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:
= a4 – a²  a² + 1
= a4 – 2a² + 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:
= a4 – 4a²  4a² + 16
= a4 – 8a² + 16
= 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:
= a4 – 9a²  9a² + 81
= a4 – 18a² + 81
= 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.