Solving comparisons modulo examples. Modulo comparisons

Comparing numbers modulo

Prepared by: Irina Zutikova

MAOU "Lyceum No. 6"

Class: 10 "a"

Scientific supervisor: Zheltova Olga Nikolaevna

Tambov

2016

• Problem
• Project goal
• Hypothesis
• Project objectives and plan for achieving them
• Comparisons and their properties
• Examples of problems and their solutions
• Used sites and literature

Problem:

Most students rarely use modulo comparison of numbers to solve non-standard and olympiad tasks.

Project goal:

Show how you can solve non-standard and olympiad tasks by comparing numbers modulo.

Hypothesis:

A deeper study of the topic “Comparing numbers modulo” will help students solve some non-standard and olympiad tasks.

Project objectives and plan for achieving them:

1. Study in detail the topic “Comparison of numbers modulo”.

2. Solve several non-standard and olympiad tasks using modulo comparison of numbers.

3.Create a memo for students on the topic “Comparing numbers modulo.”

4. Conduct a lesson on the topic “Comparing numbers modulo” in grade 10 “a”.

5.Give the class homework on the topic “Comparison by module.”

6.Compare the time to complete the task before and after studying the topic “Comparison by Module”.

7.Draw conclusions.

Before starting to study in detail the topic “Comparing numbers modulo”, I decided to compare how it is presented in various textbooks.

• Algebra and the beginnings of mathematical analysis. Advanced level. 10th grade (Yu.M. Kolyagin and others)
• Mathematics: algebra, functions, data analysis. 7th grade (L.G. Peterson and others)
• Algebra and the beginnings of mathematical analysis. Profile level. 10th grade (E.P. Nelin and others)
• Algebra and the beginnings of mathematical analysis. Profile level. 10th grade (G.K. Muravin and others)

As I found out, some textbooks do not even touch on this topic, despite the advanced level. And the topic is presented in the most clear and accessible way in the textbook by L.G. Peterson (Chapter: Introduction to the theory of divisibility), so let’s try to understand the “Comparison of numbers modulo”, relying on the theory from this textbook.

Comparisons and their properties.

Definition: If two integers a and b have the same remainders when divided by some integer m (m>0), then they say thata and b are comparable modulo m, and write:

Theorem: if and only if the difference of a and b is divisible by m.

Properties:

1. Reflexivity of comparisons.Any number a is comparable to itself modulo m (m>0; a,m are integers).
2. Symmetrical comparisons.If the number a is comparable to the number b modulo m, then the number b is comparable to the number a modulo the same (m>0; a,b,m are integers).
3. Transitivity of comparisons.If the number a is comparable to the number b modulo m, and the number b is comparable to the number c modulo the same modulo, then the number a is comparable to the number c modulo m (m>0; a,b,c,m are integers).
4. If the number a is comparable to the number b modulo m, then the number a n comparable by number b n modulo m(m>0; a,b,m-integers; n-natural number).

Examples of problems and their solutions.

1. Find the last digit of the number 3 999 .

Solution:

Because The last digit of the number is the remainder when divided by 10, then

3 999 =3 3 *3 996 =3 3 *(3 4 ) 249 =7*81 249 7(mod 10)

(Because 34=81 1(mod 10);81 n 1(mod10) (by property))

Answer: 7.

2.Prove that 2 4n -1 is divisible by 15 without a remainder. (Phystech2012)

Solution:

Because 16 1(mod 15), then

16n-1 0(mod 15) (by property); 16n= (2 4) n

2 4n -1 0(mod 15)

3.Prove that 12 2n+1 +11 n+2 Divisible by 133 without remainder.

Solution:

12 2n+1 =12*144 n 12*11 n (mod 133) (by property)

12 2n+1 +11 n+2 =12*11 n +11 n *121=11 n *(12+121) =11 n *133

Number (11n *133)divides by 133 without remainder. Therefore, (12 2n+1 +11 n+2 ) is divisible by 133 without a remainder.

4. Find the remainder of the number 2 divided by 15 2015 .

Solution:

Since 16 1(mod 15), then

2 2015 8(mod 15)

Answer:8.

5.Find the remainder of division by the 17th number 2 2015. (Phystech2015)

Solution:

2 2015 =2 3 *2 2012 =8*16 503

Since 16 -1(mod 17), then

2 2015 -8(mod 15)

8 9(mod 17)

Answer:9.

6.Prove that the number is 11 100 -1 is divisible by 100 without a remainder. (Phystech2015)

Solution:

11 100 =121 50

121 50 21 50 (mod 100) (by property)

21 50 =441 25

441 25 41 25 (mod 100) (by property)

41 25 =41*1681 12

1681 12 (-19) 12 (mod 100) (by property)

41*(-19) 12 =41*361 6

361 6 (-39) 6 (mod 100)(by property)

41*(-39) 6 =41*1521 3

1521 3 21 3 (mod100) (by property)

41*21 3 =41*21*441

441 41(mod 100) (by property)

21*41 2 =21*1681

1681 -19(mod 100) (by property)

21*(-19)=-399

399 1(mod 100) (by property)

So 11 100 1(mod 100)

11 100 -1 0(mod 100) (by property)

7. Three numbers are given: 1771,1935,2222. Find a number such that, when divided by it, the remainders of the three given numbers will be equal. (HSE2016)

Solution:

Let the unknown number be equal to a, then

2222 1935(mod a); 1935 1771(mod a); 2222 1771(mod a)

2222-1935 0(moda) (by property); 1935-17710(moda) (by property); 2222-17710(moda) (by property)

287 0(mod a); 164 0(mod a); 451 0(mod a)

287-164 0(moda) (by property); 451-2870(moda)(by property)

123 0(mod a); 164 0(mod a)

164-123 0(mod a) (by property)

41

HSE Olympiad 2016

Content.

Introduction

§1. Modulo comparison

§2. Comparison Properties

1. Module-Independent Comparison Properties
2. Module-dependent properties of comparisons

§3. Deduction system

1. Full system of deductions
2. Reduced system of deductions

§4. Euler's theorem and Fermat

1. Euler function
2. Euler's theorem and Fermat

Chapter 2. Theory of comparisons with a variable

§1. Basic concepts related to solving comparisons

1. The Roots of Comparisons
2. Equivalence of comparisons
3. Wilson's theorem

§2. First degree comparisons and their solutions

1. Selection method
2. Euler's methods
3. Euclid algorithm method
4. Continued Fraction Method

§3. Systems of comparisons of the 1st degree with one unknown

§4. Division of comparisons of higher degrees

§5. Antiderivative roots and indices

1. Deduction class order
2. Primitive roots modulo prime
3. Indexes modulo prime

Chapter 3. Application of the theory of comparisons

§1. Signs of divisibility

§2. Checking the results of arithmetic operations

§3. Conversion of an ordinary fraction to a final fraction

decimal systematic fraction

Conclusion

Literature

Introduction

In our lives we often have to deal with integers and problems related to them. In this thesis I consider the theory of comparison of integers.

Two integers whose difference is a multiple of a given natural number m are called comparable in modulus m.

The word “module” comes from the Latin modulus, which in Russian means “measure”, “magnitude”.

The statement “a is comparable to b modulo m” is usually written as ab (mod m) and is called comparison.

The definition of comparison was formulated in the book by K. Gauss “Arithmetic Studies”. This work, written in Latin, began to be printed in 1797, but the book was published only in 1801 due to the fact that the printing process at that time was extremely labor-intensive and lengthy. The first section of Gauss’s book is called: “On the comparison of numbers in general.”

Comparisons are very convenient to use in cases where it is enough to know in any research numbers accurate to multiples of a certain number.

For example, if we are interested in what digit the cube of an integer a ends with, then it is enough for us to know a only up to multiples of 10 and we can use comparisons modulo 10.

The purpose of this work is to consider the theory of comparisons and study the basic methods for solving comparisons with unknowns, as well as to study the application of the theory of comparisons to school mathematics.

The thesis consists of three chapters, with each chapter divided into paragraphs, and paragraphs into paragraphs.

The first chapter outlines general issues of the theory of comparisons. Here we consider the concept of modulo comparison, properties of comparisons, the complete and reduced system of residues, Euler's function, Euler's and Fermat's theorem.

The second chapter is devoted to the theory of comparisons with the unknown. It outlines the basic concepts associated with solving comparisons, discusses methods for solving comparisons of the first degree (selection method, Euler's method, the method of the Euclidean algorithm, the method of continued fractions, using indices), systems of comparisons of the first degree with one unknown, comparisons of higher degrees, etc. .

The third chapter contains some applications of number theory to school mathematics. The signs of divisibility, checking the results of actions, and converting ordinary fractions into systematic decimal fractions are considered.

The presentation of theoretical material is accompanied by a large number of examples that reveal the essence of the introduced concepts and definitions.

Chapter 1. General questions of the theory of comparisons

§1. Modulo comparison

Let z be the ring of integers, m be a fixed integer, and m·z be the set of all integers that are multiples of m.

Definition 1. Two integers a and b are said to be comparable modulo m if m divides a-b.

If the numbers a and b are comparable modulo m, then write a b (mod m).

Condition a b (mod m) means a-b is divisible by m.

a b (mod m)↔(a-b) m

Let us define that the comparability relation modulo m coincides with the comparability relation modulo (-m) (divisibility by m is equivalent to divisibility by –m). Therefore, without loss of generality, we can assume that m>0.

Examples.

Theorem. (a sign of comparability of spirit numbers modulo m): Two integers a and b are comparable modulo m if and only if a and b have the same remainders when divided by m.

Proof.

Let the remainders when dividing a and b by m be equal, that is, a=mq₁+r,(1)

B=mq₂+r, (2)

Where 0≤r≥m.

Subtract (2) from (1), we get a-b= m(q₁- q₂), that is, a-b m or a b (mod m).

Conversely, let a b (mod m). This means that a-b m or a-b=mt, t z (3)

Divide b by m; we get b=mq+r in (3), we will have a=m(q+t)+r, that is, when dividing a by m, the same remainder is obtained as when dividing b by m.

Examples.

5=4·(-2)+3

23=4·5+3

24=3·8+0

10=3·3+1

Definition 2. Two or more numbers that give identical remainders when divided by m are called equal remainders or comparable modulo m.

Examples.

We have: 2m+1-(m+1)²= 2m+1 - m²-2m-1=- m², and (- m²) is divided by m => our comparison is correct.

1. Prove that the following comparisons are false:

If numbers are comparable modulo m, then they have the same gcd with it.

We have: 4=2·2, 10=2·5, 25=5·5

GCD(4,10) = 2, GCD(25,10) = 5, therefore our comparison is incorrect.

§2. Comparison Properties

1. Module-independent properties of comparisons.

Many properties of comparisons are similar to the properties of equalities.

a) reflexivity: aa (mod m) (any integer a comparable to itself modulo m);

B) symmetry: if a b (mod m), then b a (mod m);

C) transitivity: if a b (mod m), and b with (mod m), then a with (mod m).

Proof.

By condition m/(a-b) and m/ (c-d). Therefore, m/(a-b)+(c-d), m/(a+c)-(b+d) => a+c b+d (mod m).

Examples.

Find the remainder when dividing at 13.

Solution: -1 (mod 13) and 1 (mod 13), then (-1)+1 0 (mod 13), that is, the remainder of the division by 13 is 0.

a-c b-d (mod m).

Proof.

By condition m/(a-b) and m/(c-d). Therefore, m/(a-b)-(c-d), m/(a-c)-(b-d) => (a-c) b-d (mod m).

1. (a consequence of properties 1, 2, 3). You can add the same integer to both sides of the comparison.

Proof.

Let a b (mod m) and k is any integer. By the property of reflexivity

k=k (mod m), and according to properties 2 and 3 we have a+k b+k (mod m).

a·c·d (mod m).

Proof.

By condition, a-b є mz, c-d є mz. Therefore a·c-b·d = (a·c - b·c)+(b·c- b·d)=(a-b)·c+b·(c-d) є mz, that is, a·c·d (mod m).

Consequence. Both sides of the comparison can be raised to the same non-negative integer power: if ab (mod m) and s is a non-negative integer, then a s b s (mod m).

Examples.

Solution: obviously 13 1 (mod 3)

2 -1 (mod 3)

5 -1 (mod 3), then

- · 1-1 0 (mod 13)

Answer: the required remainder is zero, and A is divisible by 3.

Solution:

Let us prove that 1+ 0(mod13) or 1+ 0(mod 13)

1+ =1+ 1+ =

Since 27 1 (mod 13), then 1+ 1+1·3+1·9 (mod 13).

etc.

3. Find the remainder when dividing with the remainder of a number at 24.

We have: 1 (mod 24), so

1 (mod 24)

Adding 55 to both sides of the comparison, we get:

(mod 24).

We have: (mod 24), therefore

(mod 24) for any k є N.

Hence (mod 24). Since (-8)16(mod 24), the required remainder is 16.

1. Both sides of the comparison can be multiplied by the same integer.

2. Properties of comparisons that depend on the module.

Proof.

Since a b (mod t), then (a - b) t. And since t n , then due to the transitivity of the divisibility relation(a - b n), that is, a b (mod n).

Example.

Find the remainder when 196 is divided by 7.

Solution:

Knowing that 196= , we can write 196(mod 14). Let's use the previous property, 14 7, we get 196 (mod 7), that is, 196 7.

1. Both sides of the comparison and the modulus can be multiplied by the same positive integer.

Proof.

Let a b (mod t ) and c is a positive integer. Then a-b = mt and ac-bc=mtc, or ac bc (mod mc).

Example.

Determine whether the value of an expression is an integer.

Solution:

Let's represent fractions in the form of comparisons: 4(mod 3)

1 (mod 9)

31 (mod 27)

Let's add these comparisons term by term (property 2), we get 124(mod 27) We see that 124 is not an integer divisible by 27, hence the meaning of the expressionis also not an integer.

1. Both sides of the comparison can be divided by their common factor if it is coprime to the modulus.

Proof.

If ca cb (mod m), that is, m/c(a-b) and the number With coprime to m, (c,m)=1, then m divides a-b. Hence, a b (mod t).

Example.

60 9 (mod 17), after dividing both sides of the comparison by 3 we get:

20 (mod 17).

In general, it is impossible to divide both sides of the comparison by a number that is not coprime to the modulus, since after division the numbers may be obtained that are incomparable with respect to a given modulus.

Example.

8 (mod 4), but 2 (mod 4).

1. Both sides of the comparison and the modulus can be divided by their common divisor.

Proof.

If ka kb (mod km), then k (a-b) is divided by km. Therefore, a-b is divisible by m, that is a b (mod t).

Proof.

Let P (x) = c 0 x n + c 1 x n-1 + ... + c n-1 x+ c n. By condition a b (mod t), then

a k b k (mod m) for k = 0, 1, 2, …,n. Multiplying both sides of each of the resulting n+1 comparisons by c n-k , we get:

c n-k a k c n-k b k (mod m), where k = 0, 1, 2, …,n.

Adding up the last comparisons, we get: P (a) P (b) (mod m). If a (mod m) and c i d i (mod m), 0 ≤ i ≤n, then

(mod m). Thus, in comparison modulo m, individual terms and factors can be replaced by numbers comparable modulo m.

At the same time, you should pay attention to the fact that the exponents found in comparisons cannot be replaced in this way: from

a n c(mod m) and n k(mod m) it does not follow that a k s (mod m).

Property 11 has a number of important applications. In particular, with its help it is possible to give a theoretical justification for the signs of divisibility. To illustrate, as an example, we will give the derivation of the divisibility test by 3.

Example.

Every natural number N can be represented as a systematic number: N = a 0 10 n + a 1 10 n-1 + ... + a n-1 10 + a n .

Consider the polynomial f(x) = a 0 x n + a 1 x n-1 + ... + a n-1 x+a n . Because

10 1 (mod 3), then by property 10 f (10) f(1) (mod 3) or

N = a 0 10 n + a 1 10 n-1 + ... + a n-1 10 + a n a 1 + a 2 +…+ a n-1 + a n (mod 3), i.e. for N to be divisible by 3, it is necessary and sufficient that the sum of the digits of this number is divisible by 3.

§3. Deduction systems

1. Full system of deductions.

Equal remainder numbers, or, what is the same thing, comparable modulo m, form a class of numbers modulo m.

From this definition it follows that all numbers in the class correspond to the same remainder r, and we get all the numbers in the class if, in the form mq+r, we make q run through all the integers.

Accordingly, with m different values ​​of r, we have m classes of numbers modulo m.

Any number of a class is called a residue modulo m with respect to all numbers of the same class. The residue obtained at q=0, equal to the remainder r, is called the smallest non-negative residue.

The residue ρ, the smallest in absolute value, is called the absolutely smallest residue.

Obviously, for r we have ρ=r; at r> we have ρ=r-m; finally, if m is even and r=, then any of the two numbers can be taken as ρ and -m= - .

Let us choose from each class of residues modulo T one number at a time. We get t integers: x 1,…, x m. The set (x 1,…, x t) is called complete system of deductions modulo m.

Since each class contains an infinite number of residues, it is possible to compose an infinite number of different complete systems of residues for a given module m, each of which contains t deductions.

Example.

Compile several complete systems of modulo deductions T = 5. We have classes: 0, 1, 2, 3, 4.

0 = {... -10, -5,0, 5, 10,…}

1= {... -9, -4, 1, 6, 11,…}

Let's create several complete systems of deductions, taking one deduction from each class:

0, 1, 2, 3, 4

5, 6, 2, 8, 9

10, -9, -8, -7, -6

5, -4, -3, -2, -1

etc.

The most common:

1. Complete system of least non-negative residues: 0, 1, t -1 In the example above: 0, 1, 2, 3, 4. This system of residues is simple to create: you need to write down all the non-negative remainders obtained when dividing by m.
2. Complete system of least positive residues(the smallest positive deduction is taken from each class):

1, 2, …, m. In our example: 1, 2, 3, 4, 5.

1. A complete system of absolutely minimal deductions.In the case of odd m, the absolute smallest residues are represented side by side.

- ,…, -1, 0, 1,…, ,

and in the case of even m, one of the two rows

1, …, -1, 0, 1,…, ,

, …, -1, 0, 1, …, .

In the example given: -2, -1, 0, 1, 2.

Let us now consider the basic properties of the complete system of residues.

Theorem 1 . Any collection of m integers:

x l ,x 2 ,…,x m (1)

pairwise incomparable modulo m, forms a complete system of residues modulo m.

Proof.

1. Each of the numbers in the collection (1) belongs to a certain class.
2. Any two numbers x i and x j from (1) are incomparable with each other, i.e., they belong to different classes.
3. There are m numbers in (1), i.e., the same number as there are modulo classes T.

x 1, x 2,…, x t - complete system of deductions modulo m.

Theorem 2. Let (a, m) = 1, b - arbitrary integer; then if x 1, x 2,…, x t is a complete system of residues modulo m, then the collection of numbers ax 1 + b, ax 2 + b,…, ax m + b is also a complete system of residues modulo m.

Proof.

Let's consider

Ax 1 + b, ax 2 + b,…, ax m + b (2)

1. Each of the numbers in the collection (2) belongs to a certain class.
2. Any two numbers ax i +b and ax j + b from (2) are incomparable with each other, that is, they belong to different classes.

Indeed, if in (2) there were two numbers such that

ax i +b ax j + b (mod m), (i = j), then we would get ax i ax j (mod t). Since (a, t) = 1, then the property of comparisons can reduce both parts of the comparison by A . We get x i x j (mod m).

By condition x i x j (mod t) at (i = j) , since x 1, x 2, ..., x m - a complete system of deductions.

1. The set of numbers (2) contains T numbers, that is, as many as there are classes modulo m.

So, ax 1 + b, ax 2 + b,..., ax m + b - complete system of residues modulo m.

Example.

Let m = 10, a = 3, b = 4.

Let’s take some complete system of residues modulo 10, for example: 0, 1, 2,…, 9. Let’s compose numbers of the form ax + b. We get: 4, 7, 10, 13, 16, 19, 22, 25, 28, 31. The resulting set of numbers is a complete system of residues modulo 10.

1. The given system of deductions.

Let us prove the following theorem.

Theorem 1.

Numbers of the same residue class modulo m have the same greatest common divisor with m: if a b (mod m), then (a, m) = (b, m).

Proof.

Let a b (mod m). Then a = b +mt, where t є z. From this equality it follows that (a, t) = (b, t).

Indeed, let δ be a common divisor of a and m, then aδ, m δ. Since a = b +mt, then b=a-mt, therefore bδ. Therefore, any common divisor of the numbers a and m is a common divisor of m and b.

Conversely, if m δ and b δ, then a = b +mt is divisible by δ, and therefore any common divisor of m and b is a common divisor of a and m. The theorem has been proven.

Definition 1. Greatest common modulus divisor t and any number a from this class of deductions by T called the greatest common divisor T and this class of deductions.

Definition 2. Residue class a modulo t called coprime to modulus m , if the greatest common divisor a and t equals 1 (that is, if m and any number from a are relatively prime).

Example.

Let t = 6. Residue class 2 consists of the numbers (..., -10, -4, 2, 8, 14, ...). The greatest common divisor of any of these numbers and modulus 6 is 2. Hence, (2, 6) = 2. The greatest common divisor of any number from class 5 and modulus 6 is 1. Hence, class 5 is coprime to modulus 6.

Let us choose one number from each class of residues that is coprime with modulo m. We obtain a system of deductions that is part of the complete system of deductions. They call herreduced system of residues modulo m.

Definition 3. A set of residues modulo m, taken one from each coprime with T class of residues according to this module is called a reduced system of residues.

From Definition 3 follows a method for obtaining the reduced system of modulo residues T: it is necessary to write down some complete system of residues and remove from it all residues that are not coprime with m. The remaining set of deductions is the reduced system of deductions. Obviously, an infinite number of systems of residues modulo m can be composed.

If we take as the initial system the complete system of least non-negative or absolutely least residues, then using the indicated method we obtain, respectively, a reduced system of least non-negative or absolutely least residues modulo m.

Example.

If t = 8, then 1, 3, 5, 7 is the reduced system of least non-negative residues, 1, 3, -3, -1- the reduced system of absolutely least deductions.

Theorem 2.

Let the number of classes coprime to m is equal to k.Then any collection of k integers

pairwise incomparable modulo m and coprime to m, is a reduced system of residues modulo m.

Proof

A) Each number in the population (1) belongs to a certain class.

1. All numbers from (1) are pairwise incomparable in modulus T, that is, they belong to different classes modulo m.
2. Each number from (1) is coprime with T, that is, all these numbers belong to different classes coprime to modulo m.
3. Total (1) available k numbers, that is, as many as the reduced system of residues modulo m should contain.

Therefore, the set of numbers(1) - reduced system of modulo deductions T.

§4. Euler function.

Euler's and Fermat's theorems.

1. Euler function.

Let us denote by φ(T) the number of classes of residues modulo m coprime to m, that is, the number of elements of the reduced system of residues modulo t. Function φ (t) is numeric. They call herEuler function.

Let us choose as representatives of the modulo residue classes t numbers 1, ..., t - 1, t. Then φ (t) - the number of such numbers coprime with t. In other words, φ (t) - the number of positive numbers not exceeding m and relatively prime to m.

Examples.

1. Let t = 9. The complete system of residues modulo 9 consists of the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9. Of these, the numbers 1,2,4, 5, 7, 8 are coprime to 9. So since the number of these numbers is 6, then φ (9) = 6.
2. Let t = 12. The complete system of residues consists of the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12. Of these, numbers 1, 5, 7, 11 are coprime to 12. This means

φ(12) = 4.

At t = 1, the complete system of residues consists of one class 1. The common natural divisor of the numbers 1 and 1 is 1, (1, 1) = 1. On this basis, we assume φ(1) = 1.

Let's move on to calculating the Euler function.

1) If t = p is a prime number, then φ(p) = p- 1.

Proof.

Deductions 1, 2, ..., p- 1 and only they are relatively prime with a prime number r. Therefore φ (р) = р - 1.

2) If t = p k - prime power p, then

φ(t) = (p - 1) . (1)

Proof.

Complete system of modulo deductions t = p k consists of numbers 1,..., p k - 1, p k Natural divisors T are degrees r. Therefore the number Amay have a common divisor with m other than 1, only in the case whenAdivided byr.But among the numbers 1, ... , pk -1 onronly numbers are divisiblep, 2p, ... , p2 , ... rTo, the number of which is equalrTo: p = pk-1. This means that they are coprime witht = pTorestrTo- pk-1= pk-l(p-1)numbers. This proves that

φ (pTo) = pk-1(p-1).

Theorem1.

The Euler function is multiplicative, that is, for relatively prime numbers m and n we have φ (mn) = φ(m) φ (n).

Proof.

The first requirement in the definition of a multiplicative function is fulfilled in a trivial way: the Euler function is defined for all natural numbers, and φ (1) = 1. We only need to show that iftypecoprime numbers, then

φ (tp)= φ (T) φ (p).(2)

Let us arrange the complete system of deductions modulotpin the formnXT -matrices

1 2 T

t +1 t +2 2t

………………………………

(p -1) t+1 (p -1) m+2 Fri

SinceTAndnare relatively prime, then the numberXreciprocally just withtpthen and only whenXreciprocally just withTAndXreciprocally just withn. But the numberkm+treciprocally just withTif and only iftreciprocally just withT.Therefore, numbers coprime to m are located in those columns for whichtruns through the reduced system of modulo residuesT.The number of such columns is equal to φ(T).Each column presents the complete system of modulo deductionsp.From these deductions φ(p)coprime withp.This means that the total number of numbers that are relatively prime and withTand with n, equal to φ(T)φ(n)

(φ (T)columns, in each of which φ is taken(p)numbers). These numbers, and only they, are coprime toetc.This proves that

φ (tp)= φ (T) φ (p).

Examples.

№1 . Prove the validity of the following equalities

φ(4n) =

Proof.

№2 . Solve the equation

Solution:because(m)=, That= , that is=600, =75, =3·, then x-1=1, x=2,

y-1=2, y=3

Answer: x=2, y=3

We can calculate the value of the Euler function(m), knowing the canonical representation of the number m:

m=.

Due to multiplicativity(m) we have:

(m)=.

But according to formula (1) we find that

-1), and therefore

(3)

Equality (3) can be rewritten as:

Since=m, then(4)

Formula (3) or, which is the same, (4) is what we are looking for.

Examples.

№1 . What is the amount?

Solution:,

, =18 (1- ) (1- =18 , Then= 1+1+2+2+6+6=18.

№2 . Based on the properties of the Euler number function, prove that in the sequence of natural numbers there is an infinite set of prime numbers.

Solution:Assuming that the number of prime numbers is a finite set, we assume that- the largest prime number and let a=is the product of all prime numbers, based on one of the properties of the Euler number function

Since a≥, then a is a composite number, but since its canonical representation contains all prime numbers, then=1. We have:

=1 ,

which is impossible, and thus it is proved that the set of prime numbers is infinite.

№3 .Solve the equation, where x=And=2.

Solution:We use the property of the Euler numerical function,

,

and by condition=2.

Let us express from=2 , we get, substitute in

:

(1+ -1=120, =11 =>

Then x=, x=11·13=143.

Answer:x= 143

2. Euler's theorem and Fermat.

Euler's theorem plays an important role in the theory of comparisons.

Euler's theorem.

If an integer a is coprime to m, then

(1)

Proof.Let

(2)

there is a reduced system of residues modulo m.

Ifais an integer coprime to m, then

(3)

A first degree comparison with one unknown has the form:

f(x) 0 (mod m); f(X) = Oh + and n. (1)

Solve comparison- means finding all values ​​of x that satisfy it. Two comparisons that satisfy the same values ​​of x are called equivalent.

If comparison (1) is satisfied by any x = x 1, then (according to 49) all numbers comparable to x 1, modulo m: x x 1 (mod m). This entire class of numbers is considered to be one solution. With such an agreement, the following conclusion can be drawn.

66.C alignment (1) will have as many solutions as the number of residues of the complete system that satisfy it.

Example. Comparison

6x– 4 0 (mod 8)

Among the numbers 0, 1,2, 3, 4, 5, 6, 7, two numbers satisfy the complete system of residues modulo 8: X= 2 and X= 6. Therefore, this comparison has two solutions:

x 2 (mod 8), X 6 (mod 8).

Comparison of the first degree by moving the free term (with the opposite sign) to the right side can be reduced to the form

ax b(mod m). (2)

Consider a comparison that satisfies the condition ( A, m) = 1.

According to 66, our comparison has as many solutions as there are residues of the complete system that satisfy it. But when x runs through the complete system of modulo residues T, That Oh runs through the complete system of deductions (out of 60). Therefore, for one and only one value X, taken from the complete system, Oh will be comparable to b. So,

67. When (a, m) = 1 comparison ax b(mod m)has one solution.

Let now ( a, m) = d> 1. Then, for comparison (2) to have solutions, it is necessary (out of 55) that b divided by d, otherwise comparison (2) is impossible for any integer x . Assuming therefore b multiples d, let's put a = a 1 d, b = b 1 d, m = m 1 d. Then comparison (2) will be equivalent to this (abbreviated by d): a 1 x b 1 (mod m), in which already ( A 1 , m 1) = 1, and therefore it will have one solution modulo m 1. Let X 1 – the smallest non-negative residue of this solution modulo m 1 , then all numbers are x , forming this solution are found in the form

x x 1 (mod m 1). (3)

Modulo m, numbers (3) form not one solution, but more, exactly as many solutions as there are numbers (3) in the series 0, 1, 2, ..., m – 1 least non-negative modulo residues m. But the following numbers (3) will fall here:

x 1 , x 1 + m 1 , x 1 + 2m 1 , ..., x 1 + (d – 1) m 1 ,

those. total d numbers (3); therefore comparison (2) has d decisions.

We get the theorem:

68. Let (a, m) = d. Comparison ax b ( mod m) is impossible if b is not divisible by d. When b is a multiple of d, the comparison has d solutions.

69. A method for solving comparisons of the first degree, based on the theory of continued fractions:

Expanding into a continued fraction the relation m:a,

and looking at the last two matching fractions:

according to the properties of continued fractions (according to 30 ) we have

So the comparison has a solution

to find, which is enough to calculate Pn– 1 according to the method specified in 30.

Example. Let's solve the comparison

111x= 75 (mod 321). (4)

Here (111, 321) = 3, and 75 is a multiple of 3. Therefore, the comparison has three solutions.

Dividing both sides of the comparison and the modulus by 3, we get the comparison

37x= 25 (mod 107), (5)

which we must first solve. We have

q
P 3

So, in this case n = 4, P n – 1 = 26, b= 25, and we have a solution to comparison (5) in the form

x–26 ∙ 25 99 (mod 107).

Hence, the solutions to comparison (4) are presented as follows:

X 99; 99 + 107; 99 + 2 ∙ 107 (mod 321),

Xº99; 206; 313 (mod 321).

Calculation of the inverse element by a given modulo

70.If the numbers are integers a And n are coprime, then there is a number a′, satisfying the comparison a ∙ a′ ≡ 1(mod n). Number a′ called multiplicative inverse of a modulo n and the notation used for it is a- 1 (mod n).

The calculation of reciprocal quantities modulo a certain value can be performed by solving a comparison of the first degree with one unknown, in which x number accepted a′.

To find a comparison solution

a∙x≡ 1(mod m),

Where ( a,m)= 1,

you can use the Euclid algorithm (69) or the Fermat-Euler theorem, which states that if ( a,m) = 1, then

a φ( m) ≡ 1(mod m).

x ≡ a φ( m)–1 (mod m).

Groups and their properties

Groups are one of the taxonomic classes used to classify mathematical structures with common characteristic properties. Groups have two components: many (G) And operations() defined on this set.

The concepts of set, element and membership are the basic undefined concepts of modern mathematics. Any set is defined by the elements included in it (which, in turn, can also be sets). Thus, we say that a set is defined or given if for any element we can tell whether it belongs to this set or not.

For two sets A, B records B A, B A, B∩ A, B A, B \ A, A × B respectively mean that B is a subset of the set A(i.e. any element from B is also contained in A, for example, the set of natural numbers is contained in the set of real numbers; besides, always A A), B is a proper subset of the set A(those. B A And B ≠ A), intersection of sets B And A(i.e. all such elements that lie simultaneously in A, and in B, for example, the intersection of integers and positive real numbers is the set of natural numbers), the union of sets B And A(i.e. a set consisting of elements that lie either in A, or in B), set difference B And A(i.e. the set of elements that lie in B, but do not lie in A), Cartesian product of sets A And B(i.e. a set of pairs of the form ( a, b), Where a A, b B). Via | A| always denotes the cardinality of the set A, i.e. number of elements in the set A.

An operation is a rule according to which any two elements of a set G(a And b) is matched with the third element from G: a b.

Lots of elements G with an operation is called group, if the following conditions are satisfied.

Definition 1. If two numbers are 1) a And b when divided by p give the same remainder r, then such numbers are called equiremainder or comparable in modulus p.

Statement 1. Let p some positive number. Then every number a always and, moreover, in the only way can be represented in the form

But these numbers can be obtained by setting r equal to 0, 1, 2,..., p−1. Hence sp+r=a will get all possible integer values.

Let us show that this representation is unique. Let's assume that p can be represented in two ways a=sp+r And a=s 1 p+r 1. Then

(2)

Because r 1 accepts one of the numbers 0,1, ..., p−1, then the absolute value r 1 −r less p. But from (2) it follows that r 1 −r multiple p. Hence r 1 =r And s 1 =s.

Number r called minus numbers a modulo p(in other words, the number r called the remainder of a number a on p).

Statement 2. If two numbers a And b comparable in modulus p, That a−b divided by p.

Really. If two numbers a And b comparable in modulus p, then when divided by p have the same remainder p. Then

divided by p, because the right side of equation (3) is divided by p.

Statement 3. If the difference of two numbers is divisible by p, then these numbers are comparable in modulus p.

Proof. Let us denote by r And r 1 division remainder a And b on p. Then

Examples 25≡39 (mod 7), −18≡14 (mod 4).

From the first example it follows that 25 when divided by 7 gives the same remainder as 39. Indeed, 25 = 3 7 + 4 (remainder 4). 39=3·7+4 (remainder 4). When considering the second example, you need to take into account that the remainder must be a non-negative number less than the modulus (i.e. 4). Then we can write: −18=−5·4+2 (remainder 2), 14=3·4+2 (remainder 2). Therefore, −18 when divided by 4 leaves a remainder of 2, and 14 when divided by 4 leaves a remainder of 2.

Properties of modulo comparisons

Property 1. For anyone a And p Always

there is not always a comparison

Where λ is the greatest common divisor of numbers m And p.

Proof. Let λ greatest common divisor of numbers m And p. Then

Because m(a−b) divided by k, That

Hence

And m is one of the divisors of the number p, That

Where h=pqs.

Note that we can allow comparisons based on negative modules, i.e. comparison a≡b mod( p) means in this case that the difference a−b divided by p. All properties of comparisons remain in force for negative modules.