# Investigating families of Pythagorean triples.

I was set the task of investigating families of Pythagorean triples. I started off by investigating families of Pythagorean triples where the shortest side is an odd number and c is always equal to b+1. I then extended my research into other families of triples. Finally, I came up with a formula that can calculate all Pythagorean triples.

Pythagoras’s formula is as follows:

a2 + b2 = c2

This is referring to such a triangle:

The triangle always has one right angle and c is always the longest side, also known as the hypotenuse.

An example of this is with the triple 3,4 and 5:

Here, a is 3, b is 4 and c is 5.

52 = 32 + 42

25 = 9 + 16

I will now investigate the Pythagorean triples where c is always equal to b+1. The following table shows the first ten cases, along with the areas and perimeters:

By looking at the numbers for a, I noticed that they are all odd numbers. The basic formula for odd numbers is:

2n + 1

When I applied this formula to the numbers in the table, they turned out correct. As a result,

a = 2n + 1

The difference between the numbers in column b is 8 at first, then 12, then 16, then 20, and so on, adding 4 to the difference each time. Therefore, the second difference is 4. If we divide 4 by 2, we get 2. As a result, the first part of the formula for b is:

2n2

I then noticed that by adding double n to the above formula, I could get all the numbers in column b. I finally came up with:

2n2 + 2n

This formula could then be factorised to produce:

2n ( n + 1 )

Once I had found the formula for b, finding the formula for b+1 was easy, I just needed to add one, making the formula for c:

2n2 + 2n + 1

Or

2n (n + 1 ) + 1

To find the formula for the perimeter, I had to add the formulae for a, b and c together, giving:

4n2 + 6n + 2

Or

2 ( n +1 ) ( 2n + 1 )

To find the area, I needed to add the shortest side to the medium side and divide the answer by 2. I did this to the formulae for a and b, the two shortest sides and divided the answer by 2:

2n3 + 3n2 + n

When factorised, the formula became:

n ( 2n2 + 3n + 1 )

The formulae I found for the b+1 family are shown below:

a = 2n + 1

b = 2n2 + 2n   or   2n ( n + 1 )

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

P = 4n2 + 6n + 2   or   2 ( n +1 ) ( 2n + 1 )

A = 2n3 + 3n2 + n   or   n ( 2n2 + 3n + 1 )

I then tried to prove that Pythagoras’s theorem did work for my ...