Middle Side = 2n² + 2n
Longest Side:
I know that there is only a difference of 1 between the middle side and the longest side. So:
(Middle side) + 1 = Longest side.
2n² + 2n + 1 = Longest Side
I am certain that this is the correct formula. Just in case, I will check it using the first 3 terms.
2n² + 2n +1 = 5
2 x 1² + 2 x 1 + 1 = 5
2 + 2 + 1 = 5
5 = 5
The formula works for the first term.
2n² + 2n +1 = 13
2 x 2² + 2 x 2 + 1 = 13
8 + 4 + 1 = 13
13 = 13
The formula also works for the 2nd term.
2n² + 2n +1 = 25
2 x 3² + 2 x 3 + 1 = 25
18 + 6 + 1 = 25
25 = 25
The formula works for all 3 terms. So…
Longest Side = 2n² + 2n + 1
Now, I will check that 2n + 1, 2n² + 2n and 2n² + 2n + 1 forms a Pythagorean triple (or a² + b² = c²). We have a = 2n + 1, b = 2n² + 2n, and c = 2n² + 2n + 1.
a² + b² = c²
(2n+1) ² + (2n² + 2n) ² = (2n²+2n+1) ²
4n²+4n+1+4n4+8n3+4n²=4n4+8n3+8n2+4n+1
4n4+8n3+8n²+4n+1 = 4n4+8n3+8n²+4n+1
This proves my formulas for “a”, “b” and “c” are right.
Perimeter:
To find out he formula for the perimeter is simple, all you have to do is add the 3 formulas for the side together so you get: 2n + 1 + 2n2 + 2n + 2n2 + 2n + 1 = perimeter. This simplifies down to 4n2+6n+2.
Perimeter=4n2+6n+2
Area:
The area = (a x b) divided by 2. Therefore I took my formula for ‘a’ (2n + 1) and my formula for ‘b’ (2n2 + 2n). I then did the following: -
(2n + 1)(2n2 + 2n) = area
2
Multiply this out to get
4n3 + 6n2 + 2n = area
2
Then divide 4n3 + 6n2 + 2n by 2 to get
2n3 + 3n2+ n
Summary of Formulas I have found out:
- Use 2n + 1 to get ‘a’ from ‘n’
-
Use 2n2 + 2n to get ‘b’ from ‘n’
-
Use 2n2 + 2n + 1 to get ‘c’ from ‘n’
-
Use 4n2 + 6n + 2 to get the perimeter from ‘n’
-
Use 2n3 + 3n2 + n to get the area from ‘n’
As an extra part, I am now going to find the relationships (differences) between the Short and Middle sides, the Short and Long Sides, and The Middle and Long sides.
Finding the relationship between 2 sides is quite easy, because, I already know the nth terms for each side. Because of this, all I have to do to find the relationship is to subtract one nth term from the other, leaving me with the relationship.
Middle Side – Short Side = Relationship.
(2n² + 2n) – (2n + 1) = Relationship
2n² - 1 = Relationship
Relationship between Shortest and Middle sides = 2n² – 1
The next relationship I’m going to work out is the relationship between the shortest and longest sides.
Longest Side – Short Side = Relationship.
(2n² + 2n + 1) – (2n + 1) = Relationship.
2n² = Relationship.
Relationship between the shortest and longest sides = 2n²
The next relationship I’m going to find is between the middle and longest side.
Longest Side – middle Side = Relationship.
(2n² + 2n + 1) – (2n² + 2n) = Relationship.
1 = Relationship.
Relationship between the Middle and Longest sides = 1
Using these same principles, I can work out any relationship, to prove this; I will work out the relationship between the shortest side and the perimeter and then the area.
Perimeter - Shortest Side = Relationship.
(4n² + 6n + 2) – (2n + 1) = Relationship.
4n² + 4n + 1 = Relationship.
Relationship between the Perimeter and Shortest side = 4n² + 4n + 1
Area – Shortest side= Relationship
(2n3 + 3n2 + n) – (2n + 1)= 2n3 + 3n2– n – 1
2n3 + 3n2– n – 1= Relationship
Relationship between the Area and the shortest side= 2n3 + 3n2– n – 1
Conclusion & Generalisation for Triple that start with an odd number:
As I expected my formula works.
The formula for the longest side is 2(2n2+2n+1)= 4n2+4n+2
Just to prove my formulas work, I will substitute them into a2+b2+c2
a² + b² = c²
(4n+2) ² + (4n² + 4n) ² = (4n²+4n+2) ²
16n²+16n+4+16n4+32n3+16n²=16n4+32n3+32n2+16n+4
16n4+32n3+320n²+16n+4 = 16n4+32n3+32n²+16n+4
Just as I thought the formula is 4(4n4+8n3+8n²+4n+1). It is 4, because 4 is equivalent to 2² as the numbers were double the odd numbers, and then they were squared.
Perimeter:
The perimeter is going to be exactly the same, 2 times the formula for the odd numbers, therefore the formula should be 2(4n2+6n+2)= 8n2+12n+4.
The formula to find the perimeter is 2(4n2+6n+2)= 8n2+12n+4.
Area:
The formula to find out the area is going to be 4 times as big as the formula for the odd numbers, because of 22. Therefore the formula should be 4(2n3 + 3n2+ n)=8n3+12n2+4n.
The formula for finding the area is 4(2n3 + 3n2+ n)=8n3+12n2+4n.
Summary of Formulas I have found out:
- Use 2(2n + 1) to get ‘a’ from ‘n’
-
Use 2(2n2 + 2n) to get ‘b’ from ‘n’
-
Use 2(2n2 + 2n + 1) to get ‘c’ from ‘n’
-
Use 2(4n2 + 6n + 2) to get the perimeter from ‘n’
-
Use 4(2n3 + 3n2 + n) to get the area from ‘n’
The just double from the formulas, give for odd numbers, except for the area and that is because there you got to square 2 to give you 4.
Relationships
I am now going to find the relationships (differences) between the Short and Middle sides, the Short and Long Sides, and The Middle and Long sides, for triples that start with an even number.
Middle Side – Short Side = Relationship.
(4n² + 4n) – (4n + 2) = Relationship
2n² - 1 = Relationship
Relationship between Shortest and Middle sides = 2(2n² – 1)
The next relationship I’m going to work out is the relationship between the shortest and longest sides.
Longest Side – Short Side = Relationship.
(4n² + 4n + 2) – (4n + 2) = Relationship.
4n² = Relationship.
Relationship between the shortest and longest sides = 4n²
The next relationship I’m going to find is between the middle and longest side.
Longest Side – middle Side = Relationship.
(4n² + 4n + 2) – (4n² + 4n) = Relationship.
1 = Relationship.
Relationship between the Middle and Longest sides = 2
Using these same principles, I can work out any relationship, to prove this; I will work out the relationship between the shortest side and the perimeter and then the area.
Perimeter - Shortest Side = Relationship.
(8n² + 12n + 4) – (4n + 2) = Relationship.
8n² + 8n + 2 = Relationship.
Relationship between the Perimeter and Shortest side = 2(4n² + 4n + 1)
Area – Shortest side= Relationship
(8n3 + 12n2 + 4n) – (4n + 2)= 8n3 + 12n2 – 2
8n3 + 12n2 – 2= Relationship
Relationship between the Area and the shortest side= 2(4n3 + 6n2– 1)= 8n3 + 12n2– 2
Conclusion & Generalisation for Pythagoras Triple’s that start with an even number:
- The numbers for Pythagoras triple’s that start with even numbers are just the Pythagoras triple numbers that start with odd numbers doubled. For example, 3,4,5 just doubles to form 6,8,10.
- As the numbers double, so do the formulas.
- If a Pythagoras triple starts with an even number then, all three sides, the perimeter and the area will all be even numbers.