Research on Pythagoras and his work.

Beyond Pythagoras

Introduction: Research on Pythagoras and his work

Pythagoras of Samos is often described as the first pure mathematician. He is an extremely important figure in the development of mathematics yet we know relatively little about his mathematical achievements. Unlike many later Greek mathematicians, where at least we have some of the books which they wrote, we have nothing of Pythagoras's writings. The society which he led, half religious and half scientific, followed a code of secrecy which certainly means that today Pythagoras is a mysterious figure.

We do have details of Pythagoras's life from early biographies which use important original sources yet are written by authors who attribute divine powers to him, and whose aim was to present him as a god-like figure. What we present below is an attempt to collect together the most reliable sources to reconstruct an account of Pythagoras's life. There is fairly good agreement on the main events of his life but most of the dates are disputed with different scholars giving dates which differ by 20 years. Some historians treat all this information as merely legends but, even if the reader treats it in this way, being such an early record it is of historical importance.

Pythagoras's father was Mnesarchus, while his mother was Pythais and she was a native of Samos. Mnesarchus was a merchant who came from Tyre, and there is a story that he brought corn to Samos at a time of famine and was granted citizenship of Samos as a mark of gratitude. As a child Pythagoras spent his early years in Samos but travelled widely with his father. There are accounts of Mnesarchus returning to Tyre with Pythagoras and that he was taught there by the Chaldaeans and the learned men of Syria. It seems that he also visited Italy with his father.

Little is known of Pythagoras's childhood. All accounts of his physical appearance are likely to be fictitious except the description of a striking birthmark which Pythagoras had on his thigh. It is probable that he had two brothers although some sources say that he had three. Certainly he was well educated, learning to play the lyre, learning poetry and to recite Homer. There were, among his teachers, three philosophers who were to influence Pythagoras while he was a young man. One of the most important was Pherekydes who many describe as the teacher of Pythagoras.

The other two philosophers who were to influence Pythagoras, and to introduce him to mathematical ideas, were Thales and his pupil who both lived on Miletus. In [8] it is said that Pythagoras visited in Miletus when he was between 18 and 20 years old. By this time was an old man and, although he created a strong impression on Pythagoras, he probably did not teach him a great deal. However he did contribute to Pythagoras's interest in mathematics and astronomy, and advised him to travel to Egypt to learn more of these subjects. 's pupil, Anaximander, lectured on Miletus and Pythagoras attended these lectures. Anaximander certainly was interested in geometry and and many of his ideas would influence Pythagoras's own views.

In about 535 BC Pythagoras went to Egypt. This happened a few years after the tyrant Polycrates seized control of the city of Samos. There is some evidence to suggest that Pythagoras and Polycrates were friendly at first and it is claimed [5] that Pythagoras went to Egypt with a letter of introduction written by Polycrates. In fact Polycrates had an alliance with Egypt and there were therefore strong links between Samos and Egypt at this time. The accounts of Pythagoras's time in Egypt suggest that he visited many of the temples and took part in many discussions with the priests. According to ([12] and [13]) Pythagoras was refused admission to all the temples except the one at Diospolis where he was accepted into the priesthood after completing the rites necessary for admission.

Both men and women were permitted to become members of the Society, in fact several later women Pythagoreans became famous philosophers. The outer circle of the Society were known as the akousmatics and they lived in their own houses, only coming to the Society during the day. They were allowed their own possessions and were not required to be vegetarians.

Of Pythagoras's actual work nothing is known. His school practised secrecy and communalism making it hard to distinguish between the work of Pythagoras and that of his followers. Certainly his school made outstanding contributions to mathematics, and it is possible to be fairly certain about some of Pythagoras's mathematical contributions. First we should be clear in what sense Pythagoras and the mathematikoi were studying mathematics. They were not acting as a mathematics research group does in a modern university or other institution. There were no 'open problems' for them to solve, and they were not in any sense interested in trying to formulate or solve mathematical problems.

Rather Pythagoras was interested in the principles of mathematics, the concept of number, the concept of a triangle or other mathematical figure and the abstract idea of a proof. As Brumbaugh writes in [3]:-

It is hard for us today, familiar as we are with pure mathematical abstraction and with the mental act of generalisation, to appreciate the originality of this Pythagorean contribution.

In fact today we have become so mathematically sophisticated that we fail even to recognise 2 as an abstract quantity. There is a remarkable step from 2 ships + 2 ships = 4 ships, to the abstract result 2 + 2 = 4, which applies not only to ships but to pens, people, houses etc. There is another step to see that the abstract notion of 2 is itself a thing, in some sense every bit as real as a ship or a house.

Pythagoras believed that all relations could be reduced to number relations. As wrote:-

The Pythagorean ... having been brought up in the study of mathematics, thought that things are numbers ... and that the whole cosmos is a scale and a number.

This generalisation stemmed from Pythagoras's observations in music, mathematics and astronomy. Pythagoras noticed that vibrating strings produce harmonious tones when the ratios of the lengths of the strings are whole numbers, and that these ratios could be extended to other instruments. In fact Pythagoras made remarkable contributions to the mathematical theory of music. He was a fine musician, playing the ...

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Pythagoras believed that all relations could be reduced to number relations. As wrote:-

The Pythagorean ... having been brought up in the study of mathematics, thought that things are numbers ... and that the whole cosmos is a scale and a number.

Pythagoras studied properties of numbers which would be familiar to mathematicians today, such as even and odd numbers, , etc. However to Pythagoras numbers had personalities which we hardly recognise as mathematics today [3]:-

Each number had its own personality - masculine or feminine, perfect or incomplete, beautiful or ugly. This feeling modern mathematics has deliberately eliminated, but we still find overtones of it in fiction and poetry. Ten was the very best number: it contained in itself the first four integers - one, two, three, and four [1 + 2 + 3 + 4 = 10] - and these written in dot notation formed a perfect triangle.

Of course today we particularly remember Pythagoras for his famous geometry theorem. Although the theorem, now known as Pythagoras's theorem, was known to the Babylonians 1000 years earlier he may have been the first to prove it. , the last major Greek philosopher, who lived around 450 AD wrote (see [7]):-

After [, etc.] Pythagoras transformed the study of geometry into a liberal education, examining the principles of the science from the beginning and probing the theorems in an immaterial and intellectual manner: he it was who discovered the theory of and the construction of the cosmic figures.

Again , writing of geometry, said:-

I emulate the Pythagoreans who even had a conventional phrase to express what I mean "a figure and a platform, not a figure and a sixpence", by which they implied that the geometry which is deserving of study is that which, at each new theorem, sets up a platform to ascend by, and lifts the soul on high instead of allowing it to go down among the sensible objects and so become subservient to the common needs of this mortal life.

[7] gives a list of theorems attributed to Pythagoras, or rather more generally to the Pythagoreans.

(i) The sum of the angles of a triangle is equal to two right angles. Also the Pythagoreans knew the generalisation which states that a polygon with n sides has sum of interior angles 2n - 4 right angles and sum of exterior angles equal to four right angles.

(ii) The theorem of Pythagoras - for a right angled triangle the square on the is equal to the sum of the squares on the other two sides. We should note here that to Pythagoras the square on the hypotenuse would certainly not be thought of as a number multiplied by itself, but rather as a geometrical square constructed on the side. To say that the sum of two squares is equal to a third square meant that the two squares could be cut up and reassembled to form a square identical to the third square.

(iii) Constructing figures of a given area and geometrical algebra. For example they solved equations such as a (a - x) = x2 by geometrical means.

(iv) The discovery of irrationals. This is certainly attributed to the Pythagoreans but it does seem unlikely to have been due to Pythagoras himself. This went against Pythagoras's philosophy the all things are numbers, since by a number he meant the ratio of two whole numbers. However, because of his belief that all things are numbers it would be a natural task to try to prove that the hypotenuse of an isosceles right angled triangle had a length corresponding to a number.

(v) The five regular solids. It is thought that Pythagoras himself knew how to construct the first three but it is unlikely that he would have known how to construct the other two.

(vi) In astronomy Pythagoras taught that the Earth was a sphere at the centre of the Universe. He also recognised that the orbit of the Moon was inclined to the equator of the Earth and he was one of the first to realise that Venus as an evening star was the same planet as Venus as a morning star.

Primarily, however, Pythagoras was a philosopher. In addition to his beliefs about numbers, geometry and astronomy described above, he held [2]:-

... the following philosophical and ethical teachings: ... the dependence of the dynamics of world structure on the interaction of contraries, or pairs of opposites; the viewing of the soul as a self-moving number experiencing a form of metempsychosis, or successive reincarnation in different species until its eventual purification (particularly through the intellectual life of the ethically rigorous Pythagoreans); and the understanding ...that all existing objects were fundamentally composed of form and not of material substance. Further Pythagorean doctrine ... identified the brain as the of the soul; and prescribed certain secret cultic practices.

In [3] their practical are also described:-

In their ethical practices, the Pythagorean were famous for their mutual friendship, unselfishness, and honesty.

Pythagoras's Society at Croton was not unaffected by political events despite his desire to stay out of politics. Pythagoras went to Delos in 513 BC to nurse his old teacher Pherekydes who was dying. He remained there for a few months until the death of his friend and teacher and then returned to Croton. In 510 BC Croton attacked and defeated its neighbour Sybaris and there is certainly some suggestions that Pythagoras became involved in the dispute. Then in around 508 BC the Pythagorean Society at Croton was attacked by Cylon, a noble from Croton itself. Pythagoras escaped to Metapontium and the most authors say he died there, some claiming that he committed suicide because of the attack on his Society. Iamblichus in [8] quotes one version of events:-

Cylon, a Crotoniate and leading citizen by birth, fame and riches, but otherwise a difficult, violent, disturbing and tyrannically disposed man, eagerly desired to participate in the Pythagorean way of life. He approached Pythagoras, then an old man, but was rejected because of the character defects just described. When this happened Cylon and his friends vowed to make a strong attack on Pythagoras and his followers. Thus a powerfully aggressive zeal activated Cylon and his followers to persecute the Pythagoreans to the very last man. Because of this Pythagoras left for Metapontium and there is said to have ended his days.

This seems accepted by most but Iamblichus himself does not accept this version and argues that the attack by Cylon was a minor affair and that Pythagoras returned to Croton. Certainly the Pythagorean Society thrived for many years after this and spread from Croton to many other Italian cities. Gorman [6] argues that this is a strong reason to believe that Pythagoras returned to Croton and quotes other evidence such as the widely reported age of Pythagoras as around 100 at the time of his death and the fact that many sources say that Pythagoras taught Empedokles to claim that he must have lived well after 480 BC.

The evidence is unclear as to when and where the death of Pythagoras occurred. Certainly the Pythagorean Society expanded rapidly after 500 BC, became political in nature and also spilt into a number of factions. In 460 BC the Society [2]:-

... was violently suppressed. Its meeting houses were everywhere sacked and burned; mention is made in particular of "the house of Milo" in Croton, where 50 or 60 Pythagoreans were surprised and slain. Those who survived took refuge at Thebes and other places.

The numbers 3, 4 and 5 satisfy the condition 3²+4²=5²,

3² 5²

Because 32= 3x3 =9
42= 4x4 =16
52= 5x5 =25

And so… 32+42=9+16=25=52

I now have to find out if the following sets of numbers satisfy a similar condition of (smallest number) 2+ (middle number) 2= (largest number) 2.

a) 5, 12, 13

5²+12² = 25+144 = 169 = 13².

b) 7, 24, 25

7²+24² = 49+576 = 625 +25²

Here is a table containing the results:

Perimeter and area of the right-angled triangles

a) Perimeter and area of (5,12,13) triangle:

Perimeter = 5+12+13 = 30

Area = 5 x 12 ÷ 2 = 30

Perimeter and area of (7,24,25) triangle:

Perimeter = 7+24+25 = 56

Area = 7 x 24 ÷ 2 = 84

I looked at the first table and noticed that there was only a difference of 1 between the length of the middle side and the length of the longest side.
I already know that the (smallest number) ² + (middle number) ² = (largest number) ². So I know that there will be a connection between the numbers written above. The problem is that it is obviously not:

(Middle number) ²+ (largest number) ²= (smallest number) ²

Because, 12² + 13² = 144+169 = 313
5²= 25

The difference between 25 and 313 is 288 which is far to big, so this means that the equation I want has nothing to do with 3 sides squared. I will now try 2 sides squared.

(Middle) ² + Largest number = (smallest number) ²
= 12+ 13 = 52
= 144 + 13 = 25
= 157 = 25

This does not work and neither will 13², because it is larger than 12². There is also no point in squaring the largest and the smallest or the middle number and the largest number. I will now try 1 side squared.

12² + 13 = 5

This couldn't work because 122 is already larger than 5, this also goes for 13². The only number now I can try squaring is the smallest number.

12 + 13 = 5²

25 = 25

this works with 5 being the smallest number/side but I need to know if it works with the other 2 triangles I know.

4 +5 = 3²
9 = 9

And…

24 + 25 = 7²
49 = 49

It works with both of my other triangles. So…

Middle number + Largest number = Smallest number²

If I now work backwards, I should be able to work out some other odd numbers.

E.g. 92 = Middle number + Largest number
81 = Middle number + Largest number

I know that there will be only a difference of one between the middle number and the largest number. So, the easiest way to get 2 numbers with only 1 between them is to divide 81 by 2 and then using the upper and lower bound of this number. So.

81 = 40.5
2

Lower bound = 40, Upper bound = 41.
Middle side = 40, Largest side = 41.

Key

Longest/Largest Side = Length of Longest Side.
Middle Side = Length of Middle Side.
Shortest Side = Length of Shortest Side.

Let's see if this works for a triangle that I already know.

a c

72 = Middle number + Largest number
49 = Middle number + Largest number

49 = 24.5
2

Lower bound = 24, Upper bound = 25.
Middle Side = 24, Largest Side =25.

This matches the answers I already have with 7 being the shortest side, so I think that this equation works. I now believe I can fill out a table containing the Shortest, Middle and longest sides, by using the odd numbers starting from 3. I already know that the middle and longest side with the shortest length being 3, 5,7 or 9. So I will start with the shortest side being 11.

11² = Middle number + Largest number
121 = Middle number + Largest number

a c

b
121 = 60.5
2

Lower bound = 60, Upper bound = 61.
Middle Side = 60, Largest Side =61.

13² = Middle number + Largest number
169 = Middle number + Largest number

a c

b

169 = 84.5
2

Lower bound = 84, Upper bound = 85.
Middle Side = 84, Largest Side =85.

15² = Middle number + Largest number
225 = Middle number + Largest number

a c

b

225 = 112.5
2

Lower bound = 112, Upper bound = 113.
Middle Side = 112, Largest Side =113.

17² = Middle number + Largest number
289 = Middle number + Largest number

a c

b

289 = 144.5
2

Lower bound = 144, Upper bound = 145
Middle Side = 144, Largest Side =145.

19² = Middle number + Largest number
361 = Middle number + Largest number

a c

b

361 = 180.5
2

Lower bound = 180, Upper bound = 181.
Middle Side = 180, Largest Side =181.

21² = Middle number + Largest number
441 = Middle number + Largest number

a c

441 = 220.5
2

Lower bound = 220, Upper bound = 221.
Middle Side = 220, Largest Side =221.

I now have 10 different triangles, which I think is easily enough to find a relationship between each side.

Shortest side² = Middle side + Longest side

The way I mentioned above and on the previous pages, is quite a good way of finding the middle and longest sides. An easier and faster way to work out the sides would be by using the nth term. I will now try to work out the nth term for each side (shortest middle and longest).

The formula I will work out first is for the shortest side.

3 , 5 , 7 , 9 , 11
+2 +2 +2 +2

the difference between the lengths of the shortest side is 2. This means the equation must be something to do with 2n. So,

There is only a difference of +1 between 2n and the shortest side, so this means the formula should be 2n+1. To see if I'm correct, I will now test this formula.

The Shortest side and 2n+1 columns match meaning that:

Shortest Side = 2n+1

the next formula I need to work out is the formula for the middle side.

4 , 12 , 24 , 40 , 60
+8 +12 +16 +20
+4 +4 +4

The difference of the difference of the lengths of the middle sides is 4. This means that the formula has to have something to do with 4, most likely 4n. However, because 4 are the difference of the difference, the formula must be n2. I now believe that the answer will have something to do with 4n2.

4n2 works for the first term, but, it then collapses after this, as the difference between 4n2 gets larger and larger, the thing you notice is that the difference in the 2nd term between 4n2 and the middle side is the middle side for the term before. This goes for all the other terms from the 2nd.

This means that if I subtract the previous term, then I should in theory get the correct answer.

16 - 4 = 12
36 - 12 = 24
64 - 24 = 40
Etc.

So, the equation I have so far is:

4n2 - (Previous middle side) = Middle side

All the previous term is, (n - 1), so if I put this into the above formula, then it should give me my middle side.

4n2 - 4(n - 1)2 = Middle side.

This should in theory give me my middle side. I will test my theory with the first term.

4 x 12 - 4(1-1)2 = 4
4 x 1 - 4 x 02 = 4
4 - 4 x 0 = 4
4 - 0 = 4
4 = 4

my formula works for the first term. I will now check if it works using the 2nd term.

4 x 2² - 4(2-1) ² = 12
4 x 4 - 4 x 12 = 12
16 - 4 x 1 = 12
16 - 4 = 12
12 = 12

My formula also works for the 2nd term. It's looking likely that this is the correct formula. Just to check, I will check if it works using the 3rd term.
4 x 3² - 4(3-1) ² = 24
4 x 4 - 4 x 22 = 24
36 - 4 x 4 = 24
36 - 16 = 24

20 = 24

My formula doesn't work for the 3rd term. It now looks as if "4n2 - 4(n - 1)²" is not the correct formula after all. To check, I will look to see if the formula works using the 4th term.

4 x 4² - 4(4-1) ² = 40
4 x 16 - 4 x 32 = 40
64 - 4 x 9 = 40
64 - 36 = 40
28 = 40

My formula doesn't work for the 4th term either. I can now safely say that 4n2 - 4(n-1)2 is definitely not the correct formula for the middle side.
I believe the problem with 4n2 - 4(n-1)2 was that 4n2, once you start using larger numbers, becomes far to high to bring it back down to the number that I want for the middle side. Also, 4(n-1)2 is not as small when it gets larger so it doesn't bring the 4n2 down enough, to equal the middle side.
I know that the final formula will have something to do with 4 and have to be n2. I will now try n2 + 4.

I will now look at the differences to see if I can find a pattern there.

1 , 4 , 11 , 20 , 31
+3 +7 +9 +11
+2 +2 +2

The difference of the difference here is 2, which means that the answer will involve 2 and n2.

Straight away, I can see that the difference between 2n2 and the middle number is the 2 times table. The 2 times table in the nth term is 2n. I now think that 2n² + 2n is the correct formula. I will now test it using the first 3 terms.

2 x 1² + 2 x 1 = 4
2 x 1 + 2 = 4
2 + 2 = 4
4 = 4

My formula works for the first term, so, I will now check it on the 2nd term.

2 x 2² + 2 x 2 = 12
2 x 4 + 4 = 12
8 + 4 = 12
12 = 12

My formula also works for the 2nd term. If it works for the 3rd term I can safely say that 2n2 + 2n is the correct formula.

2 x 3² + 2 x 3 = 24
2 x 9 + 6 = 24
18 + 6 = 24
24 = 24

My formula also works for the 3rd term. I am now certain that 2n2 + 2n is the correct formula for finding the middle side.

Middle Side = 2n2 + 2n

This formula also relates back to the 4 that I was writing about, because, 2 + 2 = 4 and 2 is also a factor of 4.

I now have the much easier task of finding the longest side.

I know that there is only a difference of 1 between the middle side and the longest side. So:

(Middle side) + 1 = Longest side.

2n² + 2n + 1 = Longest Side

I am 99.9% certain that this is the correct formula. Just in case, I will check it using the first 3 terms.

2n² + 2n +1 = 5
2 x 1² + 2 x 1 + 1 = 5
2 + 2 + 1 = 5
5 = 5

The formula works for the first term.

2n² + 2n +1 = 13
2 x 2² + 2 x 2 + 1 = 13
8 + 4 + 1 = 13
13 = 13

The formula also works for the 2nd term.

2n² + 2n +1 = 25
2 x 3² + 2 x 3 + 1 = 25
18 + 6 + 1 = 25
25 = 25

The formula works for all 3 terms. So…

Longest Side = 2n² + 2n + 1

Now, I will check that 2n + 1, 2n² + 2n and 2n² + 2n + 1 forms a Pythagorean triple (or a² + b² = c²). We have a = 2n + 1, b = 2n² + 2n, and c = 2n2 + 2n + 1.

c² - b² = (c - b) (c + b) = c + b = 2b + 1 = 4n² + 4n + 1 = (2n + 1)2 = a².

Or I can show it this way,

a² + b² = c²

This equals:

(2n +1) ² + (2n2 + 2n) ² = (2n2 + 2n +1) ²

If you then put these equations into brackets:

(2n + 1)(2n+1) + (2n² + 2n)(2n² + 2n) = (2n² + 2n + 1)(2n² + 2n + 1)

You need to factorise the brackets out:

4n² + 4n + 1 + 4n + 8n² + 4n² = 4n + 4n³ + 2n² + 4n³ + 4n² + 2n + 2n² +
2n +1

If I now balance out each side, and I end up with nothing, then 2n + 1, 2n² + 2n and 2n² + 2n + 1 is a Pythagorean triple.

4n² + 4n² = 4n² + 2n² + 2n²
4n = 2n + 2n
8³ = 4n³ + 4n³
4n = 4n
1 = 1

I now end up with 0 = 0, so 2n + 1, 2n² + 2n and 2n² + 2n + 1 has got to be a Pythagorean triple.

I now have the nth term for each of the 3 sides of a right-angled triangle. I can now work out, both, the nth term for the perimeter and the nth term for the area.
The perimeter of any triangle is just the length of the 3 sides added together. E.g.

1st term 3 + 4 + 5 =12

So 12 is the perimeter for the first term.

2nd term 5 + 12 + 13 = 30
3rd term 7 + 24 + 25 = 56

And so on. But, what if I didn't know these numbers, then I would have to work out it out using the nth term and this is what I want to work out. Lucky, for me, I already know the nth term for each of the sides. So, it's all a matter of putting those 3 formulas together.

Perimeter = (Shortest side) + (middle side) + (Longest Side)
= 2n + 1 + 2n² + 2n + 2n² + 2n + 1
= 2n² + 2n² + 2n + 2n + 2n + 1 + 1
= 4n² + 6n + 2

This, if I have done my calculations properly, should be the right answer. To check, I am going to use the 4th, 5th and 6th terms.

4th term:

4n² + 6n + 2 = Perimeter.
4 x 4² + 6 x 4 + 2 = 9 + 40 + 41
64 + 24 + 2 = 90
90 = 90

My formula works for the 4th term.

4n² + 6n + 2 = Perimeter.
4 x 5² + 6 x 5 + 2 = 11 + 60 + 61
100 + 30 + 2 = 132
132 = 132

And it works for the 5th term.

4n² + 6n + 2 = Perimeter.
4 x 6² + 6 x 6 + 2 = 13 + 84 + 85
144 + 36 + 2 = 182
182 = 182

My formula works for all 3 terms, so…

Perimeter = 4n² + 6n + 2

Like the perimeter, I already know that the area of a triangle is found by:

Area = ½ b h

b = Base
h = Height

Depending on which way up the right angled triangle is the shortest side or middle side can either be the base or height because it doesn't really matter which way round they go, because I'll get the same answer either way! So…

Area = ½ (Shortest Side) X (Middle Side)
= ½ (2n +1) x (2n² + 2n)
= (2n +1) (2n² + 2n)
2

I will now check this using the first 3 terms.

(2n + 1) (2n² + 2n) = ½ b h
2

(2 x 1 + 1) (2 x 12 + 2 x 1) = ½ x 3 x 4
2

3 x 4 = ½ x 12
2

12 = 6
2

6 = 6

My formula works for the first term.

(2n + 1) (2n² + 2n) = ½ b h
2

(2 x 2 + 1) (2 x 2² + 2 x 2) = ½ x 5 x 12
2

5 x 12 = ½ x 60
2

60 = 30
2

30 = 30

My formula also works for the 2nd term.

(2n + 1) (2n² + 2n) = ½ b h
2

(2 x 3 + 1) (2 x 3² + 2 x 3) = ½ x 7 x 24
2

7 x 24 = ½ x 168
2

168 = 168
2

168 = 168

My formula works for all 3 terms. So…
Area = (2n +1) (2n² + 2n)
2

If I am given a right-angled triangle I can always apply an enlargement to it (for example, I can double all lengths) and get another right-angled triangle. This means that from a given triple a, b, c we can produce many more Pythagorean triples na, nb, nc for any whole number n.
For example, starting with the triple 3, 4, 5 and taking n = 5 we get the new triple 15, 20, 25. Sometimes we can reverse this process by starting with a triple and then reducing the lengths of the sides to get another triple.
This does not always work; if we start with 3, 4, 5, for example, and halve the lengths of the sides we do not get a triple of whole numbers. However, sometimes we do; for example, by halving lengths the triple 10, 24, 26 converts into the triple 5, 12, 13.

As an extra part, I am now going to find the relationships (differences) between the Short and Middle sides, the Short and Long Sides, and The Middle and Long sides.
Finding the relationship between 2 sides is quite easy, because, I already know the nth terms for each side. Because of this, all I have to do to find the relationship is to subtract one nth term from the other, leaving me with the relationship.

Middle Side - Short Side = Relationship.
2n² + 2n - 2n + 1 = Relationship
2n² - 1 = Relationship

2n² - 1 and the Relationship between the shortest and middle side both match. So…

Relationship between Shortest and Middle sides = 2n² - 1

The next relationship I'm going to work out is the relationship between the shortest and longest sides.

Longest Side - Short Side = Relationship.
2n² + 2n + 1 - 2n + 1 = Relationship.
2n² + 1 = Relationship.

2n² + 1 and the Relationship between the shortest and longest sides both match. So…

Relationship between the shortest and longest sides = 2n² + 1

The next relationship I'm going to find is between the middle and longest side.

Longest Side - middle Side = Relationship.
2n² + 2n + 1 - 2n² + 2n = Relationship.
1 = Relationship.

I am certain that the relationship between the longest and middle side is 1.

1 and the Relationship between the middle and longest sides both match. So…

Relationship between the middle and longest sides = 1

Using these same principles, I can work out any relationship, to prove this, I will work out the relationship between the shortest side and the perimeter and then the area.

Perimeter - Shortest Side = Relationship.
4n² + 6n + 2 - 2n + 1 = Relationship.
4n² + 4n + 1 = Relationship.

I am certain that this is the write answer. So…

Relationship between the Perimeter and Shortest side = 4n² + 4n + 1

The area is even simpler because, all you have to do is knock the 2n + 1 out of the equation. So…

Relationship between the Area and Shortest side =