Pythagoras Theorem

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:

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:

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

I had noticed earlier that ‘a’ was an odd number and I wanted to see if Pythagorean triples would work when ‘a’ was an even number. To make the value of ‘a’ even I decided to double the original three sets of triples:

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I then tested this new set of numbers, by putting them trough the original theorem:

6² + 8² = 100 √100 = 10
10² + 24² = 676 √676 = 26
14² + 48² = 2,500 √2,500 ...

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I then tested this new set of numbers, by putting them trough the original theorem:

6² + 8² = 100 √100 = 10
10² + 24² = 676 √676 = 26
14² + 48² = 2,500 √2,500 = 50

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:

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:

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)

= a4 + a² + a² + 1

= a4 + 2a² + 1

This means that the formula ‘a² + b² = c²’ became:

a² + (a4 – 2a² +1)/4 = (a4 + 2a² + 1)/4

To cancel out the divide by four, all parts must have the same denominator. To stop this from changing the value of the ‘a²’ I have to multiply it by four. This means that the formula now becomes:

4a² + a4 – 2a² + 1 = a4 + 2a² + 1

= a4 + 2a² + 1 = a4 +2a² +1

By cancelling the formula down, so that it is the same on both sides, I have proven that the formulae:

b = (a² - 1)/2
c = (a² + 1)/2

Are correct, I am now going to prove the formulae for when ‘d = 2’ and ‘d = 3’ in the same way.

d = 2:

a² + b² = c²
= a² + ([a² - 4]/4)² = ([a² + 4]/4)²
= a² + ([a² - 4]/4)([a² - 4]/4) = ([a² + 4]/4) ([a² + 4]/4)
= a² + (a² - 4)(a² - 4)/16 = (a² + 4)(a² + 4)/16

Cancel out the brackets:

(a² - 4)(a² - 4)

= a4 – 4a² - 4a² + 16

= a4 – 8a² + 16

(a² + 4)(a² + 4)

= a4 + 4a² + 4a² + 16

= a4 + 8a² + 16

Change the formula:

a² + (a4 – 8a² +16)/16 = (a4 + 8a² + 16)/16

16a² + a4 – 8a² + 16 = a4 + 8a² + 16

= a4 + 8a² + 16 = a4 +8a² + 16

I have now proven that my dependent formulae for ‘d = 2’ are correct.

d = 3:

a² + b² = c²
= a² + ([a² - 9]/6)² = ([a² + 9]/6)²
= a² + ([a² - 9]/6)([a² - 9]/6) = ([a² + 9]/6) ([a² + 9]/6)
= a² + (a² - 9)(a² - 9)/36 = (a² + 9)(a² + 9)/36

Cancel out the brackets:

(a² - 9)(a² - 9)

= a4 – 9a² - 9a² + 81

= a4 – 18a² + 81

(a² + 9)(a² + 9)

= a4 + 9a² + 9a² + 81

= a4 + 18a² + 81

Change the formula:

a² + (a4 – 18a² +81)/36 = (a4 + 18a² + 81)/36

36a² + a4 – 18a² + 81 = a4 + 18a² + 81

= a4 + 18a² + 81 = a4 +18a² + 81

I have now proven that my dependent formulae for ‘d = 3’ are correct.

I now have to check that my general formulae are correct. By substituting numbers for ‘a’ and ‘d,’ I should be able to find more Pythagorean triples. I am going to substitute ‘a’ for ‘19’ and ‘d’ for ‘5.’

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.