3,4,5 are called Pythagorean triples because shortest2 + middle2 = longest2.
a2 + b2 = c2
32 + 42 = 52
9 + 16 = 25 5 X 5 = 25
- 5
3 X 3 = 9
4
4 X 4 = 16
This formula also works on 5,12,13 and 7,24,25: -
52 + 122 = 132 72 + 242 = 252
25 + 144 = 169 49 + 576 = 625
I decided to list all the Pythagorean triples but only the triples where the smallest number is an odd number. I came up with many triples but I decided to try and find patterns and observations by using only eight triples. Here are the triples.
The Original Table
Using this chart I found formulas, which can enable us to discover any of the numbers in triples that, have an odd smallest number.
In order to work out ‘a’, I had to write out the first few numbers and then find the difference between the numbers to help create a formula. I used this method to help me fnd all these numbers.
To find ‘a’: -
3 5 7
2 2 2n
Then I took 2 away from 3 and that gave me 1 which completed the formula which looks like 2n + 1.
A = 2n + 1
B = 2n2 + 2n
C = 2n2 + 2n +1
These are the original formulas and I will be relating to them often later in this.
I worked these all out using the above technique and my observations proved to be correct. These formulas are the original formulas for the triples with odd smallest sides.
By looking at the table on the previous page, I also noticed something with the ‘b’ set of numbers. I looked for many paterns and I found out that,
4 X (Triangular number) = next number in column
4 X 1 = 4
4 X 3 = 12
4 X 6 = 24
4 X 10 = 40
Numbers in red are the triangle numbers,
numbers in blue are the next number in the column and therefore,
Cancelled down = 2n (n + 1)
2n2 + 2n
The middle length also has a 2nd difference of 4. Because it has gone to a second difference in order to gain a difference that is equal, it means that the formula for this column is a quadratic.
4 12 24 40
8 12 16
4 4 2n2
4 12 24 40 2n2 + 2n
take away
2 8 18 32
2 4 6 8 2n
2n2 + 2n
This proves that ‘b’ can be found by using the above formula and by using wither of the techniques that have been shown above.
I also noticed that ‘b’ is only ever 1 less than ‘c’ so all we do is add 1 (+1) to the end of the formula for ‘b’ in order to give us the formula for ‘c’.
To work out the area of the triangles formed by the triples we have to have ‘a’ and ‘b’ as numbers. The formula is a X b / 2.
In this case, we have ‘a’ and ‘b’ so we can begin working out a formula for the area.
Area = a X b / 2
= (2n +1) X (2n2 + 2n)
= 4n3 + 4n2 + 2n2 + 2n / 2
= 4n3 + 6n2 + 2n / 2
When these become factorised and then cancelled down, we end up with,
Area = 2n3 + 3n + n
Perimeter = a + b + c therefore 2n +1 + 2n2 + 2n + 2n2 + 2n +1
= 4n2 + 6n + 2
So we have recognised the formulas that are needed to find ‘a’, ‘b’, ‘c’, the area and the perimeter of all the triples that have an odd smallest number.
Now I am going to begin looking at triples where the smallest side is an even number. When we look at these triples, we notice that they do not fit the formula that we had earlier created.
6 8 10 Doesn’t fit the pattern!!
62 + 82 = 102 Doesn’t fit the pattern!!
36 + 64 = 100 Doesn’t fit the pattern!!
Other even triples that don’t fit the formula are: -
6 8 10
12 16 20
18 24 30
24 32 40
30 40 50
36 48 60
None of these triples fit the formulas found previously and this means that once again we will need to plot the data into a table and look for similarities in order to help us find formulas for the triples with even smallest number.
I have noticed before I begin to plot the data in a table that the table for even shortest sides is double the table for odd shortest sides. Now I will show you these similarities.
The Original Table X2
The Original Table
As you can see all the figures in the top table, are double the figures in the same place of the bottom table. So therefore common sense tells us that if all the figures just become doubled, then we should just take the formulas for the original table and just double every one of the formulas.
A = (Original ‘a’) X 2
B = (Original ‘b’) X 2
C = (Original ‘c’) X 2
Therefore it would be easier to say that
a b c
Original E = e (2n + 1) e (2n2 + 2n) e (2n2 + 2n +1)
But for this formula to happen, ‘e’ must be an even number to create even triples. And looking back to the tables ‘e’ is also the difference between the two longest sides.
Now I have decided to prove all the formulas that I have found and observed. I will do this by referring back to Pythagoras’s Theorem. If this proves my formulas to work then I will be certain of these formulas working for all triples with odd and even shortest sides respectively.
a2
(2n + 1) (2n + 1) = 2n (2n + 1) = 4n2 + 2n
+ 1 (2n + 1) = 2n + 1
A = 4n2 + 4n + 1
b2
(2n2 + 2n) (2n2 + 2n) = 2n2 (2n2 + 2n) = 4n4 + 4n3
+ 2n (2n2 + 2n) = 4n3 + 4n2
B = 4n4 + 8n3 + 4n2
c2
(2n2 + 2n +1) (2n2 + 2n +1) = 2n2 (2n2 + 2n +1) = 4n4 + 4n3 + 2n2
+ 2n (2n2 + 2n +1) = 4n3 + 4n2 +2n
+ 1 (2n2 + 2n +1) = 2n2 +2n + 1
C = 4n4 + 8n3 +8n2 +4n +1
4n2 + 4n + 1
4n4 + 8n3 + 4n2
4n4 + 8n3 + 8n2 + 4n + 1
Add the formula in red to the formula in blue and the answer is the formula in green. Red is ‘a’, blue is ‘b’ and green is ‘c’, so this proves my formulas.
This shows that the formulas that I calculated at the start of this investigation do indeed satisfy Pythagoras’s Theorem of a2 + b2 = c2.
Just by playing with algebra I figured that by making ‘a’ the subject, we could prove the Original table.
a2 + b2 = c2
a2 = c2 - b2
a2 = (c - b) (c + b)
a2 = (c + b) X 1
a2 = c + b
Now I shall try and find a general formula that can help generate all the Pythagorean Triples with both odd and even shortest sides. By plenty of research I have actually found the formula but now I will give examples to show how it works.
The General Formula = (n2 – m2) + (2nm) + (n2 + m2)
A2 = (n2 – m2) 2
B2 = (2nm) 2
C2 = (n2 + m2) 2
It must be mentioned that for this General formula to work, ‘n’ must be a larger number than ‘m’ and all the numbers must be whole numbers.
(n2 - m2 ) (n2 - m2 ) = n2 (n2 - m2 ) = n4 - n2 m2
- m2 (n2 - m2) = -n2 m2 + m4
n4 - n2 m2 + m4
(2nm) (2nm) = 4n2 m2
(m2 + n2) (m2 + n2) = m2 (m2 + n2) = m4 + m2 n2
+ n2 (m2 + n2) m2 n2 + n4
m4 + 2m2 n2 + n4
n4 - n2 m2 + m4
4n2 m2
m4 + 2m2 n2 + n4
This proves that the General Formula that I researched from various sources is correct and now this formula can be used to generate as many triples as you want that can satisfy Pythagoras’s Theorem.
I will give an example that forms the 12, 16, 20 triangle by using this formula.
n = 4
m = 2
n2 - m2 = a
42 – 22 = a
16 – 4 = a
a = 12
2nm = b
2 X 4 X 2 = b
b = 16
m2 + n2 = c
22 + 42 = c
4 + 16 = c
c = 20
I believe that this is enough proof that the General formula works and works on all the triples.
In conclusion I would like to point out that this has been both an exciting but also challenging task and I am greatly satisfied with my final project although I do feel that there is room for improvement.
Thank you.
Ali
Elhamamy.
