# Beyond Pythagoras.

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Introduction

## BEYOND PYTHAGORAS

## Introduction

For this piece of work I am investigating Pythagoras. Pythagoras was a Greek mathematician. Pythagoras lived on the island of Samos and was born around 569BC. He did not write anything but he is regarded as one of the world's most important characters in maths. His most famous theorem is named after him and is called the Pythagoras Theorem. It is basically a?+b?=c?. This is what the coursework is based on.

I am going to look at the patterns, which surround this theorem and look at the different sequences that can be formed.

### The coursework

The numbers 3, 4 and 5 satisfy the condition 3?+4?=5? because

3?=3x3=9

4?=4x4=16

5?=5x5=25

And so

3?+4?=9+16=25=5?

I will now check that the following sets of numbers also satisfy the similar condition of (smallest number) ?+(middle number) ?=(largest number) ?

a) 5, 12, 13

5?=5x5=25

12?=12x12=144

25+144=169

√169 = 13

This satisfies the condition as

5?+12?=25+144=169=13?

b) 7, 24, 25

7?=7x7=49

24?=24x24=576

49+576=625

√625=25

This satisfies the condition as

7?+24?=49+576=625=25?

The numbers 3,4 and 5 can be the lengths - in appropriate units - of the side of a right-angled triangle.

5

3

The perimeter and area of this triangle are:

Perimeter = 3+4+5=12 units

Area = ?x3x4=6 units ?

The numbers 5,12,13 can also be the lengths - in appropriate units - of a right-angled triangle.

### Perimeter = 5+12+13=30

Area=?x5x12=30

This is also true for the numbers 7,24,25

### Perimeter = 7+24+25=56

### Area=?x7x24=84

I have put these results into a table to see if I can work out any patterns.

Length of shortest side | Length of middle side | Length of longest side | Perimeter | Area |

3 | 4 | 5 | 12 | 6 |

5 | 12 | 13 | 30 | 30 |

7 | 24 | 25 | 56 | 84 |

I have noticed that there seems to be a recurring pattern between the length of the middle side and the length of the longest side. They are always consecutive numbers. I shall now investigate this.

I will assume that the hypotenuse has length b + 1 where b is the length of the middle side.

Middle

To see if this statement is true I shall use the Pythagorean triple 9,40,41.

40?+41?=9?

This is not true because

40? = 1600

41? = 1681

which add together to make 3281

and

9?=81

This means that this is not a correct formula.

(Longest side) ?/(shortest side) ? = (middle side)

To check this I will use the Pythagorean triple 15,112,113.

113?/15?=112?

This is wrong as

113?=12769

15?=225

and

12769/225 = 56.751111

and as

The middle side = 113 this formula has to be wrong.

Shortest side x longest side = (middle side) ?

To check this I will use the Pythagorean triple 7,24,25

7 x 25 = 175

This is wrong because

24?= 576 and this is very different

(Longest side) ?-(middle side) ?=shortest side

To check this formula I will use the Pythagorean triple 5,12,13

13?-12?=5?

This is right as

13?=169

12?=144

169-144=25 which is also 5?.

So this formula is correct

(Middle side) ?+Smallest side=Largest side

To check this I will use the Pythagorean triple 5, 12,13.

12?+5=13

This is silly and I shouldn't really have tried it, as I should have been able to work out that this formula is incorrect just by looking at the numbers involved. 144 is bigger than 13 without adding the 5.

Largest side + Middle side = (shortest side) ?

To check this I will use the Pythagorean triple 3,4,5

5+4=3?

This is correct because 5 + 4 = 9 which is also 9.

Just to check that this formula works with all triples I shall pick 3 random numbers from the N column of the table and use the formula with them. I shall use (i) 23 (ii) 10 (iii) 16

(i) 1105+1104=47? which is also 2209. This is correct

(ii) 221+220=21? which is also 441. This is correct

(iii) 545+544=33? which is also 1089. This is correct

This seems to be a correct formula.

So the formulae I have worked out in this section are:

? (Shortest side)?+(middle side)?=(longest side)?

? (Longest side) ?-(middle side) ?=shortest side

? Largest side + Middle side = (shortest side) ?

Conclusion

Even though this is a good formula I have worked out from further research that if you use this formula it will not generate every single triple.

Internet

An interesting thing I found on the Internet, which I had never heard of before, was Pythagorean Quads (quadruples).

Pythagorean quads are the 3D version of the triples and can all be generated by the following formula, where a, b, c and d are whole numbers.

If p = a2 + b2 - c2 - d2

q = 2(ad - bc)

r = 2(ac + bd)

and s = a2 + b2 + c2 + d2

then p2 + q2 + r2 = s2.

I shall now try and work out a few of these to check if this formula is correct

I shall choose the numbers a=4, b=1, c=1 and d=1.

This means that p=16+1-1-1 = 15

q=2(4-1)=6

r=2(4+1)=10

s=16+1+1+1=19

From looking further on the Internet for confirmation I conclude that this formula does work because the quad that I worked out above is correct.

Conclusion

I started this investigation by noticing that there was a difference of 1 between the middle side and the longest side. I then went on to work out formulae relating to this. All these formulae were based on the assumption that c-b+1. however after investigating further I realised that the difference between the middle side and the longest side could be any number. So I investigated all the differences up to a difference of 5. I decided that I would stop each table at the value of b=25. I realised that this investigation could go on forever and you could investigate unlimited values of b and there are unlimited differences that I could have investigated but I had to set a limit. I also decided that I would only investigate up to c=b+10 as I felt that was a sufficient number of investigations. I wrote the extension because I thought it was interesting to see what else other people had discovered

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

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