Therefore the rule for finding the nth term of the middle side is:
2n2+2n
Next I shall find the nth term of the longest side.
4/2=2
So it’s 2n2.
So we have so far 2n2+2n. Substituting in we get for 2n2+2
If n=1 4 and the first term is 5 so a difference of +1
If n=2 12 and the first term is 13 so a difference of +1
If n=3 24 and the first term is 25 so a difference of +1
Mahmoud Elsherif Beyond Pythagoras P.5
So the nth term is:
2n2+2n+1
To see if this really works, I will expand them and see if these are the real nth terms. I will start with shortest term first.
Longest Term2= Middle Term2+ ShortestTerm2
(nth term)2= (nth term)2 + (nth term)2
(2n2+2n+1)2 (2n2+2n) 2 + (2n+1)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
(4n4+8n3+8n2+4n+1)= (4n2 +8n3+4n4)+ (4n+ 4n2+1)
Mahmoud Elsherif Beyond Pythagoras P.6
Next I will investigate the Middle Term
Middle Term2= Longest Term2- ShortestTerm2
(nth term)2= (nth term)2 + (nth term)2
(2n2+2n)2 = (2n2+2n+1) 2 - (2n+1)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
(4n2 +8n3+4n4)= (4n4+8n3+8n2+4n+1)- (4n+ 4n2+1)
Finally I will investigate the shortest Term.
Mahmoud Elsherif Beyond Pythagoras P.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
Next I shall find the prediction of the longest side next.
10,26,50,
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 10,26,50, 82,122,170
The difference is 8,16,24. Now I shall find the difference and it is n*8
First I shall find the nth term of the shortest side.
So 4/1=4, so it’s 4n. Substituting in we get for 4n
If n=1 4 and the first term is 6 so a difference is +2
If n=2 8 and the first term is 10 so a difference is +2
If n=3 12 and the first term is 14 so a difference is +2
Therefore the rule for finding the nth term of the shortest side is:
4n+2
Mahmoud Elsherif Beyond Pythagoras P.9
Next I shall find the nth term of the middle side.
8/2=4
So it’s 4n2.
Therefore the rule for finding the nth term of the middle side is:
4n2+4n
Next I shall find the nth term of the longest side.
8/2=4
So it’s 4n2.
So we have so far 4n2+4n. Substituting in we get for 4n2+4n
If n=1 8 and the first term is 10 so a difference of +2
If n=2 24 and the first term is 26 so a difference of +2
If n=3 48 and the first term is 50 so a difference of +2
Mahmoud Elsherif Beyond Pythagoras P.10
So the nth term is:
4n2+4n+2
Now I will investigate the longest term:
Longest term= Middle term+ Shortest term
(nth term)2 = (nth term)2 + (nth term)2
Longest
(4n2+4n+2) * (4n2+4n+2)
16n4+16n3+ 8n2
16n3+16n2+8n
8n2+8n+4
4+16n+32n2+32n3+16n4
Middle term
(4n2+4n)* (4n2+4n)
16n4+16n3
16n3 + 16n2
162+32n3+164
Shortest
(4n+2)*(4n+2)
16n2+8n
8n+ 4
16n+16n2+4
Longest = Middle + Shortest
(4+16n+32n2+32n3+16n4)=(16n2+32n3+16n4) +(16n+16n2+4)
Next I will investigate the Middle term:
Mahmoud Elsherif Beyond Pythagoras P.11
Middle term= Longest term- Shortest term
(nth term)2 = (nth term)2 - (nth term)2
Longest
(4n2+4n+2) * (4n2+4n+2)
16n4+16n3+ 8n2
16n3+16n2+8n
8n2+8n+4
4+16n+32n2+32n3+16n4
Middle term
(4n2+4n)* (4n2+4n)
16n4+16n3
16n3 + 16n2
162+32n3+164
Shortest
(4n+2)*(4n+2)
16n2+8n
8n+ 4
16n+16n2+4
Middle = Longest - Shortest
(16n2+32n3+16n4) = (4+16n+32n2+32n3+16n4) - (16n+16n2+4)
Finally I will investigate the shortest term
Shortest term= Longest term- Middle term
(nth term)2 = (nth term)2 - (nth term)2
Mahmoud Elsherif Beyond Pythagoras P.12
Middle term= Longest term- Shortest term
(nth term)2 = (nth term)2 - (nth term)2
Longest
(4n2+4n+2) * (4n2+4n+2)
16n4+16n3+ 8n2
16n3+16n2+8n
8n2+8n+4
4+16n+32n2+32n3+16n4
Middle term
(4n2+4n)* (4n2+4n)
16n4+16n3
16n3 + 16n2
162+32n3+164
Shortest
(4n+2)*(4n+2)
16n2+8n
8n+ 4
16n+16n2+4
Shortest = Longest - Middle
(16n+16n2+4)= (4+16n+32n2+32n3+16n4) - (16n2+32n3+16n4)
I will do this by table methods.
Mahmoud Elsherif Beyond Pythagoras P.13
So to get the second part you have to times the first part by two
To get the third part you have to times the first part by three
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 Elsherif Beyond Pythagoras P.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 Elsherif Beyond Pythagoras P.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.