# Pythagoras Theorem

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Introduction

Beyond Pythagoras

Pythagoras was a famous Greek mathematician. He produced a formula (a² + b² = c²) called the ‘Pythagoras Theorem.’ This allows you to correctly calculate the length of a right-angled triangles’ hypotenuse, when you know the length of the other two sides.

A Pythagorean triple is when a right-angled triangles three lengths (a = shortest, b = medium, c = hypotenuse) are all whole numbers- integers. At the start of this course, I was given a set of 3 Pythagorean triples. These were:

A | B | C |

3 | 4 | 5 |

5 | 12 | 13 |

7 | 24 | 25 |

I was asked to look for patterns between the numbers in each Pythagorean triple. This is what I found:

- The difference between b and c is one
- a was odd
- b was even and a multiple of four
- c was odd
- c = b + 1
- (a² + 1)/2 = c
- (a² - 1)/2 = b

The two formulae that I am going to focus on, and use to discover more Pythagorean Triples are:

- (a² + 1)/2 = c
- (a² - 1)/2 = b

Using these two rules I can use some more odd numbers to expand my table to:

A | B | C |

9 | 40 | 41 |

11 | 60 | 61 |

13 | 84 | 85 |

15 | 112 | 113 |

To test that these were possible Pythagorean triples, I put them through the original Pythagoras theory:

- 9² + 40² = 1,681 √1,681 = 41
- 11² + 60² = 3,721 √3,721 = 61
- 13² + 84² = 7,225 √7,225 = 85
- 15² + 112² = 12,769 √12,769 = 113

Middle

This proves that my second sets of numbers are correct Pythagorean Triples. I repeated what I did before and looked for patterns, this is what I found:

- the difference between b and c is 2
- a is even
- b is even
- c is even
- values for a = 8, and a = 12 are missing
- a² = 2(b + c)
- a²/2 = b + c
- b = (a² - 4)/4
- c = (a² + 4)/4

To fill in the missing values of ‘a,’ I used the last two formulae in the list. This meant that my second completed table looked like this:

A | B | C |

6 | 8 | 10 |

8 | 15 | 17 |

10 | 24 | 26 |

12 | 35 | 37 |

14 | 48 | 50 |

I then tested the 2 new triples by using the original formula:

- 8² + 15² = 289 √289 = 17
- 12² + 35² = 1,369 √1,369 = 37

I have noticed that the difference between ‘b’ and ‘c’ has changed, from 1 to 2. I decided that I wanted to investigate further and try to get the difference to 3. To get this I multiplied the original triples by three to get:

A | B | C |

9 | 12 | 15 |

15 | 36 | 39 |

21 | 72 | 75 |

I noticed that the triples for ‘a = 12’ and ‘a = 18’ were missing. I wanted to make a formula that would allow me to calculate the triples for these values of ‘a,’ whilst doing this I noticed that the difference between ‘b’ and ‘c’ was infact 4 not 3. Although this wasn’t the result that I was looking for the sets of numbers were infact all triples.

I then looked at three different formulas for ‘b’ and ‘c’ separately and this is what I found: (d id the difference between ‘b’ and ‘c’)

When d = 1:

- b = (a² - 1)/2
- c = (a² + 1)/2

When d = 2:

- b = (a² - 4)/4
- c = (a² + 4)/4

When d = 3:

- b = (a² - 9)/9
- c = (a² + 9)/9

I then decided to make a general formula for finding out the values of ‘b’ and ‘c’ when you know the value of ‘a.’ These are:

- b = (a² - d²)/2d
- c = (a² + d²)/2d

I now wanted to prove that each of my formulae were correct. To do this I am going to substitute the original formula ‘a² + b² = c²’ with the separate formulas for ‘b’ and ‘c.’

- a² + b² = c²
- = a² + ([a² - 1]/2)² = ([a² + 1]/2)²
- = a² + ([a² - 1]/2)([a² - 1]/2) = ([a² + 1]/2) ([a² + 1]/2)
- = a² + (a² - 1)(a² - 1)/4 = (a² + 1)(a² + 1)/4

To make it easier to prove the formula, I am going to multiply out the brackets:

- (a² - 1)(a² - 1)

= a4 – a² - a² + 1

= a4 – 2a² + 1

- (a² + 1)(a² + 1)

Conclusion

- b = (19² - 5²) / (2 x 5) = (361 - 25) / 10 = 33.6
- c = (19² + 5²) / (2 x 5) = (361 + 25) / 10 = 38.6

These values are incorrect as they are not integers. In this exercise, I have found that ‘a’ must be larger than ‘d’ because otherwise the answer will be negative and a triangle cannot have a negative length. To make the answers from the general formula integers, I have to multiply out the denominators. To achieve this, I am going to multiply the values of ‘a,’ ‘b’ and ‘c’ by ‘2d.’ By changing the general formula in this way, my answers should be positive integers. I now have to prove my general formula in the same way as I proved my dependent formula:

- a² + b² = c²
- = (2ad)² + (a² - d²)² = (a² + d²)²

- (2ad)(2ad) = 4a²d²
- (a² - d²)(a² - d²) = a4 – 2a²d² + d4
- (a² + d²)(a² + d²) = a4 + 2a²d² + d4

This means that the original formula ‘a² + b² = c²,’ when the respective formulas are substituted in, becomes:

- a4 + 2a²d² + d4 = a4 + 2a²d² + d4

Because both sides of the formula have cancelled down so that they are the same on both sides, I have proven that my independent general formulae are correct.

By Adam Taylor- Miss Clarke B27

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

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