The formulae I found for the b+1 family are shown below:
a = 2n + 1
b = 2n2 + 2n or 2n ( n + 1 )
c = 2n2 + 2n + 1 or 2n (n + 1 ) + 1
P = 4n2 + 6n + 2 or 2 ( n +1 ) ( 2n + 1 )
A = 2n3 + 3n2 + n or n ( 2n2 + 3n + 1 )
I then tried to prove that Pythagoras’s theorem did work for my formulae algebraically. I broke the following formula into two groups:
a2 + b2 = c2
( 2n + 1 )2 + ( 2n2 + 2n )2 = ( 2n2 + 2n + 1 )2
Left Hand Side:
( 2n + 1 )2 + ( 2n2 + 2n )2
Right Hand Side:
( 2n2 + 2n + 1 )2
I then simplified the formulae as much as I could using the following procedures:
( a + b ) 2 = ( a + b ) x ( a + b )
= a ( a + b ) + b ( a + b )
= a2 + ab + ab + b2
= a2 + b2 + 2ab
And
( a + b+ c )2 = ( a + b+ c ) x ( a + b+ c )
= a ( a + b+ c ) + b ( a + b+ c ) + c ( a + b+ c )
= a2 + b2 + c2 + 2ab + 2ac + 2bc
Left Hand Side:
( 2n + 1 )2 + ( 2n2 + 2n )2
= 4n2 + 4n + 1 + 4n4 +8n3 + 4n2
= 4n4 + 8n3 + 8n2 + 4n + 1
Right Hand Side
( 2n2 + 2n + 1 )2
= 4n4 + 8n3 + 8n2 + 4n + 1
The right hand side is the same as the left hand side which proves my formulae are correct.
Afterwards, I researched into the family of Pythagorean triples where c is equal to b+2. Many of the triples I found were multiples of the b+1 family, so I decided not to include them in my work. The following table shows the values:
In the values for a, the difference is 4, meaning that the formula starts with:
4n
By looking carefully at the formula and numbers, I noticed that by adding 4 to this formula, I would get the right numbers.
a = 4n + 4
Factorising this, gave me:
4 ( n + 1 )
I found b by halving the difference between the numbers in row b and adding 8n + 3. This gave me:
4n2 + 8n + 3
Or
( 2n + 1 ) ( 2n + 3 )
As c was 2 numbers bigger than b, all I had to do was to add 2 to b’s formula:
4n2 + 8n + 5
When factorised, it became:
( 2n + 1 ) ( 2n + 3 ) + 2
The formulas for the b+2 family are shown below:
a = 4n + 4
b = 4n2 + 8n + 3 or ( 2n + 1 ) ( 2n + 3 )
c = 4n2 + 8n + 5 or ( 2n + 1 ) ( 2n + 3 ) + 2
Once again, I used my previous techniques to show that the left hand side was equal to the right hand side:
a2 + b2 = c2
( 4n + 4 )2 + ( 4n2 + 8n + 3 ) 2 = ( 4n2 + 8n + 5 ) 2
Left Hand Side:
( 4n + 4 )2 + ( 4n2 + 8n + 3 ) 2
Right Hand Side:
( 4n2 + 8n + 5 ) 2
( a + b ) 2 = ( a + b ) x ( a + b )
= a ( a + b ) + b ( a + b )
= a2 + ab + ab + b2
= a2 + b2 + 2ab
( a + b+ c )2 = ( a + b+ c ) x ( a + b+ c )
= a ( a + b+ c ) + b ( a + b+ c ) + c ( a + b+ c )
= a2 + b2 + c2 + 2ab + 2ac + 2bc
Left Hand Side:
( 4n + 4 )2 + ( 4n2 + 8n + 3 ) 2
= 16n2 + 32n + 16 + 16n4 + 64n3 + 88n2 + 48n + 9
= 16n4 + 64n3 + 104n2 + 80n + 25
Right Hand Side:
( 4n2 + 8n + 5 ) 2
= 16n4 + 64n3 + 104n2 + 80n + 25
After I had finished exploring the b+2 triples, I decided to experiment with b+3 triples. I didn’t manage to find any that weren’t multiples. The ones I did find are shown in the table below. All of the numbers can be divided by three to produce the b+1 family:
I then moved straight on to the b+4 family, but yet again I only found multiples. The same applied for the b+5, b+6 and b+7 families. The table for b+4 is shown below. The numbers can all be divided by 4 to give the b+1 family:
The b+8 family however, had Pythagorean triples other than multiples. These are shown in the table below:
I decided not to include multiples in this case. The difference between each of the numbers in column a is 8. This means the formula for the numbers in column a starts off with:
8n
I then added 12 to the formula:
8n + 12
This is the formula for a. When factorised, it is:
4 ( 2n + 3 )
The second difference between the numbers in column b is 4, meaning the formula starts with:
4n2
By adding:
12n + 5
To the formula, b is proved correct:
4n2 + 12n + 5
Or
( 2n + 5 )( 2n + 1 )
Once again, all I have to do to find c is add 8:
4n2 + 12n + 13
Or
( 2n + 5 )( 2n + 1 ) + 8
The b+8 family formulas are:
a = 8n + 12 or 4 ( 2n + 3 )
b = 4n2 + 12n + 5 or ( 2n + 5 )( 2n + 1 )
c = 4n2 + 12n + 13 or ( 2n + 5 )( 2n + 1 ) + 8
To prove the formulae are algebraically correct, I used the previous techniques to solve it:
a2 + b2 = c2
( 8n + 12 )2 + ( 4n2 + 12n + 5 )2 = (4n2 + 12n + 13 )2
Left Hand Side:
( 8n + 12 )2 + ( 4n2 + 12n + 5 )2
Right Hand Side:
(4n2 + 12n + 13 )2
Left Hand Side:
( 8n + 12 )2 + ( 4n2 + 12n + 5 )2
= 64n2 + 192n + 144 + 16n4 + 144n2 + 25 + 120n + 40n2 + 96n3
= 16n4 + 96n3 + 248n2 + 312n + 169
Right Hand Side
(4n2 + 12n + 13 )2
= 16n4 + 144n2 + 169 + 312n + 104n2 + 96n3
= 16n4 + 96n3 + 248n2 + 312n + 169
I also progressed onto doing the b+9 family, but unfortunately, a pattern can only be found by including multiples:
Here, the formula for a is:
6n + 27
The formula for b is:
2n2 + 18n + 36
Or
2 ( n + 3 ) ( n + 6 )
And the formula for c is:
2n2 + 18n + 45
Or
2 ( n + 3 ) ( n + 6 ) + 9
I then decided to come up with a general formula for all of the Pythagorean triples. I started by using the formulae for the b+1 family:
a = 2n + 1
I figured that adding n2 and (–n)2 to the equation wouldn’t make any difference:
a = 2n + 1 + n2 – n2
I factorised this and got:
a = ( n + 1 )2 – n2
I then move on to b:
b = 2n2 + 2n
I also factorised this to give me:
b = 2n ( n + 1 )
Then I went on to c:
c = 2n2 + 2n + 1
c = n2 + n2 + 2n + 1
c = n2 + ( n + 1 )2
By looking closely at the formulae, I saw that they all had n and (n+1) in them. I named them both:
y = n
z = ( n +1 )
I then decided to try this for another one of my families, b+2.
a = 4n + 4
b = 4n2 + 8n + 3
c = 4n2 + 8n + 5
However, no matter what I did, I could not find a similar pattern. I thought the problem may be because I had not included multiples in my formulae. I then found the formulae for the b+2 family but this time including multiples. They are:
a = 2n + 4
b = n2 + 4n + 3
c = n2 + 4n + 5
I then tried to find a pattern using these formulae. However, yet again, I was unsuccessful in doing so. To try to solve this hitch, I swapped the formulae for a and b around:
a = n2 + 4n + 3
b = 2n + 4
c = n2 + 4n + 5
I then tried to find a pattern similar to the one above:
a = n2 + 4n + 3
a = n2 + 4n + 4 – 1
a = ( n + 2 )2 – 1
This was the same as the formula for a in the b+1 family, the only difference was that it was (n+2) instead of (n+1) and 1 instead of n. I then went on to b:
b = 2n + 4
b = 2 ( n + 2 )
b = 2 x 1 x ( n + 2 )
Once again, the formula was the same as the formula for b in the b+1 family, the only difference was that it was (n+2) instead of (n+1) and 1 instead of n. I finally tried c:
c = n2 + 4n + 5
c = n2 + 4n + 4 + 1
c = ( n + 2 )2 + 1
I then came up with the following formula for the b+2 family:
y = 1
z = ( n + 2 )
I then generalised the formulae for the two families above and I got:
a = z2 – y2
b = 2zy
c = z2 + y2
This method only works for numbers where z is bigger than y and both z and y are positive integers. Z must be bigger than y so that we avoid getting any negative numbers. To prove this method works, I then tried to solve the equation algebraically:
a2 + b2 = c2
( z2 – y2 )2 + ( 2zy )2 = ( z2 + y2 )2
Left Hand Side:
( z2 – y2 )2 + ( 2zy )2
Right Hand Side:
( z2 + y2 )2
Left Hand Side:
( z2 – y2 )2 + ( 2zy )2
= z4 + y4 – 2z2y2 + 4z2n2
=
Right Hand Side:
( z2 + y2 )2
= 2z2y2 + z4 + y4
As a result, the general formula for finding any Pythagorean triple, as long as z is bigger than y, and they are both positive integers is:
a = z2 – y2
b = 2zy
c = z2 + y2
