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  • Level: GCSE
  • Subject: Maths
  • Word count: 2413

Investigating families of Pythagorean triples.

Extracts from this document...

Introduction

I was set the task of investigating families of Pythagorean triples. I started off by investigating families of Pythagorean triples where the shortest side is an odd number and c is always equal to b+1. I then extended my research into other families of triples. Finally, I came up with a formula that can calculate all Pythagorean triples.

Pythagoras’s formula is as follows:

a2 + b2 = c2

This is referring to such a triangle:

image00.png

The triangle always has one right angle and c is always the longest side, also known as the hypotenuse.

An example of this is with the triple 3,4 and 5:

image01.png

Here, a is 3, b is 4 and c is 5.

52 = 32 + 42

25 = 9 + 16

I will now investigate the Pythagorean triples where c is always equal to b+1. The following table shows the first ten cases, along with the areas and perimeters:

N

a

b

c

P

A

a2

b2

c2

1

3

4

5

12

6

9

16

25

2

5

12

13

30

30

25

144

169

3

7

24

25

56

84

49

576

625

4

9

40

41

90

180

81

1600

1681

5

11

60

61

132

330

121

3600

3721

6

13

84

85

182

546

169

7056

7225

7

15

112

113

240

840

225

12544

12769

8

17

144

145

306

1224

289

20736

21025

9

19

180

181

380

1710

361

32400

32761

10

21

220

221

462

2310

441

48400

48841

By looking at the numbers for a, I noticed that they are all odd numbers. The basic formula for odd numbers is:

2n + 1

When I applied this formula to the numbers in the table, they turned out correct. As a result,

a = 2n + 1

The difference between the numbers in column b is 8 at first, then 12, then 16, then 20, and so on, adding 4 to the difference each time. Therefore, the second difference is 4. If we divide 4 by 2, we get 2.

...read more.

Middle

1936

233289

235225

In the values for a, the difference is 4, meaning that the formula starts with:

4n

By looking carefully at the formula and numbers, I noticed that by adding 4 to this formula, I would get the right numbers.

a = 4n + 4

Factorising this, gave me:

4 ( n + 1 )

I found b by halving the difference between the numbers in row b and adding 8n + 3. This gave me:

4n2 + 8n + 3

Or

( 2n + 1 ) ( 2n + 3 )

As c was 2 numbers bigger than b, all I had to do was to add 2 to b’s formula:

4n2 + 8n + 5

When factorised, it became:

( 2n + 1 ) ( 2n + 3 ) + 2

The formulas for the b+2 family are shown below:

a = 4n + 4

b = 4n2 + 8n + 3   or   ( 2n + 1 ) ( 2n + 3 )

c = 4n2 + 8n + 5   or   ( 2n + 1 ) ( 2n + 3 ) + 2

Once again, I used my previous techniques to show that the left hand side was equal to the right hand side:

a2 + b2 = c2

( 4n + 4 )2 + ( 4n2 + 8n + 3 ) 2 = ( 4n2 + 8n + 5 ) 2

Left Hand Side:

( 4n + 4 )2 + ( 4n2 + 8n + 3 ) 2

Right Hand Side:

( 4n2 + 8n + 5 ) 2

   ( a + b ) 2 = ( a + b ) x ( a + b )

= a ( a + b ) + b ( a + b )

= a2 + ab + ab + b2

= a2 + b2 + 2ab

           ( a + b+ c )2 = ( a + b+ c ) x ( a + b+ c )

        = a ( a + b+ c ) + b ( a + b+ c ) + c ( a + b+ c )

        = a2 + b2 + c2 + 2ab + 2ac + 2bc

Left Hand Side:

   ( 4n + 4 )2 + ( 4n2 + 8n + 3 ) 2

= 16n2 + 32n + 16 + 16n4 + 64n3 + 88n2 + 48n + 9

= 16n4 + 64n3 + 104n2 + 80n + 25

Right Hand Side:

   ( 4n2 + 8n + 5 ) 2

= 16n4 + 64n3 + 104n2 + 80n + 25

After I had finished exploring the b+2 triples, I decided to experiment with b+3 triples. I didn’t manage to find any that weren’t multiples. The ones I did find are shown in the table below. All of the numbers can be divided by three to produce the b+1 family:

n

a

b

c

a2

b2

c2

1

9

12

15

81

144

225

2

15

36

39

225

1296

1521

3

21

72

75

441

5184

5625

4

27

120

123

729

14400

15129

5

33

180

183

1089

32400

33489

6

39

252

255

1521

63504

65025

7

45

336

339

2025

112896

114921

8

51

432

435

2601

186624

189225

9

57

540

543

3249

291600

294849

10

63

660

663

3969

435600

439569

I then moved straight on to the b+4 family, but yet again I only found multiples. The same applied for the b+5, b+6 and b+7 families. The table for b+4 is shown below. The numbers can all be divided by 4 to give the b+1 family:

n

a

b

c

a2

b2

c2

1

12

16

20

144

256

400

2

20

48

52

400

2304

2704

3

28

96

100

784

9216

10000

4

36

160

164

1296

25600

26896

5

44

240

244

1936

57600

59536

6

52

336

340

2704

112896

115600

7

60

448

452

3600

200704

204304

8

68

576

580

4624

331776

336400

9

76

720

724

5776

518400

524176

10

84

880

884

7056

774400

781456

...read more.

Conclusion

2 + 4n + 4 – 1

        a = ( n + 2 )2 – 1

This was the same as the formula for a in the b+1 family, the only difference was that it was (n+2) instead of (n+1) and 1 instead of n. I then went on to b:

        b = 2n + 4

b = 2 ( n + 2 )

b = 2 x 1 x ( n + 2 )

Once again, the formula was the same as the formula for b in the b+1 family, the only difference was that it was (n+2) instead of (n+1) and 1 instead of n. I finally tried c:

        c = n2 + 4n + 5

c = n2 + 4n + 4 + 1

        c = ( n + 2 )2 + 1

I then came up with the following formula for the b+2 family:

        y = 1

        z = ( n + 2 )

I then generalised the formulae for the two families above and I got:

a = z2 – y2

        b = 2zy

        c = z2 + y2

This method only works for numbers where z is bigger than y and both z and y are positive integers. Z must be bigger than y so that we avoid getting any negative numbers. To prove this method works, I then tried to solve the equation algebraically:

a2 + b2 = c2

( z2 – y2 )2 + ( 2zy )2 = ( z2 + y2 )2

Left Hand Side:

( z2 – y2 )2 + ( 2zy )2

Right Hand Side:

( z2 + y2 )2

Left Hand Side:

           ( z2 – y2 )2 + ( 2zy )2

        = z4 + y4 – 2z2y2 + 4z2n2

        =

Right Hand Side:

   ( z2 + y2 )2

        = 2z2y2 + z4 + y4

As a result, the general formula for finding any Pythagorean triple, as long as z is bigger than y, and they are both positive integers is:

a = z2 – y2

        b = 2zy

        c = z2 + y2

...read more.

This student written piece of work is one of many that can be found in our GCSE Pythagorean Triples section.

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