• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month
Page
  1. 1
    1
  2. 2
    2
  3. 3
    3
  4. 4
    4
  5. 5
    5
  6. 6
    6
  7. 7
    7
  8. 8
    8
  9. 9
    9
  10. 10
    10
  11. 11
    11
  12. 12
    12
  13. 13
    13
  • 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.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related GCSE Pythagorean Triples essays

  1. Investigate the area of triangle studies including the Pythagorean Theorem and in particular Pythagorean ...

    Shortest, Middle and longest sides, by using the odd numbers starting from 3. I already know that the middle and longest side with the shortest length being 3, 5,7 or 9. So I will start with the shortest side being 11.

  2. Maths Number Patterns Investigation

    Also, 4(n-1)2 is not as small when it gets larger so it doesn�t bring the 4n2 down enough, to equal the middle side. I know that the final formula will have something to do with 4 and have to be n2. I will now try n2 + 4.

  1. Pythagoras Theorem

    = a4 + 4a� + 4a� + 16 = a4 + 8a� + 16 Change the formula: a� + (a4 - 8a� +16)/16 = (a4 + 8a� + 16)/16 o 16a� + a4 - 8a� + 16 = a4 + 8a� + 16 = a4 + 8a� + 16 =

  2. Beyond Pythagoras

    To see if I'm correct, I will now test this formula. If n=1; Shorter side=2n+1 =2(1)+1 =3 If n=2; Shorter side=2(2)+1 =5 If n=3; Shorter Side=2(3)+1 =7 Therefore all these tests prove that: Shortest Side=2n+1 The next formula I need to work out is the formula for the middle side.

  1. Math's Coursework: Pythagoras triples.

    I have now got the formula for all sides; all I need to do now is figure out the formula for Perimeter and Area. Formula of Perimeter I do not have to examine the perimeter's patterns because the perimeter is all of the sides of triangle added together.

  2. Beyond Pythagoras.

    You could just say a + b + c. however if we join all the equations found already we should be able to get a unique formula to work out the equation. I will add the formula for the shortest side with the formulas for the middle side and longest side.

  1. I am going to investigate Pythagorean triples where the shortest side is an odd ...

    +4 +24 14 48 +8 50 +2 (from number in middle side) +4 +32 18 80 +8 82 +2 (from number in middle side) +4 +40 22 120 122+2 (from number in middle side) a�+b�=c� a�+b�=c� a�+b�=c� 6�+8�=10� 14�+48�=50� 22�+120�=122� 36+64=100 196+2304=2500 484+14400=14884 100=100 2500=2500 14884=14884 Looking at the table, I have noticed that the difference has increased to +4.

  2. For this piece of work I am investigating Pythagoras. Pythagoras was a Greek mathematician.

    This must mean that the formula is: 2n�+2n I think this is the formula from looking at the first few terms but I will check it using another random number. The number will be 18. 2x18�+2x18=684 This is the same as the number given in the table so I presume that it is right.

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work