Raj Purewal Mathematics GCSE Coursework 2001-02-22 Page of

“Beyond Pythagoras”

Beyond Pythagoras

This investigation is to study Pythagoras Theorem. I will try to find patterns and formulae to help predict Pythagorean Triples.

About Pythagoras

Pythagoras was a Greek Philosopher and Mathematician who is believed to have lived in the 6th century BC. He discovered many theorems but his most famous was:

a2+b2= c2

What is a Pythagorean Triple?

To answer this I first need to explain Pythagoras Theorem. Pythagoras States that in any right-angled triangle, a2+b2=c2. a is the shortest side, b the middle length side and c the hypotenuse (the longest side).

A Pythagorean Triple is any set of integers that agrees this condition. For example 3, 4, 5 is a Pythagorean Triple because:

32+42=52

Because 32= 3x3= 9

42= 4x4= 16

52= 5x5= 25

9+16= 25

1.

The numbers 5, 12, 13 satisfy the condition:

52+122=132

Because 52= 5x5= 25

122= 12x12= 144

132= 13x13= 169

25+144= 169

The numbers 7, 24, 25

72+242= 252

Because 72= 7x7= 49

242= 24x24= 576

252= 25x25= 625

44+576= 625

2.

(a)

I found perimeter by using the formula:

Perimeter= a+b+c

P= 5+12+13

P= 30

Perimeter= a+b+c

P= 7+24+25

P= 56

I found area using the formula:

Area= (axb)÷2

Area= 0.5x5x12

Area= 30

Area= (axb)÷2

Area= 0.5x24x25

Area= 84

(b)

I next put the results for perimeter and area into the table below.

3.

Before investigating patterns I tried to find the Pythagorean triple for a shortest side of 9 to increase my depth of results. I used trial and error to find it. From the general pattern so far I knew that side b for this triple had to be roughly around 40 to 50. I also knew that c was probably one larger than b. I tried multiple triples until I found one that seemed to agree the theorem. I came up with 9, 40, 41.

I proved that this was right:

92+402= 412

Because 92= 9x9= 81

402= 40x40= 1600

412= 41x41= 1681

81+1600= 1681

Next I needed to find the perimeter and area for this triple. I used the formula:

Perimeter= a+b+c

P= 9+40+41

P= 90

I found out the area using the formula:

Area= (axb)÷2

A= (9x40)÷2

A=360÷2

A=180

Next I updated my table:

Next I set out to find a relationship between a and n. I used the Difference Method.

n a L1

1 3

>2

2 5

>2

3 7

>2

4 9

I found that the difference at level 1 (L1) had a constant of 2 every time. This meant that the formula must have 2n in it. tn= 2n alone, however, did not work.

a 3, 5, 7, 9

2n 2, 4, 6, 8

A constant at L1 means that to correct the formula I quite simply must add or subtract the appropriate number required to correct the formula. This number is 1 and must be added. I proved that this was correct:

Where n is 1

tn= 2n+1

tn= 2x1+1

tn= 2+1

tn= 3

Where n is 2

tn= 2n+1

tn= 2x2+1

tn= 4+1

tn= 5

Thus I came to the conclusion that the formula for working out the nth term for the shortest side of any Pythagorean triple with an odd number for its shortest side was:

tn= 2n+1

This was already in its simplest form and did not need factorising.

Next I needed to find a similar formula for the middle side. I again used the difference method:

n b L1

1 4

>8

2 12

>12

3 24

>16

4 40

I found that there was no constant difference at L1 so tried to find a second difference at L2.

n b L1 L2

1 4

>8

2 12 >4

>12

3 24 >4

>16

4 40

I discovered that there was a second difference at L2 of 4. This meant that the formula was quadratic. When you find a constant difference at L2 you halve the difference and the number that you are left with tells you how many quadratics there are in the formula. Halving 4 gave 2; somewhere in the formula there had to be 2n2. However 2n2 alone was obviously not correct:

b 4, 12, 24, 40

2n2 2, 8, 18, 32

The difference between 2n2 and b was not a constant number. This meant that the addition had to involve n. I quite quickly noticed that for the first triple the difference was 2 and in terms of n this was 2n. I tried adding 2n to the end of the formula and checked it for each triple. I proved that it worked:

Where n is 1

tn= 2n2+2n

tn= 2(12)+2x1

tn= 2x1+2x1

tn= 2+2

tn= 4

Where n is 2

tn= 2n2+2n

tn= 2(22)+2x2

tn= 2x4+2x2

tn= 8+4

tn= 12

I came to the conclusion that the formula must be:

tn= 2n2+2n

This worked perfectly but was not in its simplest form so I factorised it to. 2n was common to the formula so I simplified it to:

tn= 2n(n+1)

I next started to look for a formula for the hypotenuse, the longest side (c). I noticed that side c was the same as side b but +1. Therefore I tried the same formula but +1. I checked and proved that this was correct:

Where n is 1:

tn= 2n(n+1)+1

tn= 2n2+2n+1

tn= 2(12)+2x1+1

tn= 2x1+2x1+1

tn= 2+2+1

tn=5

Where n is 2:

tn= 2n(n+1)+1

tn= 2n2+2n+1

tn= 2(22)+2x2+1

tn= 2x4+2x2+1

tn= 8+4+1

tn=13

Therefore the formula for side c is:

tn= 2n(n+1)+1

This was already in its simplest form so did not require any factorising.

Next I looked for a formula for the perimeter (P). The formula for perimeter was a+b+c. The laws of algebra say that adding up the formulae found for working out sides a, b and c from n will give me the correct formula for perimeter. I did this and produced the sum:

tn=2n+1+2n2+2n+2n2+2n+1

I simplified this to:

tn=4n2+6n+2

I checked and proved that this was correct:

Where n is 1

tn= 4n2+6n+2

tn= 4(12)+6x1+2

tn= (4x1)+6+2

tn= 4+6+2

tn=12

Where n is 2

tn= 4n2+6n+2

tn= 4(22)+6x2+2

tn= (4x4)+12+2

tn= 16+12+2

tn=30

2 is common to this formula so I removed it to give me:

tn=2(2n2+3+1)

2n2+3n+1 can be factorised:

(+)(+) The signs are both the same and positive.

(2n+)(n+) 2n and n must be in front of the signs.