Math project on topic

"Comparisons modulo"

Zaripova Aisylu

Sovetsky district of Kazan

MBOU "Secondary School No. 166", 7a grade

Scientific supervisor: Antonova N.A.

Table of contents

Introduction__________________________________________________________3

    What are comparisons_____________________________________________4

    1. The concept of comparisons modulo__________________________4

      History of the emergence of the concept of comparisons modulo_____4

      Properties of comparisons_________________________________________4

    Applying comparisons to problem solving______________________________6

    1. The simplest use of modulo comparisons is to determine the divisibility of numbers___________________________6

      One comparison task_______________________________8

      Using modulo comparisons in professional activity _________________________________________9

Conclusion_________________________________________________10

List of used literature_______________________________________11

Introduction.

Topic: Modulo comparisons.

Problem: Many students are faced with problems when preparing for Olympiads, the solution of which is based on knowledge of remainders from dividing integers by a natural number. We were interested in these types of problems and possible methods for solving them. It turns out that they can be solved using modulo comparisons.

Goal: Find out the essence of modulo comparisons, the main methods of working with modulo comparisons.

Objectives: find theoretical material on this topic, consider problems that are solved using modulo comparisons, show the most common methods for solving such problems, draw conclusions.

Object of study: number theory.

Subject of research: theory of modulo comparisons.

The work relates to theoretical research and can be used in preparation for mathematics olympiads. Its content reveals the basic concepts of modulo comparisons and their main properties, and provides examples of solving problems on this topic.

I . What are comparisons?

    1. The concept of modulo comparisons.

The numbers and are said to be comparable in modulus if they are divisible by, in other words, a and b have the same remainders when divided by.

Designation

Examples:

    12 and 32 are comparable modulo 5, since 12 when divided by 5 has a remainder of 2 and 32 when divided by 2 has a remainder of 2. It is written12 ;

    101 and 17 are comparable modulo 21;

    1. History of the emergence of the concept of modulo comparisons.

The theory of divisibility was largely created by Euler. The definition of comparison was formulated in the book by K.F. Gauss “Arithmetic Studies”. This work, written in Latin, began printing 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.” It was Gauss who proposed the symbolism of modulo comparisons that has become established in mathematics.

    1. Properties of comparisons.

If

Proof:

  1. If we add the second to the first equality, we get

is the sum of two integers, therefore is an integer, therefore.

    If we subtract the second from the first equality, we get

this is the difference of two integers, which means it is an integer, therefore.

    Consider the expression:

This is the difference of products of integers, which means it is an integer, therefore.

    This is a consequence of the third property of comparisons.

Q.E.D.

5) If.

Proof: Let's find the sum of these two expressions:

is the sum of two integers, therefore it is an integer, therefore .

Q.E.D.

6) If is an integer, then

Proof: , wherep– an integer, multiply this equality by, we get: . Since is a product of integers, that is what needed to be proven.

7) If

Proof: the reasoning is similar to the proof of property 6.

8) If - coprime numbers, then

Proof: , divide this expression by, we get: - coprime numbers, which means they are divisible by an integer, i.e. =. And this means that what needed to be proven.

II . Applying comparisons to problem solving.

2.1. The simplest use of modulo comparisons is to determine the divisibility of numbers.

Example. Find the remainder of 2 2009 at 7.

Solution: Consider the powers of 2:

raising the comparison to the power of 668 and multiplying by, we get: .

Answer: 4.

Example. Prove that 7+7 2 +7 3 +…+7 4 n is divisible by 100 for anynfrom a set of integers.

Solution: Consider comparisons

etc. The cyclical nature of remainders is explained by the rules for multiplying numbers in a column. Adding up the first four comparisons, we get:

This means that this amount is divided by 100 without a remainder. Similarly, adding the following comparisons about four, we get that each such sum is divisible by 100 without a remainder. This means that the entire sum consisting of 4nterms is divisible by 100 without a remainder. Q.E.D.

Example. Determine at what valuenthe expression is divisible by 19 without a remainder.

Solution: .

Let's multiply this comparison by 20. We get.

Let's add up the comparisons, then. . Thus, the right side of the comparison is always divisible by 19 for any natural numbern, which means the original expression is divisible by 19 with naturaln.

Answer n – any natural number.

Example. What digit does the number end with?

Solution. To solve this problem, we will monitor only the last digit. Consider the powers of the number 14:

You can notice that if the exponent is odd, the value of the degree ends in 4, and if the exponent is even, it ends in 6. Then it ends in 6, i.e. is an even number. So it will end in 6.

Answer 6.

2.2. One comparison task.

The article by N. Vilenkin “Comparisons and classes of residues” presents a problem that was solved by the famous English physicist Dirac during his student years.

There is also a brief solution to this problem using modulo comparisons. But we encountered a number of similar problems. For example.

One passerby found a pile of apples near a tree on which a monkey was sitting. After counting them, he realized that if 1 apple is given to the monkey, then the number of remaining apples will be divided into n without a trace. Having given the extra apple to the monkey, he took 1/ n remaining apples and left. Later, the next passer-by approached the pile, then the next, etc. Each subsequent passer-by, having counted the apples, noticed that their number when divided by n gives the remainder 1 and, having given the extra apple to the monkey, he took 1 for himself n the remaining apples and moved on. After the last one left, n th passerby, the number of apples remaining in the pile was divided by n without a trace. How many apples were in the pile at first?

Carrying out the same reasoning as Dirac, we obtained a general formula for solving a class of similar problems: , wheren– natural number.

2.3. Application of module comparisons in professional activities.

Comparison theory is applied to coding theory, so all people who choose a profession related to computers will study, and possibly apply comparisons in their professional activities. For example, to develop public key encryption algorithms, use a whole series concepts of number theory, including modulo comparisons.

Conclusion.

The work outlines the basic concepts and properties of modulo comparisons and illustrates the use of modulo comparisons with examples. The material can be used in preparation for olympiads in mathematics and the Unified State Exam.

The following list of references allows, if necessary, to consider some more difficult moments theory of modulo comparisons and its applications.

List of used literature.

    Alfutova N.B. Algebra and number theory./N.B.Alfutova, A.V.Ustinov. M.:MCNMO, 2002, 466 p.

    Bukhshtab A.A. Number theory. /A.A.Bukhshtab. M.: Education, 1960.

    Vilenkin N. Comparisons and classes of residues./N. Vilenkin.//Quantum. – 1978.- 10.

    Fedorova N.E. Studying algebra and mathematical analysis. 10th grade.http:// www. prosv. ru/ ebooks/ Fedorova_ Algebra_10 kl/1/ xht

    ru. wikipedia. org/ wiki/Comparison_modulo.

Consider a comparison of the form x 2 ≡a(mod pα), where p– simple odd number. As was shown in paragraph 4 of §4, the solution to this comparison can be found by solving the comparison x 2 ≡a(mod p). Moreover, the comparison x 2 ≡a(mod pα) will have two solutions if a is a quadratic residue modulo p.

Example:

Solve quadratic comparison x 2 ≡86(mod 125).

125 = 5 3, 5 is a prime number. Let's check whether 86 is a square modulo 5.

The original comparison has 2 solutions.

Let's find a solution to the comparison x 2 ≡86(mod 5).

x 2 ≡1(mod 5).

This comparison could be solved in the manner indicated in the previous paragraph, but we will use the fact that the square root of 1 modulo is ±1, and the comparison has exactly two solutions. Thus, the solution to the comparison modulo 5 is

x≡±1(mod 5) or, otherwise, x=±(1+5 t 1).

Let's substitute the resulting solution into comparison modulo 5 2 =25:

x 2 ≡86(mod 25)

x 2 ≡11(mod 25)

(1+5t 1) 2 ≡11(mod 25)

1+10t 1 +25t 1 2 ≡11(mod 25)

10t 1 ≡10(mod 25)

2t 1 ≡2(mod 5)

t 1 ≡1(mod 5), or, what is the same, t 1 =1+5t 2 .

Then the solution to the comparison modulo 25 is x=±(1+5(1+5 t 2))=±(6+25 t 2). Let's substitute the resulting solution into comparison modulo 5 3 =125:

x 2 ≡86(mod 125)

(6+25t 2) 2 ≡86(mod 125)

36+12·25 t 2 +625t 2 2 ≡86(mod 125)

12·25 t 2 ≡50(mod 125)

12t 2 ≡2(mod 5)

2t 2 ≡2(mod 5)

t 2 ≡1(mod 5), or t 2 =1+5t 3 .

Then the solution to the comparison modulo 125 is x=±(6+25(1+5 t 3))=±(31+125 t 3).

Answer: x≡±31(mod 125).

Let us now consider a comparison of the form x 2 ≡a(mod 2 α). Such a comparison does not always have two solutions. For such a module the following cases are possible:

1) α=1. Then the comparison has a solution only when a≡1(mod 2), and the solution is x≡1(mod 2) (one solution).

2) α=2. Comparison has solutions only when a≡1(mod 4), and the solution is x≡±1(mod 4) (two solutions).

3) α≥3. Comparison has solutions only when a≡1(mod 8), and there will be four such solutions. Comparison x 2 ≡a(mod 2 α) for α≥3 is solved in the same way as comparisons of the form x 2 ≡a(mod pα), only solutions modulo 8 act as the initial solution: x≡±1(mod 8) and x≡±3(mod 8). They should be compared modulo 16, then modulo 32, etc. up to modulo 2 α.

Example:

Solve comparison x 2 ≡33(mod 64)

64=2 6 . Let's check whether the original comparison has a solution. 33≡1(mod 8), which means the comparison has 4 solutions.

Modulo 8 these solutions will be: x≡±1(mod 8) and x≡±3(mod 8), which can be represented as x=±(1+4 t 1). Let's substitute this expression for comparison modulo 16

x 2 ≡33(mod 16)

(1+4t 1) 2 ≡1(mod 16)

1+8t 1 +16t 1 2 ≡1(mod 16)

8t 1 ≡0 (mod 16)

t 1 ≡0 (mod 2)

Then the solution will take the form x=±(1+4 t 1)=±(1+4(0+2 t 2))=±(1+8 t 2). Let's substitute the resulting solution into comparison modulo 32:

x 2 ≡33(mod 32)

(1+8t 2) 2 ≡1(mod 32)

1+16t 2 +64t 2 2 ≡1(mod 32)

16t 2 ≡0 (mod 32)

t 2 ≡0 (mod 2)

Then the solution will take the form x=±(1+8 t 2) =±(1+8(0+2t 3)) =±(1+16 t 3). Let's substitute the resulting solution into comparison modulo 64:

x 2 ≡33(mod 64)

(1+16t 3) 2 ≡33(mod 64)

1+32t 3 +256t 3 2 ≡33(mod 64)

32t 3 ≡32 (mod 64)

t 3 ≡1 (mod 2)

Then the solution will take the form x=±(1+16 t 3) =±(1+16(1+2t 4)) =±(17+32 t 4). So, modulo 64, the original comparison has four solutions: x≡±17(mod 64)i x≡±49(mod 64).

Now let's look at the comparison general view: x 2 ≡a(mod m), (a,m)=1, - canonical expansion of the module m. According to the Theorem from paragraph 4 of §4, this comparison is equivalent to the system

If every comparison of this system is solvable, then the entire system is solvable. Having found the solution to each comparison of this system, we will obtain a system of comparisons of the first degree, solving which using the Chinese remainder theorem, we will obtain a solution to the original comparison. At the same time, the quantity various solutions the original comparison (if it is decidable) is 2 k, if α=1, 2 k+1 if α=2, 2 k+2 if α≥3.

Example:

Solve comparison x 2 ≡4(mod 21).

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 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.

At n they give the same remainders.

Equivalent formulations: a and b comparable in modulus n if their difference a - b is divisible by n, or if a can be represented as a = b + kn , Where k- some integer. For example: 32 and −10 are comparable modulo 7, since

The statement “a and b are comparable modulo n” is written as:

Modulo equality properties

The modulo comparison relation has the properties

Any two integers a And b comparable modulo 1.

In order for the numbers a And b were comparable in modulus n, it is necessary and sufficient that their difference is divisible by n.

If the numbers and are pairwise comparable in modulus n, then their sums and , as well as the products and are also comparable in modulus n.

If the numbers a And b comparable in modulus n, then their degrees a k And b k are also comparable in modulus n under any natural k.

If the numbers a And b comparable in modulus n, And n divided by m, That a And b comparable in modulus m.

In order for the numbers a And b were comparable in modulus n, presented in the form of its canonical decomposition into simple factors p i

necessary and sufficient to

The comparison relation is an equivalence relation and has many of the properties of ordinary equalities. For example, they can be added and multiplied: if

Comparisons, however, cannot, generally speaking, be divided by each other or by other numbers. Example: , however, reducing by 2, we get an erroneous comparison: . The abbreviation rules for comparisons are as follows.

You also cannot perform operations on comparisons if their moduli do not match.

Other properties:

Related definitions

Deduction classes

The set of all numbers comparable to a modulo n called deduction class a modulo n , and is usually denoted [ a] n or . Thus, the comparison is equivalent to the equality of residue classes [a] n = [b] n .

Since modulo comparison n is an equivalence relation on the set of integers, then the residue classes modulo n represent equivalence classes; their number is equal n. The set of all residue classes modulo n denoted by or.

The operations of addition and multiplication by induce corresponding operations on the set:

[a] n + [b] n = [a + b] n

With respect to these operations the set is a finite ring, and if n simple - finite field.

Deduction systems

The residue system allows you to perform arithmetic operations on a finite set of numbers without going beyond its limits. Full system of deductions modulo n is any set of n integers that are incomparable modulo n. Usually, the smallest non-negative residues are taken as a complete system of residues modulo n

0,1,...,n − 1

or the absolute smallest deductions consisting of numbers

,

in case of odd n and numbers

in case of even n .

Solving comparisons

Comparisons of the first degree

In number theory, cryptography and other fields of science, the problem of finding solutions to first-degree comparisons of the form often arises:

Solving such a comparison begins with calculating the gcd (a, m)=d. In this case, 2 cases are possible:

  • If b not a multiple d, then the comparison has no solutions.
  • If b multiple d, then the comparison has the only solution modulo m / d, or, what is the same, d modulo solutions m. In this case, as a result of reducing the original comparison by d the comparison is:

Where a 1 = a / d , b 1 = b / d And m 1 = m / d are integers, and a 1 and m 1 are relatively prime. Therefore the number a 1 can be inverted modulo m 1, that is, find such a number c, that (in other words, ). Now the solution is found by multiplying the resulting comparison by c:

Practical calculation of value c can be done in different ways: using Euler's theorem, Euclid's algorithm, the theory of continued fractions (see algorithm), etc. In particular, Euler's theorem allows you to write down the value c in the form:

Example

For comparison we have d= 2, so modulo 22 the comparison has two solutions. Let's replace 26 by 4, comparable to it modulo 22, and then reduce all 3 numbers by 2:

Since 2 is coprime to modulo 11, we can reduce the left and right sides by 2. As a result, we obtain one solution modulo 11: , equivalent to two solutions modulo 22: .

Comparisons of the second degree

Solving comparisons of the second degree comes down to finding out whether given number quadratic residue (using the quadratic reciprocity law) and subsequent calculation of the square root modulo.

Story

The Chinese remainder theorem, known for many centuries, states (in modern mathematical language) that the residue ring modulo the product of several is reciprocal prime numbers is