CUBIC CONGRUENCE EQUATIONS

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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

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Abstract

Let Nm(f (x)) denote the number of solutions of the congruence

equa-tion 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 corol-laries 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.

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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 6= 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 6≡ 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 (mod m). For proof see [1]

Example 1.2. Solve the congruence equation 8x ≡ 4 (mod 12).

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

ax2≡ b (mod m), where a, b, m ∈ Z.

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Lemma 1.1. Let p be an odd prime. Let a ∈ Z, (a, p) = 1 then x2≡ a (mod p) either has no solution or, exactly two solutions.

More detail of lemma can find in [5] Example 1.3. Solve x2_{≡ 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 x2_{≡ 5 (mod 3).}

Solution. We note that f (t) 6≡ 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) = x3≡ 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 pk where k ∈ Z+. We shall use Hensel’s lifting method. In Chapter 5 we will solve cubic congruence equations mod 2k. In Chapter 6 we will solve different kinds of cubic congruence equations modulo m where m is any composite positive integer.

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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 6= 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 pk _{| a and p}k+1

- a where p is a prime, that is pk is the highest power dividing a, then we denote it by pk|| 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 x2_{≡ 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 x2 _{≡ 4 (mod 7). This congruence has two solutions x = 2} and 5, therefore 4 is a quadratic residue.

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Definition 2.5. If p is an odd prime, a ∈ Z and (a, p) = 1 then Legendre symbola_pis defined as

(i) a_p= 1 if a is quadratic residue (mod p). (ii) a_p= −1 if a is quadratic non-residue (mod p). (iii) 0 if a ≡ 0 (mod p).

Example 2.5. Let x3 _{≡ 5 (mod 11). Then x = 3 is a solution, hence 5 is} quadratic residue therefore ₁₁5 = 1.

Definition 2.6. The discriminant ∆ of a cubic equation x3_{+ ax + b = 0 is} defined as

∆ = −4a3− 27b2_. Further information on discriminant can find in [2].

Example 2.6. Suppose that x3+ 4x + 7 ≡ 0 (mod 13). The discriminant of this congruence is ∆ = −4 · 43− 27 · 72_{= −1579.}

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Chapter 3

3 Cubic Congruence (mod p)

A cubic congruence equation be

x3≡ 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 xn ≡ 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)_6≡ 1 (mod p).

For proof of this theorem see [1].

Example 3.1. Determine the number of solutions of the congruence x3 _{≡ 6} (mod 7).

Solution. We note that (6, 7) = 1. As (3, 7 − 1) = (3, 6) = 3 and 6(6/3) ₌ 62 _{≡ 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 x3+ 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

x3+ ax + b ≡ 0 (mod p). Let t ∈ Z be such that tb2≡ a3 _{(mod p), and}

Ψp(T ) = Tk−1− 3 (p − 5)(p − 7) 22_{× 3!} T k−2_{+ · · · +} (−1)s−1(2s − 1)[p − (2s + 1)][p − (2s + 3)] · · · [p − (4s − 1)] 22(s−1)_{(2s − 1)!} T k−s_{+ · · · +} (−1)k−1(2k − 1)[p − (2k + 1)][p − (2k + 3)] · · · [p − (4k − 1)] 22(k−1)_{(2k − 1)!} . (3.1) (i) If −(4t + 27) is a quadratic non residue modulo p, then cubic congruence

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(ii) If −(4t + 27) ≡ 0 (mod p), then the congruence has one repeated root and one another means two distinct solutions x1, x2, 0 < x1, x2< p.

(iii) If −(4t+27) is a quadratic residue modulo p and if Ψp(t) ≡ 0 (mod p) then cubic congruence has three distinct solution x1, x2, x3, 0 < x1, x2, x3< p. (iv) If −(4t + 27) is a quadratic residue modulo p and if Ψp(t) 6≡ 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 x3_{− 6x + 15 ≡ 0 (mod 13).}

Solution. We get the value of t = 11, after solving the linear congruence equation t152≡ (−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 x3_{− 13x + 23 ≡ 0 (mod 7).}

Solution. We solve t232_{≡ (−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 x3− 14x + 15 ≡ 0 (mod 23).

Solution. Start to solve t152 _{≡ (−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 Ψ23(T ) = T3− 3 (23 − 5)(23 − 7) 22_{× 3!} T 2_{+ (−1)}3−1₍₅₎(23 − 7)(23 − 9)(23 − 11) 22·2_(5)! T + (−1)4−1(7)(23 − 9)(23 − 11)(23 − 13)(23 − 15) 22·3_(7)! . (3.2) We obtain that Ψ23(T ) = T3−36T2+7T −7·24−1use multiplicative inverse then Ψ23(T ) = T3−36T2+7T −7. We replace T by t so Ψ23(t) = t3−36t2+7t−7 ≡ 0 (mod 23). We have t = 6 therefore we get Ψ23(6) = 63−36·62+7·6−7 ≡ 13 6≡ 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

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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 x3+ ax + b ≡ 0 (mod p) where p > 3 is prime, k ∈ Z+

and a, b ∈ Z. The discriminant of this equation is ∆ = −4a3_{− 27b}2 _and∆ p is Legendre symbol. (i) ∆ 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 x3_{+5x+6 ≡ 0} (mod 7).

Solution. The discriminant of congruence is ∆ = −4 · 53− 27 · 62_{= −1472 and} Legendre −1472₇ = −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 x3+2x+3 ≡ 0 (mod 5).

Solution. The ∆ = −4 · 23_{− 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 −4a3_p−27b2 = 1. If vn is defined as v0 = 2, v1 = −B and vn+1= Bvn+ A₃ 3 vn−1, then Np(x3+ Ax − B) = ( 3, if v(p−(p₃))/3 ≡ 2( p 3)( A 3) (1−(p 3))/2 (mod p) 0, if v(p−(p₃))/3 ≡ −(p3)( A 3) (1−(p₃))/2 _{(mod p).} Here p₃ is Legendre while A

3 is not. For proof of this theorem see [4].

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Example 3.7. Find the number of solutions of x3− 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 · 152_{= 4901 is discriminant of the cubic congruence} and 4901₂₃ = 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

v₍₂₃₋₍23 3))/3≡ 2 23 3 (−14)(1−(233))/2(3−1)(1−( 23 3))/2 (mod 23), we get v8≡ 17 (mod 23).

By theorem 3.2 we have that v0 = 2, A = −14 and v1 = −B = −15. We calculate v2 ≡ 3 (mod 23), v3 ≡ 10 (mod 23), v4 ≡ 0 (mod 23), v5 ≡ 17 (mod 23), v6 ≡ 21 (mod 23), v7 ≡ 6 (mod 23) and v8 ≡ 17 (mod 11). The result is verified, hence prove that above congruence equation has three solu-tions.

Example 3.8. Find number of solutions of the congruence x3_{+ 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 · 23− 27 · 152_{= −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

v(13−(13 3))/3≡ − 13 3 (2)(1−(133))/2₍₃−1₎(1−(133))/2 _{(mod 13),}

and we get v4= −1 ≡ 12 (mod 13).

By theorem 3.2 we have that v0 = 2, A = 2 and v1 = −B = −15. We calculate v2= −15 · 15 + 23(3−1)3· 2 ≡ 7 (mod 13), v3= −15 · 7 − 15 · 23· 93≡ 9 (mod 13) and v4= −15 · 9 + 23· 93· 7 ≡ 12 (mod 13). Hence it is verified that above congruence equation has no solution.

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Chapter 4

4 Cubic Congruence (mod p

k

)

Let the cubic congruence be

x3≡ a (mod pk), where p be a prime, k ∈ Z+ and a ∈ Z.

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 pk−1). Then

(i) if f0(r) 6≡ 0 (mod p), then we get a unique integer t, 0 ≤ t ≤ p, such that f (r + tpk−1) ≡ 0 (mod pk−1), given as

t ≡ −f0_{(r)(f (r)/p}k−1₎ _{(mod p),} where f0_{(r) is the multiplicative inverse of f}0_{(r) modulo p.}

(ii) if f0(r) ≡ 0 (mod p) and f (r) ≡ 0 (mod pk_{), then f (r + tp}k−1_{) ≡ 0} (mod pk_{) for all integers t where 0 ≤ t ≤ p − 1.}

(iii) if f0(r) ≡ 0 (mod p) and f (r) 6≡ 0 (mod pk), then f (x) ≡ 0 (mod pk) has no solution with x ≡ r (mod pk−1).

In case (i), we note that f (x) ≡ 0 (mod pk−1) lifts a unique solution of f (x) ≡ 0 (mod pk), and in case (ii), t is arbitrary so solution lifts to p incongruent solutions modulo pk or to none at all.

For proof of lemma see [5].

Example 4.1. Solve the congruence f (x) = 2x3_{+ 7x − 4 ≡ 0 (mod 5}2_). Solution. Starting with f (x) = 2x3+ 7x − 4 ≡ 0 (mod 5), we note that x = 1 is the only solution. Here f0(1) = 13 ≡ 3 6≡ 0 (mod 5), so root is non-singular. Taking f0_{(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 52_{) is} the only solution of f (x) = 2x3_{+ 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 f0(r) 6≡ 0 (mod p), then we have a unique solution rk modulo pk, k = 2, 3, · · · such that

rk = rk−1− f (rk−1)f0(r). Here f0_{(r) is the multiplicative inverse of f}0_{(r) modulo p.}

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Example 4.2. Solve x3+ 2x + 3 ≡ 0 (mod 73).

Solution. First we note that x ≡ 6 (mod 7) is solution of x3 + 2x + 3 ≡ 0 (mod 7). Since f0(x) = 3x2+ 2, we see that f0(6) ≡ 5 6≡ 0 (mod 7). Taking f0_{(6) ≡ 3 (mod 7), we see by corollary 4.1 that the root r}

1 = 6 (mod 7) lifts to r2 = 6 − 231 · 3 ≡ 48 (mod 72). Further r2 lifts r3= 48 − 110691 · 3 ≡ 342 (mod 73_{) is only solution of x}3_{+ 2x + 3 ≡ 0 (mod 7}3_).

Theorem 4.2. Let m = 1, 2, 4, pα_{, or 2p}α_{, where is p is an odd prime. If} (a, m) = 1 then the congruence equation xn _{≡ a (mod m) has (n, φ(m))} so-lutions if aφ(m)/(n,φ(m)) _{≡ 1 (mod m) but no solutions if a}φ(m)/(n,φ(m)) _{6≡ 1} (mod m).

For proof of the theorem see [1].

Example 4.3. Find the number of solutions of x3_{≡ 7 (mod 19}2_).

Solution. Since (7, 192) = 1 the congruence has (3, φ(192)) = (3, 342) = 3 solu-tions or no solusolu-tions according as

7φ(192)/(3,φ(192)) = 7342/3≡ 1 (mod 192). The congruence x3_{≡ 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, pk) 6= 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, pk) 6= 1 then the congruence equation x3≡ a ≡ npα (mod pk) (4.1) has two cases where n an integer and k, α ∈ Z+_.

Suppose that x ≡ β + tpk−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 + tpk−1) = (β + tpk−1)3+ npα

= β3+ t3p3k−3+ 3β2tpk−1+ 3βt2p2k−2+ npα. Since 3k − 3 = k + (2k − 3) ≥ k + 1 we have that

f (β + tpk−1) ≡ (β3+ npα) + 3β2tpk−1 ≡ f (β) + 3tf0(β)pk−1

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Since throughout the congruence is divisible by pk, on dividing pk we get that c + dt ≡ 0 (mod p), (4.2) which is a linear equation, where c = f (β)/pk _{and d = f}0_{(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 f0(x) = 3x2, p and independent of n.

Case I When α ≥ k

When α ≥ k then cubic congruence is

x3≡ pα_{≡ 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 p2q where q = bk/3c.

General Expression

(a) When k ≡ 0 (mod 3), to find all possible solutions we will generate general expression as follows. Suppose that

x ≡ mpq (mod pk), (4.4) where q = bk/3c and m = 0, 1, · · · , p2q_{− 1. Put this value in (4.3) as}

(mpq)3≡ 0 (mod pk).

Take left hand side and put value of bk/3c =k−0₃ by division algorithm in above congruence equation such as

(mp(k−03 ))3≡ m3pk≡ 0 (mod pk),

shows the right side. This show that the congruence equation x3_{≡ 0 (mod p}k₎ has the solutions of form x ≡ mpq _{(mod p}k_).

(b) When k ≡ 1 (mod 3), to find all possible solutions we will general generate expression. Let us

x ≡ mpq+1 (mod pk), (4.5) where q = bk/3c and m = 0, 1, · · · , p2q_{− 1. Put this value in (4.3) we find that}

(mpq+1)3≡ 0 (mod pk). Take left hand side and put value of bk/3c =k−1

3 by division algorithm in above congruence equation such as

(mp(k−13 +1)₎3≡ m3_pk+2≡ 0 _{(mod p}k_),

the above result is equal to right hand side. This shows that the congruence equation x3≡ 0 (mod pk_{) has the solutions of form x ≡ mp}q+1 _{(mod p}k_).

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(ii) When k ≡ 2 (mod 3) the number of solutions will be p2q+1 where q = bk/3c.

General Expression

In order to find all possible solutions we will derive a general expression as follows. Suppose that

x ≡ mpq+1 (mod pk), (4.6) where q = bk/3c and m = 0, 1, · · · , p2q+1_{− 1. Put this value in (4.3)}

(mpq+1)3≡ 0 (mod pk).

Take left hand side and put value of bk/3c =k−2₃ by division algorithm in above congruence equation such as

(mp(k−23 +1)₎3≡ m3_pk+1≡ 0 _{(mod p}k_),

gives right hand side. This shows that the congruence equation x3≡ 0 (mod pk₎ has the solutions of form x ≡ mpq+1 (mod pk).

Example 4.4. Determined the solutions of congruence x3_{≡ 0 (mod 5}3_). Solution. We see that k ≡ 0 (mod 3), the congruence equation has 52·1 = 25 solutions because q = b3/3c = 1. To find all possible solutions we use equation (4.4), that is x ≡ m5 (mod 53) where m = 0, 1, · · · , 52− 1. The roots are

x ≡ 0, 5, 10, 15, · · · , 120 (mod 53).

Example 4.5. Calculate the solutions of the congruence equation x3 _{≡ 0} (mod 74_).

Solution. We note that k ≡ 1 (mod 3), the congruence equation has 72·1 = 49 solutions because q = b4/3c = 1. To determine all possible solutions by equation (4.5) that is x ≡ m72 (mod 74), where m = 0, 1, · · · , 72 − 1. The possible solutions are

x ≡ 0, 49, 98, 147, · · · , 2352 (mod 74).

Example 4.6. Find the solutions of the congruence equation x3_{≡ 0 (mod 11}5_). Solution. Since k ≡ 2 (mod 3), the congruence equation has 112·1+1 _{= 1331} solutions because q = b5/3c = 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, · · · , 114− 1. Hence solutions are

x ≡ 0, 121, 242, 363, · · · , 160930 (mod 115).

Case II When α < k

When α < k then cubic congruence is

x3≡ pα _{(mod p}k_), _(4.7) where k, α ∈ Z+_{. The congruence equation (4.7) has solutions if 3 | α otherwise} not.

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(a) when k ≡ 0 (mod 3)

Let α = k then the cubic congruence is

x3≡ pk _{(mod p}k_). _(4.8) This cubic congruence equation (4.8) has solutions of form x ≡ mpq _{(mod p}k₎ where q = bk/3c by equation (4.4) as we have discussed in above general ex-pression. Put this value of x in equation (4.8) and replace k by k + 1 we see that

f (mpq) = (mpq)3− pk_{= (mp}k 3)3− pk

≡ m3_pk_{− p}k_{≡ 0} _{(mod p}k+1_). Divide throughout the congruence by pk we obtain that

m3− 1 ≡ 0 (mod p). (4.9) The congruence equation (4.9) is solveable and has (3, Φ(p)) solutions by theo-rem 4.2. The congruence equation is cubic therefore n = 3 is fix has always 1 or 3 solutions.

(b) when k ≡ 1 (mod 3)

x3≡ pk _{(mod p}k_), _(4.10) has the solutions of form x ≡ mpq+1 _{(mod p}k_{) where q = bk/3c by equation} (4.5). Put this value of x in equation (4.8) and replace k by k + 1 we see that

f (mpq+1) = (mpq+1)3− pk _{= (mp}k−1

3 +1)3− pk

≡ m3_pk+2_{− p}k _{≡ 0} _{(mod p}k+1_). Divide throughout the congruence by pk _{we obtain that}

m3p2− 1 ≡ −1 ≡ 0 (mod p). This congruence is not solveable.

(c) when k ≡ 2 (mod 3)

x3≡ pk _{(mod p}k_). _(4.11) The congruence equation (4.8) has solutions of form x ≡ mpq+1 _{(mod p}k_{) where} q = bk/3c by equation (4.4).

Put this value of x in equation (4.8) and replace k by k + 1 we see that f (mpq+1) = (mpq+1)3− pk _{= (mp}k−2₃ +1₎3_{− p}k

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Divide throughout the congruence by pk we obtain that m3p − 1 ≡ −1 ≡ 0 (mod p). This congruence equation cannot be solved.

This shows that the congruence equation x3 ≡ pα _{(mod p}k_{) has solutions} if 3 | α. The congruence equation has p2q _{or 3p}2q _{solutions where q = bα/3c} depend on equation (4.9).

General Expression

We can construct the general expression for cubic congruence

x3≡ pα _{(mod p}k₎ _(4.12) where α, k ∈ Z+ _{to find all the solutions when α < k. Let the solutions of} congruence equation (4.12) is

x ≡ mpq+ npb (mod pk), (4.13) where q = bα/3c, b = k − 2q and n = 0, 1, · · · , p2q− 1. Here m is a solution of the cubic congruence m3≡ 1 (mod p).

Put this value of x in cubic congruence equation (4.12) such as (mpq+ npb)3≡ pα (mod pk).

Take Left hand side and by division algorithm bα/3c = (α − 0)/3 we have that (mpq+ npb)3= m3p3q+ n3p3b+ 3m2p2qpb+ 3mpqn2p2b

= m3pα+ n3p3k−2α+ 3m2p2α/3pk−2α/3+ 3mpα/3n2p(6k−4α)/3 = m3pα+ n3p3k−2α+ 3m2pk+ 3mpα/3n2p(2k−α)/3

≡ m3_pα _{(mod p}k_).

Since m3≡ 1 (mod p) therefore the congruence is (mpq+ npb)3≡ pα (mod pk).

The above result is equal right hand side of equation (4.12). This shows that the solutions of x3_{≡ p}α _{(mod p}k_{) always in the form x ≡ mp}q_{+ np}b _{(mod p}k_). Example 4.7. Determine the solutions of x3≡ 73 _{(mod 7}4_).

Solution. We start to solve the congruence equation m3_{≡ 1 (mod 7) by} equa-tion (4.9) has m ≡ 1, 2, 4 (mod 7) three soluequa-tions. The original congruence equation has 3 · 72·3/3_{= 147 solutions. To find all possible solutions we use the} general equation (4.13) that is x ≡ 7m + 72_{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

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For m = 2 we have the following solutions

x ≡ 14, 63, 112, · · · , 2366 (mod 74). The possible solutions when m = 4 are

x ≡ 28, 77, 126, · · · , 2380 (mod 74).

These are all solutions of the congruence equation x3_{≡ 7}3 _{(mod 7}4_). Example 4.8. Find the solutions of x3_{≡ 11}3 _{(mod 11}5_).

Solution. The congruence m3 _{≡ 1 (mod 11) has a unique solution m ≡ 1} (mod 11) by equation (4.9). The original cubic congruence equation has 112·3/3₌ 121 solutions. To demonstrate the solutions use equation (4.13) that is x ≡ 11m + 113_{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 115).

4.2 Cubic Congruence with linear term (mod p

k

₎

Let cubic congruence equation with linear term is

x3+ ax + b ≡ 0 (mod pk), (4.14) where p is prime, k ∈ Z+ and a, b ∈ Z. The discriminant of (4.14) is ∆ = −4a3− 27b2_. ∆ p is Legendre symbol.

(i) ∆_p= −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 pk_).

(ii) ∆ ≡ 0 (mod p) then the congruence has different solutions (mod pk_). (iii) ∆_p = 1 if ∆ is quadratic residue (mod p) and root x ≡ r (mod p)

staisfy first case of Hensel’s lemma then the congruence has 0 or 3 solutions (mod pk_).

The first two cases have taken from lemma 3.1 and case (iii) has stated in theorem 3.2.

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Example 4.9. Show that the congruence equation x3+ 2x + 3 ≡ 0 (mod 138) has unique solution.

Solution. The discriminant is ∆ = −4.23− 27.32_{= −275 ≡ 11 (mod 13). The} Legendre 11₁₃ = −1 because 11 is quadratic non-residue. We note that x = 12 is the only solution mod 13. Here f0_{(12) 6≡ 0 staisfy case (i) of Hensel’s lemma} therefore cubic congruence has a unique solution (mod 138_).

Example 4.10. Find the number of solutions of congruence equation x3_+12x+ 6 ≡ 0 (mod 56_).

Solution. Starting with ∆ = −4 · 123_{− 27 · 6}2 _{= −9099 ≡ 1 (mod 5). Now we} see that 1₅ = 1 because 1 is quadratic residue so cubic congruence has 0 or 3 solutions (mod 56_{) verify by theorem 3.2. Here A = 12, B = 6 so A · B = 72} and (72, 5) = 1 then we have

v₍₅₋₍5 3))/3≡ − 5 3 (12)(1−(53))/2₍₃−1₎(1−(53))/2 _{(mod 5).} We obtain v2= 24 ≡ 4 (mod 5).

By theorem 3.2 since v0 = 2, A = 12 and v1 = −B = −6, we find v2 = −6 · 6 + 123₍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 56_{) also has no} solution.

Example 4.11. Find the number of solutions of congruence equation x3_{+ 4x +} 6 ≡ 0 (mod 115_).

Solution. We have ∆ = −4 · 43_{− 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 115_{). 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 (4)(1−(113))/2₍₃−1₎(1−(113))/2 _{(mod 11),} we get v4≡ 1 (mod 11).

Using theorem 3.2 v0 = 2, A = 4 and v1 = −B = −6. We determine that v2= 6 · −6 + 43(3−1)3· 2 ≡ 0 (mod 11), v3= 6 · 0 − 43· (3−1)3· 6 ≡ 9 (mod 11) and v4= −6 · 9 − 43· (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 115_{also has three solutions.}

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4.2.1 Quadratic Equation (mod p3) Let us the cubic congruence equation is

f (x) = x3+ ax + b ≡ 0 (mod pk), (4.15) where p is prime, k ∈ Z+ and a 6= 0 such that a, b ∈ Z.

When the discriminant of the equation will be ∆ = −4a3− 27b2 _{= 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 f0(α) ≡ 0 (mod p) and f (α) ≡ 0 (mod p2_{) 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 pk). To find which solutions and how many will lift for next level mod p3we 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

= α3+ t3p3+ 3α2tp + 3αt2p2+ aα + atp + b = (α3+ aα + b) + (3α2+ a)tp + 3αt2p2 = f (α) + f0(α)tp + 3αt2p2

≡ 3αt2p2+ kp2t + p2m (mod p3), finally we get the quadratic equation that is

3αt2+ kt + m ≡ 0 (mod p), (4.16) where k = f0(α)/p (mod p) and m = f (α)/p2 (mod p).

This quadratic has 0 or 2 solutions (mod p).

Np(3αt2+ kt + m) =    0, if ∆_p= −1 (mod p) 2, if ∆_p= 1 (mod 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 pj_{), that p}τ _{|| f}0_{(a), and that j ≥ 2τ + 1. If b ≡ a (mod p}j−τ₎ then f (b) ≡ f (a) (mod pj_{) and p}τ _{|| f}0_{(b). Moreover there is unique t (mod p)} such that f (a + tpj−τ_{) ≡ 0 (mod p}j+1_).

In this condition, a collection of pτ _{solutions (mod p}j_{) give rise to p}τ solu-tions (mod pj+1_{), while power of p dividing f}0 _{remain constant. Since} hypoth-esis of theorem apply with a replaced by a + tpj−τ and (mod pj) replaced by

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(mod pj+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) = x3+ ax + b ≡ 0 (mod pj+1), (4.18) where p is prime, j ∈ Z+ and a, b ∈ Z.

Suppose that x ≡ α + tpj−τ (mod pj) are solutions of equation (4.18) and pτ || f0_{(α), where 0 ≤ t ≤ p − 1 and j ≥ 2τ + 1. Put value of x in cubic} congruence equation we have

f (α + tpj−τ) = (α + tpj−τ)3+ a(α + tpj−τ) + b = α3+ t3p3(j−τ )+ 3α2tpj−τ + 3αt2p2(j−τ )+ aα + atpj−τ + b, since 3j − 3τ = j + (2j − 3τ ) ≥ j + 1 therefore ≡ (α3_{+ aα + b) + (3α}2_{+ a)tp}j−τ ≡ f (α) + f0(α)tpj−τ ≡ kpj+ mtpj≡ 0 (mod pj+1).

Since both term on left side are divisible by pj_{and the right side is also we find} k + mt ≡ 0 (mod p), (4.19) where (m, p) = 1, k = f (α)/pj _{(mod p) and m = f}0_(α)/pτ _{(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 x3_{+12x+4 ≡ 0} (mod 174_).

Solution. The congruence has discriminant ∆ = −4 · 123− 27 · 42_{≡ 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) = x3_{+ 12x + 4. We note that} x = 1 is one solution. Here f0(1) = 3.12_{+ 12 6≡ 0 (mod 17) satisfy case (i)} of Hensel’s lemma. Now f0(8) = 204 ≡ 0 (mod 17) and f (8) = 612 ≡ 34 6≡ 0 (mod 172_{), 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 174_{), from which we} deduce that the congruence has a unique solution (mod 174_).

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Example 4.13. Find the number of solutions of the congruence equation x3+ 5x + 17 ≡ 0 (mod 195).

Solution. We have ∆ = −4 · 53_{− 27 · 17}2 _{≡ 0 (mod 19) so, three solutions of} congruence are,

x ≡ 12, 12, 14 (mod 19),

with one repeated root. Here f0(14) = 593 ≡ 4 6≡ 0 (mod 19) hold case (i) of Hensel’s lemma. We note that x = 12 is repeated root. We have that f0(12) = 437 ≡ 0 (mod 19) and f (12) = 1805 ≡ 0 (mod 192_{) satisfy case (ii),} so that

x ≡ 12, 31, 50, 69, · · · , 354 (mod 192),

are the solutions of x3_{+ 5x + 17 ≡ 0 (mod 19}2_{). Now we will use equation} (4.16) to find the solutions (mod 193_{). First we find value of k and m such} than k = f0(12)/19 = 437/19 ≡ 4 (mod 19), m = f (12)/192 _{= 1805/19}2 _{≡ 4} (mod 19). By quadratic equation (4.16) we see that

36t2+ 4t + 5 ≡ 0 (mod 19).

The discriminant is ∆ = 42_{− 4 · 36 · 5 = −704. Legendre symbol of discriminant} is −704₁₉ = −1 because −704 is quadratic non-residue therefore by equation (4.17) no solution will survive (mod p3_{). We conclude that congruence has a} unique solution (mod 194_{) 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 x3+ x + 10 ≡ 0 (mod 134).

Solution. Starting with ∆ = −4 · 13_{− 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 f0(4) = 49 ≡ 10 6≡ 0 (mod 13) hold case (i) of Hansel’s. We note that x = 11 is repeated root. We have that f0_{(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 132), are solutions of x3+ x + 10 ≡ 0 (mod 132).

Now we will use equation (4.16) to find solutions (mod 133). First we find value of k and m such that k = f0(11)/13 = 364/13 ≡ 2 (mod 13), m = f (11)/132= 1352/132≡ 8 (mod 13). By quadratic equation (4.16) we have

33t2+ 2t + 8 ≡ 0 (mod 13).

We have ∆ = 22_{− 4 · 33 · 8 = −1052. Legendre symbol with discriminant is} −1052

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two solution will survive (mod 132). 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 + 132t, 167 + 132t (mod 133), where 0 ≤ t ≤ 12.

By theorem 4.3 we can see that 13 || f0(141), f (141) ≡ 0 (mod 133) and 3 ≥ 2 + 1. Similarly 13 || f0_{(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 133_{. 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)/133 _{≡ 2 (mod 13) and m = f}0_{(141)/13 ≡ 12} (mod 13). Hence (12, 13) = 1 from which 12t + 2 ≡ 0 (mod 13) give that t = 2. Similarly k = f (167)/133 _{≡ 2 (mod 13) and m = f}0_{(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 134_{), from which we deduce that}

x = 479 + 132t ≡ 479, 2676, 4873, · · · , 26843 (mod 134), and

x = 2195 + 132t ≡ 2195, 4392, 6589, · · · , 28559 (mod 134),

where 0 ≤ t ≤ p − 1 are solutions of x3+ x + 10 ≡ 0 (mod 134). We conclude that (mod 134_{) has 27 solutions.}

4.3 Cubic Congruence with linear and quadratic term

(mod p

k

₎

Suppose the congruence equation is

x3+ bx2+ cx + d ≡ 0 (mod pk), (4.20) where p is prime and b, c, d ∈ Z. We substitute x = y + α in (4.20),

(y + α)3+ b(y + α)2+ c(y + α) + d ≡ 0 (mod pk), to get

y3+ (3α + b)y2+ (3α2+ 2bα + c)y + (α3+ bα2+ cα + d) ≡ 0 (mod pk). If we see at coefficient of quadratic 3α + b ≡ 0 (mod pk_{) 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 pk_{) 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 x3_{+ cx + d ≡ 0} (mod pk_{). We can use lemma 3.1 and theorem 3.2 to find the solutions of this} cubic congruence equation.

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Example 4.15. Simplify the congruence equation x3 + 4x2 + 5x + 8 ≡ 0 (mod 53) into x3+cx+d ≡ 0 (mod 53) 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

= (y3+ α3+ 3y2α + 3yα2) + 4(y2+ α2+ 2yα) + 5(y + α) + 8. We obtain that

y3+ (3α + 4)y2+ (3α2+ 8α + 5)y + (α3+ 4α2+ 5α + 8) ≡ 0 (mod 53). (4.21) We take cofficient of quadratic term to find value of α that is

3α + 4 ≡ 0 (mod 53)

gives α = 82 by Hensel’s lemma. Put this value α = 82 in congruence equation (4.21) we obtain

y3+ 250y2+ 20833y + 578682 ≡ 0 (mod 53). Further, we simplify this equation we have that

y3+ 83y + 57 ≡ 0 (mod 53).

The discriminant ∆ = −4 · 833_{− 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 53_{) verify by theorem 3.2 such as}

v₍₅₋₍5 3))/3≡ − 5 3 (83)(1−(53))/2(3−1)(1−( 5 3))/2 (mod 5),

and we get v2≡ 1 (mod 5).

By theorem 3.2 we have that v0 = 2, A = 83 and v1 = −B = −57. We calculate v2 = −57 · 1 + 833(3−1)3· 2 ≡ 1 (mod 5). Hence it is verified that above congruence y3_{+ 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

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The coefficient of x2, x and the constant term can be divided by at least p, p2 and p3 respectively. To get rid of the quadratic term we divide coefficient of x2 by 3 and then put in original polynomial such as

fx − α 3 =x −α 3 3 − αx −α 3 2 + (ap2+ bp2+ abp2)x −α 3 − mnp3_, where α = p + ap + bp and after simplifying we will obtain

f (x) = x3−cp 2_x 3 −

dp3 27,

where c, d ∈ Z. Now for more simplicity we take this polynomial (mod p3_). f (x) = x3−cp 2_x 3 − dp3 27 ≡ x3− np2x ≡ 0 (mod p3), where n ∈ Z.

In this thesis we just focus on the cubic polynomial of form

f (x) = x3− np2_{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 p2_{) where 0 ≤ t ≤ p − 1. Let k = 3 and put x = pt in equation (4.22) as}

(pt)3− np2_{(pt) ≡ 0} _{(mod p}3_). _(4.23) This shows that always all solutions will survive from mod p2to mod p3and that the number of solutions will be p2_{. 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 3p2 solu-tions.

(ii) If (n

p) = −1 then one solutions will survive and modulo p

4_{has p}2_solutions. To find which solutions will survive for (mod p4) we can construct a general expression. Suppose that the cubic congruence is

f (x) = x3− np2_{x ≡ 0} _{(mod p}4_). _(4.24) Let x ≡ α + p2t be the solutions of the above congruence. By putting this value in (4.23) we obtain

f (α + p2t) = (α + p2t)3− np2(α + p2t)

= α3+ t6p6+ 3α2tp2+ 3αp4t2− np2_{α − np}4_t ≡ f (α) + f0(α)p2t (mod p4),

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and we find that

kp3+ mp3t ≡ 0 (mod p4).

Since throughout the congruence divisible by p3. Moreover, on dividing through by p3 we find that

k + mt ≡ 0 (mod p), (4.25) where k = f (α)/p3 _{and m = f}0_(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 p5_{) can be found by the above expression. The} num-ber of solutions after (mod p5_{) 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 56_{) 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) = x3− 65x2+ 1175x − 4375 ≡ 0 (mod 56). 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 53_{) , we see} that f (x) = x3−700 3 x + 20000 27 ≡ 0 (mod 5 3 ), hence cubic congruence is

f (x) = x3+ 100x ≡ 0 (mod 53)

We focus on this modified congruence and find number of solutions (mod 56_). The discriminant ∆ = −4a3_{= −4000000 ≡ 0 (mod 5) so this congruence has}

x ≡ 0, 0, 0 (mod 5),

three solutions. Here f0(0) = 100 ≡ 0 (mod 52_{) and f (0) ≡ 0 (mod 5}3_{) hold} case (ii) of Hensel’s lemma, hence

x ≡ 0, 5, 10, 15, 20 (mod 52),

are the solutions of x3+ 100x ≡ 0 (mod 52). 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 53). The number of solutions are 52 (mod 53). The Legendre symbol is (−4₅ ) = 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)/53 _{= 5 and m = f}0_{(5)/5 = 35.} _{We have that (35, 5) = 5 and} k further divisible by 5 therefore 5 + 52_{t will survivors of} _{(mod 5}4_{) where}

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0 ≤ t ≤ 4. Similarly k = f (20)/53= 80 and m = f0(20)/5 = 260. We see that (260, 5) = 5 and k divisible by 5 therefore 20 + 52t will survivors of (mod 54) where 0 ≤ t ≤ 4.

To find which solution will survive for (mod 55) we use same technique that we have applied on (mod 54). The survivors (mod 55) are 0 + 53t,55 + 53t and 70 + 53_{t where 0 ≤ t ≤ 4. In fact we note that 5}2 _{|| f}0_{(0) and 5}5 _|| f (0), so theorem 4.3 shows that number of solutions will remain constant after (mod 55_{). By equation (4.19) we find that k = f (55)/5}5 _{≡ 2 (mod 5) and} m = f0(55)/52 _{≡ 0 (mod 5) and verify that there is precisely one value of t} (mod 5), namely t = 0, for which 55 + 0 · 53 _{= 55 survives for} _{(mod p}6_). Similarly k = f (70)/55_{≡ 2 (mod 5) and m = f}0_(70)/52_{≡ 2 (mod 5), moreover} we can verify that there is precisely one value of t (mod 5), namely t = 4, for which 70+4·53_{= 570 is survive for (mod 5}6_{). We conclude that 0+5}4_t,55+54_t and 570+54_{t are solutions (mod 5}6_{) where 0 ≤ t ≤ 4 and every solution further} split into 5 new ones. The orignal congruence has 3 · 52 _{solutions. The solutions} of the equation is described as a tree showed in figure 1.

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Example 4.17. Simplify the congruence equation f (x) = (x − 7)(x − 35)(x − 56) ≡ 0 (mod 75) into general form x3− np2_{x ≡ 0 (mod p}k_{) and find the} number of solutions of simplified congruence equation.

Solution. The cubic congruence equation is

f (x) = x3− 98x2+ 2597x − 13720 ≡ 0 (mod 75). 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 73_{) , so that}

f (x) = x3−1813 3 x +

37730

27 ≡ 0 (mod 7 3_), and finally the cubic congruence is

f (x) = x3+ 196x ≡ 0 (mod 75).

We find the number of solutions this modified congruence. Starting with ∆ = −4a3_{= −30118144 ≡ 0 (mod 7). Three solutions of congruence are}

x ≡ 0, 0, 0 (mod 7).

We note that f0(0) = 196 ≡ 0 (mod 72) and f (0) ≡ 0 (mod 73) staisfy case (ii) of Hensel’s lemma, so that

x ≡ 0, 7, 14, 21, 28, 35, 42 (mod 72),

are solutions of f (x) = x3_{+ 196x ≡ 0 (mod 7}2_{). By theorem 4.3 we see that} 72_{|| f}0_{(0) and 7}5_{||f (0) shows that the number of solutions will remain constant} after (mod 75_{). 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 73₎ where 0 ≤ t ≤ 6. The number of solutions are 72 _{(mod 7}3_{). The Legender} symbol is (4₇) = −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 + 72_{t survivors} (mod 74_{). Similarly zero uniquely lift for (mod 7}5_{) and survivors are 0 + 7}3_t (mod 75). The number of solutions (mod 75) are 72and remain same after this level by theorem 4.3.

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Chapter 5

5 Cubic congruence (mod 2

k

)

Let the cubic congruence equation be

x3≡ a (mod 2k), (5.1) where 2 is prime, k ∈ Z+ and a, ∈ Z.

Theorem 5.1. Suppose that k ≥ 3 and let a is odd. If n is odd then the congruence xn _{≡ a (mod 2}k_{) has exactly one solution. If n is even, then select} β such that (n, 2α−2_{) = 2}β_{.The congruence x}n _{≡ a (mod 2}k_{) has 2}β+1 _solution if a ≡ 1 (mod 2β+2_{) or no solution a 6≡ 1 (mod 2}β+2₎

For proof of theorem see [1].

Example 5.1. Find the number of solutions of x3_{≡ 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) = x3_{− 7 ≡ 0} (mod 2) has one solution that is x ≡ 1 (mod 2). Since f0(x) = 3x2_{, we note} that f0(1) = 3 ≡ 1 6≡ 0 (mod 2), the first case of Hensel’s lemma is satisfied so there is unique solution modulo 22 _{of the form 1 + 2t, where t find as}

t = −f0_{(1)(f (1)/2)} _{(mod 2).}

Here f0_{(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 22₎ is unique solution of f (x) ≡ 0 (mod 22_{). Furthermore, by corollary 4.1 we see} that f (3) = 20, we conclude that x = 3 − 20 · 1 = −17 ≡ 7 (mod 23_{) is the} unique solution of f (x) ≡ 0 (mod 23_).

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

x3≡ a ≡ 2j_{· b} _{(mod 2}k_), _(5.2) where k ∈ Z+

and a, b ∈ Z. If we substitute x = yc in 5.2 then we get y3_c3_{≡ 2}j_b (mod 2k_{). Here b is an odd number and we resolve the equation into y}3 _{≡ 2}j (mod 2k_{) 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 y3_{≡ 2}j (mod 2k).

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5.1 Case I when j ≥ k

When j ≥ k, where j, k ∈ Z then the solutions of the congruence x3 _{≡ 2}j _{≡ 0} (mod 2k_{) 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 22q solutions, where q = bk/3c.

(b) When k ≡ 2 (mod 3), then the cubic congruence has 22q+1 _solutions, where q = bk/3c.

General Expression

(a) We can construct the general expression for cubic congruence

x3≡ 0 (mod 2k), (5.3) where k ∈ Z+ _{to find all solutions when k ≡ 0, 1 (mod 3). Let us}

x ≡ m2b (mod 2k), (5.4) here b = q + r, hence q = bk/3c and r is raminder where m = 0, 1, · · · , 2q+r− 1. Put value of x in equation (5.3) we see

(m2b)3≡ 0 (mod 2k).

Take left hand side and substitute value of bk/3c = k−r₃ by division algorithm in above congruence equation that is

(m2b)3= m323(k−r3 +r)= m32k+2r≡ 0 (mod 2k),

is equal right hand side. The congruence x3≡ 2j _{(mod 2}k_{) has the solutions of} form x ≡ m2b (mod 2k).

(b) When k ≡ 2 (mod 3) then we can construct the general expression for cubic congruence equation

x3≡ 2j _{(mod 2}k₎ _(5.5) to find all possible solutions. Let us

x ≡ m2b (mod 2k), (5.6) here b = q + 1, where q = bk/3c and m = 0, 1, · · · , 22q+1_{− 1. Put value of x in} equation (5.5)

(m2b)3≡ 0 (mod 2k). We take left hand side and put value of bk/3c = k−2

3 by division algorithm in above congruence equation such as

(m2b)3= m323(k−23 +1)= m32k+1≡ 0 (mod 2k),

is equal right hand side of above congruence. This shows that the congruence x3_{≡ 2}j _{(mod 2}k_{) has the solutions of form x ≡ m2}b _{(mod 2}k_).

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Example 5.2. Determine the solutions of x3≡ 28 _{(mod 2}6_).

Solution. We note that k ≡ 0 (mod 3), the congruence has 22·2_{= 16 solutions} because q = b6/3c = 2.

To find all solutions use equation (5.4) that is x ≡ m22+0 _{(mod 2}6_{), where} m = 0, 1, · · · , 22·2_{− 1. The solutions are}

x ≡ 0, 4, 8, 12, · · · , 60 (mod 26).

Example 5.3. Find the solutions of x3_{≡ 2}6 _{(mod 2}4_).

Solution. We have that k ≡ 1 (mod 3) so congruence has 22·1 = 4 solutions because q = b4/3c = 1. Further we use equation (5.4) that is x ≡ m21+1 (mod 24_{), where m = 0, 1, · · · , 2}2_{− 1. The congruence has}

x ≡ 0, 4, 8, 12 (mod 24), possible solutions.

Example 5.4. Determine the solutions of x3_{≡ 2}7 _{(mod 2}11_).

Solution. Starting with k ≡ 2 (mod 3), therefore the congruence has 22·3+1 = 128 solutions because q = b11/3c = 3. The equation 5.6 become x ≡ m23+1 (mod 211), where m = 0, 1, · · · , 23·2+1− 1. The possible solutions are

x ≡ 0, 16, 32, · · · , 2032 (mod 211).

5.2 Case II when j < k

When j < k, number of solutions of congruence equation x3_{≡ 2}j _{(mod 2}k_{) will} depend on j. The congruence equation has 22j/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

x3≡ 2k _{(mod 2}k_). _(5.7) This equation (5.7) has solutions of form x ≡ m2b (mod 2k) where b = q + r, q = bk/3c 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 (m2b) = (m2b)3− 2k_{= (m2}k

3₎3− 2k

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Divide throughout the congruence equation by 2k we obtain that m3− 1 ≡ 0 (mod 2).

This congruence equation is solveable and has always one solution by theorem 5.1.

(b) when k ≡ 1 (mod 3)

x3≡ 2k _{(mod 2}k_). _(5.8) The equation (5.8) has solutions of form x ≡ m2b _{(mod 2}k_{) where b = q + r,} q = bk/3c 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 (m2b) = (m2b)3− 2k _{= (m2}k−1₃ +1₎3_{− 2}k

≡ m3₂k+2_{− 2}k _{≡ 0} _{(mod 2}k+1_). Divide throughout the congruence equation by 2k _{we get that}

m322− 1 ≡ −1 ≡ 0 (mod 2). This congruence equation cannot be solved.

(c) when k ≡ 2 (mod 3)

x3≡ 2k _{(mod 2}k_). _(5.9) The solutions of form x ≡ m2b (mod 2k) where b = q + r, q = bk/3c 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 (m2b) = (m2b)3− 2k = (m2k−23 +1₎3− 2k

≡ m32k+1− 2k ≡ 0 (mod 2k+1). Divide throughout the congruence by 2k _{we obtain that}

m32 − 1 ≡ −1 ≡ 0 (mod 2).

This congruence equation is not solveable. Hence these result shows that when 3 | j then congruence equation x3≡ 2j _{(mod 2}k_{) has solutions otherwise not.}

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General Expression

We can generate the general expression for cubic congruence equation

x3≡ 2j _{(mod 2}k_), _(5.10) where j, k ∈ Z+ _{to find all solutions when j < k. Let}

x ≡ 2q+ m2b (mod 2k), (5.11) where q = bj/3c, b = k − 2q and m = 0, 1, · · · , 22q_{− 1.}

Put this value of x in cubic congruence equation (5.10) such as (2q+ m2b)3≡ 2j _{(mod 2}k_).

By division algorithm bj/3c = (j − 0)/3 We take Left hand side (2q+ m2b)3= 23q+ m323b+ 3m22q2b+ 3m22q22b

= 2j+ m323k−2j+ 3m22j/32k−2j/3+ 3m22j/32(6k−4j)/3 = 2j+ m323k−2j+ 3m2k+ 3m222k−j

≡ 2j _{(mod 2}k_).

The above result is equal right hand side of equation (5.11). This shows that the solutions of x3_{≡ 2}j _{(mod 2}k_{) always in the form x ≡ 2}a_{+ m2}b_.

Example 5.5. Find the number of solutions to x3_{≡ 2}3 _{(mod 2}4_).

Solution. Since j = 3, we see that 3 | j so congruence has 22·3/3 _{= 2}2 _{= 4} solutions.

We find these 4 solution by Hensel’s lemma how they look like. Starting with f (x) = x3− 8 (mod 2), we note that x = 0 is the only solution. Since f0(x) = 3x2, we see that f0(0) ≡ 0 (mod 2) and f (0) ≡ 0 (mod 4), satisfying case (ii) of Hensel’s so that

x ≡ 0, 2 (mod 22),

are the solutions x3 _{≡ 2}3 _{(mod 2}2_). _{We see that f}0_{(0) ≡ 0 (mod 2), and} f (0) ≡ 0 (mod 8), so that we have the roots x = 2 and x = 4 (mod 8). Now f0(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 23), are the solutions x3_{≡ 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) 6≡ 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 24) we conclude that x = 2 and x = 10 are solutions

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(mod 16). Similarly f (4) 6≡ 0 (mod 24) so congruence has no solution at x = 4. At last f (6) 6≡ 0 (mod 24) and so the second case of Hensel’s lemma is satisfied. Therefore we obtain two roots, 6, 14 (mod 24), from which we deduce that

x ≡ 2, 6, 10, 14 (mod 24),

are the solutions of x3_{≡ 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 x3_{≡ 2}6 _{(mod 2}7_).

Solution. We use the equation (5.11) that is x ≡ 22_{+ 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 27), are solutions of x3≡ 26 _{(mod 2}7_{). The number of solutions for this congruence} are 16.

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Chapter 6

6 Cubic Congruence (mod m)

Let us assume that the cubic congruence equation is of the form x3≡ 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} con-gruence f (x) ≡ 0 (mod m). If m = m1m2 where (m1, m2) = 1, then N (m) = N (m1)N (m2). If m = Πpα is canonical factorization of m, then N (m) = ΠN (pα_).

For proof of this theorem see [1].

Example 6.1. Determine the number of solutions of the congruence equation x3_{≡ 8 (mod 1200).}

Solution. If we factor the integer 1200 into a product of primes, we get 1200 = 24_{· 3 · 5}2_{. Now we split original congruence as follows}

x3≡ 8 (mod 3). (6.1) x3≡ 8 (mod 24). (6.2) x3≡ 8 (mod 52). (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 22·3/3 _{= 4} solutions. We find these four solutions by general expression 5.2 that is x ≡ 2 + m22_{, m = 0, 1, · · · , 2}2_{− 1. We conclude that}

x ≡ 0, 2, 4, 6 (mod 24), are solutions of x3_{≡ 8 (mod 2}4_).

Furthermore, we take equation (6.3) and solve by theorem 4.2, according as (8, 52) = 1 so congruence has (3, φ(52)) = (3, 20) = 1 solutions or no solutions verify as

8φ(52)/(3,φ(52))= 820/1≡ 1 (mod 52).

The congruence x3 _{≡ 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 f0(x) = 3x2_{, we see that f}0_{(2) ≡ 2 6≡ 0 (mod 5).} Taking f0_{(2) ≡ 3 (mod 7), we see that the root r}

1 = 2 (mod 5) lifts to r2 = 2 − 0 · 3 ≡ 2 (mod 52_{) is only solution of x}3_{≡ 8 (mod 5}2_).

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6.1 Cubic Congruence with linear term (mod m)

Let us the cubic congruence equation with linear term is of form x3+ 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 x3_{+ 8x + 9 ≡ 0 (mod 190333).}

Solution. We factor the integer 190333 = 13 · 114 _{in product of primes. Now we} split orignal congruence as follows

x3+ 8x + 9 ≡ 0 (mod 13). (6.4) x3+ 8x + 9 ≡ 0 (mod 114). (6.5) Start to solve equation (6.4), the discriminant is ∆ = −4·83−27·92_{= −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

v₍₁₃₋₍13 3))/3≡ 2 13 3 (8)(1−(133))/2(3−1)(1−( 13 3))/2 (mod 13), we get v4≡ 2 (mod 13).

By theorem 3.2 since v0 = 2, A = 8 and v1 = −B = −9. We calculate v2= 9 · −9 + 83(3−1)3· 2 ≡ 0 (mod 13), v3= 9 · 0 − 83· (3−1)3· −9 ≡ 7 (mod 13) and v4= 9 · 7 − 43· (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 · 83− 27 · 92_{= −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 f0(2) ≡ 9 6≡ 0 (mod 11) hold case (i) of Hensel’s lemma. We note that x = 10 is repeated root. We have that f0(10) = 308 ≡ 0 (mod 11) and f (10) = 1089 ≡ 0 (mod 112_{) satisfy case (ii) of} Hensel’s lemma. So that

x ≡ 10, 21, 32, · · · , 120 (mod 112),

are solutions of x3_{+ 8x + 9 ≡ 0 (mod 11}2_{). By theorem 4.3 we have 11 || 308} means that 11 | 308 but 112

- 308. So τ = 1, j ≥ 3 that is f (10) = 1089 6≡ 0 (mod 113_{) shows theorem can not apply at beginning .}

Now we will use quadratic equation (4.16) to find solutions (mod 113_). First we find value of k and m such that k = f0(10)/11 = 308/11 ≡ 6 (mod 11), m = f (10)/112_{= 1089/11}2_{≡ 9 (mod 11). The quadratic equation is}

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The discriminant is ∆ = 62− 4 · 30 · 9 = −1044. Legendre symbol with discrim-inant is −1044₁₁ = 1 because −1044 is quadratic residue therefore two solution will survive (mod 112) 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 + 132t, 120 + 132t (mod 113), where 0 ≤ t ≤ 10.

Moreover we can see that 11 || f0(43), so that f (43) ≡ 0 (mod 113_). Sim-ilarly 11 || f0(120), so that f (120) ≡ 0 (mod 113_{) 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)/113 _{≡ 5} (mod 11) and m = f0(43)/11 ≡ 10 (mod 11). Hence (10, 11) = 1 from which 10t + 5 ≡ 0 (mod 11) give that t = 5. Similarly k = f (120)/113 _{≡ 1 (mod 11)} and m = f0(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 114_{). The} solutions are

x ≡ 43, 120, 164, 241, · · · , 1330 (mod 114)

Thus by theorem 6.1, the congruence x3+ 8x + 9 ≡ 0 (mod 190333) has 23 · 3 = 69 solutions.

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7 Bibliography

References

[1] Ivan Niven, An introduction to the theory of numbers (Fifth Edition), Her-bert 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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