Beyond Pythagoras - Pythagorean Triples
Beyond Pythagoras
Pythagorean Triples:
Three integers a, b, and c that satisfy a2 + b2 = c2 are called Pythagorean Triples.
a2 + b2 = c2
The numbers 3, 4 and 5 satisfy the condition because:
32 + 42 = 52
32 = 3 x 3 = 9
42 = 4 x 4 =16
52 = 5 x 5 = 25
32 + 42 = 9 +16 = 25 = 52
. Each of the following sets of numbers satisfy a similar condition of
(smallest number)2 + (middle number)2 = (largest number)2
a) 5, 12, 13.
52 + 122 = 132
52 = 5 x 5 = 25
122 = 12 x 12 = 144
132 = 13 x 13 = 169
52 + 122 = 25 +144 = 169 = 132
b) 7, 24, 25
72 + 242 = 252
72 = 7 x 7 = 49
242 = 24 x 24 = 576
252 = 25 x 25 = 625
72 + 242 = 49 +576 = 625 = 252
The numbers 3, 4 and 5 can be the lengths - in appropriate units - of the sides of a right-angled triangle.
3 5
4
The perimeter and area of this triangle are :
Perimeter = 3 + 4 + 5 = 12 units
Area = 1/2 x 3 x 4 = 6 square units
The numbers 5, 12, and 13 can also be the lengths - in appropriate units - of a right-angled triangle :
5 13
12
The perimeter and area of this triangle are :
Perimeter = 5 + 12 + 13 = 30
Area = 1/2 x 5 x 12 = 30
(c) This is also true for the numbers 7, 24 and 25:
7 25
24
The perimeter and area for this triangle are :
Perimeter = 7 + 24 + 25 = 56
Area = 1/2 x 7 x 24 = 84
Below is a table showing many Pythagorean triples:
Length of shortest side
(a)
Length of middle side
(b)
Length of longest side
(c)
Perimeter
Area
3
4
5
2
6
5
2
3
30
30
7
24
25
56
84
9
40
41
90
80
1
60
61
32
330
3
84
85
82
546
5
12
13
240
840
7
44
45
306
224
9
80
81
380
710
21
220
221
462
2310
23
264
265
552
3036
I have noticed that the first column goes up by two every time, there is a constant difference of two. I have also noticed that the second column is a quadratic sequence, with it going up in fours and then a difference of four at the end. The third column is simply the number in the second column plus one.
Generalisation:
(a)=
Term no.
2
3
4
5
Sequence
3
5
7
9
1
Difference
2
2
2
2
I have found the nth term formula to be 2n + 1
(b)=
Term no.
2
3
4
5
Sequence
4
2
24
40
60
2n2
2
8
8
32
50
R2 - R3
2
4
6
8
0
Difference
2
2
2
2
I have found the nth term formula to be 2n2 + 2n
This can also be expressed in the form of triangle numbers because:
(b)
Multiples
4
4 x 1
2
4 x 3
24
4 x 6
40
4 x 10
The triangle formula will be simply be 4t
(c)=
To find the nth term formula for this column is very simple. You take the formula for column (b) and just add one at the end because that is always the difference between column (b) and (c).
The formula will be 2n2 + 2n +1.
This can also be expressed in the form of triangle numbers.
The triangle formula for column (c) will be 4t + 1
Proof:
a2 + b2 = c2
(2n +1) 2 + (2n2 + 2n) 2 = (2n2 + 2n +1) 2
4n +2 + 4n4 + 4n = 4n4 + 4n2 +2
4n4 + 4n2 + 2 = 4n4 + 4n2 + 2
Left hand side = Right hand side
Perimeter =
To find the perimeter, you have to (a)+ (b) + (c) = perimeter
...
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This can also be expressed in the form of triangle numbers.
The triangle formula for column (c) will be 4t + 1
Proof:
a2 + b2 = c2
(2n +1) 2 + (2n2 + 2n) 2 = (2n2 + 2n +1) 2
4n +2 + 4n4 + 4n = 4n4 + 4n2 +2
4n4 + 4n2 + 2 = 4n4 + 4n2 + 2
Left hand side = Right hand side
Perimeter =
To find the perimeter, you have to (a)+ (b) + (c) = perimeter
2n2 + 2n + 1
2n2 + 2n +
2n + 1
4n2 + 6n + 2
Area =
To find the area, you have to
= (a) x (b)
2
= (2n2 + 2n) (2n + 1)
2
= 4n3 + 2n2 + 4n2 + 2n
2
= 4n3 + 6n2 + 2n
2
=2n3 +3n2 + n
Here is a family of triples, this time with even units.
Length of shortest side
(a)
Length of middle side
(b)
Length of longest side
(c)
Perimeter
Area
6
8
0
24
40
0
24
26
60
300
4
48
50
12
200
8
80
82
80
3280
22
20
22
264
7320
From looking at the table, I have noticed that the first column has a constant difference of 4. I have also noticed that column (b) consists of the triangle numbers like column (b) did for the odd family of triples.
(b)
Multiples
8
8 x 1
24
8 x 3
48
8 x 6
80
8 x 10
This time in column (c), the number is obtained by adding 2 to the number in column (b). Something else that is evident from looking at the table, is that the numbers from the odd family of triples are enlarged by a scale factor of 2 to receive the set of triples that are present in the even set of triples.
Generalisation:
(a)=
Term no.
2
3
4
5
Sequence
6
0
4
8
22
Difference
4
4
4
4
The nth term is 4n + 2
This is obvious as the family or triples are simply double the first one, which in turn doubles all generalisations.
(b)=
Term no
2
3
4
5
Sequence
8
24
48
80
20
4n2
4
6
36
64
00
R2-R3
4
8
2
6
20
Difference
4
4
4
4
The nth term is 4n2 + 4n
This is obvious as the family or triples are simply double the first one, which in turn doubles all generalisations.
(c)=
To find the formula for this column is very simple. You take the formula for column (b) and just add two at the end because that is always the difference between column (b) and (c). The formula will look like this:
4n2 + 4n + 2
This is obvious as the family or triples are simply double the first one, which in turn doubles all generalisations.
Proof:
a2 + b2 = c2
(4n +2) 2 + (4n2 + 4n) 2 = (4n2 + 4n +2) 2
6n + 4 + 16n4 + 16n = 16n4 + 16n2 + 4
6n4 + 16n2 + 4 = 16n4 + 16n2 + 4
LHS = RHS
This is obvious as the family or triples are simply double the first one, which in turn doubles all generalisations and proofs.
Perimeter =
To find the perimeter, you have to:
(a) + (b) + (c) = perimeter
4n2 + 4n + 2
4n2 + 4n +
4n + 2
8n2 + 12n + 4
This is obvious as the family or triples are simply double the first one, which in turn doubles all generalisations.
Area =
To find the area, you have to:
1/2 x (a) x (b)
= (4n2 + 4n) (4n + 2)
2
= 16n3 + 16n2 + 8n2 + 8n
2
= 16n + 24n2 + 8n
2
=8n3 + 12n2 + 4n
This is obvious as the family or triples are simply double the first one, which in turn all generalisations.
Here is another family of triples. It is the odd family of triples enlarged by a scale factor of 3.
Length of shortest side
(a)
Length of middle side
(b)
Length of longest side
(c)
Perimeter
Area
9
2
5
36
54
5
36
39
90
270
21
72
75
68
756
27
20
23
270
620
33
80
83
396
2970
39
252
255
546
4914
45
336
339
720
7560
51
432
435
918
1016
From the above table I have noticed that there is a constant difference of 6 in the first column. I have also noticed that the third column is now obtained by adding 3 to column (b) this time, not 2. Column (b) can be again made using triangle numbers.
Generalisation:
(a) =
Term no.
2
3
4
5
Sequence
9
5
21
27
33
Difference
6
6
6
6
The nth term for this is 6n + 3
(b)=
Term no
2
3
4
5
Sequence
2
36
72
20
80
6n2
6
24
54
96
50
R2-R3
6
2
8
24
30
Difference
6
6
6
6
The nth term for this is 6n2 + 6n
(c)=
The formula for this term can be found using the nth term formula for column (b).
It will simply be 6n2 + 6n + 3
Proof
a2 + b2 = c2 ?
(6n + 3) 2 + (6n2 + 6n) 2 = (6n2 + 6n + 3) 2
36n + 9 + 36n4 + 36n = 36n4 + 36n2 + 9
36n4 + 36n2 + 9 = 36n4 + 36n2 + 9
LHS = RHS
Perimeter =
To find the perimeter, you have to (a) + (b) + (c) = perimeter
6n2 + 6n + 3
6n2 + 6n +
6n + 3
2n2 + 18n + 6
Area =
To find the area, you have to:
1/2 x (a) x (b)
= (6n2 + 6n) (6n + 3)
2
= 36n3 + 36n2 + 18n2 + 18n
2
= 16n3 + 54n2 + 18n
2
Here is a table of a family of triples, this time with the even family of triples enlarged by a scale factor of 3.
Length of shortest side
(a)
Length of middle side
(b)
Length of longest side
(c)
Perimeter
Area
8
24
30
72
216
30
72
78
80
080
42
44
50
336
3024
54
240
246
540
6480
66
360
366
792
1880
In this family of triples, I have noticed that the first column has a constant difference of 12, twice the constant difference of the odd family of triples that were enlarged by a scale factor of 3. To acquire the number in column (c), you now have to add 6 to the number in column (b), which is also twice the number of that in the odd family of triples that were also enlarged by a scale factor 3.
Generalisation:
(a)=
Term no.
2
3
4
5
Sequence
8
30
42
54
66
Difference
2
2
2
2
The nth term for this is 12n + 6
(b)=
Term no
2
3
4
5
Sequence
24
72
44
240
360
2n2
2
48
08
92
300
R2-R3
2
24
36
48
60
Difference
2
2
2
2
The nth term for this is 12n2 + 12n
(c)=
The formula for this term can be found using the nth term formula for column (b).
It will simply be 12n2 + 12n + 6
Proof
a2 + b2 = c2 ?
(12n + 6) 2 + (12n2 + 12n) 2 = (12n2 + 12n + 6) 2
44n + 36 + 144n4 + 144n = 144n4 + 144n2 + 36
44n4 + 144n2 + 36 = 144n4 + 144n2 + 36
LHS = RHS
Perimeter =
To find the perimeter, you have to:
(a) + (b) + (c) = perimeter
so...
2n2 + 12n + 6
2n2 + 12n +
12n + 6
24n2 + 36n + 12
Area =
To find the area, you have to:
1/2 x (a) x (b)
(12n2 + 12n) (12n + 6)
2
= 144n3 + 72n2 + 144n2 + 72n
2
= 144n3 + 216n2 + 72n
2
Here is a table of a family of triples, this time with the odd family of triples enlarged by a scale factor of 4.
Length of shortest side
(a)
Length of middle side
(b)
Length of longest side
(c)
Perimeter
Area
2
6
20
48
96
20
48
52
20
480
28
96
00
224
344
36
60
64
360
2880
44
240
244
528
5280
In this family of triples, I have noticed that the first column has a constant difference of 8. To acquire the number in column (c), you have to add 4 to the number in column (b), which is also half the number of the constant difference. The numbers in column (b) can also be generated using triangle numbers.
Generalisation:
(a)=
Term no.
2
3
4
5
Sequence
2
20
28
36
44
Difference
8
8
8
8
The nth term for this is 8n + 4
(b)=
Term no
2
3
4
5
Sequence
6
48
96
60
240
8n2
8
32
72
28
200
R2-R3
8
6
24
32
40
Difference
8
8
8
8
The nth term for this is 8n2 + 8n
(c)=
The formula for this term can be found using the nth term formula for column (b).
It will simply be 8n2 + 8n + 4, because 4 is the difference between column (b) numbers and column (c) numbers.
Proof
a2 + b2 = c2
(8n + 4) 2 + (8n2 + 8n) 2 = (8n2 + 8n + 4) 2
64n + 16 + 64n4 + 64n = 64n4 + 64n2 + 16
64n4 + 64n2 + 16 = 64n4 + 64n2 + 16
LHS = RHS
Perimeter =
To find the perimeter, you have to:
(a) + (b) + (c) = perimeter
so...
8n2 + 8n + 4
8n2 + 8n +
8n + 4
6n2 + 24n + 8
Area =
To find the area, you have to:
1/2 x (a) x (b)
(8n2 + 8n) (8n + 4)
2
= 64n3 + 32n2 + 64n2 + 32n
2
= 64n3 + 96n2 + 32n
2
Here is a table of a family of triples, this time with the even family of triples enlarged by a scale factor of 4.
Length of shortest side
(a)
Length of middle side
(b)
Length of longest side
(c)
Perimeter
Area
24
32
40
96
384
40
96
04
240
920
56
92
200
448
5376
72
320
328
720
1520
88
480
488
056
21120
In this family of triples, I have noticed that the first column has a constant difference of 16, which is double the constant difference in the odd family of triples when it was enlarged by a scale factor of 4. To acquire the number in column (c), you have to add 8 to the number in column (b), which is double the difference of the odd family of triples. Again, the numbers in column (b) can also be generated using triangle numbers.
Generalisation:
(a)=
Term no.
2
3
4
5
Sequence
24
40
56
72
88
Difference
6
6
6
6
The nth term for this is 16n + 8
(b)=
Term no
2
3
4
5
Sequence
32
96
92
320
480
6n2
6
64
44
256
400
R2-R3
6
32
48
64
80
Difference
6
6
6
6
The nth term for this is 16n2 + 16n
(c)=
The formula for this term can be found using the nth term formula for column (b).
It will simply be 16n2 + 16n + 8, because 8 is the difference between column (b) numbers and column (c) numbers.
Proof
a2 + b2 = c2
(16n + 8) 2 + (16n2 + 16n) 2 = (16n2 + 16n + 8) 2
256n + 64 + 256n4 + 256n = 256n4 + 256n2 + 64
256n4 + 256n2 + 64 = 256n4 + 256n2 + 64
LHS = RHS
Perimeter =
To find the perimeter, you have to:
(a) + (b) + (c) = perimeter
so...
6n2 + 16n + 8
6n2 + 16n +
16n + 8
32n2 + 48n + 16
Area =
To find the area, you have to:
1/2 x (a) x (b)
so...
(16n2 + 16n) (16n + 8)
2
= 256n3 + 256n2 + 128n2 + 128n
2
= 256n3 + 384n2 + 128n
2
After investigating enlarging family of triples by a certain scale factor I can predict that the triples obtained by, for example, enlarging the odd family of triples by 7, will give the even family of triples enlarged by 7, when doubled. This is the same for all scale factors. I have also found that in the original odd family of triples, adding the middle and longest side and the square rooting the answer will give the shortest side. For example:
3,4,5 5,12,13
3 = (4 + 5) 5 = (12 + 13)
A formula for generating all Pythagorean Triples is:
a = t2 - s2 , b = 2st , c = t2 + s2
in which s and t are integer positive numbers, neither both even or both odd and in which s < t.
by choosing values of s and t properly, Pythagorean Triples like these fore example are formed :
s
t
(a)
(b)
(c)
2
3
4
5
2
3
5
2
3
3
8
6
0
Proving the formula
The only triple that consists of consecutive integers is 3,4,5.
Here are some properties that I found apply to all Pythagorean Triples :
) Either (a) or (b) is divisible by 3.
2) Either (a) or (b) is divisible by 4.
3) Either (a), (b) or (c) is divisible by 5.
4) The product of (a), (b) and (c) is divisible by 60.
5)
6)
7)
8)
