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
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
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
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
