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. So

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, then 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 traveled 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.

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

This academic year, in mathematics lessons, we studied one of the most popular theorems in 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 numbersa , 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 the statement of the theorem, the converse of 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 a common divisor of 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 of whose sides are 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 a very interesting topic and finding answers to my questions became 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

"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.

Objective: 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 USE test and measuring materials. 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

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.

https://pandia.ru/text/80/406/images/image007_0.gif" width="240" height="139 src=">

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 (the 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 by 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 problems.

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 classical 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.

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.

Wikimedia Foundation. 2010 .

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 ... ...

Triplets of positive integers x, y, z satisfying the equation x2+y 2=z2. All solutions of this equation, and, consequently, all P. p., are expressed by the formulas x=a 2 b2, y=2ab, z=a2+b2, where a, b are arbitrary positive integers (a>b). P. h ... Mathematical Encyclopedia

Triplets of natural numbers such that a triangle, the lengths of the sides to which are proportional (or equal) to these numbers, is rectangular, for example. triple of numbers: 3, 4, 5… Natural science. encyclopedic Dictionary

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

Curly numbers are the general name of numbers associated with a particular geometric figure. This historical concept goes back to the Pythagoreans. There are the following types of curly numbers: Linear numbers are numbers that do not decompose into factors, that is, their ... ... Wikipedia

- The “pi paradox” is a joke on the topic of mathematics, which was in circulation among students until the 80s (in fact, before the mass distribution of microcalculators) and was associated with the limited accuracy of calculating trigonometric functions and ... ... Wikipedia

- (Greek arithmetika, from arithmys number) the science of numbers, primarily natural (positive integer) numbers and (rational) fractions, and operations on them. Possession of a sufficiently developed concept of a natural number and the ability to ... ... Great Soviet Encyclopedia

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.…

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 the smaller factors from the pairs of factors that form 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. (one)

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, we need several of these solutions, 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 triplets 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 \u003d 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.