2n+1
Mahmoud Elsherif Beyond Pythagoras P.4
Next I shall find the nth term of the middle side.
4/2=2
So it’s 2n2.
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
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)
Now I will investigate the other Pythagorean triples because my theory says use the bases then you times them. For Example a=b+ (b+1)= c so a=3 b=4 c=5
So it’s 5= 4+(2+1) So it’s 52= 42+ 32
25=16+9
Then you times the bases 3,4, 5 by two to get a=b+(b+1) divide by b/2 in brackets.
So it will become 10= 8+ (8+2) divide b/2 in the brackets
102=82+62
Mahmoud Elsherif Beyond Pythagoras P.13
100=64+36
Then you times the bases 3,4,5 by three to get a=b+(b+3)
So it will become 15=12+(12+3) divide b/2 in the brackets
So it will become 152= 122+92
Then it will become 225= 144+81
Finally you times the bases 3,4,5 by four to get a=b+(b+4)
So it will become 20=16+ (16+4) divide b/2 in the brackets
So it will be 202= 162+ 122
400= 256 + 144
Now to find the formula of this.
I will do this by table methods.
So to get the second part you have to times the first part by two
Mahmoud Elsherif Beyond Pythagoras P.14
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)
My formula is right
A= B +(B+ X)
You get for the top part of every table you divide it by two the B that is. B/2
Same with the first except in the second you divide it by three. B/3 For 5,12,13.
Then if it’s the next slot then you have to plus the number divide by plus one from the previous one to get next one. For 7,24,25 divide 24/4 +x.
That is my conclusion.
So my theory is correct with B. I can prove it by this
Mahmoud Elsherif Beyond Pythagoras P.15
13= 12+ (12+1) divide b/3 in the brackets
132=122+52
25= 24+ (24+2) divide b/3 in the brackets
252=242+102
25= 24+ (24+1) divide b/4 in the brackets
252=242+72
50= 48+ (48+2) divide b/4 in the brackets
502=482+142
I have gotten all the evidence right here to prove my theory.
Both of these methods I have used the table and the dividing b have also proven the theory and gave me all the evidence I needed to finish this.