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

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

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.

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

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

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.