Beyond Pythagoras
Introduction
Believed to have been born in 582BC and died in 500BC approximately, Pythagoras was a Greek philosopher and mathematician. He discovered some of the most influential theories of number, geometry and proportion which are still frequently used in modern mathematics.
Pythagorean triples, on which this investigation is based, are sets of three numbers as seen in the above diagram. Here it is the simplest triple; 3,4,5. They will be written in the following form throughout this paper; a,b,c. There are other Pythagorean triples such as 5,12,13; which is the 2nd odd triple and 361,65160,65161; which is the 180th odd triple.
His theorem is this: the square of the hypotenuse (longest side (c (5 (25)))) of a right-angled triangle is equal to the sum of the squares of the other two sides (a (3 (9)) and b (4 (16))). This can be expressed using the following equation:
a2+b2=c2
Aim
The aim of the investigation is to explore and mathematically express the relationships between the number of the triple (n), the length of each side (a; the shortest), (b; the intermediate), (c; the longest), the perimeter (P) and area (A) of a right-angled triangle.
Please note: Throughout this investigation, the sum on the bottom line is the final answer to each set of sums. For example:
y=2+2
y=4
In this example, the bottom line (y=4) should be interpreted as the answer.
The main rule of the investigation is that the values of the above (n, a, b, c, P and A) must all be positive integers (i.e. whole numbers of greater value than zero). If any of the numbers concerned do not match these criteria (i.e. they are decimals), then they and their respective triple will be excluded from the sequence of triples and will not be considered in this investigation, as this would cause the investigation to be infinitely complex. For example, if in the first term of the sequence, a equalled 1, then b and c would not be integers (1·3 recurring and 1.6 recurring respectively), as an integer has to be a whole number by definition. Therefore, the first odd Pythagorean triple to satisfy these criteria is the 3,4,5 triple, and it is subsequently the first term of the odd triples sequence.
The first sequence of triples that will be investigated are those with the first number (i.e. 3 of 3,4,5) as an odd number. I will refer to these triples as "odd triples" throughout the investigation for convenience.
The second sequence of triples that will be investigated are those with the first number (i.e. 6 of 6,8,10, although this triple should not be counted as a "true" triple, as we shall see later) as an even number. I will refer to these triples as "even triples" throughout the investigation for convenience.
The first section of the investigation is to collect odd triples so that they can be analysed and turned into formulae relating to the number of the term of the sequence (n). The first set of triples was to be where a equalled an odd number. I wanted to collect the first 8 odd triples using trial and error, but realised that using this time to devise a method to find b and c from a would be more effective. This would be achieved by using a to calculate the other two sides, and this would be simple because a is already known (it is an odd number starting from and including 3). The formula was calculated in the following way.
* I found that each of the three triples of which I had discovered by trial and error had a factor in common with each other, and this was that the difference between b and c was always 1. This meant that if I could find the relationship between a and the midway point between b and c (b+0·5 or c-0·5), I could subtract or add 0·5 to this midway point to find b and c respectively.
* I began by finding the relationship between a and this midway point; the relationship between 3 and 4·5. I tried multiplying a by 1·5 and ended up with 4·5. This worked for the first triple, but when it came to the 5,12,13 triple it did not. I then realised the connection. If you square a and then divide it by 2, you will end up with the midway number (4·5). I tested this on the next two triples and it was successful, so I had found one of the most time saving formulae in this investigation.
* To find the remaining five triples before I analysed all eight of them, I could now utilise this formula.
I discovered that if you square a and divide it by 2, you end up with the base number for the next two sides, b and c. By subtracting and adding 0·5 to the answer, you can find b and c respectively. For example, in the first Pythagorean triple:
b= a2-0·5 c= a2+0·5
2 2
b= 32-0·5 c= 32+0·5
2 2
b= 4·5-0·5 c= 4·5+0·5
b= 4 c= 5
These ...
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* To find the remaining five triples before I analysed all eight of them, I could now utilise this formula.
I discovered that if you square a and divide it by 2, you end up with the base number for the next two sides, b and c. By subtracting and adding 0·5 to the answer, you can find b and c respectively. For example, in the first Pythagorean triple:
b= a2-0·5 c= a2+0·5
2 2
b= 32-0·5 c= 32+0·5
2 2
b= 4·5-0·5 c= 4·5+0·5
b= 4 c= 5
These formulae can be summed up as the relationship between a and b and a and c.
I then constructed a table of the first 8 Pythagorean triples whereby a equalled an odd number by using the above formula. In this table, I also included the perimeter and area of the triangles in relation to a, b, and c.
n
a
b
c
P
A
=a
=a2/2-0·5
=a2/2+0·5
=a+b+c
=1/2ab
x
x
x
x
x
3
4
5
2
6
2
5
2
3
30
30
3
7
24
25
56
84
4
9
40
41
90
80
5
1
60
61
32
330
6
3
84
85
82
546
7
5
12
13
240
840
8
7
44
45
306
224
The formulas displayed in the second row of the 2nd, 3rd and 4th columns of the table are displaying relationships of a towards a, b and c. What were now required were the formulae that displayed a relationship between the number of the term of the sequence (n) and the value of the measures of the triangle (a, b, c, P and A). It is ironic that the formula displaying the relationship of a towards b and c will be needed later on in the investigation. For example, if the user wanted to know the length of a, then he/she could calculate it by implementing the number of the term of the sequence (n) into the correct formula.
A note on the displaying of sequences in this investigation. The leftmost column of values will be titled with the name of the sequence of values being investigated. If the values under L1 are all identical, then this shows a linear relationship between the values in the leftmost column. If the values under L2 are all identical, then this shows an exponential relationship between the values in the leftmost column. If the values under L3 are all identical, then this shows a furthermore exponential relationship between the values in the leftmost column.
Relationship between n and a
The formula for calculating side a from the number of the term of the sequence (n) was worked out in the following way:
The terms of a that had been previously collected were written down and the relationships between them were present in their most basic form at L1 as +2.
a L1
3
>2
5
>2
7
>2
9
>2
1
>2
3
From previous work, it is known that if the lowest level is L1, then the value here is the first part of the formula. Because the extracted value (2) is from L1, it shows that the relationship between the terms of a is linear. In maths, linearity is displayed as n1 or just n for short.
tn=2n
However, this does not yet work and requires further refinement. For example, the first term is not:
tn=2n
tn=2x1
t1=2 it is t1=3
The difference between 2 and 3 is 1, so the formula may be:
t1=2n+1
t1=2x1+1
t1=3 THE FORMULA IS CORRECT FOR THE FIRST TERM
This can be checked by implementing it in the next term of the sequence.
t2=2n+1 t3=2n+1 t4=2n+1
t2=2x2+1 t3=2x3+1 t4=2x4+1
t2=5 t3=7 t4=9
Note that in the above, the number which replaces n has been written in bold.
The formula has now been proven and it works for all of the terms of the sequence.
When writing formulae, they are conventionally factorised. This is the process of finding the highest common factor of all of the values in a formula to write them in a more compact form. However, the highest common factor in this formula is 1, and so it does not have to be factorised because it would then not be in its simplest form. Just to show the expanded formula:
tn=1(2n+1)
This is more complex than if not factorised. Therefore, the formula of the relationship between n and a for odd triples is:
tn=2n+1
Relationship between n and b
The formula for calculating side b from the number of the term of the sequence (n) was worked out in the following way:
The terms of b that had been previously collected were written down and the relationships between them were present in their most basic form at L2 as +4.
b L1 L2
4
>8
2 >4
>12
24 >4
>16
40 >4
>20
60 >4
>24
84 >4
>28
12 >4
>32
44
From previous work, it is known that if the lowest level is L2, you halve the number on this level (4/2=2) and this forms the first part of the formula. Because the extracted value (4) is from L2, it shows that the relationship between the terms of b is exponential. In maths, exponents are displayed as indices (n2).
It can be noted that the L1 (level 1) relationship between the terms was steadily increasing in value (by 4 each time), and L2 said this. This suggests that the first component of the formula will be quadratic, because the L1 relationship is exponential.
tn=2n2
However, this does not yet work and requires further refinement. For example, the first term is not:
t1=2n2
t1=2(1x1)
t1=2 it is t1=4
The difference between 2 and 4 is 2, so the formula may be:
t1=2n2+2
t1=2x12+2
t1=4 THIS IS CORRECT FOR THE FIRST TERM
This can be checked by implementing it in the next terms of the sequence:
t2=2n2+2 t3=2n2+2 t4=2n2+2
t2=2x22+2 t3=2x32+2 t4=2x42+2
t2=10 t3=20 t4=34
Note that in the above formula, the number which replaces n has been written in bold.
The formula does not work, as the terms except for the first are not true using this formula. This problem can be solved by comparing the above values for t2, t3 and t4 with the true values. Because this formula does not work, the +2 at the end of tn=2n2+2 can be disregarded.
b difference current formula
t1=4 0 t1=4
t2=12 2 t2=10
t3=24 4 t3=20
t4=40 6 t4=34
It can be seen that as each term passes, the difference increases by two. As the comparison is between n and b, n must be used to display variables, because it is the only value that is already known. The above differences can be compared to n so that they can be expressed in the final formula.
n relationship difference
+2n 2
2 +2n 4
3 +2n 6
4 +2n 8
This table can be summarised in that as the difference increases in each term, b stays in linear relationship to n, and therefore the last component of the formula is +2n. So the formula is:
tn=2n2+2n
This can be checked by implementing it in the next terms of the sequence:
t2=2n2+2n t3=2n2+2n t4=2n2+2n
t2=(2x22)+(2x2) t3=(2x32)+(2x3) t4=(2x42)+(2x4)
t2=8+4 t3=18+6 t4=32+8
t2=12 t3=24 t4=40
The formula has now been proven and it works for all of the terms of the sequence.
When writing formulae, they are conventionally factorised. This is the process of finding the highest common factor of all of the values in a formula to write them in a more compact form. For this formula, the highest common factor is 2n (2n is common to both "2n2+2n"). This means that the formula can be compacted, using brackets, like this:
tn=2n2+2n
tn=2(n2+n)
tn=2n(n+1)
To check that the formula has been factorised properly, the reverse, expanding the formula should be done. To do this, the value outside the bracket (2n) is multiplied by everything inside the bracket in turn.
tn=2n(n+1)
tn=2(n2+n)
tn=2n2+2n
The formula has been proved to work and factorised. The formula of the relationship between n and b for odd triples is:
tn=2n(n+1)
Relationship between n and c
To calculate the formula of the relationship between n and c, the above method was not required. Looking at the relationship between b and c in odd triples, it is true to say that b+1=c in all cases of each term. Therefore, if the formula of the relationship between n and b is tn=2n(n+1), this is expanded to tn=2n2+2n, then the formula for c is tn=2n2+2n+1. This can be checked by implementing it in the terms of the sequence:
t1=2n2+2n+1 t2=2n2+2n+1 t3=2n2+2n+1 t4=2n2+2n+1
t1=(2x12)+(2x1)+1 t2=(2x22)+(2x2)+1 t3=(2x32)+(2x3)+1 t4=(2x42)+(2x4)+1
t1=2+2+1 t2=8+4+1 t3=18+6+1 t4=32+8+1
t1=5 t2=13 t3=25 t4=41
Note that in the above, the number which replaces n has been written in bold.
The formula has now been proven and it works for all of the terms of the sequence.
When writing formulae, they are conventionally factorised. This is the process of finding the highest common factor that satisfies all of the values in a formula to write them in a more compact form. In this formula, the highest common factor is 2n (2n is common to both "2n2+2n+1"). This means that the formula can be compacted, using brackets, like this:
tn=2n2+2n+1
tn=2(n2+n)+1
tn=2n(n+1)+1
To check that the formula has been factorised properly, the reverse, expanding the formula should be done. To do this, the value outside the bracket (2n) is multiplied by everything inside the bracket in turn.
tn=2n(n+1)+1
tn=2(n2+n)+1
tn=2n2+2n+1
The formula has been proved to work and factorised. The formula of the relationship between n and c for odd triples is:
tn=2n(n+1)+1
Relationship between n and P
To calculate the relationship between n and P, the above method for calculating the relationship between n and a and n and b was not required. Instead, the relationship between n and P could be calculated algebraically. Because P=a+b+c, a, b and c can be replaced with their respective formulae and they can then be added together to give the relationship between n and P.
The formula was written down and a, b and c were substituted with their respective formulae:
P=a+b+c
P=a+b+c=(2n+1)+(2n(n+1))+(2n(n+1)+1)
The individual formulae were expanded:
P=(2n+1)+(2n(n+1))+(2n(n+1)+1)=(2n+1)+(2n2+2n)+(2n2+2n+1)
The formulae were added together:
P=(2n+1)+(2n2+2n)+(2n2+2n+1)=4n2+6n+2
tn=4n2+6n+2
The above is the formula of the relationship between n and P. To prove that this is true, the formula can be implemented on the terms of the sequence:
t1=4n2+6n+2 t2=4n2+6n+2 t3=4n2+6n+2 t4=4n2+6n+2
t1=(4x12)+(6x1)+2 t2=(4x22)+(6x2)+2 t3=(4x32)+(6x3)+2 t4=(4x42)+(6x4)+2
t1=4+6+2 t2=16+12+2 t3=36+18+2 t4=64+24+2
t1=12 t2=30 t3=56 t4=90
Note that in the above, the number which replaces n has been written in bold.
The formula has now been proven and it works for all of the terms of the sequence.
When writing formulae, they are conventionally factorised. This is the process of finding the highest common factor of all of the values in a formula to write them in a more compact form. For this formula, the highest common factor is 2 (2 is common to "4n2+6n+2"). This means that the formula can be compacted, using brackets, like this:
tn=4n2+6n+2
tn=4n2+6n+2(1)
tn=4n2+2(3n+1)
tn=2(2n2+3n+1)
To check that the formula has been factorised properly, the reverse, expanding the formula should be done. To do this, the value outside the bracket (2) is multiplied by everything inside the bracket in turn:
tn=2(2n2+3n+1)
tn=4n2+2(3n+1)
tn=4n2+6n+2(1)
tn=4n2+6n+2
The formula has been proven to work and has been factorised. The formula of the relationship between n and P for odd triples is:
tn=2(2n2+3n+1)
Relationship between n and A
To calculate the formula of the relationship between n and A, the substitution method used above can be used. Because the formula of the area of a triangle is A=ab/2, the letters here can be substituted for the formulae to which they correspond.
The letters were substituted with their respective formulae:
A=ab/2=(2n+1)(2n(n+1))/2
The formulae were expanded:
A=(2n+1)(2n(n+1))/2=(2n+1)(2n2+2n)/2
The formulae were multiplied together:
A=(2n+1)(2n2+2n)/2=4n3+4n2+2n2+2n/2
The answer was simplified:
A=4n3+4n2+2n2+2n/2=4n3+6n2+2n/2
To simplify it further, it was divided by two:
A=4n3+6n2+2n/2=2n3+3n2+n
A=2n3+3n2+n
tn=2n3+3n2+n
The above is the formula of the relationship between n and A. To prove that this is true, the formula can be implemented in the terms of the sequence:
t1=2n3+3n2+n t2=2n3+3n2+n t3=2n3+3n2+n t4=2n3+3n2+n
t1=(2x13)+(3x12)+1 t2=(2x23)+(3x22)+2 t3=(2x33)+(3x32)+3 t4=(2x43)+(3x42)+4
t1=2+3+1 t2=16+12+2 t3=54+27+3 t4=128+48+4
t1=6 t2=30 t3=84 t4=180
Note that in the above, the number which replaces n has been written in bold.
The formula has now been proven and it works for all of the terms of the sequence.
When writing formulae, they are conventionally factorised. This is the process of finding the highest common factor of all of the values in a formula to write them in a more compact form. For this formula, the highest common factor is n (n is common to "2n3+3n2+n"). This means that the formula can be compacted, using brackets, like this:
tn=2n3+3n2+n
tn=2n3+3n2+n(1)
tn=2n3+n(3n+1)
tn=n(2n2+3n+1)
To check that the formula has been factorised properly, the reverse, expanding the formula, should be done. To do this, the value outside the bracket (n) is multiplied by everything inside the bracket in turn:
tn=n(2n2+3n+1)
tn=2n3+n(3n+1)
tn=2n3+3n2+n(1)
tn=2n3+3n2+n
The formula has been proven to work and has been factorised. The formula of the relationship between n and A for odd triples is:
tn=n(2n2+3n+1)
Summary
All of the formulae of the relationships between n and a, b, c, P and A have been calculated and are displayed in the following table:
n
a
b
c
P
A
tn=n
tn=2n+1
tn=2n(n+1)
tn=2n(n+1)+1
tn=2(2n2+3n+1)
tn=n(2n2+3n+1)
x
x
x
x
x
3
4
5
2
6
2
5
2
3
30
30
3
7
24
25
56
84
4
9
40
41
90
80
5
1
60
61
32
330
6
3
84
85
82
546
7
5
12
13
240
840
8
7
44
45
306
224
To prove that the formulae are correct, algebra can be used. For example, Pythagoras' theorem states that a2+b2=c2, and so by substituting a, b and c for their respective formulae the same should be true; that by adding the square of the formula of the relationship between n and a to the square of the formula of the relationship between n and b, this will equate the square of the formula of the relationship between n and c.
a2+b2=c2=(2n+1)2+(2n(n+1))2=(2n(n+1)+1)2
a2 was calculated:
a2=(2n+1)2
a2=(2n+1)(2n+1)
a2=4n2+2n+2n+1
a2=4n2+4n+1
b2 was calculated:
b2=(2n(n+1))2
b2=(2n2+2n)2
b2=(2n2+2n)(2n2+2n)
b2=4n4+4n3+4n3+4n2
b2=4n4+8n3+4n2
c2 was calculated:
c2=(2n(n+1)+1)2
c2=(2n2+2n+1)2
c2=(2n2+2n+1)(2n2+2n+1)
c2=4n4+4n3+2n2+4n3+4n2+2n+2n2+2n+1
c2=4n4+8n3+8n2+4n+1
Using the above, it is now possible to see if the formulae are true:
a2+b2=c2
(4n2+4n+1)+(4n4+8n3+4n2)=4n4+8n3+8n2+4n+1
4n2+4n+1+4n4+8n3+4n2=4n4+8n3+8n2+4n+1
4n4+8n3+8n2+4n+1=4n4+8n3+8n2+4n+1
The above calculation proves that the formulae are true because they obey Pythagoras' theorem.
The whole process was repeated for even numbers.
When all of the formulas had been calculated, a spreadsheet was built to work out the terms of the sequence after 8 which utilised the discovered formulae.
The values of b and c in the even triples can be discussed. The "true" triples are the ones that satisfy the statement n-1<b<n+1: the value of b is greater than the previous term and less than the next term and the same is true for c. Also to be a "true" triple, the difference between b and c must be 2 in any one term. These two criteria directly affect each other.
PROVE THE FORMULAE.
EVEN
The first aim of the investigation is to collect data so that it can be analysed. I wanted to calculate the first 8 Pythagorean triples using trial and error at first, but realised that using this time to devise a method to find these numbers would be more effective. After much trial and error, I came to that if you square a and divide it by 2, you end up with the base number for the next two sides, b and c. By subtracting and adding 0·5 to the answer, you can find b and c respectively. For example, in the first Pythagorean triple:
1
