Beyond Pythagoras

The differences show that ‘a’ and ‘n’ are changing at the same rate – increasing by 2 each time, thus proving that the formula includes 2n. The difference between ‘a’ and ‘n’ is always 1; therefore, our formula needs to have a +1 included.

From this, I know the formula for ‘a’, in terms of ‘n’, to be;

a  =  2n + 1

I personally cannot see an obvious link between ‘n’ and ‘b’ or ‘c’, so I will use the known formula, a² +b² = c² to discover ‘b’ and ‘c’ in terms of ‘n’.

3² + 4² = 5² and 5² + 12² = 13² both have a hypotenuse one greater than one side (in this case known as ‘b’). Looking at my table of triples I can see that there are more odd triples that follow this pattern. Using this, I should be able to discover ‘b’ and ‘c’ in terms of ‘a’.

3² + 4²        = 5²

a² + b²        = c²

a² + b²        = (b + 1)²

a² + b²        = b² + 1 + 2b

a²         = 2b + 1

2b        = (a² - 1)

b         = (a² - 1)

                2

Remove extra variable ‘c’ from formula

Multiply out brackets

Simplify formula

Rearrange formula

Rearrange formula

From this, I know the formula for ‘b’, in terms of ‘a’, to be;

b = (a² - 1)

     2

From this formula for ‘b’ in terms of ‘a’ I can discover ‘c’ in terms of ‘a’;

This table shows that if you add 1 before dividing by two, instead of subtracting 1 and then dividing, then you get ‘c’ as the result instead of ‘b’.

c = (a² + 1)

      2

Using the two formulae on the above page (‘b’ and ‘c’ in terms of ‘a’) with my formula for ‘a’ in terms of ‘n’, I can calculate ‘b’ and ‘c’ in terms of ‘n’ by putting

a = 2n + 1 into the formulae.

b         = (a² - 1)

                2

b         = ((2n + 1) ² - 1)

                        2

b         = 4n² + 2n + 2n +1 - 1

     2

b         = 4n² + 4n

        2

b        = 2n² + 2n

Insert a = 2n + 1

Multiply out brackets

Sort like numbers/letters

Cancel down to simplest form

And the same with ‘c’;

c         = (a² - 1)

                2

c         = ((2n + 1) ² + 1)

                        2

c         = ((2n + 1) ² + 1)

                        2

c         = 4n² + 2n + 2n +1 + 1

     2

c         = 4n² + 4n +2

        2

c        = 2n² + 2n + 1

Insert a = 2n + 1

Multiply out brackets

Sort like numbers/letters

Cancel down to simplest form

b = 2n² + 2n

c = 2n² + 2n + 1

Now that I have formulae for ‘a’, ‘b’ and ‘c’ in terms of ‘n’; I can put these into my simple area and perimeter formulae and will then be able to calculate both area and perimeter in terms of ‘n’.

Perimeter

Perimeter = a + b + c

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

P = 2n + 1 + 2n² + 2n + 2n² + 2n + 1

P = 6n + 2 + 4n²

Perimeter = 4n² + 6n + 2

Area

Area = a x b

           2

A = (2n+1) x (2n² + 2n)

                2

        

A = 4n³ + 4n² + 2n² +2n

                2

A = 4n³ + 6n² + 2n

        2

A = 2n³ + 3n² + n

Area = 2n³ + 3n² + n

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Even Triples

Now that I have done all my odd Pythagorean triples formulae, I will do the same with the even Pythagorean triples.

First, I need to work out ‘a’ in terms of ‘n’:

The differences between the numbers show that ‘a’ is changing twice as fast as ‘n’. Therefore, 2n must be part of the formula.

The differences show that ‘a’ and ‘n’ are changing at the same rate – increasing by 2 each time, thus proving that the formula includes 2n. The difference between ‘a’ and ‘n’ is always 4; therefore, our formula needs to have a +4 included.

From this, I know the formula for ‘a’, in terms of ‘n’, to be;

a  =  2n + 4

I personally cannot see an obvious link between ‘n’ and ‘b’ or ‘c’, so I will use the known formula, a² +b² = c² to discover ‘b’ and ‘c’ in terms of ‘n’.

6² + 8² = 10² and 8² + 15² = 17² both have a hypotenuse two greater than one side (in this case known as ‘b’). Looking at my table of triples I can see that there are more even triples that follow this pattern. Using this, I should be able to discover ‘b’ and ‘c’ in terms of ‘a’.

6² + 8²        = 10²

a² + b²        = c²

a² + b²        = (b + 2)²

a² + b²        = b² + 4 + 4b

a²         = 4b + 4

4b        = (a² - 4)

b         = (a² - 4)

                4

Remove extra variable ‘c’ from formula

Multiply out brackets

Simplify formula

Rearrange formula

Rearrange formula

From this, I know the formula for ‘b’, in terms of ‘a’, to be;

b = (a² - 4)

     4

From this formula for ‘b’ in terms of ‘a’ I can discover ‘c’ in terms of ‘a’;

This table shows that if you add 4 before dividing by four, instead of subtracting 4 and then dividing, then you get ‘c’ as the result instead of ‘b’.

c = (a² + 4)

      4

Using the two formulae on the above page (‘b’ and ‘c’ in terms of ‘a’) with my formula for ‘a’ in terms of ‘n’, I can calculate ‘b’ and ‘c’ in terms of ‘n’ by putting

a = 2n + 4 into the formulae.

b         = (a² - 4)

                4

b         = ((2n + 4) ² - 4)

                        4

b         = 4n² + 8n + 8n +16 - 4

 4

b         = 4n² + 16n + 12

                4

b        = n² + 4n + 3

Insert a = 2n + 4

Multiply out brackets

Sort like numbers/letters

Cancel down to simplest form

And the same with ‘c’;

c         = (a² + 4)

                4

c         = ((2n + 4) ² + 4)

                        4

c         = 4n² + 8n + 8n +16 + 4

 4

c         = 4n² + 16n + 20

                4

c        = n² + 4n + 5

Insert a = 2n + 4

Multiply out brackets

Sort like numbers/letters

Cancel down to simplest form

b = 2n² + 2n

c = 2n² + 2n + 1

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Now that I have formulae for ‘a’, ‘b’ and ‘c’ in terms of ‘n’; I can put these into my simple area and perimeter formulae and will then be able to calculate both area and perimeter in terms of ‘n’.

Perimeter

Perimeter = a + b + c

P = (2n + 4) + (n² + 4n +3) + (n² + 4n + 5)

P = 2n + 4 + n² + 4n +3 + n² + 4n + 5

P = 10n + 12 + 2n²

Perimeter = 2n² + 10n + 12

Area

Area = a x b

           2

A = (2n+4) x (n² + 4n +3)

                2

        

A = 2n³ + 8n² + 6n + 12 + 4n² + 16n

                2

A = 2n³ + 12n² + 22n + 12

        2

A = n³ + 6n² + 11n + 6

Area = n³ + 6n² + 11n + 6

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Formulae Summary

These are all the formulae I have discovered in my investigation. I can now workout any odd or even Pythagorean triple and its perimeter and area knowing only the number, n. For example, the 26 odd triple is: a = 53, b = 1404, c = 1405, perimeter = 2862. area = 37206

Odd Pythagorean Triples

a                =        2n + 1

b                =        2n² + 2n

c                =        2n² + 2n + 1

Perimeter        =        4n² + 6n + 2

Area                =        2n³ + 3n² + n

Even Pythagorean Triples

a                =        2n + 4

b                =        n² + 4n + 3

c                =        n² + 4n + 5

Perimeter        =        2n² + 10n + 12

Area                =        n³ + 6n² + 11n + 6