α)p 2 t (mod p 4 ), 25

and we find that kp 3 + mp 3 t 0 (mod p 4 ). Since throughout the congruence divisible by p 3. Moreover, on dividing through by p 3 we find that k + mt 0 (mod p), (4.25) where k = f(α)/p 3 and m = f (α)/p. If above k and m be divided by p then solutions will survive and split into p new ones. The survivor of (mod p 5 ) can be found by the above expression. The number of solutions after (mod p 5 ) will remain constant by theorem 4.3. Note that zero will always survive for every level. Example 4.16. Simplify the congruence equation f(x) = (x 5)(x 25)(x 35) 0 (mod 5 6 ) into general form x 3 np 2 x 0 (mod p k ) and find the solutions of simplified congruence equation. Solution. The cubic congruence is f(x) = x 3 65x 2 + 1175x 4375 0 (mod 5 6 ). Replace x by x = x + 65 3 in above equation to get rid of the quadratic term. Then for the more simplicity take this congruence equation (mod 5 3 ), we see that f(x) = x 3 700 3 x + 20000 0 (mod 5 3 ), 27 hence cubic congruence is f(x) = x 3 + 100x 0 (mod 5 3 ) We focus on this modified congruence and find number of solutions (mod 5 6 ). The discriminant = 4a 3 = 4000000 0 (mod 5) so this congruence has x 0, 0, 0 (mod 5), three solutions. Here f (0) = 100 0 (mod 5 2 ) and f(0) 0 (mod 5 3 ) hold case (ii) of Hensel s lemma, hence x 0, 5, 10, 15, 20 (mod 5 2 ), are the solutions of x 3 + 100x 0 (mod 5 2 ). By equation (4.23) all solutions will survive for next level therefore 0 + 5t,5 + 5t,10 + 5t,15 + 5t and 20 + 5t are solutions (mod 5 3 ). The number of solutions are 5 2 (mod 5 3 ). The Legendre symbol is ( 4 5 ) = 1 (mod p) so three solutions will survive for (mod 54 ) becuase in section 4.4 we have discussed it. One survivor is zero that will lift for every level. To find which solution will survive we use equation (4.25), we note that k = f(5)/5 3 = 5 and m = f (5)/5 = 35. We have that (35, 5) = 5 and k further divisible by 5 therefore 5 + 5 2 t will survivors of (mod 5 4 ) where 26

0 t 4. Similarly k = f(20)/5 3 = 80 and m = f (20)/5 = 260. We see that (260, 5) = 5 and k divisible by 5 therefore 20 + 5 2 t will survivors of (mod 5 4 ) where 0 t 4. To find which solution will survive for (mod 5 5 ) we use same technique that we have applied on (mod 5 4 ). The survivors (mod 5 5 ) are 0 + 5 3 t,55 + 5 3 t and 70 + 5 3 t where 0 t 4. In fact we note that 5 2 f (0) and 5 5 f(0), so theorem 4.3 shows that number of solutions will remain constant after (mod 5 5 ). By equation (4.19) we find that k = f(55)/5 5 2 (mod 5) and m = f (55)/5 2 0 (mod 5) and verify that there is precisely one value of t (mod 5), namely t = 0, for which 55 + 0 5 3 = 55 survives for (mod p 6 ). Similarly k = f(70)/5 5 2 (mod 5) and m = f (70)/5 2 2 (mod 5), moreover we can verify that there is precisely one value of t (mod 5), namely t = 4, for which 70+4 5 3 = 570 is survive for (mod 5 6 ). We conclude that 0+5 4 t,55+5 4 t and 570+5 4 t are solutions (mod 5 6 ) where 0 t 4 and every solution further split into 5 new ones. The orignal congruence has 3 5 2 solutions. The solutions of the equation is described as a tree showed in figure 1. Figure 1: Tree 27

Example 4.17. Simplify the congruence equation f(x) = (x 7)(x 35)(x 56) 0 (mod 7 5 ) into general form x 3 np 2 x 0 (mod p k ) and find the number of solutions of simplified congruence equation. Solution. The cubic congruence equation is f(x) = x 3 98x 2 + 2597x 13720 0 (mod 7 5 ). Substitute x for x = x + 98 3 in above polynomial to get rid of the quadratic term. Further for more simplicity take this polynomial (mod 7 3 ), so that f(x) = x 3 1813 3 x + 37730 0 (mod 7 3 ), 27 and finally the cubic congruence is f(x) = x 3 + 196x 0 (mod 7 5 ). We find the number of solutions this modified congruence. Starting with = 4a 3 = 30118144 0 (mod 7). Three solutions of congruence are x 0, 0, 0 (mod 7). We note that f (0) = 196 0 (mod 7 2 ) and f(0) 0 (mod 7 3 ) staisfy case (ii) of Hensel s lemma, so that x 0, 7, 14, 21, 28, 35, 42 (mod 7 2 ), are solutions of f(x) = x 3 + 196x 0 (mod 7 2 ). By theorem 4.3 we see that 7 2 f (0) and 7 5 f(0) shows that the number of solutions will remain constant after (mod 7 5 ). Due to equation 4.24 all solutions will survive for next level and 0 + 7t,7 + 7t,14 + 7t,21 + 7t,28 + 7t, 35 + 7t and 42 + 7t are solutions (mod 7 3 ) where 0 t 6. The number of solutions are 7 2 (mod 7 3 ). The Legender symbol is ( 4 7 ) = 1 (mod 7) so unique solutions will survive for (mod 74 ) that we have discussed in section 4.4 and that will be zero therefore 0 + 7 2 t survivors (mod 7 4 ). Similarly zero uniquely lift for (mod 7 5 ) and survivors are 0 + 7 3 t (mod 7 5 ). The number of solutions (mod 7 5 ) are 7 2 and remain same after this level by theorem 4.3. 28

Chapter 5 5 Cubic congruence (mod 2 k ) Let the cubic congruence equation be where 2 is prime, k Z + and a, Z. x 3 a (mod 2 k ), (5.1) Theorem 5.1. Suppose that k 3 and let a is odd. If n is odd then the congruence x n a (mod 2 k ) has exactly one solution. If n is even, then select β such that (n, 2 α 2 ) = 2 β.the congruence x n a (mod 2 k ) has 2 β+1 solution if a 1 (mod 2 β+2 ) or no solution a 1 (mod 2 β+2 ) For proof of theorem see [1]. Example 5.1. Find the number of solutions of x 3 7 (mod 2 3 ). Solution. Since n = 3 and a = 7 is odd, this equation has exactly one solution by theorem 5.1. Now we see solution by Hensel s lemma. The congruence f(x) = x 3 7 0 (mod 2) has one solution that is x 1 (mod 2). Since f (x) = 3x 2, we note that f (1) = 3 1 0 (mod 2), the first case of Hensel s lemma is satisfied so there is unique solution modulo 2 2 of the form 1 + 2t, where t find as t = f (1)(f(1)/2) (mod 2). Here f (1) = 1 (mod 2), so we have that f(1)/2 = 6/2 = 3. It give us that t = 1 3 1 (mod 2). We derived that x 1 + 2 1 = 3 (mod 2 2 ) is unique solution of f(x) 0 (mod 2 2 ). Furthermore, by corollary 4.1 we see that f(3) = 20, we conclude that x = 3 20 1 = 17 7 (mod 2 3 ) is the unique solution of f(x) 0 (mod 2 3 ). Special case Among the refrences there is no generalization of theorem 5.1 that describes the case when a is even. Now we are going to solve the special case which is not recovered by theorem 5.1. Let the cubic congruence with special case when a is even x 3 a 2 j b (mod 2 k ), (5.2) where k Z + and a, b Z. If we substitute x = yc in 5.2 then we get y 3 c 3 2 j b (mod 2 k ). Here b is an odd number and we resolve the equation into y 3 2 j (mod 2 k ) and c 3 b (mod 2 k ). We conclude that b is an odd and n = 3, so by theorem (5.1) this equation has a unique solution therefore number of solutions of congruence will independent of b. There exist two cases for congruence y 3 2 j (mod 2 k ). 29

5.1 Case I when j k When j k, where j, k Z then the solutions of the congruence x 3 2 j 0 (mod 2 k ) will depend on k, therefore if j vary, it will not affect the number of solutions. It includes two sub-cases given as (a) When k 0, 1 (mod 3), then the cubic congruence has 2 2q solutions, where q = k/3. (b) When k 2 (mod 3), then the cubic congruence has 2 2q+1 solutions, where q = k/3. General Expression (a) We can construct the general expression for cubic congruence x 3 0 (mod 2 k ), (5.3) where k Z + to find all solutions when k 0, 1 (mod 3). Let us x m2 b (mod 2 k ), (5.4) here b = q + r, hence q = k/3 and r is raminder where m = 0, 1,, 2 q+r 1. Put value of x in equation (5.3) we see (m2 b ) 3 0 (mod 2 k ). Take left hand side and substitute value of k/3 = k r 3 by division algorithm in above congruence equation that is (m2 b ) 3 = m 3 k r 3( 2 3 +r) = m 3 2 k+2r 0 (mod 2 k ), is equal right hand side. The congruence x 3 2 j (mod 2 k ) has the solutions of form x m2 b (mod 2 k ). (b) When k 2 (mod 3) then we can construct the general expression for cubic congruence equation to find all possible solutions. Let us x 3 2 j (mod 2 k ) (5.5) x m2 b (mod 2 k ), (5.6) here b = q + 1, where q = k/3 and m = 0, 1,, 2 2q+1 1. Put value of x in equation (5.5) (m2 b ) 3 0 (mod 2 k ). We take left hand side and put value of k/3 = k 2 3 by division algorithm in above congruence equation such as (m2 b ) 3 = m 3 k 2 3( 2 3 +1) = m 3 2 k+1 0 (mod 2 k ), is equal right hand side of above congruence. This shows that the congruence x 3 2 j (mod 2 k ) has the solutions of form x m2 b (mod 2 k ). 30

Example 5.2. Determine the solutions of x 3 2 8 (mod 2 6 ). Solution. We note that k 0 (mod 3), the congruence has 2 2 2 = 16 solutions because q = 6/3 = 2. To find all solutions use equation (5.4) that is x m2 2+0 (mod 2 6 ), where m = 0, 1,, 2 2 2 1. The solutions are x 0, 4, 8, 12,, 60 (mod 2 6 ). Example 5.3. Find the solutions of x 3 2 6 (mod 2 4 ). Solution. We have that k 1 (mod 3) so congruence has 2 2 1 = 4 solutions because q = 4/3 = 1. Further we use equation (5.4) that is x m2 1+1 (mod 2 4 ), where m = 0, 1,, 2 2 1. The congruence has possible solutions. x 0, 4, 8, 12 (mod 2 4 ), Example 5.4. Determine the solutions of x 3 2 7 (mod 2 11 ). Solution. Starting with k 2 (mod 3), therefore the congruence has 2 2 3+1 = 128 solutions because q = 11/3 = 3. The equation 5.6 become x m2 3+1 (mod 2 11 ), where m = 0, 1,, 2 3 2+1 1. The possible solutions are x 0, 16, 32,, 2032 (mod 2 11 ). 5.2 Case II when j < k When j < k, number of solutions of congruence equation x 3 2 j (mod 2 k ) will depend on j. The congruence equation has 2 2j/3 solutions if 3 j otherwise no solution. The solutions will be 4 or multiple of 4. (a) when k 0 (mod 3) Let j = k then the cubic congruence is x 3 2 k (mod 2 k ). (5.7) This equation (5.7) has solutions of form x m2 b (mod 2 k ) where b = q + r, q = k/3 and r is remainder by equation (5.4) as we have discussed in above general expression. Put this value of x in equation (5.7) and replace k by k + 1 we see that f(m2 b ) = (m2 b ) 3 2 k = (m2 k 3 ) 3 2 k m 3 2 k 2 k 0 (mod 2 k+1 ). 31

Divide throughout the congruence equation by 2 k we obtain that m 3 1 0 (mod 2). This congruence equation is solveable and has always one solution by theorem 5.1. (b) when k 1 (mod 3) Let j = k then the cubic congruence is x 3 2 k (mod 2 k ). (5.8) The equation (5.8) has solutions of form x m2 b (mod 2 k ) where b = q + r, q = k/3 and r is remainder by equation (5.4) as we have discussed in above general expression. Put this value of x in equation (5.8) and take k + 1 instead of k we note that f(m2 b ) = (m2 b ) 3 2 k = (m2 k 1 3 +1 ) 3 2 k m 3 2 k+2 2 k 0 (mod 2 k+1 ). Divide throughout the congruence equation by 2 k we get that m 3 2 2 1 1 0 (mod 2). This congruence equation cannot be solved. (c) when k 2 (mod 3) Let j = k then the cubic congruence is x 3 2 k (mod 2 k ). (5.9) The solutions of form x m2 b (mod 2 k ) where b = q + r, q = k/3 and r is remainder of the equation (5.9) by equation (5.4). Put this value of x in equation (5.9) and take k + 1 instead of k we see that f(m2 b ) = (m2 b ) 3 2 k = (m2 k 2 3 +1 ) 3 2 k m 3 2 k+1 2 k 0 (mod 2 k+1 ). Divide throughout the congruence by 2 k we obtain that m 3 2 1 1 0 (mod 2). This congruence equation is not solveable. Hence these result shows that when 3 j then congruence equation x 3 2 j (mod 2 k ) has solutions otherwise not. 32

General Expression We can generate the general expression for cubic congruence equation where j, k Z + to find all solutions when j < k. Let x 3 2 j (mod 2 k ), (5.10) x 2 q + m2 b (mod 2 k ), (5.11) where q = j/3, b = k 2q and m = 0, 1,, 2 2q 1. Put this value of x in cubic congruence equation (5.10) such as (2 q + m2 b ) 3 2 j (mod 2 k ). By division algorithm j/3 = (j 0)/3 We take Left hand side (2 q + m2 b ) 3 = 2 3q + m 3 2 3b + 3m2 2q 2 b + 3m 2 2 q 2 2b = 2 j + m 3 2 3k 2j + 3m2 2j/3 2 k 2j/3 + 3m 2 2 j/3 2 (6k 4j)/3 = 2 j + m 3 2 3k 2j + 3m2 k + 3m 2 2 2k j 2 j (mod 2 k ). The above result is equal right hand side of equation (5.11). This shows that the solutions of x 3 2 j (mod 2 k ) always in the form x 2 a + m2 b. Example 5.5. Find the number of solutions to x 3 2 3 (mod 2 4 ). Solution. Since j = 3, we see that 3 j so congruence has 2 2 3/3 = 2 2 = 4 solutions. We find these 4 solution by Hensel s lemma how they look like. Starting with f(x) = x 3 8 (mod 2), we note that x = 0 is the only solution. Since f (x) = 3x 2, we see that f (0) 0 (mod 2) and f(0) 0 (mod 4), satisfying case (ii) of Hensel s so that x 0, 2 (mod 2 2 ), are the solutions x 3 2 3 (mod 2 2 ). We see that f (0) 0 (mod 2), and f(0) 0 (mod 8), so that we have the roots x = 2 and x = 4 (mod 8). Now f (2) 0 (mod 2), and f(2) 0 (mod 8) the second case of Hensel s is staisfied so that we have roots 2 + 4t (mod 8) where 0 t 1. We get that x 0, 2, 4, 6 (mod 2 3 ), are the solutions x 3 2 3 (mod 2 3 ). We are now in a position to determine which, if any, of the root s 0, 2, 4, 6 (mod 8) can be lifted to root s (mod 16). We find that f(0) 0 (mod 16) third case of Hensel s lemma is satisfied so 0 will not lift solution for next level. We see that f(2) 0 (mod 2 4 ) we conclude that x = 2 and x = 10 are solutions 33

(mod 16). Similarly f(4) 0 (mod 2 4 ) so congruence has no solution at x = 4. At last f(6) 0 (mod 2 4 ) and so the second case of Hensel s lemma is satisfied. Therefore we obtain two roots, 6, 14 (mod 2 4 ), from which we deduce that x 2, 6, 10, 14 (mod 2 4 ), are the solutions of x 3 2 3 (mod 2 4 ). The solutions of the equation is described as a tree showed in figure 2. Figure 2: Tree Hensel Example 5.6. Find the solution of x 3 2 6 (mod 2 7 ). Solution. We use the equation (5.11) that is x 2 2 + m2 3, m = 0, 1,, 2 4 1 to find all solutions. We conclude that x 4, 12, 20, 28, 36, 44, 52, 60, 68, 76, 84, 92, 100, 108, 116, 124 (mod 2 7 ), are solutions of x 3 2 6 (mod 2 7 ). The number of solutions for this congruence are 16. 34

Chapter 6 6 Cubic Congruence (mod m) Let us assume that the cubic congruence equation is of the form x 3 a (mod m), where a Z and m Z + is any composite integer. Theorem 6.1. Suppose f(x) be a fixed polynomial with integral coefficients, and for any m Z + let N(m) represent the number of solutions of the congruence f(x) 0 (mod m). If m = m 1 m 2 where (m 1, m 2 ) = 1, then N(m) = N(m 1 )N(m 2 ). If m = Πp α is canonical factorization of m, then N(m) = ΠN(p α ). For proof of this theorem see [1]. Example 6.1. Determine the number of solutions of the congruence equation x 3 8 (mod 1200). Solution. If we factor the integer 1200 into a product of primes, we get 1200 = 2 4 3 5 2. Now we split original congruence as follows x 3 8 (mod 3). (6.1) x 3 8 (mod 2 4 ). (6.2) x 3 8 (mod 5 2 ). (6.3) First we solve equation (6.1) using theorem 3.1 we note that (8, 3) = 1. As (3, 3 1) = (3, 2) = 1 and 8 (2/1) = 8 2 1 (mod 3). We deduce that the congruence has a unique solution. The unique solution of congruence is x = 2 (mod 3). Further, we solve equation (6.2) by case 5.2 the congruence has 2 2 3/3 = 4 solutions. We find these four solutions by general expression 5.2 that is x 2 + m2 2, m = 0, 1,, 2 2 1. We conclude that x 0, 2, 4, 6 (mod 2 4 ), are solutions of x 3 8 (mod 2 4 ). Furthermore, we take equation (6.3) and solve by theorem 4.2, according as (8, 5 2 ) = 1 so congruence has (3, φ(5 2 )) = (3, 20) = 1 solutions or no solutions verify as 8 φ(52 )/(3,φ(5 2 )) = 8 20/1 1 (mod 5 2 ). The congruence x 3 8 (mod 5 2 ) has a unique solution. Now we find this solution by corollary 4.1. Starting with x 2 (mod 5) a unique solution of above congruence. Here f (x) = 3x 2, we see that f (2) 2 0 (mod 5). Taking f (2) 3 (mod 7), we see that the root r 1 = 2 (mod 5) lifts to r 2 = 2 0 3 2 (mod 5 2 ) is only solution of x 3 8 (mod 5 2 ). Thus by theorem 6.1, the number of solutions modulo 1200 are 1 4 1 = 4. 35

6.1 Cubic Congruence with linear term (mod m) Let us the cubic congruence equation with linear term is of form x 3 + ax + b 0 (mod m), where a, b Z and m Z + is composite integer. Example 6.2. Determine the number of solutions of the congruence equation x 3 + 8x + 9 0 (mod 190333). Solution. We factor the integer 190333 = 13 11 4 in product of primes. Now we split orignal congruence as follows x 3 + 8x + 9 0 (mod 13). (6.4) x 3 + 8x + 9 0 (mod 11 4 ). (6.5) Start to solve equation (6.4), the discriminant is = 4 8 3 27 9 2 = 4235 3 (mod 13), further we take Legendre symbol of discriminant that is ( 3 13) = 1 because 3 is quadratic residue therefore cubic congruence has 0 or 3 solutions we verify by theorem 3.2. Here A = 166, B = 2 so A B = 332 then we see that ( ) 13 13 v (13 ( 13 3 ))/3 (1 ( 2 (8) 3 ))/2 (3 1 13 (1 ( ) 3 ))/2 (mod 13), 3 we get v 4 2 (mod 13). By theorem 3.2 since v 0 = 2, A = 8 and v 1 = B = 9. We calculate v 2 = 9 9 + 8 3 (3 1 ) 3 2 0 (mod 13), v 3 = 9 0 8 3 (3 1 ) 3 9 7 (mod 13) and v 4 = 9 7 4 3 (3 1 ) 3 0 2 (mod 13).Result is verify and the congruence has x = 6, 8, 12 three solutions. Next we solve equation (6.5) we have that = 4 8 3 27 9 2 = 4235 0 (mod 11), so this equation has three solutions that are x 10, 10, 2 (mod 11), with one repeated root. We see that f (2) 9 0 (mod 11) hold case (i) of Hensel s lemma. We note that x = 10 is repeated root. We have that f (10) = 308 0 (mod 11) and f(10) = 1089 0 (mod 11 2 ) satisfy case (ii) of Hensel s lemma. So that x 10, 21, 32,, 120 (mod 11 2 ), are solutions of x 3 + 8x + 9 0 (mod 11 2 ). By theorem 4.3 we have 11 308 means that 11 308 but 11 2 308. So τ = 1, j 3 that is f(10) = 1089 0 (mod 11 3 ) shows theorem can not apply at beginning. Now we will use quadratic equation (4.16) to find solutions (mod 11 3 ). First we find value of k and m such that k = f (10)/11 = 308/11 6 (mod 11), m = f(10)/11 2 = 1089/11 2 9 (mod 11). The quadratic equation is 30t 2 + 6t + 9 0 (mod 11). 36

The discriminant is = 6 2 4 30 9 = 1044. Legendre symbol with discriminant is ( ) 1044 11 = 1 because 1044 is quadratic residue therefore two solution will survive (mod 11 2 ) by equation (4.17). We solve this quadratic we get t = 3, 10 which gives that the 10 + 3 11 = 43, 10 + 10 11 = 120 these two roots give 2 11 + 1 solutions of the form 43 + 13 2 t, 120 + 13 2 t (mod 11 3 ), where 0 t 10. Moreover we can see that 11 f (43), so that f(43) 0 (mod 11 3 ). Similarly 11 f (120), so that f(120) 0 (mod 11 3 ) by theorem 4.3 now there will be unique t that will lift for every next level and number of solutions will remain constant. We have found for each k 3 there are precisely 23 solutions. We find this unique t by equation (4.19). We find that k = f(43)/11 3 5 (mod 11) and m = f (43)/11 10 (mod 11). Hence (10, 11) = 1 from which 10t + 5 0 (mod 11) give that t = 5. Similarly k = f(120)/11 3 1 (mod 11) and m = f (120)/11 1 (mod 11). We see (1, 11) = 1 from which t + 1 0 (mod 11) give t = 10. This shows that t = 5, 10 will lift for (mod 11 4 ). The solutions are x 43, 120, 164, 241,, 1330 (mod 11 4 ) Thus by theorem 6.1, the congruence x 3 + 8x + 9 0 (mod 190333) has 23 3 = 69 solutions. 37

7 Bibliography References [1] Ivan Niven, An introduction to the theory of numbers (Fifth Edition), Herbert S. Zuckerman, Hugh L. Montgomery, Published by John Wiley and sons Inc, 1991. [2] K.T.Leung, Polynomials and Equations, I.A.C Mok, S.N.Suen, Hong Kong University Press, 1992. [3] Paulo Ribenboim, Fermat s Last Theorem for Amateurs, Published by Springer, 1997. [4] Zhi-HOng Sun, Cubic and quartic congruences modulo a prime (Journal of Number Theory), Published by Academic Press, 2003. [5] Kenneth H.Rosen, Elementary Number Theory and its applications (5th edition), Published by Pearson Addison Wesley, 2005.

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