# Beyond Pythagoras

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Introduction

Mahmoud ElsherifBeyond PythagorasP.1

Pythagoras Theorem is a2+b2= c2 ‘a’ is being the shortest side, ‘b’ being the middle side and ‘c’ being the longest side (hypotenuse) of a right angled triangle.

The numbers 3,4,5 satisfy this condition and so

32+ 42=52

Because 32= 3*3=9

42=4*4=16

52=5*5=25

32+ 42=52

9+16=25

25=25

This proves Pythagoras Theorem goes with the right angled triangle with the numbers 3,4,5. Next I shall prove that Pythagoras’s Theorem applies to 5,12,13 right angled triangle.

52+122=132

Because 52= 5*5=25

122= 12*12=144

132= 13*13=169

Mahmoud ElsherifBeyond PythagorasP.2

This satisfies the Theorem of Pythagoras’s goes with these numbers 5,12,13. Finally I shall prove that Pythagoras’s Theorem applies to 7,24,25 right angled triangle.

72+ 242=252

Because 72= 7*7=49

242= 24*24= 576

252=25*25=625.

So

a2+b2=c2

72+242=252

49+576=625

This proves Pythagoras Theorem goes with the right angle triangle with the sides 7,24,25

Shortest Side | Middle Side | Longest Side |

3 | 4 | 5 |

5 | 12 | 13 |

7 | 24 | 25 |

9 | 40 | 41 |

11 | 60 | 61 |

13 | 84 | 85 |

Mahmoud ElsherifBeyond PythagorasP.3

I shall find the prediction of the shortest side first.

3,5,7

It goes up in 2 so in my conclusion so it will become

3,5,7,9,11,13

Now I will find the difference between them.

The difference is 2

Next I shall find the prediction of the middle side next.

4,12,24

It goes up by 4,8,12. So in my conclusion I think it will become 4,8,12,16,20,24

So it will be 4,12,24, 40, 60 84.

The difference is 4,8,12. Now I shall find the difference and it is n*4

Middle

Finally I will investigate the shortest Term.

Mahmoud ElsherifBeyond PythagorasP.7

Shortest Term2= Longest Term2- Middle Term2

(nth term)2= (nth term)2 + (nth term)2

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

Shortest

(2n+1)* (2n+1)

4n2+ 2n

2n + 1

4n+ 4n2+1

Middle

(2n2+2n) * (2n2+2n)

4n4+4n3

4n3+ 4n2

4n2 +8n3+4n4

Longest

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

4n4+4n3+2n2

4n3+ 4n2+2n

2n2+2n+1

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

(4n+ 4n2+1)= (4n4+8n3+8n2+4n+1)-(4n2 +8n3+4n4)

Now I have finished that I will start having even numbers, to see if Pythagoras’s Theorem works. I shall do this all over again but with an even short side.

Shortest Side | Middle Side | Longest Side |

6 | 8 | 10 |

10 | 24 | 26 |

14 | 48 | 50 |

18 | 80 | 82 |

22 | 120 | 122 |

26 | 168 | 170 |

Mahmoud Elsherif Beyond Pythagoras P.8

I shall find the prediction of the shortest side first.

6,10,14

It goes up in 4 so in my conclusion so it will become

6,10,14,18,22,26.

Now I will find the difference between them.

The difference is 4

Next I shall find the prediction of the middle side next.

8,24,48

It goes up by 8,16,24. So in my conclusion I think it will become 8,16,24,32,40,48

So it will be 8,24,48, 80, 120, 168.

The difference is 8,16,24. Now I shall find the difference and it is n*8

Conclusion

9 | 12 | 15 |

15 | 36 | 39 |

21 | 72 | 75 |

27 | 120 | 123 |

33 | 180 | 183 |

39 | 252 | 255 |

Finally to get the fourth part you have to times the first part by four.

Shortest Side | Middle Side | Longest Side |

12 | 16 | 20 |

20 | 48 | 52 |

28 | 96 | 100 |

36 | 160 | 164 |

44 | 240 | 244 |

52 | 336 | 340 |

So my theory is correct with the times.

For the shortest formula it is x (2n+1)

For the middle formula it is x (2n2+2n)

For the longest formula it is x (2n2+2n+1)

1(2n+1) 2* 1 =2 +1=3

2 (2n+1) 4n +2 = 4*1= 4+2=6

Mahmoud ElsherifBeyond PythagorasP.14

3 (2n+1) 6n+3= 6*1 + 3= 9

4 (2n+1) 8n+4= 8*1+4= 12

M(2n +1)

1 (2n2+2n) 2*12= 2*1=2, 2+2=4

2(2n2+2n) 4n2+4n = 4*12= 4, 4*1=4. 4+4=8

3(2n2+2n) 6n2+6n= 6*12= 6, 6*1=6. 6+6= 12

4(2n2+2n) 8n2+8n= 8*12= 8, 8*1=8. 8+8= 16

M(2n2+2n)

1 (2n2+2n+1) 2*12= 2*1=2, 2+2+1=5

2 (2n2+2n+1) 4n2+4n = 4*12= 4, 4*1=4. 4+4+2=10

3 (2n2+2n+1) 6n2+6n = 6*12= 6, 6*1=6. 6+6+3=15

4 (2n2+2n+1) 8n2+8n = 8*12= 8, 8*1=8. 8+8+4=20

M (2n2+2n+1)

N=1

Now to investigate the Pythagoreans Triples

1st Family= c, b, b+1 generate this family using (b+1)2 = b2+c2

(4+1)2= 42+ 32

52=42+ 32

25=16+ 9

2nd Family=c, b, b+2 generate this family using (b+2)2 = b2+c2

(8+2)2= 82+ 62

102=82+ 62

100=64+ 36

3rd Family= c, b, b+3 generate this family using (b+3)2 = b2+c2

(12+3)2= 122+ 92

152=122+ 92

225= 144+81

Mahmoud ElsherifBeyond PythagorasP.15

4th Family= c,b,b+4 generate this family using(b+4)2 = b2+c2

(16+4)2= 162+ 122

202=162+ 122

400= 256+144

I have noticed you have to make b the subject and the formula is this.

Xth family= c, b, b,+ x generates this rule using (b+ x)2= b2+c2

(B+X)2= b2+c2

X2= b2+c2

X= b+c

This is the formula I have proven this by my methods by using the families of the Pythagorean triples.

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

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