The only factors of 1 are 1 and 1 so it factorises to (2n+1)(n+1)
2 needs to be added to the front of these brackets before the formula for perimeter is complete.
Therefore the formula for P is:
tn=2(2n+1)(n+1)
Next I needed to find a formula for area (A). As in the case of perimeter the laws of algebra say that if I insert my formulae for a and b to the formula for area ((axb)÷2) I will come up with the correct formula for A. My calculation turned out to be:
(2n+1)(2n2 +2n)
2
The answer to this is:
2n3+3n2+n
I checked and proved that this was correct:
Where n is 1
tn= 2n3+3n2+n
tn= 2(13)+3x(12)+1
tn= 2x1+3x1+1
tn= 2+3+1
tn= 6
Where n is 2
tn= 2n3+3n2+n
tn= 2(23)+3x(22)+2
tn= 2x8+3x4+2
tn= 16+12+2
tn= 30
I immediately noticed that n was common to the formula and on taking n out of the formula I was left with:
2n2+3n+1
This is now a quadratic equation and can be factorised:
(+)(+) The signs are both the same and positive.
(2n+)(n+) I tried two factors of 2n2, 2n and n.
(2n+1)(n+1) Two factors of 2 are 1 and 1.
I need to add n in front of this to find the formula for A from n, therefore the formula for A from n is:
tn=n(2n+1)(n+1)
I have made a table of my findings.
If I had worked out the formulae correctly they would agree to Pythagoras theorem. For example, the formula for a2+ the formula for b2 must equal the formula for c2. I proved this by the following calculation:
a2+b2=c2
(2n+1)2+(2n(n+1))2= (2n(n+1)+1)2–Brackets need to be multiplied out
(2n+1)2+(2n2+2n)2= (2n2+2n+1)2
a2= (2n+1)2= (2n+1)(2n+1)
= 4n2+2n+2n+1= 4n2+4n+1
b2= (2n2+2n)2= (2n2+2n)(2n2+2n)
= 4n4+4n3+4n3+4n2= 4n4+8n3+4n2
c2= (2n(n+1)+1)2=(2n2+2n+1)( 2n2+2n+1)
= 4n4+4n3+2n2+4n3+4n2+2n+2n2+2n+1=4n4+8n3+8n2+4n+1.
(4n2+4n+1)+(4n4+8n3+4n2)= 4n4+8n3+8n2+4n+1√ CORRECT
This rather large calculation proves that my formulae can be applied to Pythagoras’ Theorem.
I decided to continue my investigation finding similar formulae only from a, for example, a formula to find b from just a. I first tried to find a formula to find b from a.
I had to use trial and error. The difference between b and a was 1, 7, 17, 31. The difference was not constant. This meant that the formula must have involved a. It was also probably quadratic. a2 gave values over twice larger than b:
a2 9, 25, 49, 81
b 4, 12, 24, 40
Using trail and error I came up with:
b= a2÷2-0.5
I checked and proved that this was correct:
Where a is 3
b= a2÷2-0.5
b= 32÷2-0.5
b= 9÷2-0.5
b= 4.5-0.5
b= 4
Where a is 5
b= a2÷2-0.5
b= 52÷2-0.5
b= 25÷2-0.5
b= 12.5-0.5
b= 12
After I found the formula for b from a, the formula for c from a was very simple to find. I already knew that side c was exactly 1 more than b. Therefore the formula to find c from a is exactly the same as to find b from a, only +1 at the end of it:
b= a2÷2-0.5+1
This simplifies to:
b= a2÷2+0.5
I checked and proved that this was correct:
Where a is 3
c= a2÷2+0.5
c= 32÷2+0.5
c= 9÷2+0.5
c= 4.5+0.5
c= 5
Where a is 5
c= a2÷2+0.5
c= 52÷2+0.5
c= 25÷2+0.5
c= 12.5+0.5
c= 13
Next I needed a formula to find P from a. The difference between P and a went 9, 25, 49, 81. The difference was not constant meaning that a must have been involved and it was probably quadratic. However a2 alone is too small:
a2 9, 25, 49, 81
P 12, 30, 56, 90
Showing this against a made it very obvious what remaining component was required:
a 3, 5, 7, 9
a2 9, 25, 49, 81
+3 +5 +7 +9
∨ ∨ ∨ ∨
P 12, 30, 56, 90
I had to add a to the end before completing the formula. I checked and proved that what I had found was correct:
Where a is 3
P= a2+a
P= 32+3
P= 9+3
P= 12
Where a is 5
P= a2+a
P= 52+5
P= 25+5
P= 30
Therefore the formula to find P from a is:
P= a2+a
I next had to find a formula to find A from a. This was much more difficult than the others. Again I used trial and error. The difference between A and a went 3, 25, 77, 171. There was no constant pattern. This meant that a must be involved in the formula and it was probably quadratic or even cubic. I compared both a2 and a3 to A:
a2 9, 25, 49, 81
A 6, 30, 84, 180
a3 27, 125, 343, 729
Apart from one exception a2 was too small and a3 was too big. It couldn’t possibly be a2 because in the first triple, 3, 4, 5, it was too small but in the other triples it was too big. A formula can’t possibly change for different triples. It must have involved a3. The difference between a3 and A was not constant. This meant that a was involved in the subtraction or division. The difference between a3 and A went 19, 95, 259, 549. I compared this to a:
D 19, 95, 259, 549
a 3, 5, 7, 9
D stands for difference between a3 and A
Not every difference could be expressed in terms of a (i.e. 2a, 6a etc…). This meant that the suffix to the formula had to involve both a subtraction and a division, one of which had to be in terms of a. I tried multiple combinations until I found:
A= a3-a÷4
I checked and proved that what I had found did work:
Where a is 3
A= a3-a÷4
A= 33-3÷4
A= 27-3÷4
A= 24÷4
A= 6
Where a is 5
A= a3-a÷4
A= 53-5÷4
A= 125-5÷4
A= 120÷4
A= 30
I made a table of my findings:
Anomalies
Whilst searching for triples I came across the triple 119, 120, 169. This cannot be found using my formulae in terms of n. Side c is not 1 larger than Side b and b is 1 larger than a. Also it doesn’t agree to my formulae in terms of a. I cannot explain this but will investigate it later in my investigation.
Other Patterns
A quite obvious pattern is that c= b+1. Another pattern is a2= b+c or a=√b+c. I will try to explain this later in the investigation.
Even Numbers
I decided to continue my investigation by investigating similar patterns but instead for Pythagorean triples with an even number for its shortest side. Firstly I had to collect some triples. I started off using the method of trial and error but this became a long and tedious process. I began to look for an alternative method to gather results. I came across an ancient Babylonian formula that went as follows:
- P>Q>0; P and Q have no common divisors other than 1; P and Q are not both odd; P and Q are Both Positive Integers.
-
X= P2-Q2
- Y=2xPxQ
-
c=p2+22
- X and Y correspond to a and b but not necessarily in that order.
This was a very good method but still took quite a lot of time. I decided to design a spreadsheet to do the working for me, all I had to do was put values of P and Q that agreed the laws stated. Part of the spreadsheet is below:
This shows the formulae I used:
This saved myself valuable time and I quickly found the triples I required. It soon became apparent that the triples for even numbered shortest sides were a lot more complicated than for odd numbers. I realised that there were no triples possible for a shortest side of 2 or 4. The first triple was 6, 8, 10. This was double of 3, 4, 5. It also became apparent that there would be many multiples of previous triples; for example for the shortest side of 12 there were two possibilities; 12, 16, 20 and 12, 35, 37. 12, 16, 20 is just double of 6, 8, 10. I decided to ignore any multiples such as 12, 35, 37 and stick with true triples that cannot be halved to produce a different triple with an even number. I noticed that all true triples had a difference of 2 between c and b. I soon developed the table below using my spreadsheet:
As before, I started by trying to find formulae that would find a, b, c, P and A from n.
I began to search for a relationship between a from n. I used the difference method:
n a L1
1 6
>2
2 8
>2
3 10
>2
4 12
>2
5 14
>2
6 16
>2
7 18
There was a constant difference at L1 of 2. This meant that the formula had to contain 2n. I compared 2n to a.
2n a
2 >+4 6
4 >+4 8
6 >+4 10
8 >+4 12
10 >+4 14
12 >+4 16
14 >+4 18
It became obvious that all I had to do to complete my formula was add 4 to the 2n I already had found. I checked and proved that what I had found did work:
Where n is 1
tn= 2n+4
tn= 2x1+4
tn= 3+4
tn= 7
Where n is 2
tn= 2n+4
tn= 2x2+4
tn= 4+4
tn= 8
Therefore the formula to find a from n for Pythagorean triples with an even number for its shortest side is:
tn= 2n+4
This is not in its simplest form. 2 is common to the formula. I factorised it to:
tn= 2(n+2)
I next began looking for a relationship between n and b. I used the difference method:
n b L1
1 8
>7
2 15
>9
3 24
>11
4 35
>13
5 48
>15
6 63
>17
7 80
There was no constant at L1 so I looked for a constant second difference:
n b L1 L2
1 8
>7
2 15 >2
>9
3 24 >2
>11
4 35 >2
>13
5 48 >2
>15
6 63 >2
>17
7 80
There was a constant difference of 2 at L2. This meant the formula was quadratic. Halving 2 gives 1, this means that the formula contains n2 (1n2) somewhere. I next compared n2 to b:
n2 b
1 >+7 8
4 >+11 15
9 >+15 24
16 >+19 35
25 >+23 48
36 >+27 63
49 >+31 80
The difference between n2 and b was not constant. This meant that n must have been involved in the addition, however the difference between n2 and b could not be expressed in terms of n. This meant that both an integer in terms of n and a whole number must be involved in the addition. I noticed the addition had to be fairly large. I used trial and error to find the addition. I came up with +4n+3. I checked and proved that what I had found was correct:
Where n is 1
tn= n2+4n+3
tn= 12+4x1+3
tn= 1+4+3
tn= 8
Where n is 2
tn= n2+4n+3
tn= 22+4x2+3
tn= 4+8+3
tn= 15
Therefore the formula to find b from n for Pythagorean triples with and even number for its shortest side is:
tn= n2+4n+3
This is a quadratic equation and can be factorised:
(+)(+) The signs are both the same and positive.
(n+)(n+) I tried two factors of n2, n and n.
(n+1)(n+3) Two factors of 3 are 1 and 3.
Therefore the formula to find b from n for Pythagorean triples with and even number for its shortest side in its simplest form is:
tn= (n+1)(n+3)
I next started looking for a formula to find side b from n. This was much easier than the others. I noticed that side c was identical to side b but +2. The laws of algebra say that the formula should be identical but +2. I proved that this was correct:
Where n is 1
tn= (n+1)(n+3)+2
tn= n2+4n+3+2
tn= 12+4x1+3+2
tn= 1+4+3+2
tn= 10
Where n is 2
tn= (n+1)(n+3)+2
tn= n2+4n+3+2
tn= 22+4x2+3+2
tn= 4+8+3+2
tn= 17
Therefore the formula to find c from n for Pythagorean triples with and even number for its shortest side is:
tn= (n+1)(n+3)+2
This is already in its simplest form and does not require factorising.
Next I looked for a formula to find perimeter (P) from n. The formula to find perimeter is:
P=a+b+c
The laws of algebra say that if I add the formulae to find a, b and c from n I will find the right formula. This is exactly what I did. I used unfactorised versions of the formulae to find a, b and c from n:
a= tn= 2n+4
b= tn= n2+4n+3
c= tn= n2+4n+3+2
My calculation was:
tn= 2n+4+n2+4n+3+n2+4n+3+2
After collecting similar terms you are left with:
tn= 2n2+10n+12
I proved that this was correct:
Where n is 1
tn= 2n2+10n+12
tn= 2x(12)+10x1+12
tn= 2x1+10x1+12
tn= 2+10+12
tn= 24
Where n is 2
tn= 2n2+10n+12
tn= 2x(22)+10x2+12
tn= 2x4+10x2+12
tn= 8+20+12
tn= 40
2 is common to this formula:
tn= 2(n2+5n+6)
The n2+5n+6 is a quadratic equation and can be factorised:
(+)(+) The signs are both the same and positive.
(n+)(n+) I tried two factors of n2, n and n.
(n+3)(n+2) Two factors of 6 that add up to make 5 are 3 and 2.
Therefore the formula to find P from n for Pythagorean triples with and even number for its shortest side is:
tn= 2(n+3)(n+2)
Next I began to look for a relationship between A and n. The formula for area (A) is:
A= (axb)÷2
Again the laws of algebra state that if I multiply the formulae to find a and b from n together and divide the answer by 2, I will find the correct formula to find A from n. This made my calculation much easier and I would still come up with the same answer. This was the calculation I was faced with; again I used unfactorised formulae to make the calculation easier:
(2n+4) (n2+4n+3) ÷2
(2n3+8n2+6n+4n2+16n+12) ÷2
(2n3+122+22n+12) ÷2
n3+6n2+11n+6
I proved that this was correct:
Where n is 1
tn= n3+6n2+11n+6
tn= 13+6x12+11x1+6
tn= 1+6x1+11x1+6
tn= 1+6+11+6
tn= 24
Where n is 2
tn= n3+6n2+11n+6
tn= 23+6x22+11x2+6
tn= 8+6x4+11x2+6
tn= 8+24+22+6
tn= 60
I noticed that this was not in its simplest form but didn’t know how to factorise cubic equations. I noticed that if I altered the area formula A= (axb)÷2 to A= ½axb I would still get the right answer but in a more simple form. Half of a is n+2. I immediately noticed if I multiplied this by the fully factorised version of b, I would get the calculation:
(n+2)(n+1)(n+3)
On multiplying the brackets out you get n3+6n2+11n+6 as the answer. What this meant was that (a+2)(a+1)(a+3) is the simplest form of n3+6n2+11n+6.
I know had found formulae for each of a, b, c, P and A from n. I made a table of my findings:
If I had worked out the formulae correctly they would agree to Pythagoras theorem. For example, the formula for a2+ the formula for b2 must equal the formula for c2. I proved this by the following calculation:
a2+b2=c2
(2n+4)2+((n+1)(n+3))2= ((n+1)(n+3)+2)2–Brackets need to be multiplied out
(2n+4)2+(n2+4n+3)2= (2n2+2n+1)2
a2= (2n+4)2= (2n+4)(2n+4)
= 4n2+8n+8n+16= 4n2+16n+16
b2= (n2+4n+3)2= (n2+4n+3)( n2+4n+3)
= n4+4n3+3n2+4n3+16n2+12n+3n2+12n+9= n4+8n3+22n2+24n+9
c2= (n2+4n+5)2=( n2+4n+5)( n2+4n+5)
= n4+4n3+5n2+4n3+16n2+20n+5n2+20n+25 = n4+8n3+26n2+40n+25.
(4n2+16n+16)+( n4+8n3+22n2+24n+9)= n4+8n3+26n2+40n+25√ CORRECT
This rather large calculation proves that my formulae can be applied to Pythagoras’ Theorem.
As I did when investigating triples with an odd number for its shortest side, I continued my investigation by looking for formulae in terms of a.
I first began to look for a formula to find b from a. The difference between b and a went 2, 7, 14, 23, 34, 47, 62. There was no constant difference meaning that a was involved in the formula and it was probably quadratic. I compared a2 to b:
a2 36, 64, 100, 144
b 8, 15, 24, 35
Using trail and error I came up with:
b= a2÷4-1
I checked and proved that this was correct:
Where a is 6
b= a2÷4-1
b= 62÷4-1
b= 36÷4-1
b= 9-1
b= 8
Where a is 8
b= a2÷4-1
b= 82÷4-1
b= 64÷4-1
b= 16-1
b= 15
After I found the formula for b from a, the formula for c from a was very simple to find. I already knew that side c was exactly 2 more than b. Therefore the formula to find c from a is exactly the same as to find b from a, only +2 at the end of it:
b= a2÷4-1+2
This simplifies to:
b= a2÷4+1
I checked and proved that this was correct:
Where a is 6
b= a2÷4+1
b= 62÷4+1
b= 36÷4+1
b= 9+1
b= 10
Where a is 8
b= a2÷4+1
b= 82÷4+1
b= 64÷4+1
b= 16+1
b= 17
I next began to look for a formula for P in terms of a. The difference between P and a was 18, 32, 50, 72. The difference was not constant. This meant that a must have been involved in the difference and the formula was probably quadratic. I compared a2 to P:
a2 36, 64, 100, 144
∨ ∨ ∨ ∨
+12 +24 +40 +60
P 24, 40, 60, 84
a 6 8 10 12
The difference between a2 and P was not constant but could be expressed in terms of a. In the first triple the difference between a2 and P could be expressed as 2a, for the second triple it could be expressed as 3a, in the next 4a and the next 5a. I had found the pattern but could not express this as a formula without including a. Another pattern I noticed was that 24 was 4a, 40 was 5a, 60 was 6a and 84 was 7a. Again I couldn’t express this as a formula without including n. I tried comparing a3 to P:
a3 216, 512, 1000, 1728
∨ ∨ ∨ ∨
+192 +472 +940 +1644
P 24, 40, 60, 84
a 6 8 10 12
The difference between a3 and P was not constant and couldn’t be expressed in terms of a. Then the formula hit me. A quadratic or cubic was not required. Halving a gives you 3, 4, 5, 6 etc… Adding 1 to this number gives you 4, 5, 6, 7etc… These are the numbers I required to complete a simple pattern. Multiplying these numbers to a gives you P for each triple. Therefore the formula to find P from a for true Pythagorean triples with an even number for its shortest side is:
P= (½a+1)xa
This was already in its simplest form. I checked and proved that this was correct:
Where a is 6
P= (½a+1)xa
P= (½x6+1)x6
P= (3+1)x6
P= 4x6
P= 24
Where a is 8
P= (½a+1)xa
P= (½x8+1)x8
P= (4+1)x8
P= 5x8
P= 40
I next required a formula to find A from a. The difference between A and a went 18, 52, 110, 198, 322. There was no constant difference this meant that the formula had to involve a and was probably quadratic or cubic. I compared a2 and a3 to A:
a3 216, 512, 1000, 1728
∨ ∨ ∨ ∨
+192 +472 +940 +1644
A 24, 60, 120, 210
∧ ∧ ∧ ∧
-12 -4 +20 +66
a2 36, 64, 100, 144
a 6, 8, 10, 12
The formula couldn’t possibly be based around a2. For the first 2 triples a2 was too big but for the others it was too small. The formula can’t possibly decrease a2 for some triples and increase it for others. Therefore the formula I was looking either included a or a3. I compared a and a3 to A again:
a3 216, 512, 1000, 1728
∨ ∨ ∨ ∨
-192 -472 -940 -1644
A 24, 60, 120, 210
∧ ∧ ∧ ∧
+18 +52 +110 +198
a 6, 8, 10, 12
Neither of the differences, a and A, and a2 and A, were constant and also not all triples for either of them could be expressed in terms of a. This meant that I was looking for an integer in terms of a and an integer in the suffix of the formula. I tried comparing 2a to A:
A 24, 60, 120, 210
∧ ∧ ∧ ∧
+12 +44 +100 +186
2a 12, 16, 20, 24
There was no constant difference between A and 2a and the difference could not be expressed in terms of a. I tried 3a:
A 24, 60, 120, 210
∧ ∧ ∧ ∧
+6 +36 +90 +174
2a 18, 24, 30, 36
Again there was no constant difference between A and 3a and the difference could not be expressed in terms of a. I tried 2a3:
2a3 432, 1024, 2000, 3456
∨ ∨ ∨ ∨
+192 +964 +1060 +3246
A 24, 60, 120, 210
I tried many combinations until I finally found:
A= (a3÷8)-½a
This was already in its simplest form. I checked and proved that this was correct:
Where a is 6
A= (a3÷8)-½a
A= (63÷8)-½x6
A= (216÷8)-3
A= 27-3
A= 24
Where a is 8
A= (a3÷8)-½a
A= (83÷8)-½x8
A= (512÷8)-4
A= 64
A= 60
I now had all of the formulae to find b, c, P and A from a for triples with an odd number for its shortest side.
Anomalies
Whilst searching for triples I came across the triple 20, 21, 29 and 696, 697, 985. This cannot be found using my formulae in terms of n. Side c is not 2 larger than Side b and b is 1 larger than a. Also it doesn’t agree to my formulae in terms of a. I cannot explain this but will investigate it later in my investigation.
Other Patterns
A quite obvious pattern is that c= b+2. The pattern a2= b+c which I found for odd numbered shortest side triples does not work for triples with an even number for its shortest side. Instead a2=2(b+c) works. I will try to explain this later in the investigation. Whilst trying to find formulae to find A and P from a I found the following formulae:
P= a+2n2+4n
P= (n+3)xa
A= (½a)(a+n2+2n+1)
A= (a3÷8)-(n+2)
a2= b+c
I decided to investigate why a2= b+c only worked for odd numbered shortest side triples and not even numbered. The alternative was identical but c and b had to be doubled before hand. I decided to compare what the differences between the two. The most obvious difference was that for odd numbered triples the difference between b and c was 1, whilst for even numbered triples the difference was 2. I noticed that this difference was the same number required to be multiplied to b and c before a2= b+c worked. This was when I realised that the formula was actually:
a2= (c-b)(b+c)
c-b for odd numbered shortest side triples was 1, therefore the c-b can be ignored, but for even numbered shortest side triples c-b= 2 so 2 is required in front of b+c. This formula works for all triples. In the triple 9, 12, 15 (triple of 3, 4, 5) the difference between b and c is 3. 3x(12+15) is 81, which is also 92. Then the next stage of this suddenly appeared to me; on multiplying a2= (c-b)(b+c) you get a2=cb+c2-b2-cb, this simplifies to a2=c2-b2 or a2=b2+c2, Pythagoras Theorem! I suspect this is how Pythagoras himself discovered his theorem.
The Perfect Pythagorean Triples
The anomalies to my formulae which I found earlier, 20, 21, 29, 119, 120, 169, 696, 697, 985. These are so called “Perfect Triples”. A Perfect Triple is a triple in which side b is exactly one larger than side b. Perfect Triples are a very strange phenomenon. The first Perfect triple is 3, 4, 5; the next 20, 21, 29. If you multiply the hypotenuse of the last triple by the special number (√2+ 1)2 (approx 5.828) you get near perfectly the hypotenuse of the next Perfect Triple. For example:
20, 21, 29- hypotenuse is 29x(√2+ 1)2= 169.024
The next triple is 119, 120, 169- the hypotenuse is 169, 0.024 off from calculated.
