# Beyond Pythagoras

Extracts from this document...

Introduction

William Kerslake Mathematics GCSE Coursework

Beyond Pythagoras

## Aim

I am conducting this investigation to discover formulae that will allow me to calculate many Pythagorean triples. I will first find formulae for odd Pythagorean triples and then even ones. For each, I will find formulae for ‘a’, ‘b’, ‘c’, perimeter and area, all in terms of ‘n’. After this I hope to discover a general formula for all Pythagorean triples, although it is unlikely that I will.

Contents

Page Number | |

1 | Aim and Contents |

2 Odd and Even | Tables of Pythagorean triples |

3 Odd | Formula for ‘a’ in terms of ‘n’ |

4 Odd | Formula for ‘b’ and ‘c’ in terms of ‘a’ |

5 Odd | Formula for ‘b’ and ‘c’ in terms of ‘n’ |

6 Odd | Perimeter and Area in terms of ‘n’ |

7 Even | Formula for ‘a’ in terms of ‘n’ |

8 Even | Formula for ‘b’ and ‘c’ in terms of ‘a’ |

9 Even | Formula for ‘b’ and ‘c’ in terms of ‘n’ |

10 Even | Perimeter and Area in terms of ‘n’ |

11 Odd and Even | Summary – All Formulae |

Odd Pythagorean Triples

Table of first 10 odd triples

n | a | b | c | Perimeter | Area |

1 | 3 | 4 | 5 | 12 | 6 |

2 | 5 | 12 | 13 | 30 | 30 |

3 | 7 | 24 | 25 | 56 | 84 |

4 | 9 | 40 | 41 | 90 | 180 |

5 | 11 | 60 | 61 | 132 | 330 |

6 | 13 | 84 | 85 | 182 | 546 |

7 | 15 | 112 | 113 | 240 | 840 |

8 | 17 | 144 | 145 | 306 | 1224 |

9 | 19 | 180 | 181 | 380 | 1710 |

10 | 21 | 220 | 221 | 462 | 2310 |

Middle

Difference

1

1

3

2

1

2

5

2

1

3

7

2

1

4

9

2

The differences between the numbers show that ‘a’ is changing twice as fast as ‘n’. Therefore, 2n must be part of the formula.

Difference | 2n | Difference between n and a | a | Difference |

2 | 2 | 1 | 3 | 2 |

2 | 4 | 1 | 5 | 2 |

2 | 6 | 1 | 7 | 2 |

2 | 8 | 1 | 9 | 2 |

The differences show that ‘a’ and ‘n’ are changing at the same rate – increasing by 2 each time, thus proving that the formula includes 2n. The difference between ‘a’ and ‘n’ is always 1; therefore, our formula needs to have a +1 included.

From this, I know the formula for ‘a’, in terms of ‘n’, to be;

a = 2n + 1

I personally cannot see an obvious link between ‘n’ and ‘b’ or ‘c’, so I will use the known formula, a² +b² = c² to discover ‘b’ and ‘c’ in terms of ‘n’.

3² + 4² = 5² and 5² + 12² = 13² both have a hypotenuse one greater than one side (in this case known as ‘b’). Looking at my table of triples I can see that there are more odd triples that follow this pattern. Using this, I should be able to discover ‘b’ and ‘c’ in terms of ‘a’.

3² + 4² = 5²

a² + b² = c²

a² + b² = (b + 1)²

a² + b² = b² + 1 + 2b

a² = 2b + 1

2b = (a² - 1)

b = (a² - 1)

2

Remove extra variable ‘c’ from formula

Multiply out brackets

Simplify formula

Rearrange formula

Rearrange formula

From this, I know the formula for ‘b’, in terms of ‘a’, to be;

b = (a² - 1)

2

From this formula for ‘b’ in terms of ‘a’ I can discover ‘c’ in terms of ‘a’;

a | b | (a² - 1)/2 | a²-1 | a² | a²+1 | (a² + 1)/2 | c |

3 | 4 | 4 | 8 | 9 | 10 | 5 | 5 |

5 | 12 | 12 | 24 | 25 | 26 | 13 | 13 |

7 | 24 | 24 | 48 | 49 | 50 | 25 | 25 |

9 | 40 | 40 | 80 | 81 | 82 | 41 | 41 |

Conclusion

4

b = 4n² + 8n + 8n +16 - 4

4

b = 4n² + 16n + 12

4

b = n² + 4n + 3

Insert a = 2n + 4

Multiply out brackets

Sort like numbers/letters

Cancel down to simplest form

And the same with ‘c’;

c = (a² + 4)

4

c = ((2n + 4) ² + 4)

4

c = 4n² + 8n + 8n +16 + 4

4

c = 4n² + 16n + 20

4

c = n² + 4n + 5

Insert a = 2n + 4

Multiply out brackets

Sort like numbers/letters

Cancel down to simplest form

b = 2n² + 2n

c = 2n² + 2n + 1

Now that I have formulae for ‘a’, ‘b’ and ‘c’ in terms of ‘n’; I can put these into my simple area and perimeter formulae and will then be able to calculate both area and perimeter in terms of ‘n’.

Perimeter

Perimeter = a + b + c

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

P = 2n + 4 + n² + 4n +3 + n² + 4n + 5

P = 10n + 12 + 2n²

Perimeter = 2n² + 10n + 12

Area

Area = a x b

2

A = (2n+4) x (n² + 4n +3)

2

A = 2n³ + 8n² + 6n + 12 + 4n² + 16n

2

A = 2n³ + 12n² + 22n + 12

2

A = n³ + 6n² + 11n + 6

## Area = n³ + 6n² + 11n + 6

## Formulae Summary

These are all the formulae I have discovered in my investigation. I can now workout any odd or even Pythagorean triple and its perimeter and area knowing only the number, n. For example, the 26 odd triple is: a = 53, b = 1404, c = 1405, perimeter = 2862. area = 37206

## Odd Pythagorean Triples

a = 2n + 1

b = 2n² + 2n

c = 2n² + 2n + 1

Perimeter = 4n² + 6n + 2

Area = 2n³ + 3n² + n

## Even Pythagorean Triples

a = 2n + 4

b = n² + 4n + 3

c = n² + 4n + 5

Perimeter = 2n² + 10n + 12

Area = n³ + 6n² + 11n + 6

/

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