Beyond Pythagoras

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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:


Because 32= 3x3= 9  

42= 4x4= 16                  

52= 5x5= 25

9+16= 25


The numbers 5, 12, 13 satisfy the condition:


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



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



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


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



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        5


3        7


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


2        12


3        24


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


2        12                >4


3        24                >4


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


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


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:


I simplified this to:


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


Where n is 2

tn= 4n2+6n+2

tn= 4(22)+6x2+2

tn= (4x4)+12+2

tn= 16+12+2


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


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.

Join now!

The only factors of 1 are 1 and 1 so it factorises to (2n+1)(n+1)

2 needs to be added to the front of these brackets before the formula for perimeter is complete.

Therefore the formula for P is:


Next I needed to find a formula for area (A). As in the case of perimeter the laws of algebra say that if I insert my formulae for a and b to the formula for area ((axb)÷2) I will come up with the correct formula for A. My calculation turned out to be:

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


The answer to this is:


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