Pythagorean numbers. Pythagorean triples of numbers (Creative work of the student). Using Pythagorean triples in solving

Belotelov V.A. Pythagorean triplets and their number // Nesterov Encyclopedia

This article is a response to one professor - shchipach. Look, professor, how they do it in our village.

Nizhny Novgorod region, Zavolzhye.

Requires knowledge of the algorithm for solving Diophantine equations (ARDE) and knowledge of progressions of polynomials.

IF is a prime number.

SP is a composite number.

Let there be an odd number N. For any odd number except one, you can create an equation.

p 2 + N = 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 a prime number, this equation is unique. If the number N is composite, then similar equations can be constructed using 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 wondered if it was possible, by clinging to this difference between IF and MF, to find a method for identifying them.

Let's introduce some notation;

Let's change the lower equation, -

N = in 2 – a 2 = (in – a)(in + a).

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

The numbers N were summarized in a matrix, -

It was for this task that we had to deal with progressions of polynomials and their matrices. Everything turned out to be in vain - the IFs held their defense powerfully. Let's enter a column in Table 1 where b - a = 1 (q - p = 1).

And 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 = 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 = ℓ 2, where

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

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

This article is a response to one professor - shchipach.

I asked for help; I needed a series of numbers that I couldn’t find on the Internet. I ran into questions like “why?”, “show me the method.” In particular, there was a problem asking whether the series is infinite Pythagorean triples, “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 triplets, -

x 2 = y 2 + z 2. (1)

Let's pass it through the ARD.

Three situations are possible:

I. x – 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 – even number,

y is an odd number,

z is an odd number.

x > y > z.

Let's start in order from I.

Let's introduce new variables

Let's substitute into equation (1).

Let's reduce by a 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 variable while simultaneously introducing 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 = (2ƒ + 2k) 2 + (2k + 1) 2

Let's square it -

(x – y) 2 + 2(2ƒ + 2k)(x – y) + (2ƒ + 2k) 2 = (2ƒ + 2k) 2 + (2k + 1) 2,

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

The ARDE gives, through parameters, the relationship between the leading terms of the equation, so we get equation (3).

It’s not a good idea to select solutions. But, firstly, there is nowhere to go, and secondly, several of these solutions are needed, and we can restore an endless series of solutions.

When 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.

It can take a long time to select, but in the end the series will take the form -

x – y = 1, 9, 25, 49, 81, ….

Let's consider option II.

Let's introduce new variables into equation (1)

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

Let's reduce by a smaller variable 2 β, -

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

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

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

2α – 2γ = x – z and substitute it 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 = 2, 8, 18, 32, 50, ….

Let's draw a trapezoid -

Let's write the formula.

where n=1, 2,... ∞.

We will not describe case III - there are no solutions there.

For condition II, the set of triplets 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, there are 9 columns of triplets, with five triplets 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 = 81.

For the quantities x we ​​write a trapezoid, -

Let's write the formula -

For the quantities y we write a trapezoid, -

Let's write the formula -

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

Let's write the formula -

Where n = 1 ÷ ∞.

As promised, a series of triples at x – y = 81 flies to ∞.

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

Let's write down the values ​​of x from the top rows from the last five columns and construct a trapezoid.

It didn’t work out, but the pattern should be quadratic. To keep everything in order, it turned out that it was necessary to combine columns I and II.

In case II, the values ​​y and z are swapped again.

We managed to merge for one reason - the cards worked well in this task - luck.

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

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

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

Run to the closet for the chest.

Total: In addition to one, each odd number on the number line participates in the formation of Pythagorean triplets equal to the number of pairs of generator factors given number N, including the 1 x N factor.

The number N = ℓ 2, where ℓ is the frequency factor, forms one Pythagorean triple; if ℓ is the frequency frequency, then there is no triple on the factors ℓxℓ.

Let's construct matrices for the values ​​x, y.

Let's start working with the matrix for x. To do this, let’s stretch the coordinate grid from the task of identifying the IF and MF onto it.

The numbering of vertical rows is normalized by the expression

We will remove the first column, because

The matrix will take the form -

Let us describe the vertical rows, -

Let us describe the coefficients for "a", -

Let us describe the free terms -

Let's create a general formula for "x" -

If we carry out similar work for "y", we get -

You can approach this result from the other side.

Let's take the equation -

a 2 + N = in 2.

Let's transform it a little -

N = in 2 – a 2 .

Let's square it -

N 2 = in 4 – 2 in 2 a 2 + a 4.

To the left and right sides of the equation we add in value 4в 2 а 2, -

N 2 + 4b 2 a 2 = b 4 + 2b 2 a 2 + a 4.

And finally, -

(in 2 + a 2) 2 = (2va) 2 + N 2.

Pythagorean triplets are composed as follows:

Let's 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 ​​a – a, while the vertical columns of Table 3 are numbered with values ​​x – y.

x – y = (c – a) 2,

x = y + (c – a) 2.

Let's make three equations.

(y + 1 2) 2 = y 2 + 117 2,

(y + 3 2) 2 = y 2 + 117 2,

(y + 9 2) 2 = 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 coprime numbers, so one triple was obtained with a coefficient of 9.

Let us depict what is written above in general symbols -

Everything in this work, including an example of 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 identifying the IF and MF.

A lot has been done in this direction, and today the following idea has come to light: an identification equation, and one that can determine the factors, does not exist.

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

There is a formula

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

For any degree n given equation there is a number N having 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 work, the numbers N were considered for the equation x 2 = y 2 + z 2 when they are in place of z in the equation. When N is in place of x, this is a different problem.

Sincerely, Belotelov V.A.

Educational: study a number of Pythagorean triples, develop an algorithm for their use in different situations, create a reminder on their use.

Educational: formation of a conscious attitude towards learning, development of cognitive activity, culture of educational work.

Developmental: development of geometric, algebraic and numerical intuition, intelligence, observation, memory.

Lesson progress

I. Organizational moment

II. Explanation of new material

Teacher: The mystery of the attractive power of Pythagorean triplets has long worried humanity. The unique properties of Pythagorean triplets explain them special role in nature, music, mathematics. The Pythagorean spell, the Pythagorean theorem, remains in the brains of millions, if not billions, of people. This is a fundamental theorem that every schoolchild is forced to memorize. Although Pythagorean theorem can be understood by ten-year-olds, it is an inspiring start to a problem that the greatest minds in the history of mathematics have failed to solve, Fermat's theorem. Pythagoras from the island of Samos (see. Appendix 1 , slide 4) was one of the most influential and yet mysterious figures in mathematics. Because no reliable accounts of his life and work survive, his life has become shrouded in myth and legend, and historians may find it difficult to separate fact from fiction. There is no doubt, however, that Pythagoras developed the idea of ​​the logic of numbers and that it is to him that we owe the first golden age of mathematics. Thanks to his genius, numbers ceased to be used only for counting and calculations and were appreciated for the first time. Pythagoras studied the properties of certain classes of numbers, the relationships between them and the figures that form numbers. Pythagoras realized that numbers exist independently of the material world, and therefore the study of numbers is not affected by the inaccuracy of our senses. This meant that Pythagoras gained the ability to discover truths independent of anyone else's opinion or prejudice. Truths more absolute than any previous knowledge. Based on the literature studied regarding Pythagorean triples, we will be interested in the possibility of using Pythagorean triples in solving trigonometry problems. Therefore, we will set ourselves the goal: to study a number of Pythagorean triplets, develop an algorithm for their use, compile a memo on their use, and conduct research on their use in various situations.

Triangle ( slide 14), whose sides are equal to Pythagorean numbers, is rectangular. Moreover, any such triangle is Heronian, i.e. one in which all sides and area are integers. The simplest of them is the Egyptian triangle with sides (3, 4, 5).

Let's create a series of Pythagorean triples by multiplying the numbers (3, 4, 5) by 2, by 3, by 4. We will obtain a series of Pythagorean triples, sort them in ascending order of the maximum number, and select primitive ones.

(3, 4, 5), (6, 8, 10), (5, 12, 13) , (9, 12, 13), (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).

III. Lesson progress

1. Let's spin around the tasks:

1) Using relations between trigonometric functions of the same argument, find if

it is known that .

2) Find the value of the trigonometric functions of the angle?, if it is known that:

3) System of training tasks on the topic “Addition formulas”

knowing that sin = 8/17, cos = 4/5, and are the angles of the first quarter, find the value of the expression:

knowing that and are the angles of the second quarter, sin = 4/5, cos = – 15/17, find: .

4) System of training tasks on the topic “Double angle formulas”

a) Let sin = 5/13 be the angle of the second quarter. Find sin2, cos2, tg2, ctg2.

b) It is known that tg? = 3/4, – third quarter angle. Find sin2, cos2, tg2, ctg2.

c) It is known that , 0< < . Найдите sin, cos, tg, ctg.

d) It is known that , < < 2. Найдите sin, cos, tg.

e) Find tan( + ) if it is known that cos = 3/5, cos = 7/25, where and are the angles of the first quarter.

f) Find , – third quarter angle.

We solve the problem in the traditional way using basic trigonometric identities, and then we solve the same problems in a more rational way. To do this, we use an algorithm for solving problems using Pythagorean triples. Let's create a guide to solving problems using Pythagorean triples. To do this, recall the definition of sine, cosine, tangent and cotangent, acute angle right triangle, we depict it, depending on the conditions of the problem, we correctly arrange the Pythagorean triples on the sides of the right triangle ( rice. 1). We write down the ratio and arrange the signs. The algorithm has been developed.

Figure 1

Algorithm for solving problems

Review (study) theoretical material.

Know primitive Pythagorean triples by heart and, if necessary, be able to construct new ones.

Apply the Pythagorean theorem for points with rational coordinates.

Know the definition of sine, cosine, tangent and cotangent of an acute angle of a right triangle, be able to draw a right triangle and, depending on the conditions of the problem, correctly place Pythagorean triples on the sides of the triangle.

Know the signs of sine, cosine, tangent and cotangent depending on their location in coordinate plane.

Necessary requirements:

1. know what signs sine, cosine, tangent, cotangent have in each of the quarters of the coordinate plane;
2. know the definition of sine, cosine, tangent and cotangent of an acute angle of a right triangle;
3. know and be able to apply the Pythagorean theorem;
4. know the basics trigonometric identities, addition formulas, double angle formulas, half argument formulas;
5. know reduction formulas.

Taking into account the above, let's fill out the table ( table 1). It must be completed following the definition of sine, cosine, tangent and cotangent or using the Pythagorean theorem for points with rational coordinates. In this case, it is always necessary to remember the signs of sine, cosine, tangent and cotangent, depending on their location in the coordinate plane.

Table 1

		Triples of numbers		sin	cos	tg	ctg
		(3, 4, 5)	I hour
		(6, 8, 10)	Part II			-	-
		(5, 12, 13)	Part III	-	-
		(8, 15, 17)	IV hour	-		-	-
		(9, 40, 41)	I hour

For successful work You can use the instructions for using Pythagorean triples.

Table 2

(3, 4, 5), (6, 8, 10), (5, 12, 13) , (9, 12, 13), (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), …

2. Let's decide together.

1) Problem: find cos, tg and ctg, if sin = 5/13, if - the angle of the second quarter.

Properties

Since Eq. x 2 + y 2 = z 2 homogeneous, when multiplying 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, coprime numbers.

Examples

Some Pythagorean triples (sorted in ascending order of maximum number, primitive ones 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)…

Story

Pythagorean triplets have been known for a very long time. Found in the architecture of ancient Mesopotamian tombstones isosceles triangle, composed of two rectangular ones with sides of 9, 12 and 15 cubits. The pyramids of Pharaoh Snofru (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.

X All-Russian Symposium on Applied and Industrial Mathematics. St. Petersburg, May 19, 2009

Report: Algorithm for solving Diophantine equations.

The paper discusses the method of studying Diophantine equations and presents the solutions solved by this method: - great theorem Farm; - search for Pythagorean triples, etc. http://referats.protoplex.ru/referats_show/6954.html

Links

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

Wikimedia Foundation. 2010.

See what “Pythagorean triplets” are in other dictionaries:

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The study of the properties of natural numbers led the Pythagoreans to another “eternal” problem of theoretical arithmetic (number theory) - a problem whose germs appeared long before Pythagoras in Ancient Egypt and Ancient Babylon, and a general solution has not been found to this day. Let's start with the problem, which in modern terms can be formulated as follows: solve an indefinite equation in natural numbers

Today this task is called Pythagorean problem, and its solutions - triplets of natural numbers satisfying equation (1.2.1) - are called Pythagorean triplets. Due to the obvious connection of the Pythagorean theorem with the Pythagorean problem, the latter can be given a geometric formulation: find all right triangles with integer legs x, y and an integer hypotenuse z.

Particular solutions to the Pythagorean problem were known in ancient times. In a papyrus from the time of Pharaoh Amenemhat I (c. 2000 BC), kept in the Egyptian Museum in Berlin, we find a right triangle with the aspect ratio (). According to the greatest German historian of mathematics M. Cantor (1829 - 1920), there was a special profession in Ancient Egypt harpedonaptians- “rope pullers”, who, during the solemn ceremony of laying temples and pyramids, marked right angles using a rope with 12 (= 3 + 4 + 5) equally spaced knots. The way the harpedonaptes construct a right angle is obvious from Figure 36.

It must be said that another expert categorically disagrees with Kantor ancient mathematics- van der Waerden, although the very proportions of ancient Egyptian architecture testify in favor of Kantor. Be that as it may, today a right triangle with the ratio of its sides is called Egyptian.

As noted on p. 76, a clay tablet dating back to the ancient Babylonian era and containing 15 lines of Pythagorean triplets has been preserved. In addition to the trivial triple obtained from the Egyptian (3, 4, 5) by multiplying by 15 (45, 60, 75), there are also very complex Pythagorean triplets, such as (3367, 3456, 4825) and even (12709, 13500, 18541)! There is no doubt that these numbers were found not by simple search, but according to certain uniform rules.

And yet the question of general decision equation (1.2.1) in natural numbers was posed and solved only by the Pythagoreans. General setting whatever mathematical problem was alien to both the ancient Egyptians and the ancient Babylonians. Only with Pythagoras does the development of mathematics as a deductive science begin, and one of the first steps on this path was the solution of the problem of Pythagorean triplets. The ancient tradition associates the first solutions of equation (1.2.1) with the names of Pythagoras and Plato. Let's try to reconstruct these solutions.

It is clear that Pythagoras thought of equation (1.2.1) not in analytical form, but in the form of a square number, within which it was necessary to find the square numbers and. It was natural to represent the number as a square with side y one less side z the original square, i.e. Then, as is easy to see from Figure 37 (just see!), the remaining square number must satisfy the equality . Thus we arrive at the system linear equations

By adding and subtracting these equations, we find the solution to equation (1.2.1):

It is easy to verify that the resulting solution gives natural numbers only for odd . Thus, we finally have

Etc. Tradition associates this decision with the name of Pythagoras.

Note that system (1.2.2) can also be obtained formally from equation (1.2.1). In fact,

from where, assuming , we arrive at (1.2.2).

It is clear that the Pythagorean solution was found under a fairly strict constraint () and does not contain all Pythagorean triples. The next step can be put , then , since only in this case will square number. This is how the system arises, which will also be a Pythagorean triple. Now the main

Theorem. If p And q mutually prime numbers different parities, then all primitive Pythagorean triples are found by the formulas

Next, we will consider known methods for generating effective Pythagorean triples. Pythagoras' students were the first to invent a simple way of generating 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 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 triples are generated:

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

As we see, these formulas cannot give all possible primitive triplets.

Consider the following polynomial, which can be expanded 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 triplets in which the average number differs from the largest number by exactly one, that is, not all possible triplets are generated either. Here the first threes are equal to: (5,12,13), (7,24,25), (9,40,41), (11,60,61).

To determine how to generate all primitive triplets, their properties should be examined. Firstly, if ( a,b,c) is a primitive triple, then a And b, b And c, A And c- must be relatively simple. Let a And b are divided into d. Then a 2 + b 2 - also divisible by d. Respectively, c 2 and c must be divided by d. That is, this is not a primitive three.

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, like 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 0 or 1 and cannot have a remainder 2. Thus, a And b cannot be unpaired, that is a 2 + b 2 = 4k 2 +4k+4l 2 +4l+1 and the remainder of the division With 2 by 4 must be 1, which means that With must be unpaired.

Such requirements for the elements of a 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— mutually prime with different pairings. These dependencies first became known from the works of Euclid, who lived 2300 r. back.

Let us prove the validity of dependencies (2). Let A- paired, then b And c- unpaired. Then c + b i c − b- paired. They can be represented as c + b = 2u And c − b = 2v, Where u,v- 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- mutually simple. Let u And v- divided into d. Then ( c + b) And ( c − b) are divided into d. And so c And b must be divided by d, and this contradicts the condition for the Pythagorean triple.

Because uv = (a/2) 2 and u And v are relatively prime, 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 relatively prime. If m And n- unpaired, then b = m 2 − n 2 and c = m 2 + n 2 would be paired, which is impossible, since c And b- mutually simple.

Thus, any primitive Pythagorean triple must satisfy conditions (2). At the same time, the numbers m And n are called generating numbers primitive triplets. For example, let us 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 mutually prime and of different pairs.

You can prove the opposite, that the numbers m, n using formulas (2) they 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 in this case a,b,c are mutually prime numbers by contradiction. Let these numbers be divisible 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 , but this is impossible, since p≠ 2. Therefore m, n- mutually prime and a,b,c- are also relatively simple.

Table 1 shows all primitive Pythagorean triples generated using 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 divisible by 3;
• one of the numbers a,b,c divisible by 5;
• number A divisible by 4;
• work a· b divisible by 12.

In 1971, American mathematicians Teigan and Hedwin proposed such little-known parameters of a right triangle as its 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 it is the additional distance that must be passed along the legs of the triangle from one vertex to the opposite, if not going along its diagonal.

Through the excess and growth of 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 are products of a certain number d. This number d has the name of growth and refers to h as follows: 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.

Using pairs ( k,h) you can generate all Pythagorean triangles, including non-primitive and generalized ones, 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 relatively prime and if h =± q 2 at q- unpaired.
Moreover, this will be precisely a Pythagorean triple if k> √2· h/d And h > 0.

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

• 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 by ( k,h) = (3,2).

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

Setting triplets using 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- its perimeter. This follows from the equality eP = 4S, which follows from the Pythagorean theorem.

For a right triangle e equal to the diameter of the circle inscribed in the triangle. This follows from the fact that the hypotenuse With = (A − r)+(b − r) = a + b − 2r, Where r- 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 sequence Pythagorean triangles with growth h. From Table 2, which presents several options for triplets generated by pairs h, k, it is clear that with increasing k the sizes of the sides of the triangle increase. Thus, unlike classical numbering, numbering in pairs h, k has greater 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 inequality 2√ h ≤ d ≤ 2h, in which the lower limit is reached at p= 1, and the top one - at q= 1. Therefore the value d relative to 2√ h is a measure of how much a number h distant from the square of a certain number.