# Beyond Pythagoras

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Introduction

Beyond Pythagoras

I have been asked to investigate Pythagorean triplets where the shortest side is an odd number and all the three sides are positive integers. A pythagorean triple is a set of integers (a,b,c) that specifies the lengths of a right angle triangle a²+b²=c² in which ‘a’ is the shortest side ‘b’ is the middle side and ‘c’ is the hypotenuse.

The first set of triples (3,4,5) which has already been proved to satisfy Pythagoras’s theory.

I have also been given two other pythagorean triples (5,12,13) and (7,24,25) I will now prove these to satisfy Pythagoras’s theory

a² = 5² =25

b² = 12² =144

c² = 13² =169

a²+b² =25+144 =169

a²+b²=c² so Pythagoras’s theory holds for (5,12,13) because they satisfy the condition of a²+b²=c² in a right angled triangle.

a² = 7² =49

b² = 24² =576

c² = 25² =625

a²+b² =49+576 =625

a²+b²=c² so Pythagoras’s theory holds for (7,24,25) because they satisfy the condition of a²+b²=c² in a right angled triangle.

I am now going to put my results in to a table so that I can predict more values:

a | b | c |

3 | 4 | 5 |

5 | 12 | 13 |

7 | 24 | 25 |

I will now predict the next two values in the table so I can work out a general formula for this pythagorean family.

Middle

So now with 5 triplets that I have proved I will now find a general formula to find other pythagorean triplets in this family.

By looking at the table it’s obvious that the formulae for ‘a’ is a=2n+1

I’m now going to find the general formulae for the ‘b’ and ‘c’ value.

(b+1) ² = (2n+1)² + b

b²+2b+1 = 4n²+4n+1+b²

b²+2b-b² = 4n²+4n+1-1

2b = 4n²+4n

b = 2n²+2n

c = 2n²+2n+1

I now have the formulae:

a =2n+1

b = 2n²+2n

c = 2n²+2n+1

I am now going to prove that these formulae work

a² = (2n+1)² )

b² = (2n²+2n)² )

c² = (2n²+2n+1)² =

a² = (2n+1) (2n+1)

4n²+2n+2n+1

4n²+4n+1

b² = (2n²+2n) (2n²+2n)

4n +4n³+4n³+4n²

a²+b² = 4n + 8n³+8n²+2n+1

c² = (2n²+2n+1) (2n²+2n+1)

4n + 4n³+2n²+ 4n³+4n²+2n+2n²+2n+1

4n + 8n³+8n²+2n+1

I will give an example of how to use the formulae for the 10th as the nth term:

a =2n+1 =21

b = 2n²+2n = 220

c = 2n²+2n+1 = 221

the 10th pythagorean triple in this pythagorean family is (21,220,221)

I am now going to investigate the 2nd pythagorean family. I will get this by doubling the values of the first family.

(3,4,5) = (6,8,10)

(5,12,13) = (10,24,26)

(7,24,25) = (14,48,50)

(9,40,41) = (18,80,82)

(11,60,61) = (22,120,122)

Conclusion

c² = 123² =15,129

a²+b² =729+14,400 =15,129

a²+b²=c² so Pythagoras’s theory holds for (27,120,123) because they satisfy the condition of a²+b²=c² in a right angled triangle.

a² = 33² =1089

b² = 180² =32,400

c² = 183² =33,489

a²+b² =441+5184 =5625

a²+b²=c² so Pythagoras’s theory holds for (33,180,183) because they satisfy the condition of a²+b²=c² in a right angled triangle.

I am now going to find a general formula for the 3rd pythagorean family that I will name the b+3 family.

From the values I have I have used the differencing method to find out that a = 6n+3 I will now work out the ‘b’ and ‘c’ values.

(b+3) ² = (6n+3)² + b

b²+6b+1 = 36n²+36n+9+b²

b²+6b-b² = 36n²+36n+9-9

6b = 36n²+36n

b = 6n²+6n

c = 6n²+6n+3

I now have the formulae:

a = 6n+3

b = 6n²+6n

c = 6n²+6n+3

I am now going to prove that these formulae work:

a² = (6n+3)² )

b² = (6n²+6n)² )

c² = (6n²+6n+3)² =

a² = (6n+3) (6n+3)

36n²+18n+18n+9

36n²+36n+9

b² = (6n²+6n) (6n²+6n)

36n +36n³+36n³+36n²

36n +72n³+36n²

a²+b² = 36n +72n³+72n²+36n+9

c² = (6n²+6n+3) (6n²+6n+3)

36n + 36n³+18n²+36n³+36n²+18n+18n²+18n+9

36n + 72n³+82n²+36n+9

I will give an example of how to use the formulae for the 10th as the nth term:

a = 6n+3 =63

b = 6n²+6n = 3660

c = 6n²+6n+3= 3663

the 10th pythagorean triple in this pythagorean family is (21,220,221)

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

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