Pythagorean triples of numbers (Creative work of the student). Pythagorean numbers Three identical hypotenuses in Pythagorean triples

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 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 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. The method of constructing a right angle with harpedonapts is obvious 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. The general formulation of any 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 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 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 a system of 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

A convenient and very accurate method used by land surveyors to draw perpendicular lines on the ground is as follows. Let it be required to draw a perpendicular to the line MN through point A (Fig. 13). Lay off from A in the direction of AM three times some distance a. Then three knots are tied on the cord, the distances between which are 4a and 5a. Attaching the extreme knots to points A and B, pull the cord over the middle knot. The cord will be located in a triangle, in which the angle A is a right one.

This ancient method, apparently used thousands of years ago by the builders of the Egyptian pyramids, is based on the fact that each triangle, the sides of which are related as 3:4:5, according to the well-known Pythagorean theorem, is right-angled, since

3 2 + 4 2 = 5 2 .

In addition to the numbers 3, 4, 5, there is, as is known, an uncountable set of positive integers a, b, c, satisfying the relation

A 2 + b 2 \u003d c 2.

They are called Pythagorean numbers. According to the Pythagorean theorem, such numbers can serve as the lengths of the sides of some right triangle; therefore, a and b are called "legs", and c is called the "hypotenuse".

It is clear that if a, b, c is a triple of Pythagorean numbers, then pa, pb, pc, where p is an integer factor, are Pythagorean numbers. Conversely, if the Pythagorean numbers have a common factor, then by this common factor you can reduce them all, and again you get a triple of Pythagorean numbers. Therefore, we will first study only triples of coprime Pythagorean numbers (the rest are obtained from them by multiplying by an integer factor p).

Let us show that in each of such triplets a, b, c one of the "legs" must be even and the other odd. Let's argue "on the contrary". If both "legs" a and b are even, then the number a 2 + b 2 will be even, and hence the "hypotenuse". This, however, contradicts the fact that the numbers a, b, c do not have common factors, since three even numbers have a common factor of 2. Thus, at least one of the "legs" a, b is odd.

There remains one more possibility: both "legs" are odd, and the "hypotenuse" is even. It is easy to prove that this cannot be. Indeed, if the "legs" have the form

2x + 1 and 2y + 1,

then the sum of their squares is

4x 2 + 4x + 1 + 4y 2 + 4y + 1 \u003d 4 (x 2 + x + y 2 + y) + 2,

i.e., it is a number that, when divided by 4, gives a remainder of 2. Meanwhile, the square of any even number must be divisible by 4 without a remainder. So the sum of the squares of two odd numbers cannot be the square of an even number; in other words, our three numbers are not Pythagorean.

So, from the "legs" a, b, one is even and the other is odd. Therefore, the number a 2 + b 2 is odd, which means that the "hypotenuse" c is also odd.

Suppose, for definiteness, that odd is "leg" a, and even b. From equality

a 2 + b 2 = c 2

we easily get:

A 2 \u003d c 2 - b 2 \u003d (c + b) (c - b).

The factors c + b and c - b on the right side are coprime. Indeed, if these numbers had a common prime factor other than one, then the sum would also be divisible by this factor.

(c + b) + (c - b) = 2c,

and difference

(c + b) - (c - b) = 2b,

and work

(c + b) (c - b) \u003d a 2,

i.e. the numbers 2c, 2b and a would have a common factor. Since a is odd, this factor is different from two, and therefore the numbers a, b, c have the same common factor, which, however, cannot be. The resulting contradiction shows that the numbers c + b and c - b are coprime.

But if the product of coprime numbers is an exact square, then each of them is a square, i.e.

Solving this system, we find:

C \u003d (m 2 + n 2) / 2, b \u003d (m 2 - n 2) / 2, and 2 \u003d (c + b) (c - b) \u003d m 2 n 2, a \u003d mn.

So, the considered Pythagorean numbers have the form

A \u003d mn, b \u003d (m 2 - n 2) / 2, c \u003d (m 2 + n 2) / 2.

where m and n are some coprime odd numbers. The reader can easily verify the opposite: for any odd type, the written formulas give three Pythagorean numbers a, b, c.

Here are some triplets of Pythagorean numbers obtained with various types:

For m = 3, n = 1 3 2 + 4 2 = 5 2 for m = 5, n = 1 5 2 + 12 2 = 13 2 for m = 7, n = 1 7 2 + 24 2 = 25 2 for m = 9, n = 1 9 2 + 40 2 = 41 2 at m = 11, n = 1 11 2 + 60 2 = 61 2 at m = 13, n = 1 13 2 + 84 2 = 85 2 at m = 5 , n = 3 15 2 + 8 2 = 17 2 for m = 7, n = 3 21 2 + 20 2 = 29 2 for m = 11, n = 3 33 2 + 56 2 = 65 2 for m = 13, n = 3 39 2 + 80 2 = 89 2 at m = 7, n = 5 35 2 + 12 2 = 37 2 at m = 9, n = 5 45 2 + 28 2 = 53 2 at m = 11, n = 5 55 2 + 48 2 = 73 2 at m = 13, n = 5 65 2 + 72 2 = 97 2 at m = 9, n = 7 63 2 + 16 2 = 65 2 at m = 11, n = 7 77 2 + 36 2 = 85 2

(All other triples of Pythagorean numbers either have common factors or contain numbers greater than one hundred.)

An important example of a Diophantine equation is given by the Pythagorean theorem, which relates the lengths x and y of the legs of a right triangle to the length z of its hypotenuse:

Of course, you have come across one of the wonderful solutions of this equation in natural numbers, namely the Pythagorean triple of numbers x=3, y=4, z=5. Are there any other triplets?

It turns out that there are infinitely many Pythagorean triples, and all of them were found a long time ago. They can be obtained by well-known formulas, which you will learn about from this paragraph.

If the Diophantine equations of the first and second degree have already been solved, then the question of solving the equations is more high degrees still remains open, despite the efforts of the greatest mathematicians. At present, for example, Fermat's famous conjecture that for any integer value n2 the equation

has no solutions in integers.

For solving certain types of Diophantine equations, the so-called complex numbers. What it is? Let the letter i denote some object that satisfies the condition i 2 \u003d -1(it is clear that no real number satisfies this condition). Consider expressions of the form α+iβ, where α and β are real numbers. Such expressions will be called complex numbers, having defined the operations of addition and multiplication over them, as well as over binomials, but with the only difference that the expression i 2 everywhere we will replace the number -1:

7.1. Many of the three

Prove that if x0, y0, z0- Pythagorean triple, then triples y 0 , x 0 , z 0 And x 0 k, y 0 k, z 0 k for any value of the natural parameter k are also Pythagorean.

7.2. Private formulas

Check that for any natural values m>n trinity of the form

is Pythagorean. Is it any Pythagorean triple x, y, z can be represented in this form, if you allow to rearrange the numbers x and y in the triple?

7.3. Irreducible triplets

A Pythagorean triple of numbers that do not have a common divisor greater than 1 will be called irreducible. Prove that a Pythagorean triple is irreducible only if any two of the numbers in the triple are coprime.

7.4. Property of irreducible triples

Prove that in any irreducible Pythagorean triple x, y, z the number z and exactly one of the numbers x or y are odd.

7.5. All irreducible triples

Prove that a triple of numbers x, y, z is an irreducible Pythagorean triple if and only if it coincides with the triple up to the order of the first two numbers 2mn, m 2 - n 2, m 2 + n 2, where m>n- coprime natural numbers of different parity.

7.6. General formulas

Prove that all solutions of the equation

in natural numbers are given up to the order of the unknown x and y by the formulas

where m>n and k are natural parameters (in order to avoid duplication of any triples, it is enough to choose numbers of type coprime and, moreover, of different parity).

7.7. First 10 triplets

Find all Pythagorean triples x, y, z satisfying the condition x

7.8. Properties of Pythagorean triplets

Prove that for any Pythagorean triple x, y, z statements are true:

a) at least one of the numbers x or y is a multiple of 3;

b) at least one of the numbers x or y is a multiple of 4;

c) at least one of the numbers x, y or z is a multiple of 5.

7.9. Application of complex numbers

The modulus of a complex number α + iβ called a non-negative number

Check that for any complex numbers α + iβ And γ + iδ property is executed

Using the properties of complex numbers and their moduli, prove that any two integers m and n satisfy the equality

i.e., they give a solution to the equation

integers (compare with Problem 7.5).

7.10. Non-Pythagorean triples

Using the properties of complex numbers and their moduli (see Problem 7.9), find formulas for any integer solutions of the equation:

a) x 2 + y 2 \u003d z 3; b) x 2 + y 2 \u003d z 4.

Solutions

7.1. If x 0 2 + y 0 2 = z 0 2 , then y 0 2 + x 0 2 = z 0 2 , and for any natural value of k we have

Q.E.D.

7.2. From equalities

we conclude that the triple indicated in the problem satisfies the equation x 2 + y 2 = z 2 in natural numbers. However, not every Pythagorean triple x, y, z can be represented in this form; for example, the triple 9, 12, 15 is Pythagorean, but the number 15 cannot be represented as the sum of the squares of any two natural numbers m and n.

7.3. If any two numbers from the Pythagorean triple x, y, z have a common divisor d, then it will also be a divisor of the third number (so, in the case x = x 1 d, y = y 1 d we have z 2 \u003d x 2 + y 2 \u003d (x 1 2 + y 1 2) d 2, whence z 2 is divisible by d 2 and z is divisible by d). Therefore, for a Pythagorean triple to be irreducible, it is necessary that any two of the numbers in the triple be coprime,

7.4. Note that one of the numbers x or y, say x, of an irreducible Pythagorean triple x, y, z is odd because otherwise the numbers x and y would not be coprime (see problem 7.3). If the other number y is also odd, then both numbers

give a remainder of 1 when divided by 4, and the number z 2 \u003d x 2 + y 2 gives a remainder of 2 when divided by 4, that is, it is divisible by 2, but not divisible by 4, which cannot be. Thus, the number y must be even, and the number z must therefore be odd.

7.5. Let the Pythagorean triple x, y, z is irreducible and, for definiteness, the number x is even, while the numbers y, z are odd (see Problem 7.4). Then

where are the numbers are whole. Let us prove that the numbers a and b are coprime. Indeed, if they had a common divisor greater than 1, then the numbers would have the same divisor z = a + b, y = a - b, i.e., the triple would not be irreducible (see Problem 7.3). Now, expanding the numbers a and b into products prime factors, we note that any prime factor must be included in the product 4ab = x2 only to an even degree, and if it is included in the expansion of the number a, then it is not included in the expansion of the number b and vice versa. Therefore, any prime factor is included in the expansion of the number a or b separately only to an even degree, which means that these numbers themselves are squares of integers. Let's put then we get the equalities

moreover, the natural parameters m>n are coprime (due to the coprimeness of the numbers a and b) and have different parity (due to the odd number z \u003d m 2 + n 2).

Let now natural numbers m>n of different parity be coprime. Then the troika x \u003d 2mn, y \u003d m 2 - n 2, z \u003d m 2 + n 2, according to Problem 7.2, is Pythagorean. Let us prove that it is irreducible. To do this, it suffices to check that the numbers y and z do not have common divisors (see Problem 7.3). In fact, both of these numbers are odd, since the type numbers have different parities. If the numbers y and z have some simple common divisor (then it must be odd), then each of the numbers and and with them and each of the numbers m and n has the same divisor, which contradicts their mutual simplicity.

7.6. By virtue of the assertions formulated in Problems 7.1 and 7.2, these formulas define only Pythagorean triples. On the other hand, any Pythagorean triple x, y, z after its reduction by the greatest common divisor k, the pair of numbers x and y becomes irreducible (see Problem 7.3) and, therefore, can be represented up to the order of the numbers x and y in the form described in Problem 7.5. Therefore, any Pythagorean triple is given by the indicated formulas for some values ​​of the parameters.

7.7. From inequality z and the formulas of Problem 7.6, we obtain the estimate m 2 i.e. m≤5. Assuming m = 2, n = 1 And k = 1, 2, 3, 4, 5, we get triplets 3, 4, 5; 6, 8, 10; 9, 12, 15; 12,16,20; 15, 20, 25. Assuming m=3, n=2 And k = 1, 2, we get triplets 5, 12, 13; 10, 24, 26. Assuming m = 4, n = 1, 3 And k = 1, we get triplets 8, 15, 17; 7, 24, 25. Finally, assuming m=5, n=2 And k = 1, we get three 20, 21, 29.

During the classes

I. Organizational moment

II. Explanation of new material

Teacher: The mystery of the attractive power of Pythagorean triples has long worried humanity. 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 the Pythagorean theorem can be understood by ten-year-olds, it is the inspiring beginning of the problem that failed the greatest minds in the history of mathematics, Fermat's theorem. Pythagoras from the island of Samos (cf. Attachment 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, develop an algorithm for their application, compile a memo on their use, and 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 the sine, cosine, tangent and cotangent, the acute angle of a right triangle, depict it, depending on the conditions of the problem on the sides of the right triangle, we correctly arrange the Pythagorean triples ( rice. one). 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 the 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 by 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.

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:

History

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

see also

Links

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

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