Pythagorean triples and their number. Modern science-intensive technologies Prime numbers as part of Pythagorean triples

"Regional center of education"

Methodical development

Using Pythagorean triples in solving

geometric problems and trigonometric tasks USE

Kaluga, 2016

I Introduction

The Pythagorean theorem is one of the main and, one might even say, the most important theorem of geometry. Its significance lies in the fact that most of the theorems of geometry can be deduced from it or with its help. The Pythagorean theorem is also remarkable in that in itself it is not at all obvious. For example, the properties of an isosceles triangle can be seen directly on the drawing. But no matter how you look at a right triangle, you will never see that there is such a simple ratio between its sides: a2+b2=c2. However, it was not Pythagoras who discovered the theorem that bears his name. It was known even earlier, but perhaps only as a fact derived from measurements. Presumably, Pythagoras knew this, but found proof.

There are an infinite number of natural numbers a, b, c, satisfying the relation a2+b2=c2.. They are called Pythagorean numbers. According to the Pythagorean theorem, such numbers can serve as the lengths of the sides of some right-angled triangle - we will call them Pythagorean triangles.

Goal of the work: to study the possibility and effectiveness of using Pythagorean triples for solving problems of a school mathematics course, USE assignments.

Based on the purpose of the work, the following tasks:

To study the history and classification of Pythagorean triples. Analyze tasks using Pythagorean triples that are available in school textbooks and found in the control and measuring materials of the exam. Evaluate the effectiveness of using Pythagorean triples and their properties for solving problems.

Object of study: Pythagorean triples of numbers.

Subject of study: tasks of the school course of trigonometry and geometry, in which Pythagorean triples are used.

The relevance of research. Pythagorean triples are often used in geometry and trigonometry, knowing them will eliminate errors in calculations and save time.

II. Main part. Solving problems using Pythagorean triples.

2.1. Table of triples of Pythagorean numbers (according to Perelman)

Pythagorean numbers have the form a= m n, , where m and n are some coprime odd numbers.

Pythagorean numbers have a number of interesting features:

One of the "legs" must be a multiple of three.

One of the "legs" must be a multiple of four.

One of the Pythagorean numbers must be a multiple of five.

The book "Entertaining Algebra" contains a table of Pythagorean triples containing numbers up to one hundred, which do not have common factors.

		32+42=52
		52+122=132
		72+242=252
		92+402=412
		112+602=612
		132+842=852
		152+82=172
		212 +202=292
		332+562=652
		392+802=892
		352+122=372
		452+282=532
		552+482=732
		652+722=972
		632+162=652
		772+362=852

2.2. Shustrov's classification of Pythagorean triples.

Shustrov discovered the following pattern: if all Pythagorean triangles are divided into groups, then the following formulas are valid for the odd leg x, even y and hypotenuse z:

x \u003d (2N-1) (2n + 2N-1); y = 2n (n+2N-1); z = 2n (n+2N-1)+(2N-1) 2, where N is the number of the family and n is the ordinal number of the triangle in the family.

Substituting in the formula in place of N and n any positive integers, starting from one, you can get all the main Pythagorean triples of numbers, as well as multiples of a certain type. You can make a table of all Pythagorean triples for each family.

2.3. Planimetry tasks

Let's consider problems from various textbooks on geometry and find out how often Pythagorean triples are found in these tasks. Trivial problems of finding the third element in the table of Pythagorean triples will not be considered, although they are also found in textbooks. Let us show how to reduce the solution of a problem whose data is not expressed by natural numbers to Pythagorean triples.

Consider tasks from a geometry textbook for grades 7-9.

№ 000. Find the hypotenuse of a right triangle A=, b=.

Solution. Multiply the lengths of the legs by 7, we get two elements from the Pythagorean triple 3 and 4. The missing element is 5, which we divide by 7. Answer.

№ 000. In rectangle ABCD find BC if CD=1.5, AC=2.5.

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Solution. Let's solve right triangle ACD. We multiply the lengths by 2, we get two elements from the Pythagorean triple 3 and 5, the missing element is 4, which we divide by 2. Answer: 2.

When solving the next number, check the ratio a2+b2=c2 it is completely optional, it is enough to use Pythagorean numbers and their properties.

№ 000. Find out if a triangle is right-angled if its sides are expressed by numbers:

a) 6,8,10 (Pythagorean triple 3,4.5) - yes;

One of the legs of a right triangle must be divisible by 4. Answer: no.

c) 9,12,15 (Pythagorean triple 3,4.5) - yes;

d) 10,24,26 (Pythagorean triple 5,12.13) - yes;

One of the Pythagorean numbers must be a multiple of five. Answer: no.

g) 15, 20, 25 (Pythagorean triple 3,4.5) - yes.

Of the thirty-nine tasks in this section (Pythagorean theorem), twenty-two are solved orally using Pythagorean numbers and knowledge of their properties.

Consider problem #000 (from the "Additional Tasks" section):

Find the area of ​​quadrilateral ABCD where AB=5 cm, BC=13 cm, CD=9 cm, DA=15 cm, AC=12 cm.

The task is to check the ratio a2+b2=c2 and prove that the given quadrilateral consists of two right triangles (the inverse theorem). And the knowledge of Pythagorean triples: 3, 4, 5 and 5, 12, 13, eliminates the need for calculations.

Let's give solutions to several problems from a textbook on geometry for grades 7-9.

Problem 156 (h). The legs of a right triangle are 9 and 40. Find the median drawn to the hypotenuse.

Solution . The median drawn to the hypotenuse is equal to half of it. The Pythagorean triple is 9.40 and 41. Therefore, the median is 20.5.

Problem 156 (i). The sides of the triangle are: A= 13 cm, b= 20 cm and height hс = 12 cm. Find the base With.

Task (KIM USE). Find the radius of a circle inscribed in an acute triangle ABC if the height BH is 12 and it is known that sin A=,sin C \u003d left "\u003e

Solution. We solve rectangular ∆ ASC: sin A=, BH=12, hence AB=13,AK=5 (Pythagorean triple 5,12,13). Solve rectangular ∆ BCH: BH =12, sin C===https://pandia.ru/text/80/406/images/image015_0.gif" width="12" height="13">3=9 (Pythagorean triple 3,4,5).The radius is found by the formula r === 4. Answer.4.

2.4. Pythagorean triples in trigonometry

The main trigonometric identity is a special case of the Pythagorean theorem: sin2a + cos2a = 1; (a/c) 2 + (b/c)2 =1. Therefore, some trigonometric tasks are easily solved orally using Pythagorean triples.

Problems in which it is required to find the values ​​of other trigonometric functions from a given value of a function can be solved without squaring and extracting a square root. All tasks of this type in the school textbook of algebra (10-11) Mordkovich (No. 000-No. 000) can be solved orally, knowing only a few Pythagorean triples: 3,4,5 ; 5,12,13 ; 8,15,17 ; 7,24,25 . Let's consider the solutions of two tasks.

No. 000 a). sin t = 4/5, π/2< t < π.

Solution. Pythagorean triple: 3, 4, 5. Therefore, cos t = -3/5; tg t = -4/3,

No. 000 b). tg t = 2.4, π< t < 3π/2.

Solution. tg t \u003d 2.4 \u003d 24/10 \u003d 12/5. Pythagorean triple 5,12,13. Given the signs, we get sin t = -12/13, cos t = -5/13, ctg t = 5/12.

3. Control and measuring materials of the exam

a) cos (arcsin 3/5)=4/5 (3, 4, 5)

b) sin (arccos 5/13)=12/13 (5, 12, 13)

c) tg (arcsin 0.6)=0.75 (6, 8, 10)

d) ctg (arccos 9/41) = 9/40 (9, 40, 41)

e) 4/3 tg (π–arcsin (–3/5))= 4/3 tg (π+arcsin 3/5)= 4/3 tg arcsin 3/5=4/3 3/4=1

e) check the validity of the equality:

arcsin 4/5 + arcsin 5/13 + arcsin 16/65 = π/2.

Solution. arcsin 4/5 + arcsin 5/13 + arcsin 16/65 = π/2

arcsin 4/5 + arcsin 5/13 = π/2 - arcsin 16/65

sin (arcsin 4/5 + arcsin 5/13) = sin (arccos 16/65)

sin (arcsin 4/5) cos (arcsin 5/13) + cos (arcsin 4/5) sin (arcsin 5/13) = 63/65

4/5 12/13 + 3/5 5/13 = 63/65

III. Conclusion

In geometric problems, one often has to solve right triangles, sometimes several times. After analyzing the tasks of school textbooks and USE materials, we can conclude that triplets are mainly used: 3, 4, 5; 5, 12, 13; 7, 24, 25; 9, 40, 41; 8,15,17; which are easy to remember. When solving some trigonometric tasks, the classic solution using trigonometric formulas and a large number of calculations takes time, and knowledge of Pythagorean triples will eliminate errors in calculations and save time for solving more difficult problems on the exam.

Bibliographic list

1. Algebra and the beginnings of analysis. 10-11 grades. At 2 hours. Part 2. A task book for educational institutions / [and others]; ed. . - 8th ed., Sr. - M. : Mnemosyne, 2007. - 315 p. : ill.

2. Perelman algebra. - D.: VAP, 1994. - 200 p.

3. Roganovsky: Proc. For 7-9 cells. with a deep the study of mathematics general education. school from Russian lang. learning, - 3rd ed. - Mn.; Nar. Asveta, 2000. - 574 p.: ill.

4. Mathematics: Reader on history, methodology, didactics. / Comp. . - M.: Publishing house of URAO, 2001. - 384 p.

5. Journal "Mathematics at School" No. 1, 1965.

6. Control and measuring materials of the exam.

7. Geometry, 7-9: Proc. for educational institutions /, etc. - 13th ed. - M .: Education, 2003. – 384 p. : ill.

8. Geometry: Proc. for 10-11 cells. avg. school /, etc. - 2nd ed. - M .: Education, 1993, - 207 p.: ill.

Perelman algebra. - D.: VAP, 1994. - 200 p.

Journal "Mathematics at School" No. 1, 1965.

Geometry, 7-9: Proc. for educational institutions /, etc. - 13th ed. - M .: Education, 2003. – 384 p. : ill.

Roganovsky: Proc. For 7-9 cells. with a deep the study of mathematics general education. school from Russian lang. learning, - 3rd ed. - Mn.; Nar. Asveta, 2000. - 574 p.: ill.

Algebra and the beginnings of analysis. 10-11 grades. At 2 hours. Part 2. A task book for educational institutions / [and others]; ed. . - 8th ed., Sr. - M. : Mnemosyne, 2007. - 315 p. : ill., p.18.

Belotelov V.A. Pythagorean triples and their number // Encyclopedia of the Nesterovs

This article is an answer to one professor - a pincher. Look, professor, how they do it in our village.

Nizhny Novgorod region, Zavolzhye.

Knowledge of the algorithm for solving Diophantine equations (ADDE) and knowledge of polynomial progressions is required.

IF is a prime number.

MF is a composite number.

Let there be an odd number N. For any odd number other than one, you can write an equation.

p 2 + N \u003d q 2,

where р + q = N, q – р = 1.

For example, for the numbers 21 and 23, the equations would be, -

10 2 + 21 = 11 2 , 11 2 + 23 = 12 2 .

If N is prime, this equation is unique. If the number N is composite, then it is possible to compose similar equations for the number of pairs of factors representing this number, including 1 x N.

Let's take the number N = 45, -

1 x 45 = 45, 3 x 15 = 45, 5 x 9 = 45.

I dreamed, but is it possible, clinging to this difference between the IF and the MF, to find a method for their identification.

Let us introduce the notation;

Let's change the lower equation, -

N \u003d in 2 - a 2 \u003d (b - a) (b + a).

Let us group the values ​​of N according to the criterion in - a, i.e. let's make a table.

The numbers N were summarized in a matrix, -

It was for this task that I had to deal with the progressions of polynomials and their matrices. Everything turned out to be in vain - the PCh defenses are powerfully held. Let's enter a column in table 1, where in - a \u003d 1 (q - p \u003d 1).

Once again. Table 2 was obtained as a result of an attempt to solve the problem of identifying the IF and MF. It follows from the table that for any number N, there are as many equations of the form a 2 + N \u003d in 2, into how many pairs of factors the number N can be divided, including the factor 1 x N. In addition to the numbers N \u003d ℓ 2, where

ℓ - FC. For N = ℓ 2 , where ℓ is IF, there is a unique equation p 2 + N = q 2 . What additional proof can we talk about if the table lists smaller factors from pairs of factors forming N, from one to ∞. We will place Table 2 in a chest, and hide the chest in a closet.

Let's return to the topic stated in the title of the article.

This article is an answer to one professor - a pincher.

I asked for help - I needed a series of numbers that I could not find on the Internet. I ran into questions like, - "what for?", "But show me the method." In particular, there was a question of whether the series of Pythagorean triples is infinite, "how to prove it?". He didn't help me. Look, professor, how they do it in our village.

Let's take the formula of Pythagorean triples, -

x 2 \u003d y 2 + z 2. (1)

Let's pass through ARDU.

Three situations are possible:

I. x is an odd number,

y is an even number

z is an even number.

And there is a condition x > y > z.

II. x is an odd number

y is an even number

z is an odd number.

x > z > y.

III.x - an even number,

y is an odd number

z is an odd number.

x > y > z.

Let's start with I.

Let's introduce new variables

Substitute into equation (1).

Let us cancel by the smaller variable 2γ.

(2α - 2γ + 2k + 1) 2 = (2β - 2γ + 2k) 2 + (2k + 1) 2 .

Let us reduce the variable 2β – 2γ by a smaller one with the simultaneous introduction of a new parameter ƒ, -

(2α - 2β + 2ƒ + 2k + 1) 2 = (2ƒ + 2k) 2 + (2k + 1) 2 (2)

Then, 2α - 2β = x - y - 1.

Equation (2) will take the form, –

(x - y + 2ƒ + 2k) 2 \u003d (2ƒ + 2k) 2 + (2k + 1) 2

Let's square it -

(x - y) 2 + 2 (2ƒ + 2k) (x - y) + (2ƒ + 2k) 2 \u003d (2ƒ + 2k) 2 + (2k + 1) 2,

(x - y) 2 + 2(2ƒ + 2k)(x - y) - (2k + 1) 2 = 0. (3)

ARDU gives through the parameters the relationship between the senior terms of the equation, so we got equation (3).

It is not solid to deal with the selection of solutions. But, firstly, there is nowhere to go, and secondly, several of these solutions are needed, and we can restore an infinite number of solutions.

For ƒ = 1, k = 1, we have x – y = 1.

With ƒ = 12, k = 16, we have x - y = 9.

With ƒ = 4, k = 32, we have x - y = 25.

You can pick it up for a long time, but in the end the series will take the form -

x - y \u003d 1, 9, 25, 49, 81, ....

Consider option II.

Let us introduce new variables into equation (1)

(2α + 2k + 1) 2 = (2β + 2k) 2 + (2γ + 2k + 1) 2 .

We reduce by a smaller variable 2 β, -

(2α - 2β + 2k + 1) 2 = (2α - 2β + 2k+1) 2 + (2k) 2 .

Let us reduce by the smaller variable 2α – 2β, –

(2α - 2γ + 2ƒ + 2k + 1) 2 = (2ƒ + 2k + 1) 2 + (2k) 2 . (4)

2α - 2γ = x - z and substitute into equation (4).

(x - z + 2ƒ + 2k + 1) 2 = (2ƒ + 2k + 1) 2 + (2k) 2

(x - z) 2 + 2 (2ƒ + 2k + 1) (x - z) + (2ƒ + 2k + 1) 2 = (2ƒ + 2k + 1) 2 + (2k) 2 (x - z) 2 + 2(2ƒ + 2k + 1) (x - z) - (2k) 2 = 0

With ƒ = 3, k = 4, we have x - z = 2.

With ƒ = 8, k = 14, we have x - z = 8.

With ƒ = 3, k = 24, we have x - z = 18.

x - z \u003d 2, 8, 18, 32, 50, ....

Let's draw a trapezoid -

Let's write a formula.

where n=1, 2,...∞.

Case III will not be described - there are no solutions there.

For condition II, the set of triples will be as follows:

Equation (1) is presented as x 2 = z 2 + y 2 for clarity.

For condition I, the set of triples will be as follows:

In total, 9 columns of triples are painted, five triples in each. And each of the presented columns can be written up to ∞.

As an example, consider the triples of the last column, where x - y \u003d 81.

For the values ​​of x, we write a trapezoid, -

Let's write the formula

For the values ​​\u200b\u200bof we write a trapezoid, -

Let's write the formula

For the values ​​of z, we write a trapezoid, -

Let's write the formula

Where n = 1 ÷ ∞.

As promised, a series of triplets with x - y = 81 flies to ∞.

There was an attempt for cases I and II to construct matrices for x, y, z.

Write out the last five columns of x from the top rows and build a trapezoid.

It did not work, and the pattern should be quadratic. To make everything in openwork, it turned out that it was necessary to combine columns I and II.

In case II, the quantities y, z are again interchanged.

We managed to merge for one reason - the cards fit well in this task - we were lucky.

Now you can write matrices for x, y, z.

Let's take from the last five columns of the x value from the top rows and build a trapezoid.

Everything is fine, you can build matrices, and let's start with a matrix for z.

I run to the closet for a chest.

Total: In addition to one, each odd number of the numerical axis participates in the formation of Pythagorean triples by an equal number of pairs of factors forming this number N, including the factor 1 x N.

The number N \u003d ℓ 2, where ℓ - IF, forms one Pythagorean triple, if ℓ is MF, then there is no triple on the factors ℓхℓ.

Let's build matrices for x, y.

Let's start with the matrix for x. To do this, we will pull on it the coordinate grid from the problem of identifying the IF and MF.

The numbering of vertical rows is normalized by the expression

Let's remove the first column, because

The matrix will take the form -

Let's describe the vertical rows, -

Let us describe the coefficients at "a", -

Let's describe the free members, -

Let's make a general formula for "x", -

If we do a similar job for "y", we get -

You can approach this result from the other side.

Let's take the equation,

and 2 + N = in 2 .

Let's change it a bit -

N \u003d in 2 - a 2.

Let's square it -

N 2 \u003d in 4 - 2v 2 a 2 + a 4.

To the left and right sides of the equation, add in magnitude 4v 2 a 2, -

N 2 + 4v 2 a 2 \u003d in 4 + 2v 2 a 2 + a 4.

And finally -

(in 2 + a 2) 2 \u003d (2va) 2 + N 2.

Pythagorean triples are composed as follows:

Consider an example with the number N = 117.

1 x 117 = 117, 3 x 39 = 117, 9 x 13 = 117.

The vertical columns of Table 2 are numbered with values ​​in - a, while the vertical columns of Table 3 are numbered with values ​​x - y.

x - y \u003d (c - a) 2,

x \u003d y + (c - a) 2.

Let's make three equations.

(y + 1 2) 2 \u003d y 2 + 117 2,

(y + 3 2) 2 \u003d y 2 + 117 2,

(y + 9 2) 2 \u003d y 2 + 117 2.

x 1 = 6845, y 1 = 6844, z 1 = 117.

x 2 = 765, y 2 = 756, z 2 = 117 (x 2 = 85, y 2 = 84, z 2 = 13).

x 3 = 125, y 3 = 44, z 3 = 117.

Factors 3 and 39 are not relatively prime numbers, so one triple turned out with a factor of 9.

Let us depict the above written in general symbols, -

In this work, everything, including an example for calculating Pythagorean triples with the number

N = 117, tied to the smaller factor in - a. Explicit discrimination in relation to the factor in + a. Let's correct this injustice - we will compose three equations with a factor in + a.

Let's return to the question of identification of IF and MF.

A lot of things have been done in this direction, and today the following thought has come through the hands - there is no identification equation, and there is no such thing as to determine the factors.

Suppose we have found the relation F = a, b (N).

There is a formula

You can get rid of in the formula F from in and you get a homogeneous equation of the nth degree with respect to a, i.e. F = a(N).

For any degree n of this equation, there is a number N with m pairs of factors, for m > n.

And as a consequence, a homogeneous equation of degree n must have m roots.

Yes, this cannot be.

In this paper, the numbers N were considered for the equation x 2 = y 2 + z 2 when they are in the equation at the place z. When N is in place of x, this is another task.

Sincerely, Belotelov V.A.

Next, we consider the well-known methods for generating effective Pythagorean triples. The students of Pythagoras were the first to devise a simple way to generate Pythagorean triples, using a formula whose parts represent a Pythagorean triple:

m 2 + ((m 2 − 1)/2) 2 = ((m 2 + 1)/2) 2 ,

Where m- unpaired, m>2. Really,

4m 2 + m 4 − 2m 2 + 1
m 2 + ((m 2 − 1)/2) 2 = ————————— = ((m 2 + 1)/2) 2 .
4

A similar formula was proposed by the ancient Greek philosopher Plato:

(2m) 2 + (m 2 − 1) 2 = (m 2 + 1) 2 ,

Where m- any number. For m= 2,3,4,5 the following triplets are generated:

(16,9,25), (36,64,100), (64,225,289), (100,576,676).

As you can see, these formulas cannot give all possible primitive triples.

Consider the following polynomial, which is decomposed into a sum of polynomials:

(2m 2 + 2m + 1) 2 = 4m 4 + 8m 3 + 8m 2 + 4m + 1 =
=4m 4 + 8m 3 + 4m 2 + 4m 2 + 4m + 1 = (2m(m+1)) 2 + (2m +1) 2 .

Hence the following formulas for obtaining primitive triples:

a = 2m +1 , b = 2m(m+1) = 2m 2 + 2m , c = 2m 2 + 2m + 1.

These formulas generate triples in which the average number differs from the largest by exactly one, that is, not all possible triples are also generated. Here the first triples are: (5,12,13), (7,24,25), (9,40,41), (11,60,61).

To determine how to generate all primitive triples, one must examine their properties. First, if ( a,b,c) is a primitive triple, then a And b, b And c, A And c— must be coprime. Let a And b are divided into d. Then a 2 + b 2 is also divisible by d. Respectively, c 2 and c should be divided into d. That is, it is not a primitive triple.

Secondly, among the numbers a, b one must be paired and the other unpaired. Indeed, if a And b- paired, then With will be paired, and the numbers can be divided by at least 2. If they are both unpaired, then they can be represented as 2 k+1 i 2 l+1, where k,l- some numbers. Then a 2 + b 2 = 4k 2 +4k+1+4l 2 +4l+1, that is, With 2 , as well as a 2 + b 2 has a remainder of 2 when divided by 4.

Let With- any number, that is With = 4k+i (i=0,…,3). Then With 2 = (4k+i) 2 has a remainder of 0 or 1 and cannot have a remainder of 2. Thus, a And b cannot be unpaired, that is a 2 + b 2 = 4k 2 +4k+4l 2 +4l+1 and remainder With 2 by 4 should be 1, which means that With should be unpaired.

Such requirements for the elements of the Pythagorean triple are satisfied by the following numbers:

a = 2mn, b = m 2 − n 2 , c = m 2 + n 2 , m > n, (2)

Where m And n are coprime with different pairings. For the first time, these dependencies became known from the works of Euclid, who lived 2300 r. back.

Let us prove the validity of dependencies (2). Let A- double, then b And c- unpaired. Then c + b i c − b- couples. They can be represented as c + b = 2u And c − b = 2v, Where u,v are some integers. That's why

a 2 = With 2 − b 2 = (c + b)(c − b) = 2u 2 v = 4UV

And therefore ( a/2) 2 = UV.

It can be proven by contradiction that u And v are coprime. Let u And v- are divided into d. Then ( c + b) And ( c − b) are divided into d. And therefore c And b should be divided into d, and this contradicts the condition for the Pythagorean triple.

Because UV = (a/2) 2 and u And v coprime, it is easy to prove that u And v must be squares of some numbers.

So there are positive integers m And n, such that u = m 2 and v = n 2. Then

A 2 = 4UV = 4m 2 n 2 so
A = 2mn; b = u − v = m 2 − n 2 ; c = u + v = m 2 + n 2 .

Because b> 0, then m > n.

It remains to show that m And n have different pairings. If m And n- paired, then u And v must be paired, but this is impossible, since they are coprime. If m And n- unpaired, then b = m 2 − n 2 and c = m 2 + n 2 would be paired, which is impossible because c And b are coprime.

Thus, any primitive Pythagorean triple must satisfy conditions (2). At the same time, the numbers m And n called generating numbers primitive triplets. For example, let's have a primitive Pythagorean triple (120,119,169). In this case

A= 120 = 2 12 5, b= 119 = 144 − 25, and c = 144+25=169,

Where m = 12, n= 5 - generating numbers, 12 > 5; 12 and 5 are coprime and of different pairings.

It can be proved that the numbers m, n formulas (2) give a primitive Pythagorean triple (a,b,c). Really,

A 2 + b 2 = (2mn) 2 + (m 2 − n 2) 2 = 4m 2 n 2 + (m 4 − 2m 2 n 2 + n 4) =
= (m 4 + 2m 2 n 2 + n 4) = (m 2 + n 2) 2 = c 2 ,

That is ( a,b,c) is a Pythagorean triple. Let us prove that while a,b,c are coprime numbers by contradiction. Let these numbers be divided by p> 1. Since m And n have different pairings, then b And c- unpaired, that is p≠ 2. Since R divides b And c, That R must divide 2 m 2 and 2 n 2 , which is impossible because p≠ 2. Therefore m, n are coprime and a,b,c are also coprime.

Table 1 shows all primitive Pythagorean triples generated by formulas (2) for m≤10.

Table 1. Primitive Pythagorean triples for m≤10

m	n	a	b	c	m	n	a	b	c
2	1	4	3	5	8	1	16	63	65
3	2	12	5	13	8	3	48	55	73
4	1	8	15	17	8	5	80	39	89
4	3	24	7	25	8	7	112	15	113
5	2	20	21	29	9	2	36	77	85
5	4	40	9	41	9	4	72	65	97
6	1	12	35	37	9	8	144	17	145
6	5	60	11	61	10	1	20	99	101
7	2	28	45	53	10	3	60	91	109
7	4	56	33	65	10	7	140	51	149
7	6	84	13	85	10	9	180	19	181

Analysis of this table shows the presence of the following series of patterns:

• or a, or b are divided by 3;
• one of the numbers a,b,c is divisible by 5;
• number A is divisible by 4;
• work a· b is divisible by 12.

In 1971, the American mathematicians Teigan and Hedwin proposed such little-known parameters of a right-angled triangle as its height (height) to generate triplets h = c− b and excess (success) e = a + b − c. In Fig.1. these quantities are shown on a certain right triangle.

Figure 1. Right triangle and its growth and excess

The name "excess" is derived from the fact that this is the additional distance that must be passed along the legs of the triangle from one vertex to the opposite, if you do not go along its diagonal.

Through excess and growth, the sides of the Pythagorean triangle can be expressed as:

e 2 e 2
a = h + e, b = e + ——, c = h + e + ——, (3)
2h 2h

Not all combinations h And e may correspond to Pythagorean triangles. For a given h possible values e is the product of some number d. This number d is called growth and refers to h in the following way: d is the smallest positive integer whose square is divisible by 2 h. Because e multiple d, then it is written as e = kd, Where k is a positive integer.

With the help of pairs ( k,h) you can generate all Pythagorean triangles, including non-primitive and generalized, as follows:

(dk) 2 (dk) 2
a = h + dk, b = dk + ——, c = h + dk + ——, (4)
2h 2h

Moreover, a triple is primitive if k And h are coprime and if h =± q 2 at q- unpaired.
Moreover, it will be exactly a Pythagorean triple if k> √2 h/d And h > 0.

To find k And h from ( a,b,c) do the following:

• h = c − b;
• write down h How h = pq 2 , where p> 0 and such that is not a square;
• d = 2pq If p- unpaired and d = pq, if p is paired;
• k = (a − h)/d.

For example, for the triple (8,15,17) we have h= 17−15 = 2 1, so p= 2 and q = 1, d= 2, and k= (8 − 2)/2 = 3. So this triple is given as ( k,h) = (3,2).

For the triple (459,1260,1341) we have h= 1341 − 1260 = 81, so p = 1, q= 9 and d= 18, hence k= (459 − 81)/18 = 21, so the code of this triple is ( k,h) = (21, 81).

Specifying triples with h And k has a number of interesting properties. Parameter k equals

k = 4S/(dP), (5)

Where S = ab/2 is the area of ​​the triangle, and P = a + b + c is its perimeter. This follows from the equality eP = 4S, which comes from the Pythagorean theorem.

For a right triangle e equals the diameter of the circle inscribed in the triangle. This comes from the fact that the hypotenuse With = (A − r)+(b − r) = a + b − 2r, Where r is the radius of the circle. From here h = c − b = A − 2r And e = a − h = 2r.

For h> 0 and k > 0, k is the ordinal number of triplets a-b-c in a sequence of Pythagorean triangles with increasing h. From table 2, which shows several options for triplets generated by pairs h, k, it can be seen that with increasing k the sides of the triangle increase. Thus, unlike classical numbering, numbering in pairs h, k has a higher order in sequences of triplets.

Table 2. Pythagorean triples generated by pairs h, k.

h	k	a	b	c	h	k	a	b	c
2	1	4	3	5	3	1	9	12	15
2	2	6	8	10	3	2	15	36	39
2	3	8	15	17	3	3	21	72	75
2	4	10	24	26	3	4	27	120	123
2	5	12	35	37	3	5	33	180	183

For h > 0, d satisfies the inequality 2√ h ≤ d ≤ 2h, in which the lower bound is reached at p= 1, and the upper one, at q= 1. Therefore, the value d with respect to 2√ h is a measure of how much h far from the square of some number.

Properties

Since the equation x 2 + y 2 = z 2 homogeneous, when multiplied x , y And z for the same number you get another Pythagorean triple. The Pythagorean triple is called primitive, if it cannot be obtained in this way, that is - relatively prime numbers.

Examples

Some Pythagorean triples (sorted in ascending order of maximum number, primitive ones are highlighted):

(3, 4, 5), (6, 8, 10), (5, 12, 13), (9, 12, 15), (8, 15, 17), (12, 16, 20), (15, 20, 25), (7, 24, 25), (10, 24, 26), (20, 21, 29), (18, 24, 30), (16, 30, 34), (21, 28, 35), (12, 35, 37), (15, 36, 39), (24, 32, 40), (9, 40, 41), (14, 48, 50), (30, 40, 50)…

Based on the properties of Fibonacci numbers, you can make them, for example, such Pythagorean triples:

.

Story

Pythagorean triples have been known for a very long time. In the architecture of ancient Mesopotamian tombstones, an isosceles triangle is found, made up of two rectangular ones with sides of 9, 12 and 15 cubits. The pyramids of Pharaoh Snefru (XXVII century BC) were built using triangles with sides of 20, 21 and 29, as well as 18, 24 and 30 tens of Egyptian cubits.

see also

Links

• E. A. Gorin Powers of prime numbers in Pythagorean triples // Mathematical education. - 2008. - V. 12. - S. 105-125.

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See what "Pythagorean numbers" are in other dictionaries:

Triples of natural numbers such that a triangle whose side lengths are proportional (or equal) to these numbers is right-angled, e.g. triple of numbers: 3, 4, 5… Big Encyclopedic Dictionary

Triples of natural numbers such that a triangle whose side lengths are proportional (or equal) to these numbers is rectangular, for example, a triple of numbers: 3, 4, 5. * * * PYTHAGORAN NUMBERS PYTHAGORAN NUMBERS, triples of natural numbers such that ... ... encyclopedic Dictionary

Triples of natural numbers such that a triangle whose side lengths are proportional (or equal) to these numbers is a right triangle. According to the theorem, the inverse of the Pythagorean theorem (see Pythagorean theorem), for this it is enough that they ... ...

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In mathematics, Pythagorean numbers (Pythagorean triple) is a tuple of three integers that satisfy the Pythagorean relation: x2 + y2 = z2. Contents 1 Properties 2 Examples ... Wikipedia

Curly numbers are the general name of numbers associated with a particular geometric figure. This historical concept goes back to the Pythagoreans. Presumably, the expression “Square or cube” arose from curly numbers. Contents ... ... Wikipedia

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Books

• Archimedean summer, or the history of the community of young mathematicians. Binary number system, Bobrov Sergey Pavlovich. Binary number system, "Tower of Hanoi", knight's move, magic squares, arithmetic triangle, curly numbers, combinations, concept of probabilities, Möbius strip and Klein bottle.…

» Honored Professor of Mathematics at the University of Warwick, a well-known popularizer of science Ian Stewart, dedicated to the role of numbers in the history of mankind and the relevance of their study in our time.

Pythagorean hypotenuse

Pythagorean triangles have a right angle and integer sides. In the simplest of them, the longest side has a length of 5, the rest are 3 and 4. There are 5 regular polyhedra in total. A fifth-degree equation cannot be solved with fifth-degree roots - or any other roots. Lattices in the plane and in three-dimensional space do not have a five-lobe rotational symmetry; therefore, such symmetries are also absent in crystals. However, they can be in lattices in four-dimensional space and in interesting structures known as quasicrystals.

Hypotenuse of the smallest Pythagorean triple

The Pythagorean theorem states that the longest side of a right triangle (the notorious hypotenuse) correlates with the other two sides of this triangle in a very simple and beautiful way: the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Traditionally, we call this theorem after Pythagoras, but in fact its history is rather vague. Clay tablets suggest that the ancient Babylonians knew the Pythagorean theorem long before Pythagoras himself; the glory of the discoverer was brought to him by the mathematical cult of the Pythagoreans, whose supporters believed that the universe was based on numerical patterns. Ancient authors attributed to the Pythagoreans - and hence to Pythagoras - a variety of mathematical theorems, but in fact we have no idea what kind of mathematics Pythagoras himself was engaged in. We don't even know if the Pythagoreans could prove the Pythagorean Theorem, or if they simply believed it was true. Or, more likely, they had convincing data about its truth, which nevertheless would not have been enough for what we consider proof today.

Evidence of Pythagoras

The first known proof of the Pythagorean theorem is found in Euclid's Elements. This is a rather complicated proof using a drawing that Victorian schoolchildren would immediately recognize as "Pythagorean pants"; the drawing really resembles underpants drying on a rope. Literally hundreds of other proofs are known, most of which make the assertion more obvious.

// Rice. 33. Pythagorean pants

One of the simplest proofs is a kind of mathematical puzzle. Take any right triangle, make four copies of it and collect them inside the square. With one laying, we see a square on the hypotenuse; with the other - squares on the other two sides of the triangle. It is clear that the areas in both cases are equal.

// Rice. 34. Left: square on the hypotenuse (plus four triangles). Right: the sum of the squares on the other two sides (plus the same four triangles). Now eliminate the triangles

The dissection of Perigal is another puzzle piece of evidence.

// Rice. 35. Dissection of Perigal

There is also a proof of the theorem using stacking squares on the plane. Perhaps this is how the Pythagoreans or their unknown predecessors discovered this theorem. If you look at how the oblique square overlaps the other two squares, you can see how to cut the large square into pieces and then put them together into two smaller squares. You can also see right-angled triangles, the sides of which give the dimensions of the three squares involved.

// Rice. 36. Proof by paving

There are interesting proofs using similar triangles in trigonometry. At least fifty different proofs are known.

Pythagorean triplets

In number theory, the Pythagorean theorem became the source of a fruitful idea: to find integer solutions to algebraic equations. A Pythagorean triple is a set of integers a, b and c such that

Geometrically, such a triple defines a right triangle with integer sides.

The smallest hypotenuse of a Pythagorean triple is 5.

The other two sides of this triangle are 3 and 4. Here

32 + 42 = 9 + 16 = 25 = 52.

The next largest hypotenuse is 10 because

62 + 82 = 36 + 64 = 100 = 102.

However, this is essentially the same triangle with doubled sides. The next largest and truly different hypotenuse is 13, for which

52 + 122 = 25 + 144 = 169 = 132.

Euclid knew that there were an infinite number of different variations of Pythagorean triples, and he gave what might be called a formula for finding them all. Later, Diophantus of Alexandria offered a simple recipe, basically the same as Euclidean.

Take any two natural numbers and calculate:

their double product;

difference of their squares;

the sum of their squares.

The three resulting numbers will be the sides of the Pythagorean triangle.

Take, for example, the numbers 2 and 1. Calculate:

double product: 2 × 2 × 1 = 4;

difference of squares: 22 - 12 = 3;

sum of squares: 22 + 12 = 5,

and we got the famous 3-4-5 triangle. If we take the numbers 3 and 2 instead, we get:

double product: 2 × 3 × 2 = 12;

difference of squares: 32 - 22 = 5;

sum of squares: 32 + 22 = 13,

and we get the next famous triangle 5 - 12 - 13. Let's try to take the numbers 42 and 23 and get:

double product: 2 × 42 × 23 = 1932;

difference of squares: 422 - 232 = 1235;

sum of squares: 422 + 232 = 2293,

no one has ever heard of the triangle 1235–1932–2293.

But these numbers work too:

12352 + 19322 = 1525225 + 3732624 = 5257849 = 22932.

There is another feature in the Diophantine rule that has already been hinted at: having received three numbers, we can take another arbitrary number and multiply them all by it. Thus, a 3-4-5 triangle can be turned into a 6-8-10 triangle by multiplying all sides by 2, or into a 15-20-25 triangle by multiplying everything by 5.

If we switch to the language of algebra, the rule takes the following form: let u, v and k be natural numbers. Then a right triangle with sides

2kuv and k (u2 - v2) has a hypotenuse

There are other ways of presenting the main idea, but they all boil down to the one described above. This method allows you to get all Pythagorean triples.

Regular polyhedra

There are exactly five regular polyhedra. A regular polyhedron (or polyhedron) is a three-dimensional figure with a finite number of flat faces. Facets converge with each other on lines called edges; edges meet at points called vertices.

The culmination of the Euclidean "Principles" is the proof that there can be only five regular polyhedra, that is, polyhedra in which each face is a regular polygon (equal sides, equal angles), all faces are identical, and all vertices are surrounded by an equal number of equally spaced faces. Here are five regular polyhedra:

tetrahedron with four triangular faces, four vertices and six edges;

cube, or hexahedron, with 6 square faces, 8 vertices and 12 edges;

octahedron with 8 triangular faces, 6 vertices and 12 edges;

dodecahedron with 12 pentagonal faces, 20 vertices and 30 edges;

icosahedron with 20 triangular faces, 12 vertices and 30 edges.

// Rice. 37. Five regular polyhedra

Regular polyhedra can also be found in nature. In 1904, Ernst Haeckel published drawings of tiny organisms known as radiolarians; many of them are shaped like the same five regular polyhedra. Perhaps, however, he slightly corrected nature, and the drawings do not fully reflect the shape of specific living beings. The first three structures are also observed in crystals. You will not find a dodecahedron and an icosahedron in crystals, although irregular dodecahedrons and icosahedrons sometimes come across there. True dodecahedrons can appear as quasicrystals, which are like crystals in every way, except that their atoms do not form a periodic lattice.

// Rice. 38. Drawings by Haeckel: radiolarians in the form of regular polyhedra

// Rice. 39. Developments of Regular Polyhedra

It can be interesting to make models of regular polyhedra out of paper by first cutting out a set of interconnected faces - this is called a polyhedron sweep; the scan is folded along the edges and the corresponding edges are glued together. It is useful to add an additional area for glue to one of the edges of each such pair, as shown in Fig. 39. If there is no such platform, you can use adhesive tape.

Equation of the fifth degree

There is no algebraic formula for solving equations of the 5th degree.

In general, the equation of the fifth degree looks like this:

ax5 + bx4 + cx3 + dx2 + ex + f = 0.

The problem is to find a formula for solving such an equation (it can have up to five solutions). Experience with quadratic and cubic equations, as well as with equations of the fourth degree, suggests that such a formula should also exist for equations of the fifth degree, and, in theory, the roots of the fifth, third and second degree should appear in it. Again, one can safely assume that such a formula, if it exists, will turn out to be very, very complicated.

This assumption ultimately turned out to be wrong. Indeed, no such formula exists; at least there is no formula consisting of the coefficients a, b, c, d, e and f, composed using addition, subtraction, multiplication and division, as well as taking roots. Thus, there is something very special about the number 5. The reasons for this unusual behavior of the five are very deep, and it took a lot of time to figure them out.

The first sign of a problem was that no matter how hard mathematicians tried to find such a formula, no matter how smart they were, they always failed. For some time, everyone believed that the reasons lie in the incredible complexity of the formula. It was believed that no one simply could understand this algebra properly. However, over time, some mathematicians began to doubt that such a formula even existed, and in 1823 Niels Hendrik Abel was able to prove the opposite. There is no such formula. Shortly thereafter, Évariste Galois found a way to determine whether an equation of one degree or another - 5th, 6th, 7th, generally any - is solvable using this kind of formula.

The conclusion from all this is simple: the number 5 is special. You can solve algebraic equations (using nth roots for different values ​​of n) for powers of 1, 2, 3, and 4, but not for powers of 5. This is where the obvious pattern ends.

No one is surprised that equations of powers greater than 5 behave even worse; in particular, the same difficulty is connected with them: there are no general formulas for their solution. This does not mean that the equations have no solutions; it does not mean also that it is impossible to find very precise numerical values ​​of these solutions. It's all about the limitations of traditional algebra tools. This is reminiscent of the impossibility of trisecting an angle with a ruler and a compass. There is an answer, but the listed methods are not sufficient and do not allow you to determine what it is.

Crystallographic limitation

Crystals in two and three dimensions do not have 5-beam rotational symmetry.

The atoms in a crystal form a lattice, that is, a structure that repeats periodically in several independent directions. For example, the pattern on the wallpaper is repeated along the length of the roll; in addition, it is usually repeated in the horizontal direction, sometimes with a shift from one piece of wallpaper to the next. Essentially, the wallpaper is a two-dimensional crystal.

There are 17 varieties of wallpaper patterns on the plane (see chapter 17). They differ in the types of symmetry, that is, in the ways of rigidly shifting the pattern so that it lies exactly on itself in its original position. The types of symmetry include, in particular, various variants of rotational symmetry, where the pattern should be rotated through a certain angle around a certain point - the center of symmetry.

The order of symmetry of rotation is how many times you can rotate the body to a full circle so that all the details of the picture return to their original positions. For example, a 90° rotation is 4th order rotational symmetry*. The list of possible types of rotational symmetry in the crystal lattice again points to the unusualness of the number 5: it is not there. There are variants with rotational symmetry of 2nd, 3rd, 4th and 6th orders, but no wallpaper pattern has 5th order rotational symmetry. There is also no rotational symmetry of order greater than 6 in crystals, but the first violation of the sequence still occurs at the number 5.

The same happens with crystallographic systems in three-dimensional space. Here the lattice repeats itself in three independent directions. There are 219 different types of symmetry, or 230 if we consider the mirror reflection of the pattern as a separate version of it - moreover, in this case there is no mirror symmetry. Again, rotational symmetries of orders 2, 3, 4, and 6 are observed, but not 5. This fact is called the crystallographic constraint.

In four-dimensional space, lattices with 5th order symmetry exist; in general, for lattices of sufficiently high dimension, any predetermined order of rotational symmetry is possible.

// Rice. 40. Crystal lattice of table salt. Dark balls represent sodium atoms, light balls represent chlorine atoms.

Quasicrystals

While 5th order rotational symmetry is not possible in 2D and 3D lattices, it can exist in slightly less regular structures known as quasicrystals. Using Kepler's sketches, Roger Penrose discovered flat systems with a more general type of fivefold symmetry. They are called quasicrystals.

Quasicrystals exist in nature. In 1984, Daniel Shechtman discovered that an alloy of aluminum and manganese can form quasi-crystals; Initially, crystallographers greeted his message with some skepticism, but later the discovery was confirmed, and in 2011 Shechtman was awarded the Nobel Prize in Chemistry. In 2009, a team of scientists led by Luca Bindi discovered quasi-crystals in a mineral from the Russian Koryak Highlands - a compound of aluminum, copper and iron. Today this mineral is called icosahedrite. By measuring the content of various oxygen isotopes in the mineral with a mass spectrometer, scientists showed that this mineral did not originate on Earth. It formed about 4.5 billion years ago, at a time when the solar system was just emerging, and spent most of its time in the asteroid belt, orbiting the sun, until some kind of disturbance changed its orbit and brought it eventually to Earth.

// Rice. 41. Left: one of two quasi-crystalline lattices with exact fivefold symmetry. Right: Atomic model of an icosahedral aluminum-palladium-manganese quasicrystal