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Level: GCSE
Subject: Maths
Word count: 1272

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Pyhtagorean Theorem

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Introduction

Jaymin Patel 10Q Page of

Pythagoras’ theorem states that a²+b²=c²

As you can see, a is the longest side, b is the middle side and c is the longest side (hypotenuse). The point of this coursework is to find b when a is an odd number and all of the sides are positive integers. Then after that I will go looking when a is a positive number.

The numbers 3, 4 and 5 work in Pythagoras’ theorem,

3²+4²=5²

because 3²=3×3=9

4²=4×4=16

5²=5×5=25

and so 3²+4²=9+16=25=5²

The numbers 5, 12 and 13 also work,

5²+12²=13²

because 5²=5×5=25

12²=12×12=144

13²=13×13=169

and so 5²+12²=25+144=169=13²

The numbers 7, 24 and 25 also work,

7²+24²=25²

because 7²=7×7=49

24²=24×24=576

25²=25×25=625

and so 7²+24²=49+576=625=25²

3, 4 and 5

Perimeter= 3+4+5=12

Area= ½ ×3×4=6

5, 12 and 13

Perimeter= 5+12+13=30

Area= ½ ×5×12=30

7, 24 and 25

Perimeter= 7+24+25=56

Area= ½ ×7×24=84

From the first three terms I have realised that: -

a increases by 2 each time

a is equal to the formula ×2+1 from n

b is always even

c is always odd

c is always +1 of b

b=(a×n) + n

I have also added 2 more terms using what I think are my formulae.

‘n’	‘a’	‘b’	‘c’	Perimeter	Area
1	3	4	5	12	6
2	5	12	13	30	30
3	7	24	25	56	84
4	9	40	41	90	180
5	11	60	61	132	330

Here are my formulas.

Middle

‘n’	‘a’	‘b’	‘c’	area	perimeter
1	3	4	5	6	12
2	5	12	13	30	30
3	7	24	25	84	56
4	9	40	41	180	90
5	11	60	61	330	132
6	13	84	85	546	182
7	15	112	113	840	240
8	17	144	145	1224	306
9	19	180	181	1710	360
10	21	220	221	2310	462

Now I have to prove that my formulae for a, b and c work. To do this I must incorporate 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

4n⁴+8n³+8n²+4n+1 = 4n⁴+8n³+8n²+4n+1

This proves that my formulae for a, b and c are all correct!

Now for part 2, I will try to solve the problem of Pythagorean triples with even numbers for a.

The numbers 6, 8 and 10 work for Pythagoras’ theorem,

6²+8²=10²

because 6²=6×6=36

8²=8×8=64

10²=10×10=100

and so 6²+8²=36+64=100=10²

The numbers 10, 24 and 26 also work,

10²+24²=26²

because 10²=10×10=100

24²=24×24=576

26²=26×26=676

and so 10²+24²=100+576=676=26²

Conclusion

So 2n+n²-1+n²+1

= 2n+2n²

= 2n²+2n

Here is how I got to my b and c formulae.

Getting b

a²+b²=c²

=c²= (b+2) ²

=(2n) ²+b²=(b+2) ²

=4n²+b²=b²+4b+4

=4n²-4=4b

=n²-1=b

=b=n²-1

Getting c

By looking at b you can see that c is always +2 than b.

So c is n²-1+2

=n²+1

Here is a table with all of the results for even numbers.

‘n’	‘a’	‘b’	‘c’	Perimeter	Area
1	2	0	2	4	0
2	4	3	5	12	6
3	6	8	10	24	24
4	8	15	17	40	60
5	10	24	26	60	120
6	12	35	37	84	210
7	14	48	50	112	336
8	16	63	65	144	504
9	18	80	82	180	720
10	20	99	101	220	990

Now I have to prove my formulae for a, b and c. To do this I have to incorporate a²+b²=c²

a²+b²=c²

= (2n)²+(n²-1) ²=(n²+1) ²

= (2n)(2n)=4n²

(n²-1)(n²-1)=n⁴-n²-n²+1

(n²+1)(n²+1)=n⁴+n²+n²+1

= 4n²+n⁴-n²-n²+1=n⁴+n²+n²+1

= n⁴+2n³+1=n⁴+2n³+1

This proves that my formulas for a, b and c are all correct.

This student written piece of work is one of many that can be found in our GCSE Beyond Pythagoras section.

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