Comparisons by simple modulo. Modulo comparisons

Content.

Introduction

§1. Modulo comparison

§2. Comparison Properties

Module-Independent Comparison Properties
Module-dependent properties of comparisons

§3. Deduction system

Full system of deductions
Reduced system of deductions

§4. Euler's theorem and Fermat

Euler function
Euler's theorem and Fermat

Chapter 2. Theory of comparisons with a variable

§1. Basic concepts related to solving comparisons

The Roots of Comparisons
Equivalence of comparisons
Wilson's theorem

§2. First degree comparisons and their solutions

Selection method
Euler's methods
Euclid algorithm method
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

Deduction class order
Primitive roots modulo prime
Indexes modulo prime

Chapter 3. Application of the theory of comparisons

§1. Signs of divisibility

§2. Checking the results of arithmetic operations

§3. Converting 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 some studies 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, considers 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.

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

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 at 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).

(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.

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.

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.

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).

Generally speaking, it is impossible to divide both sides of a 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).

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, it should be noted 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 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

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:

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.
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.

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.

Each of the numbers in the collection (1) belongs to a certain class.
Any two numbers x i and x j from (1) are incomparable with each other, i.e., they belong to different classes.
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)

Each of the numbers in the collection (2) belongs to a certain class.
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.

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.

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 is 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.

All numbers from (1) are pairwise incomparable in modulus T, that is, they belong to different classes modulo m.
Each number from (1) is coprime with T, that is, all these numbers belong to different classes coprime to modulo m.
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.

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.

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.
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 m = 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 - power of a prime number 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- Rk-1= pk-l(p-1)numbers. This proves that

φ (RTo) = 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 modulotpasPXT -matrices

1 2 T

t +1 t +2 2t

………………………………

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

Because theTAndPare relatively prime, then the numberXreciprocally just withtpthen and only whenXreciprocally just withTAndXreciprocally just withP. 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 deductions moduloP.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:

Because the=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

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)

Consider a comparison of the form x 2 ≡a(mod pα), where p– a 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 a general comparison: 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. Moreover, the number of different solutions to the original comparison (if it is solvable) 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).

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 mod 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 a unique 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 implemented 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 as:

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 a given number is a quadratic residue (using the quadratic reciprocity law) and then calculating 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 coprime numbers is

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

Applying comparisons to problem solving______________________________6

The simplest use of modulo comparisons is to determine the divisibility of numbers___________________________6
One comparison task_______________________________8
The use of modulo comparisons in professional activities________________________________________________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?

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;

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 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.” It was Gauss who proposed the symbolism of modulo comparisons that has become established in mathematics.

Proof:

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

is the sum of two integers, therefore it 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 proved.

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 only monitor 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/ 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, a number of number theory concepts, including modulo comparisons, are used to develop public key encryption algorithms.

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 given list of references allows, if necessary, to consider some more complex aspects of the 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. Study of algebra and mathematical analysis. Grade 10.http:// www. prosv. ru/ ebooks/ Fedorova_ Algebra_10 kl/1/ xht

ru. wikipedia. org/ wiki/Comparison_modulo.

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: a bunch of (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 many 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, either 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| the power of the set is always denoted 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.

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