# Beyond Pythagoras

Beyond Pythagoras

Introduction

Believed to have been born in 582BC and died in 500BC approximately, Pythagoras was a Greek philosopher and mathematician. He discovered some of the most influential theories of number, geometry and proportion which are still frequently used in modern mathematics.

Pythagorean triples, on which this investigation is based, are sets of three numbers as seen in the above diagram. Here it is the simplest triple; 3,4,5. They will be written in the following form throughout this paper; a,b,c. There are other Pythagorean triples such as 5,12,13; which is the 2nd odd triple and 361,65160,65161; which is the 180th odd triple.

His theorem is this: the square of the hypotenuse (longest side (c (5 (25)))) of a right-angled triangle is equal to the sum of the squares of the other two sides (a (3 (9)) and b (4 (16))). This can be expressed using the following equation:

a2+b2=c2

Aim

The aim of the investigation is to explore and mathematically express the relationships between the number of the triple (n), the length of each side (a; the shortest), (b; the intermediate), (c; the longest), the perimeter (P) and area (A) of a right-angled triangle.

Please note: Throughout this investigation, the sum on the bottom line is the final answer to each set of sums. For example:

y=2+2

y=4

In this example, the bottom line (y=4) should be interpreted as the answer.

The main rule of the investigation is that the values of the above (n, a, b, c, P and A) must all be positive integers (i.e. whole numbers of greater value than zero). If any of the numbers concerned do not match these criteria (i.e. they are decimals), then they and their respective triple will be excluded from the sequence of triples and will not be considered in this investigation, as this would cause the investigation to be infinitely complex. For example, if in the first term of the sequence, a equalled 1, then b and c would not be integers (1·3 recurring and 1.6 recurring respectively), as an integer has to be a whole number by definition. Therefore, the first odd Pythagorean triple to satisfy these criteria is the 3,4,5 triple, and it is subsequently the first term of the odd triples sequence.

The first sequence of triples that will be investigated are those with the first number (i.e. 3 of 3,4,5) as an odd number. I will refer to these triples as "odd triples" throughout the investigation for convenience.

The second sequence of triples that will be investigated are those with the first number (i.e. 6 of 6,8,10, although this triple should not be counted as a "true" triple, as we shall see later) as an even number. I will refer to these triples as "even triples" throughout the investigation for convenience.

The first section of the investigation is to collect odd triples so that they can be analysed and turned into formulae relating to the number of the term of the sequence (n). The first set of triples was to be where a equalled an odd number. I wanted to collect the first 8 odd triples using trial and error, but realised that using this time to devise a method to find b and c from a would be more effective. This would be achieved by using a to calculate the other two sides, and this would be simple because a is already known (it is an odd number starting from and including 3). The formula was calculated in the following way.

* I found that each of the three triples of which I had discovered by trial and error had a factor in common with each other, and this was that the difference between b and c was always 1. This meant that if I could find the relationship between a and the midway point between b and c (b+0·5 or c-0·5), I could subtract or add 0·5 to this midway point to find b and c respectively.

* I began by finding the relationship between a and this midway point; the relationship between 3 and 4·5. I tried multiplying a by 1·5 and ended up with 4·5. This worked for the first triple, but when it came to the 5,12,13 triple it did not. I then realised the connection. If you square a and then divide it by 2, you will end up with the midway number (4·5). I tested this on the next two triples and it was successful, so I had found one of the most time saving formulae in this investigation.

* To find the remaining five triples before I analysed all eight of them, I could now utilise this formula.

I discovered that if you square a and divide it by 2, you end up with the base number for the next two sides, b and c. By subtracting and adding 0·5 to the answer, you can find b and c respectively. For example, in the first Pythagorean triple:

b= a2-0·5 c= a2+0·5

2 2

b= 32-0·5 c= 32+0·5

2 2

b= 4·5-0·5 c= 4·5+0·5

b= 4 c= 5

These ...