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Introduction

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:

 A B C 3 4 5 5 12 13 7 24 25

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:

 A B C 9 40 41 11 60 61 13 84 85 15 112 113

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

Middle

10² + 24² = 676                                          √676 = 2614² + 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:

 A B C 6 8 10 8 15 17 10 24 26 12 35 37 14 48 50

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:

 A B C 9 12 15 15 36 39 21 72 75

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)

Conclusion

• 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
• (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.

By Adam Taylor-   Miss Clarke B27

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

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