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 for the generation of triples right triangle, as his height (height) 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 triplets, 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.

educational: to study a number of Pythagorean triples, develop an algorithm for their application in different situations, make a memo on their use.

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

Educational: development of geometric, algebraic and numerical intuition, ingenuity, observation, memory.

During the classes

I. Organizational moment

II. Explanation of new material

Teacher: The mystery of the attractive power of Pythagorean triples has long worried mankind. The unique properties of Pythagorean triples explain their special role in nature, music, and mathematics. The Pythagorean spell, the Pythagorean theorem, remains in the brains of millions, if not billions, of people. This is a fundamental theorem, which every schoolchild is forced to memorize. Although comprehensible to ten-year-olds, the Pythagorean theorem is the inspiring start to the problem that the greatest minds in the history of mathematics have failed to solve: Fermat's Theorem. Pythagoras from the island of Samos (cf. Annex 1 , slide 4) was one of the most influential yet enigmatic figures in mathematics. Since there are no reliable records of his life and work, his life has become shrouded in myths and legends, and historians 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 were no longer used only for counting and calculations and were first appreciated. 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 inaccuracy of our senses does not affect the study of numbers. This meant that Pythagoras gained the ability to discover truths independent of anyone's opinion or prejudice. Truths are more absolute than any previous knowledge. Based on the studied literature concerning 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 triples, to develop an algorithm for their application, to compile a memo on their use, to conduct a study on their application 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 make a series of Pythagorean triples by multiplying the numbers (3, 4, 5) by 2, by 3, by 4. We will get a series of Pythagorean triples, sort them in ascending order of the maximum number, 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. During the classes

1. Let's spin around the tasks:

1) Using the relationships 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) The 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) The 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? \u003d 3/4, - the angle of the third quarter. 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 tg( + ) if it is known that cos = 3/5, cos = 7/25, where and are the angles of the first quarter.

f) Find , is the angle of the third quarter.

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. We compose a memo for solving problems using Pythagorean triples. To do this, we recall the definition of sine, cosine, tangent and cotangent, acute angle right-angled triangle, depict it, depending on the conditions of the problem on the sides of the right-angled triangle, we correctly arrange the Pythagorean triples ( rice. 1). We write down the ratio and arrange the signs. The algorithm has been developed.

Picture 1

Problem solving algorithm

Repeat (study) theoretical material.

Know by heart the primitive Pythagorean triples 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 condition of the problem, correctly arrange Pythagorean triples on the sides of the triangle.

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

Required 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 basic trigonometric identities, addition formulas, double angle formulas, half argument formulas;
5. know the formulas of reduction.

Based on the above, fill in the table ( Table 1). It must be filled in following the definition of sine, cosine, tangent and cotangent or using the Pythagorean theorem for points with rational coordinates. In this case, it is constantly necessary to remember the signs of the sine, cosine, tangent and cotangent, depending on their location in the coordinate plane.

Table 1

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

For successful work, you can use the memo of 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. We decide together.

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

Pythagorean triples of numbers

creative work

student 8 ”A” class

MAOU "Gymnasium No. 1"

Oktyabrsky district of Saratov

Panfilova Vladimir

Supervisor - teacher of mathematics of the highest category

Grishina Irina Vladimirovna

Content

Introduction……………………………………………………………………………………3

Theoretical part of the work

Finding the basic Pythagorean triangle

(formulas of the ancient Hindus)…………………………………………………………………4

Practical part of the work

Composing Pythagorean triples in various ways……………………........ 6

An important property of Pythagorean triangles………………………………………...8

Conclusion………………………………………………………………………………....9

Literature………………………………………………………………………………...10

Introduction

In that academic year in mathematics lessons, we studied one of the most popular theorems of geometry - the Pythagorean theorem. The Pythagorean theorem is applied in geometry at every step, it has found wide application in practice and everyday life. But, besides the theorem itself, we also studied the theorem inverse to the Pythagorean theorem. In connection with the study of this theorem, we have become acquainted with Pythagorean triples of numbers, i.e. with sets of 3 natural numbers a , b Andc , for which the relation is valid: = + . Such sets include, for example, the following triplets:

3,4,5; 5,12,13; 7,24,25; 8,15,17; 20,21,29; 9,40,41; 12,35,37

I immediately had questions: how many Pythagorean triples can you come up with? And how to compose them?

In our geometry textbook, after presenting the theorem converse to the Pythagorean theorem, an important remark was made: it can be proved that the legsA Andb and hypotenuseWith right-angled triangles, the lengths of whose sides are expressed in natural numbers, can be found by the formulas:

A = 2km b = k( - )c = k( + , (1)

Wherek , m , n are any natural numbers, andm > n .

Naturally, the question arises - how to prove these formulas? And is it only by these formulas that Pythagorean triples can be formed?

In my work, I have attempted to answer the questions that have arisen in my mind.

Theoretical part of the work

Finding the main Pythagorean triangle (formulas of the ancient Hindus)

Let us first prove formulas (1):

Let us denote the lengths of the legs throughX Andat , and the length of the hypotenuse throughz . By the Pythagorean theorem, we have the equality:+ = .(2)

This equation is called the Pythagorean equation. The study of Pythagorean triangles is reduced to solving equation (2) in natural numbers.

If each side of some Pythagorean triangle is increased by the same number of times, then we get a new right-angled triangle similar to the given one with sides expressed in natural numbers, i.e. again the Pythagorean triangle.

Among all similar triangles, there is the smallest one, it is easy to guess that this will be a triangle whose sidesX Andat expressed in coprime numbers

(gcd (x,y )=1).

We call such a Pythagorean trianglemain .

Finding the main Pythagorean triangles.

Let the triangle (x , y , z ) is the main Pythagorean triangle. NumbersX Andat are coprime, and therefore cannot both be even. Let us prove that they cannot both be odd. For this, we note thatThe square of an odd number when divided by 8 gives a remainder of 1. Indeed, any odd natural number can be represented as2 k -1 , Wherek belongsN .

From here: = -4 k +1 = 4 k ( k -1)+1.

Numbers( k -1) Andk are consecutive, one of them must be even. Then the expressionk ( k -1) divided by2 , 4 k ( k -1) is divisible by 8, which means when divided by 8, the remainder is 1.

The sum of the squares of two odd numbers gives a remainder of 2 when divided by 8, therefore, the sum of the squares of two odd numbers is an even number, but not a multiple of 4, and therefore this numbercannot be the square of a natural number.

So equality (2) cannot hold ifx Andat both are odd.

Thus, if the Pythagorean triangle (x, y, z ) - the main one, then among the numbersX Andat one must be even and the other must be odd. Let the number y be even. NumbersX Andz odd (oddz follows from equality (2)).

From the equation+ = we get that= ( z + x )( z - x ) (3).

Numbersz + x Andz - x as the sum and difference of two odd numbers are even numbers, and therefore (4):

z + x = 2 a , z - x = 2 b , WhereA Andb belongN .

z + x =2 a , z - x = 2 b ,

z = a+b , x = a - b. (5)

From these equalities it follows thata Andb are relatively prime numbers.

We prove this by arguing from the contrary.

Let GCD (a , b )= d , Whered >1 .

Thend z Andx , and hence the numbersz + x Andz - x . Then, based on equality (3) would be a divisor . In this cased would be common divisor numbersat AndX , but the numbersat AndX must be coprime.

Numberat is known to be even, soy = 2s , WhereWith - natural number. Equality (3) based on equality (4) takes the following form: =2a*2 b , or =ab.

It is known from arithmetic thatif the product of two coprime numbers is the square of a natural number, then each of those numbers is also the square of a natural number.

Means,a = Andb = , Wherem Andn are coprime numbers, because they are divisors of coprime numbersA Andb .

Based on equality (5) we have:

z = + , x = - , = ab = * = ; c = mn

Theny = 2 mn .

Numbersm Andn , because are coprime, cannot be even at the same time. But they cannot be odd at the same time, because in this casex = - would be even, which is impossible. So one of the numbersm orn is even and the other is odd. Obviously,y = 2 mn is divisible by 4. Therefore, in every main Pythagorean triangle, at least one of the legs is divisible by 4. It follows that there are no Pythagorean triangles, all sides of which would be prime numbers.

The results obtained can be expressed as the following theorem:

All major triangles in whichat is an even number, are obtained from the formula

x = - , y =2 mn , z = + ( m > n ), Wherem Andn - all pairs of coprime numbers, one of which is even and the other odd (it doesn't matter which one). Every basic Pythagorean triple (x, y, z ), Whereat – even, is determined uniquely in this way.

Numbersm Andn cannot be both even or both odd, because in these cases

x = would be even, which is impossible. So one of the numbersm orn even and the other oddy = 2 mn divisible by 4).

Practical part of the work

Composing Pythagorean triples in various ways

In Hindu formulasm Andn - coprime, but can be numbers of arbitrary parity and it is quite difficult to make Pythagorean triples using them. Therefore, let's try to find a different approach to compiling Pythagorean triples.

= - = ( z - y )( z + y ), WhereX - odd,y - even,z – odd

v = z - y , u = z + y

= UV , Whereu - odd,v – odd (coprime)

Because the product of two odd coprime numbers is the square of a natural number, thenu = , v = , Wherek Andl are coprime, odd numbers.

z - y = z + y = k 2 , whence, adding the equalities and subtracting from one another, we get:

2 z = + 2 y = - that is

z= y= x = cl

k

l

x

y

z

37

9

1

9

40

41 (szeros)*(100…0 (szeros) +1)+1 =200…0 (s-1zeros) 200…0 (s-1zeros) 1

An important property of Pythagorean triangles

Theorem

In the main Pythagorean triangle, one of the legs is necessarily divisible by 4, one of the legs is necessarily divisible by 3, and the area of ​​the Pythagorean triangle is necessarily a multiple of 6.

Proof

As we know, in any Pythagorean triangle at least one of the legs is divisible by 4.

Let us prove that one of the legs is also divisible by 3.

To prove this, suppose that in the Pythagorean triangle (x , y , z x ory multiple of 3.

Now we prove that the area of ​​the Pythagorean triangle is divisible by 6.

Any Pythagorean triangle has an area expressed as a natural multiple of 6. This follows from the fact that at least one of the legs is divisible by 3 and at least one of the legs is divisible by 4. The area of ​​the triangle, determined by the half-product of the legs, must be expressed by a multiple of 6 .

Conclusion

In work

- proven formulas of the ancient Hindus

- conducted a study on the number of Pythagorean triples (there are infinitely many of them)

- methods for finding Pythagorean triples are indicated

- Studied some properties of Pythagorean triangles

For me it was very interesting topic and finding answers to my questions has become a very interesting activity. In the future, I plan to consider the connection of Pythagorean triples with the Fibonacci sequence and Fermat's theorem and learn many more properties of Pythagorean triangles.

Literature

L.S. Atanasyan "Geometry. 7-9 grades" M .: Education, 2012.

V. Serpinsky “Pythagorean triangles” M.: Uchpedgiz, 1959.

Saratov

2014

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

Today this task is called problem of Pythagoras, and its solutions - triples 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 integer hypotenuse z.

Particular solutions of the Pythagorean problem were known in ancient times. In a papyrus from the time of Pharaoh Amenemhet I (c. 2000 BC), stored in the Egyptian Museum in Berlin, we find a right triangle with an aspect ratio (). According to the largest German historian of mathematics M. Kantor (1829 - 1920), in ancient Egypt there was a special profession harpedonapts- "rope tensioners", who, during the solemn ceremony of laying temples and pyramids, marked right angles with a rope having 12 (= 3 + 4 + 5) equally spaced knots. Construction method right angle harpedonapt is evident from Figure 36.

It must be said that another connoisseur of ancient mathematics, van der Waerden, categorically disagrees with Cantor, although the very proportions of ancient Egyptian architecture testify in favor of Cantor. Be that as it may, today a right triangle with an aspect ratio 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 triples, such as (3367, 3456, 4825) and even (12709, 13500, 18541)! There is no doubt that these numbers were found not by simple enumeration, but by some uniform rules.

Nevertheless, the question of the general solution of equation (1.2.1) in natural numbers was raised 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 formation of mathematics as a deductive science begin, and one of the first steps along this path was the solution of the problem of Pythagorean triples. 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 an analytical form, but in the form of a square number , inside which it was necessary to find the square numbers and . It was natural to represent the number in the form of a square with a side y one less side z original square, i.e. . Then, as it is easy to see from Figure 37 (just see!), for the remaining square number, the equality must be satisfied. Thus we arrive at the system linear equations

Adding and subtracting these equations, we find the solution of equation (1.2.1):

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

And so on. Tradition connects this decision with the name of Pythagoras.

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

whence, assuming , we arrive at (1.2.2).

It is clear that the Pythagorean solution was found under a rather rigid constraint () and contains far from all Pythagorean triples. The next step is to put , then , since only in this case will be a square number. So the system arises will also be a Pythagorean triple. Now the main

Theorem. If p And q coprime numbers of different parity, then all primitive Pythagorean triples are found by the formulas

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)…

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.

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

Report: Algorithm for solving Diophantine equations.

The paper considers the method of studying Diophantine equations and presents the solutions solved by this method: - grand theorem Farm; - search for Pythagorean triplets, 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. - S. 105-125.

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