Beyond Pythagoras

* 'b' increases by +4 each term

* ('a' x 'n') + n = 'b'

From these observations I have worked out the next two terms.

I will now put the first five terms in a table format.

Term Number 'n'

Shortest Side 'a'

Middle Side 'b'

Longest Side 'c'

Perimeter

Area

3

4

5

2

6

2

5

2

3

30

30

3

7

24

25

56

84

4

9

40

41

90

80

5

1

60

61

32

330

I have worked out formulas for

. How to get 'a' from 'n'

2. How to get 'b' from 'n'

3. How to get 'c' from 'n'

4. How to get the perimeter from 'n'

5. How to get the area from 'n'

My formulas are

. 2n + 1

2. 2n2 + 2n

3. 2n2 + 2n + 1

4. 4n2 + 6n + 2

5. 2n3 + 3n2 + n

To get these formulas I did the following

. Take side 'a' for the first five terms 3, 5, 7, 9, 11. From these numbers you can see that the formula is 2n + 1 because these are consecutive odd numbers (2n + 1 is the general formula for consecutive odd numbers) You may be able to see the formula if you draw a graph

2. From looking at my table of results, I noticed that 'an + n = b'. So I took my formula for 'a' (2n + 1) multiplied it by 'n' to get '2n2 + n'. I then added my other 'n' to get '2n2 + 2n'. This is a parabola as you can see from the equation and also the graph

3. Side 'c' is just the formula for side 'b' +1

4. The perimeter = a + b + c. Therefore I took my formula for 'a' (2n + 1), my formula for 'b' (2n2 + 2n) and my formula for 'c' (2n2 + 2n + 1). I then did the following: -

2n + 1 + 2n2 + 2n + 2n2 + 2n + 1 = perimeter

Rearranges to equal

4n2 + 6n + 2 = perimeter

5. 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

To prove my formulas for 'a', 'b' and 'c' are correct. I decided incorporate my formulas into a2 + b2 = c2: -

a2 + b2= c2

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

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

4n2 + 2n + 2n + 1 + 4n4 + 4n2 + 4n3 + 4n3= 4n4 + 8n3 + 8n2 + 4n + 1

4n2 + 4n + 1 + 4n4 + 8n3 + 4n2= 4n4 + 8n3 + 8n2 + 4n + 1

4n4 + 8n3 + 8n2 + 4n + 1 = 4n4 + 8n3 + 8n2 + 4n + 1

This proves that my 'a', 'b' and 'c' formulas are correct

Extension

Polynomials

Here are the formulas for the perimeter and the area

Perimeter = 4n2 + 6n + 2

Area = 2n3 + 3n2 + n

From my table of results I know that the perimeter = area at term number 2. Therefore

(n-2) is my factor

I would like to find out at what other places (if any) the perimeter is equal to the area

To find this out I have been to the library and looked at some A-level textbooks and learnt 'Polynomials'

4n2 + 6n + 2 = 2n3 + 3n2 + n

6n + 2 = 2n3 - n2 + n

2 = 2n3 - n2 - 5n

0 = 2n3 - n2 - 5n - 2

2n2 + 3n + 1 b

n - 2 ) 2n3 - n2 - 5n - 2

2n3 - 4n2 b

3n2 - 5n - 2

3n2 - 6n b

n - 2

n - 2 b

0

This tells us that the only term where the perimeter = area is term number 2

Therefore when f(x) = 2n3 - n2 - 5n - 2 is divided by n - 2 there is no remainder and a quotient 2n2 + 3n + 1. The result can be written as

f(x) = 2n3 - n2 - 5n - 2= (n - 2)(2n2 + 3n + 1)

If 2 was substituted for the x in this identity so that n - 2 = 0, the quotient is eliminated giving f (2) = + 2

Now I will complete the square on '4n2 + 6n + 2' to see what the solution to this is.

4n2 + 6n + 2

4(n +3)2 - 9 + 2

4(n +3)2 - 7

4(n +3)2 = 7

(n +3)2 = 1.75

n + 3 = 1.322875656

n + 3 = -1.322875656

n = -1.677124344

n = -4.322875656

Arithmatic Progression

I would like to know whether or not the Pythagorean triple 3,4,5 is the basis of all triples just some of them.

To find this out I have been to the library and looked at some A-level textbooks and learnt 'Arithmatic Progression'

3, 4, 5 is a Pythagorean triple

The pattern is plus one

If a = 3 and d = difference (which is +1) then

3 = a

4 = a + d

5 = a +2d

a, a +d, a + 2d

Therefore if you incorporate this into Pythagoras theorem

a2 + (a + d)2 = (a + 2d)2

a2 + (a + d)(a + d) = (a + 2d)2

a2 + a2 + ad + ad + d2 = (a + 2d)2

2a2 + 2ad + d2 = (a + 2d)2

2a2 + 2ad + d2 = (a + 2d)(a + 2d)

2a2 + 2ad + d2 = a2 + 2ad + 2ad + 4d2

2a2 + 2ad + d2 = 4d2 + a2 + 4ad

If you equate these equations to 0 you get

a2 - 3d2 - 2ad = 0

Change a to x

x2 - 3d2 - 2dx = 0

Factorise this equation to get

(x + d)(x - 3d)

Therefore

x = -d

x = 3d

x = -d is impossible as you cannot have a negative dimension

a, a+d, a + 2d

Is the same as

3d, 4d, 5d

This tells us that the only Pythagorean triples are 3, 4, 5 or multiples of 3, 4, 5 e.g. 6, 8, 10 or 12, 16, 20 etc.