Download A Derivation of Formulas Used to Generate Pythagorean Triples

PROBLEM STATEMENT
In a triangle with a right angle, there are 2 legs and the hypotenuse of a triangle.
The hypotenuse of a triangle is the side of a right triangle that is opposite the 90° angle.
The legs of a triangle are the other 2 shorter, perpendicular sides. The Pythagorean
Theorem can be used to figure out the legs and the hypotenuse of a right triangle. The
Pythagorean Theorem states that the square of the hypotenuse of a right triangle is equal
to the sum of the squares of the other two sides.
a2 + b2 = c2
hypotenuse
(c)
leg
(b)
leg (a)
Figure 1. A right triangle with the legs and hypotenuse labeled.
A triple is a set of 3 numbers. The average student knows usually 3 of these
Pythagorean triples. They usually know the 3,4,5 triple, the 5,12,13 triple, and the
8,15,17 triple. This research deals with the exploration of Pythagorean triples.
How difficult it is to get these Pythagorean Triples? In Table 1, there is a list of
numbers that the calculator chose randomly. In Table 2, there is a list of selected. The
numbers listed under a, are the lengths of the legs, the numbers listed under b are the
length of the legs, and the letter c represents the length of the hypotenuse.
Amanda Magli
Page 1
Table 1. Selected Numbers Used to Find Pythagorean Triples
a
b
c
4
3
2
√105
14
4
√21
8
6
2
51
65
17
√1092
44
33
√4185
12
33
28
4
14
2
√1835
51
5
12
√624
53
4
8
13
7
5
√12
8
15
√48
10
12
√45
3
52
79
√195
32
81
49
64
√2772
75
36
√8820
√60
33
24
70
80
41
1
√1160
6
80
15
√65
√35
4
13
√421
8
11
√108
9
√13
√5305
√10466
22
46
√8497
√3490
91
54
√6714
√2080
94
16
√1093
49
√7501
√6425
√1825
25
63
√52
√6464
√394
“Does It Form a
Triple”?
NO
NO
NO
NO
NO
NO
NO
NO
NO
NO
NO
NO
NO
NO
NO
NO
NO
NO
NO
NO
NO
NO
NO
NO
NO
NO
NO
NO
NO
NO
NO
NO
Amanda Magli
Page 2
Table 2. The Numbers the Calculator Randomly Picked to Find Pythagorean Triples
a
b
c
9
33
69
88
11
98
√1053
92
79
√648
62
12
√1332
81
√801
22
61
√3813
62
33
80
√4416
46
9
√2673
√4187
1
1
83
25
10
22
√3080
49
9
35
81
2
√767
18
√4059
√3952
72
13
√228
54
39
70
37
√1904
69
54
67
55
21
√3900
45
28
98
40
√3360
80
46
√4620
45
86
25
47
21
36
13
√3480
27
92
72
√476
√4845
59
√6248
40
64
72
38
23
√205
45
33
94
37
56
√2997
√2610
√9661
√9113
45
√14365
63
√12953
√9266
33
88
√2169
46
√16165
49
62
√10121
77
92
√3114
√13796
71
√4325
√522
63
66
59
√730
√15353
√5809
24
73
81
93
41
√5321
√11745
√1448
36
23
78
71
√14020
√1538
58
“Does This Form a
Ttriple”?
NO
NO
NO
NO
NO
NO
NO
NO
NO
NO
NO
NO
NO
NO
NO
NO
NO
NO
NO
NO
NO
NO
NO
NO
NO
NO
NO
NO
NO
NO
NO
NO
NO
NO
YES
NO
NO
NO
NO
NO
NO
NO
NO
NO
NO
Amanda Magli
Page 3
Table 1 shows us that we found no numbers that formed a Pythagorean Triple.
From Table 2, one set of numbers formed a Pythagorean triple. Based on this
information, we can conjecture that it is not easy to pick out Pythagorean triples
randomly or by guess and check. This paper explores Pythagorean triples and to derive a
formula to get triples more efficiently.
The questions that we will investigate during this report are:
•
Is there a formula to figure out different Pythagorean Triples?
•
How many Pythagorean triples are there?
These two questions will be answered throughout this paper.
RELATED AND ORIGINAL RESEARCH
When the Smallest Leg is Odd
The tables in this section show the steps of deriving a formula to generate
Pythagorean triples using several given triples. Each table shows us, step by step, the
process and hints of how to get a Pythagorean triple when the smallest leg is odd.
Amanda Magli
Page 4
Table. 3 The Perimeter of 10 Selected Right Triangles
Pythagorean Triple
Perimeter
3,4,5
3+4+5 =12
5,12,13
5+12+13 = 30
7,24,25
7+24+25 = 56
9,40,41
9+40+41 = 90
11,60,61
11+60+61 = 132
13,84,85
13+84+85 = 182
15,112,113
15+112+113 = 240
17, 144,145
17+144+145 = 306
19,180,181
19+180+181 = 380
21,220,221
21+220+221 = 462
…..
……….
n ?,?
On column 1 of Table 3, several Pythagorean triples are presented. Many patterns are
formed with these Pythagorean Triplets. In the second column, the sums of the elements
are shown. Each integer of the Pythagorean triple is added together to get a sum of the 3
numbers.
Table 4. Alternative Method for Finding the Perimeter of Right Triangles
Pythagorean Triple
Perimeter
3,4,5
5,12,13
7,24,25
9,40,41
11,60,61
13,84,85
15,112,113
17, 144,145
19,180,181
21,220, 221
…….
n ?, ?
3+4+5 =12
5+12+13 = 30
7+24+25 = 56
9+40+41 = 90
11+60+61 = 132
13+84+85 = 182
15+112+113 = 240
17+144+145 = 306
19 + 180+181 = 380
21 + 220 + 221 = 462
…….
Alternative Method for
Finding the Perimeter
3 ·4
5 ·6
7 ·8
9 ·10
11 ·12
13 ·14
15 ·16
17 ·18
19 · 20
21 ·22
………..
n(n+1)
Amanda Magli
Page 5
The third column of Table 4 shows an alternative method for finding the perimeter. The
smallest element is involved as a factor in this formula for the perimeter. For example,
the first Pythagorean triple is 3, 4, 5. We take the smallest integer (the 3) and multiply
that by 4. Then the next largest triple we take is the smallest integer in the triple and
multiply that by the next consecutive integer. So, for 5, 12, 13 we multiply 5 by 6 to get
the perimeter. Now, since we know this, we can substitute n. Since we know that to find
the sum of the elements we multiply the smallest integer by itself plus 1. We conjecture
that the formula for finding the alternative sum of the elements is:
n(n+1)
Now after looking at the alternative method for finding the perimeter we are going to
look at the method for finding the sum of the two larger elements.
Amanda Magli
Page 6
Table 5. The Method for Finding the Sum of the Two Larger Elements
Pythagorean
Perimeter
Alternative
The Smallest
Method for
Triple
Method for
Number
Finding the
Finding the
Squared
Sum of the
Perimeter
Two Larger
Elements
3,4,5
3+4+5 =12
3·4
9
12-3 = 9
5,12,13
5+12+13 = 30
5 ·6
25
30-5 = 25
7,24,25
7+24+25 = 56
7· 8
49
56-7 = 49
9,40,41
9+40+41 = 90
9 ·10
81
90-9 = 81
11,60,61
11+60+61 = 132 11 ·12
121
132-11 =
121
13,84,85
13+84+85 = 182 13 ·14
169
182-13 =
169
15,112,113
15+112+113 =
15 ·16
225
240-15 =
240
225
17, 144,145
17+144+145 =
17 ·18
289
306 – 17 =
306
289
19,180,181
19+180+181 =
19 ·20
361
380 – 19 =
380
361
21,220,221
21+220+221 =
21 ·22
441
462 - 21 =
462
441
……
……
……..
…..
……….
n ?,?
n(n+1)
n2
n(n+1) –n =
n2
The 4th column of Table 5 shows the smallest integer of each triple squared. As we can
see it relates to the 5th column which it demonstrates a method for finding the sum of the
two larger elements. In this case, let’s take the perimeter and subtract the smallest integer
in the triple to get the sum of the two larger integers. For example, in the 7,24,25 triple
we take 56
7 × 8 = 56
Then we subtract 7
56-7 = 49
Notice that the smallest element squared is equal to the sum of the two larger elements.
Therefore we get the equation.
Amanda Magli
Page 7
n(n+1)-n = n2
After finding the sum of the two larger elements we will now look at what the even
element will be.
Table 6. Finding the Even Element of the Pythagorean Triples
Pythagorean
Triple
Perimeter
Alternative
Method for
Finding the
Perimeter
The
Smallest
Number
Squared
3,4,5
3+4+5 =12
3 ·4
9
5,12,13
5+12+13 =
30
7+24+25 =
56
9+40+41 =
90
11+60+61 =
132
13+84+85 =
182
15+112+113
= 240
17+144+145
= 306
19+180+181
= 380
21+220+221
= 462
…….
5 ·6
25
7 ·8
49
9 ·10
81
11 ·12
121
13 ·14
225
15 ·16
289
17· 18
361
19 ·20
441
21 ·22
…..
……….
n(n+1)
….
n2
7,24,25
9,40,41
11,60,61
13,84,85
15,112,113
17, 144,145
19,180,181
21,220,221
……..
n ?,?
Method
Even
for
Element
Finding
the Sum
of the
Two
Larger
Elements
12-3 = 9 9 − 1
=4
2
30-5 =
25 − 1
= 12
25
2
56-7 =
49 − 2
= 24
49
2
90-9 =
81 − 1
= 40
81
2
132-11 = 121 − 1
= 60
121
2
182-13 = 169 − 1
= 84
169
2
240-15 = 225 − 1
= 112
225
2
306 – 17 289 − 1
= 144
= 289
2
380 – 19 361 − 1
= 180
= 361
2
462 _ 21 441 − 1
= 220
= 441
2
……
……
n(n+1) – n 2 − 1
n = n2
2
Amanda Magli
Page 8
The 6th column of Table 6 shows how to get the even element. We take the sum of the
two larger elements, subtract 1, and then divide by 2. We add or subtract in this case to
identify the remaining element. For example, we take 9, (the answer when we subtracted
the sum of the elements by the smallest integer) and we subtract 1. Let’s then divide it by
2 to get 4, the even element in this triple. Also by knowing this we can figure out a
formula. The n formula is:
n2 − 1
2
As you can see, this is how we got the even element. Now we are going to try to find the
greatest element (hypotenuse).
Amanda Magli
Page 9
Table 7. Finding the Greatest Element (Hypotenuse)
Pythagorean
Triple
Perimeter
Alternative
Method for
Finding the
Perimeter
The
Smallest
Number
Squared
3,4,5
3+4+5 =12
3 ·4
9
Method for
Finding the
Sum of the
Two Larger
Elements
12-3 = 9
5,12,13
5+12+13 =
30
7+24+25 =
56
9+40+41 =
90
11+60+61 =
132
13+84+85 =
182
15+112+113
= 240
17+144+145
= 306
19+180+181
= 380
5 ·6
25
30-5 = 25
7 ·8
49
56-7 = 49
9 ·10
81
90-9 = 81
11 ·12
121
13 ·14
289
15 ·16
361
17· 18
441
19 ·20
…..
21+220+221
= 462
……..
21 ·22
….
……..
n(n+1)
n2
7,24,25
9,40,41
11,60,61
13,84,85
15,112,113
17, 144,145
19,180,181
21,220,221
……
n ?,?
Even
Element
Greatest
Element
(Hypotenuse)
132-11 =
121
182-13 =
169
240-15 =
225
306 – 17 =
289
380 – 19 =
361
9 −1
=4
2
25 − 1
= 12
2
49 − 2
= 24
2
81 − 1
= 40
2
121 − 1
= 60
2
169 − 1
= 84
2
225 − 1
= 112
2
289 − 1
= 144
2
361 − 1
= 180
2
9 +1
=5
2
25 + 1
= 13
2
49 + 1
= 25
2
81 + 1
= 41
2
121 + 1
= 61
2
169 + 1
= 85
2
225 + 1
= 113
2
289 + 1
= 145
2
361 + 1
= 181
2
462 -21 =
441
………
n(n+1) –n =
n2
441 − 1
= 220
2
……
n2 − 1
2
441 + 1
= 221
2
………
n2 + 1
2
The last column in Table 7 is the column that shows the greatest element, which is the
hypotenuse. We take the sum of the two larger elements, add 1 and divide by 2 to get the
Amanda Magli
Page 10
greatest element. For example we take 25 which is the sum of the two larger elements of
the 5,12,13 triple.
Add 1 to 25:
25 + 1
Then divide that by 2
25 + 1
= 13
2
The formula that can be used to find the greatest integer in a Pythagorean triple is:
n2 + 1
2
Now we can find out the formula for the legs in the 1st column. If n is the smallest
integer then the even integer is
n2 − 1
2
The greatest integer (the hypotenuse) is
n2 + 1
2
We conjecture that these formulas generate a Pythagorean triple.
Now, we can look at the last line of the table. It shows how we got that n(n+1) – n = n2.
This was found by taking the smallest integer, n, and then multiplying that by (n +1).
This gave us the sum of the 3 integers in Pythagorean Triples. We then subtracted the
smallest integer from that product to get n2. This was the method for finding the sum of
Amanda Magli
Page 11
n2 − 1
is the
the two larger elements. By using this information we found out that
2
n2 + 1
is the formula for deriving the greatest
2
formula for deriving the even element and
element (the hypotenuse). After getting those two pieces of information, we can fill in
the last row of the first column. The smallest integer is represented by n, the even
element is represented by
n2 − 1
n2 + 1
, and the greatest element is represented by
.
2
2
We are going to check that these 3 formulas work before completing the final row. First
we are going to take the 3 formulas and put them into the Pythagorean Theorem.
x2 + y2 = z2
Let
x=n
Let
y=
n2 − 1
2
and let
n2 + 1
z=
2
Substitute:
2
 n2 − 1   n2 + 1 
(n ) + 
 =

 2   2 
2
2
 n2 − 1 
 n2 + 1 
We now are going to multiply 
 and 

 2 
 2 
2
?
2
Amanda Magli
Page 12
 n 2 − 1  n 2 − 1   n 2 + 1  n 2 + 1 
n2 + 

= 

?
 2  2   2  2 
This equals
 n 4 − 2n + 1  n 4 + 2n 2 + 1
n2 + 
=
4
4


?
Then make the expression on the left one fraction
4n 2 + n 4 − 2n 2 + 1 n 4 + 2n 2 + 1
=
4
4
?
Combine like terms:
n 4 + 2n 2 + 1 n 4 + 2n 2 + 1
=
4
4
The expressions n,
n2 − 1 n2 + 1
,
actually do generate triples. Our conjecture is true.
2
2
Now we can fill in the last row of the table.
Table 8. The Last Column of Each Table
n,
n2 − 1 n2 + 1
,
2
2
n2 − 1
n+
2
2
n +1
+
2
n(n+1)
n2
n(n+1) –n
= n2
n2 − 1
2
n2 + 1
2
If you look at this column, we might think to ourselves how do we know that the answer
we get won’t be a fraction. We know that the hypotenuse and the even element won’t be
a fraction because when we subtract or add 1 to an odd number, we will always get an
Amanda Magli
Page 13
even number as the result. Even numbers are always divisible by 2 so therefore we will
never get a fraction.
Here is an example using this formula: Let n = 15
The even element is
152 − 1
= 112
2
The hypotenuse is
152 + 1
= 113
2
We now know that a Pythagorean triple is 15, 112 , and 113.
Another example is if n = 23.
The even element is
232 − 1
= 264
2
The hypotenuse is
232 + 1
= 265
2
This Pythagorean triple is therefore, 23, 264 and 265.
Now since we have looked at the smallest element being odd, now we are going to look
at the smallest element being even.
Amanda Magli
Page 14
When the Smallest Element is Even
The next series of tables shows that if the smallest integer in a Pythagorean triple
is even than a Pythagorean triple can also be formed.
Table 9. Pythagorean Triples and the Associated Perimeter
n
3
4
5
6
7
8
9
…..
n, ?, ?
Associated Triple
6,8,10
8,15,17
10,24,26
12, 35,37
14,48,50
16,63,65
18,80,82
…….
2n,?,?
Perimeter
24
40
60
84
112
144
180
…….
In Table 8 there are the triples and the perimeters of the triples. We took the smallest
even numbered element of these sets of triples and designated it 2n. The n is derived by
taking the 2n and dividing it by 2. For example, in the 6, 8, 10 triple, 6 is the smallest
even numbered element. When n = 3, 6 is that smallest even numbered element of that
Pythagorean triple.
n=
2n
2
The perimeter was found by adding all of the numbers of each triple. An example is,
6 + 8 + 10 = 24
Notice that the first numbers of successive triples differ by 2. After deriving the
perimeter we will figure out a formula to form the perimeter in a different way.
Amanda Magli
Page 15
Table 10. An Alternate Formula for Finding the Perimeter of the Triangle
n
Associated Triple
Perimeter
3
4
5
6
7
8
9
……..
n
6,8,10
8,15,17
10,24,26
12, 35,37
14, 48,50
16, 63,65
18,80,82
…..
2n, ?, ?
24
40
60
84
112
144
180
…..
….
An Alternative
Way of Generating
the Perimeter
6·4
8·5
10 · 6
12 · 7
14 · 8
16 · 9
18 · 10
….
2n(n+1)
Table 9 shows the value of n, the associated triple, the perimeter of the triple, and an
alternate formula for the perimeter of the formula without adding up the 3 numbers
together. The formula that is used to express these values is
2n(n+1).
An example of this is the triple 6,8,10.
n=3
This is because the number 6 in this triple is equal to 2n. Therefore the formula to get the
perimeter of the triple is
(6)(3 + 1)
After finding an alternate formula for find the perimeter of a triangle, we are now going
to look at the difference between the perimeter and the smallest side.
Amanda Magli
Page 16
Table 1. Difference Between the Perimeter and the Smallest Side
n
2n2
Associated
Triple
Perimeter
3
18
6,8,10
4
32
5
Difference
Between
Perimeter
and Smallest
Side
24
An
Alternative
Way of
Generating
the
Perimeter
6·4
8,15,17
40
8·5
40 - 8 = 32
50
10,24,26
60
10 · 6
60 -10 = 50
6
72
12, 35,37
84
12 · 7
84 – 12 = 72
7
98
14, 48,50
112
14 · 8
8
128
16, 63,65
144
16 · 9
9
162
18,80,82
180
18 · 10
……..
…..
…..
…..
….
112 – 14 =
98
144 - 16 =
128
180 – 18 =
162
…
n
2n2
2n, ?, ?
….
2n(n+1)
2n2
24 - 6 = 18
In Table 11, the 6th column shows the smallest element subtracted from the perimeter of
each triangle.
Perimeter – Smallest element = Sum of 2 larger elements.
In the 6,8,10 triple we take the perimeter
6 + 8 + 10 = 24
We subtract the smallest element from it
24 -6 =18
We can conjecture that the sum of the 2 larger elements is 2n2. For example, in the
8,15,17 triple, it is 42 multiplied by 2 equals 32 which is the difference between the
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Page 17
perimeter and the smallest side. Now we are going to look at the formula for the middle
element.
Table 12. The Formula for Generating the Middle Element
n
2n2
Associated Perimeter
Triple
3
18
6,8,10
24
An
Alternative
Way of
Generating
the
Perimeter
6·4
4
32
8,15,17
40
8·5
5
50
10,24,26
60
10 · 6
6
72
12, 35,37
84
12 · 7
7
98
14, 48,50
112
14 · 8
8
128
16, 63,65
144
16 · 9
9
162
18,80,82
180
18 · 10
……..
…..
…..
…..
n
2n2
2n, ?, ?
….
Formula for
the Middle
Element
….
Difference
Between
Perimeter
and
Smallest
Side
24 - 6 =
18
40 - 8 =
32
60 -10 =
50
84 – 12 =
72
112 – 14 =
98
144 - 16 =
128
180 – 18 =
162
…
2n(n+1)
2n2
?
18 − 2
=8
2
32 − 2
= 15
2
50 − 2
= 24
2
72 − 2
= 35
2
98 − 2
= 48
2
128 − 2
= 63
2
162 − 2
= 80
2
……
In column 7 of table 12, we find the formula for the middle element. We take
2n 2 − 2
2
This is the same as saying
n2 - 1
For example, for the 8, 15, 17 triple we take
Amanda Magli
Page 18
32 − 2
= 15
2
Now we are going to show how to get the largest element.
Table 13. The Formula for Generating the Largest Element
n
2n2
3
18
Associated Perimeter An
Triple
Alternative
Way of
Generating
the
Perimeter
6,8,10
24
6·4
4
32
8,15,17
40
8·5
5
50
10,24,26
60
6
72
12, 35,37
7
98
8
Difference Formula for
Between
the Middle
Perimeter Element
and
Smallest
Side
24 - 6 =
18 − 2
=8
18
2
Formula for
the Largest
Element
40 - 8 =
32
32 − 2
= 15
2
32 + 2
= 17
2
10 · 6
60 -10 =
50
50 − 2
= 24
2
50 + 2
= 26
2
84
12 · 7
84 – 12 =
72
72 − 2
= 35
2
72 + 2
= 37
2
14, 48,50
112
14 · 8
112 – 14
= 98
98 − 2
= 48
2
98 + 2
= 50
2
128
16, 63,65
144
16 · 9
144 - 16 =
128
128 − 2
= 63
2
128 + 2
= 65
2
9
162
18,80,82
180
18 · 10
180 – 18
= 162
162 − 2
= 80
2
162 + 2
= 82
2
……..
…..
…..
…..
….
…
……
……
n
2n2
2n, ?, ?
….
2n(n+1)
2n2
n2 - 1
n2 + 1
18 + 2
= 10
2
Amanda Magli
Page 19
In column 8 of Table 13, we find the formula for the largest element. We take the
difference between the perimeter and the smallest side and add 2. We then divide two to
get the largest element. For example, in the 10, 24, 26 triple we take
50 + 2
= 26
2
We know that
2n 2 + 2
= n2 + 1
2
We conjecture that n2 + 1 is the formula for the largest element.
Now we are going to check to see if these 3 formulas work by substituting them into the
Pythagorean Theorem.
x2 + y2 = z2
Let
x = 2n
Let
y = n2 – 1
Let
z = n2 + 1
Substitute:
(2n)2 + (n2 – 1)2 = (n2 + 1)2 ?
We are now going to multiply (2n)2, (n2 – 1)2 and (n2 + 1)2
(2n)(2n) + (n2 – 1)(n2 – 1) = (n2 + 1)(n2 + 1) ?
This equals
Amanda Magli
Page 20
4n2 + n4 – n2 – n2 + 1 = n4 + n2 + n2 + 1 ?
Combine like terms
n4 + 2n2 + 1 = n4 + 2n2 + 1
Our conjecture is true. These 3 expressions do generate Pythagorean triples.
The Euclidean Method
We now are going to find a third formula for finding several Pythagorean triples. We
first are going to set up a column for the sum of the elements like we have done in the
previous tables.
Table 14. Sum of Elements of Selected Triples
Triple
4, 3, 5
8, 15, 17
12, 5, 13
24, 7, 25
40, 9, 41
60, 11, 61
72, 65, 97
120, 119, 169
240, 161, 289
……..
?, ?, ?
Sum of Elements
12
40
30
56
90
132
234
408
690
….
In Table 14 columns 1 and 2, we have a list of triples in which the even element is first.
In column 2 there is the sum of the elements. For example, in the 8, 15,17 triple we add
each number together to get the sum of the elements.
Now we are going to look at the sum of the odd elements.
Amanda Magli
Page 21
Table 15. Sum of the Odd Elements
Triple
Sum of Elements
Sum of Odd
Elements
4, 3, 5
8, 15, 17
12, 5, 13
24, 7, 25
40, 9, 41
60, 11, 61
72, 65, 97
120, 119, 169
240, 161, 289
……..
? ?, ?
12
40
30
56
90
132
234
408
690
….
8
32
18
32
50
72
162
288
450
………
Sum of Odd
Elements Expressed
as Twice a Perfect
Square
2· 4
2 · 16
2· 9
2 · 16
2 · 25
2 · 36
2 · 81
2 · 144
2 · 225
……
In Table 15, column 3 we have the sum of the odd elements. The first (even) element is
subtracted from the sum of the triple to generate the sum of the odd elements. For
example in the 4,3,5 triple we add the two odd elements, 3 and 5, to get the sum of the
odd elements. In this case the sum of the odd integers is 8. We can also relate column 3
with column 4. Notice that the sum of the odd integers is twice a perfect square. For
example in the 40, 9, 41 triple the sum of the odd elements is 50. If you divide that by 2,
the answer you will get will be a perfect square, in this case 25.
Now we are going to look at the difference between the two odd elements.
Amanda Magli
Page 22
Table 16. Difference Between the Two Odd Elements
Triple
Sum of
Elements
Sum of Odd
Elements
4, 3, 5
8, 15, 17
12, 5, 13
24, 7, 25
40, 9, 41
60, 11, 61
72, 65, 97
120, 119, 169
240, 161, 289
……..
? ?, ?
12
40
30
56
90
132
234
408
690
….
8
32
18
32
50
72
162
288
450
………
Sum of Odd
Elements
Expressed as
Twice a Perfect
Square
2· 4
2 · 16
2· 9
2 · 16
2 · 25
2 · 36
2 · 81
2 · 144
2 · 225
……
Difference
Between the
Two Odd
Elements
2
2
8
18
32
50
32
50
128
….
In Table 16 column 5 we have the difference between odd elements. We took the odd
elements from each triple and subtracted them. For example, in the 12, 5, 13 triple, we
took
13 -5 = 8
Column 4 also relates to column 5. Twice a perfect square comes up with a number in
the difference between the two odd elements column. For example, in the 24, 7, 25
triple, the difference between the two odd elements is 18. If you divide 18 by 2 you get 9
which is a perfect square.
The next column that we are going to look at is the column that explains the average
value of the odd elements.
Amanda Magli
Page 23
Table 17. Average Value of the Odd Elements
Triple
Sum of
Elements
Sum of Odd
Elements
Sum of Odd
Elements
Expressed as
Twice a
Perfect
Square
Difference
Between the
Two Odd
Elements
4, 3, 5
8, 15, 17
12, 5, 13
24, 7, 25
40, 9, 41
60, 11, 61
72, 65, 97
120, 119,
169
240, 161,
289
……..
? ?, ?
12
40
30
56
90
132
234
408
8
32
18
32
50
72
162
288
2·
2·
2·
2·
2·
2·
2·
2·
2
2
8
18
32
50
32
50
Average
Value of the
Two Odd
Elements
Expressed as
a Perfect
Square
4 = 22
16 = 42
9 = 32
16 = 42
25 = 52
36 = 62
81 = 92
144 = 122
690
450
2 · 225
128
225 = 152
….
………
……
….
….
u2
4
16
9
16
25
36
81
144
In column 6 table 17, we have the average value of the odd elements. For example, if we
look at the 12, 5, 13 triple we add the two odd numbers in that triple.
5 + 13 = 18
Then we divide by 2 to get the average
18 ÷ 2 = 9
The number 9 is the same as 32. Nine is a perfect square. We can represent the average
value of the odd elements with u2.
Now we are going to look at half of the difference of the odd elements.
Amanda Magli
Page 24
Table 18. Half of the Difference of the Odd Elements
Triple
Sum of
Elements
Sum of
Odd
Elements
Sum of
Odd
Elements
Expressed
as Twice a
Perfect
Square
Difference
Between
the Two
Odd
Elements
Half of the
Difference
of the Odd
Elements
Expressed
as a
Perfect
Square
2
2
8
18
32
50
32
50
Average
Value of
the Two
Odd
Elements
Expressed
as a
Perfect
Square
4 = 22
16 = 42
9 = 32
16 = 42
25 = 52
36 = 62
81 = 92
144 = 122
4, 3, 5
8, 15, 17
12, 5, 13
24, 7, 25
40, 9, 41
60, 11, 61
72, 65, 97
120, 119,
169
240, 161,
289
……..
? ?, ?
12
40
30
56
90
132
234
408
8
32
18
32
50
72
162
288
2·
2·
2·
2·
2·
2·
2·
2·
690
450
2 · 225
128
225 = 152
64 = 82
….
………
……
….
….
u2
……..
v2
4
16
9
16
25
36
81
144
1 = 12
1 = 12
4 = 22
9 = 32
16 = 42
25 = 52
16 = 42
25 = 52
In column 7 table 18, we explore half of the difference of the odd elements. For example,
in the 24, 7, 25 triple the difference of the odd elements is 18. Therefore half of this
difference gives you 9.
9 = 32
Nine is a perfect square.
We can represent the half of the difference of the odd elements with v2.
It is now time to look at the hypotenuse.
Amanda Magli
Page 25
Table 19. The Formula for Generating the Hypotenuse
Triple Sum of
Sum of
Sum of
Elements Odd
Odd
Elements Elements
Expressed
as Twice
a Perfect
Square
Difference
Between
the Two
Odd
Elements
Average
Value of
the Two
Odd
Elements
Expressed
as a
Perfect
Square
12
40
8
32
2· 4
2 · 16
2
2
4 = 22
16 = 42
Half of
Hypotenuse
the
Difference
of the
Odd
Elements
Expressed
as a
Perfect
Square
1 = 12
22 + 12
1 = 12
42 + 12
4, 3, 5
8, 15,
17
12, 5,
13
24, 7,
25
40, 9,
41
60,
11, 61
72,
65, 97
120,
119,
169
240,
161,
289
……..
? ?, ?
30
18
2· 9
8
9 = 32
4 = 22
32 + 22
56
32
2 · 16
18
16 = 42
9 = 32
42 + 32
90
50
2 · 25
32
25 = 52
16 = 42
52 + 42
132
72
2 · 36
50
36 = 62
25 = 52
62 + 52
234
162
2 · 81
32
81 = 92
16 = 42
92 + 42
408
288
2 · 144
50
144 = 122
25 = 52
122 + 52
690
450
2 · 225
128
225 = 152
64 = 82
152 + 82
….
………
……
….
….
u2
……
v2
……
u2 + v2
Column 8 of Table 19 shows the hypotenuse of a triangle. As we can see the hypotenuse
is the sum of average value of the odd elements and half of the difference of the odd
elements. For example, in the 40, 9, 41 triple the average value of the odd elements is 52
and half of the difference of the odd elements is 42.
52 + 42 = 41 (the hypotenuse)
Amanda Magli
Page 26
Since we know this, we can label the hypotenuse as u2 + v2 since u2 represents the
average value of the odd elements and v2 represents half of the difference of the odd
elements.
The odd leg is the next side of the triangle we are going to look at.
Table 20. The Formula for Generating the Odd Leg
Half of
the
Difference
of the Odd
Elements
Expressed
as a
Perfect
Square
1 = 12
1 = 12
Hypotenuse
Odd Leg
2
2
Average
Value of
the Two
Odd
Elements
Expressed
as a
Perfect
Square
4 = 22
16 = 42
22 + 12
42 + 12
22 - 12
42 - 12
2· 9
8
9 = 32
4 = 22
32 + 22
32 - 22
32
2 · 16
18
16 = 42
9 = 32
42 + 32
42 - 32
90
50
2 · 25
32
25 = 52
16 = 42
52 + 42
52 - 42
60, 11, 61
132
72
2 · 36
50
36 = 62
25 = 52
62 + 52
62 - 52
72, 65, 97
234
162
2 · 81
32
81 = 92
16 = 42
92 + 42
92 - 42
120, 119,
169
408
288
2 · 144
50
144 = 122
25 = 52
122 + 52
122 -52
240, 161,
289
690
450
2 · 225
128
225 = 152
64 = 82
152 + 82
152- 82
……..
? ?, ?
….
………
……
….
….
u2
…….
v2
……
u2 + v2
…
u2 - v2
Triple
Sum of
Elements
Sum of
Odd
Elements
Sum of
Odd
Elements
Expressed
as Twice a
Perfect
Square
Difference
Between
the Two
Odd
Elements
4, 3, 5
8, 15, 17
12
40
8
32
2· 4
2 · 16
12, 5, 13
30
18
24, 7, 25
56
40, 9, 41
In Column 9 of Table 20, look at the odd leg of each triangle. The odd leg of each
triangle is half of the difference of the odd elements subtracted from the average value of
Amanda Magli
Page 27
the odd elements. For example, in the 8, 15, 17 triple the average value of the odd
elements is 42 and half of the difference of the odd elements is 12. Now we subtract
them.
42 – 12 = 15 (odd leg)
We can label the odd leg as u2 – v2 since it is half of the difference of the odd elements
subtracted from the average value of the sum of the odd elements.
We are now going to look at the even leg.
Amanda Magli
Page 28
Table 21. The Formula for Generating the Even Leg
Hypotenuse
Odd
Leg
Even
Leg
4 = 22
Half of
the
Difference
of the
Odd
Elements
Expressed
as a
Perfect
Square
1 = 12
22 + 12
22 - 12
2
16 = 42
1 = 12
42 + 12
42 - 12
2· 9
8
9 = 32
4 = 22
32 + 22
32 - 22
32
2 · 16
18
16 = 42
9 = 32
42 + 32
42 - 32
90
50
2 · 25
32
25 = 52
16 = 42
52 + 42
52 - 42
132
72
2 · 36
50
36 = 62
25 = 52
62 + 52
62 - 52
234
162
2 · 81
32
81 = 92
16 = 42
92 + 42
92 - 42
408
288
2 · 144
50
144 = 122
25 = 52
122 + 52
122 -52
2
(2)(1)
2
(4)(1)
2
(3)(2)
2
(4)(3)
2
(5)(4)
2
(6)(5)
2
(9)(4)
2
(12)(5)
690
450
2 · 225
128
225 = 152
64 = 82
152 + 82
152- 82
….
………
……
….
….
u2
……
u2 + v2
…
u2 - v2
Triple
Sum of
Elements
Sum of
Odd
Elements
Sum of
Odd
Elements
Expressed
as Twice
a Perfect
Square
Difference
Between
the Two
Odd
Elements
Average
Value of
the Two
Odd
Elements
Expressed
as a
Perfect
Square
4, 3, 5
12
8
2· 4
2
8, 15,
17
12, 5,
13
24, 7,
25
40, 9,
41
60, 11,
61
72, 65,
97
120,
119,
169
240,
161,
289
……..
2uv,
40
32
2 · 16
30
18
56
2
v
2
(15)(8)
….
2uv
u2+/-v2
Look at the 10th column of Table 21. Notice that the even leg is twice u times v. For
example, for the 24, 7, 25 triple the hypotenuse is 42 + 32. The 42 is represented by u2 and
the 32 is represented by the v2. If we leave out the squared for the hypotenuse we get the
numbers 4 and 3. The number 4 represents u and the number 3 represents v. The even
leg is therefore, twice the product of u · v. In this case
2 (4)(3) = 24
The number 24 is the even leg in this triple.
Amanda Magli
Page 29
We have now discovered the Euclidean Method. We are going to show that this works in
the Pythagorean Theorem.
Let
x = u2 – v2
Let
y = 2uv
Let
z = u2 + v2
Substitute
(u2 – v2)2 + (2uv)2 = (u2 + v2) ?
We are now going to multiply (u2 – v2)2, (2uv)2, and (u2 + v2)
(u2 – v2)(u2 – v2) + (2uv)(2uv) = (u2 + v2)(u2 + v2) ?
This equals
u4- 2u2v2 + v4 + 4u2v2 = u4 + 2u2v2 + v4 ?
Combine like terms
u4 +2u2v2 +v4 = u4 +2u2v2 +v4
Our conjecture is true. These 3 expressions do generate Pythagorean triples. Now we are
going to pick random numbers and plug them into the Euclidean method.
Amanda Magli
Page 30
Table 22. Examples of Triples Generated Using the Euclidean Method
u
3
4
5
7
8
8
9
7
10
8
2
v
2
1
4
3
2
4
1
6
3
6
1
x
5
15
9
40
60
48
80
13
91
28
3
y
12
8
40
42
32
64
18
84
60
96
4
z
13
17
41
58
68
80
82
85
109
100
5
In Table 22, column 1 and 2 are the numbers being substituted for u and v. We
picked a random number for u and a random number for v to be substituted into the
Euclidean method. We then found out what x is by substituting the u and v numbers into
the formula u2 – v2. We found out what y is by substituting in the u and v numbers for
the formula 2uv. Finally we found out what z is by substituting in the u and v numbers
into the formula u2 + v2. Here is an example of what is explained above.
Let
u=3
Let
v=2
We substitute these numbers into the x formula
32 - 22 = 5
We substitute these numbers into the y formula
2(3)(2) = 12
Finally we substitute these numbers into the z formula
Amanda Magli
Page 31
32 + 22 = 13
Now we are going to put these 3 numbers, 5, 12, 13 into the Pythagorean Theorem.
52 + 122 = 132
When we solve this equation, we see that the left side equals the right side so therefore 5,
12, 13 is a Pythagorean triple.
Now we are going to look at the product of x, y and z
Table 23. Product of x, y, and z
x
5
15
9
40
60
48
80
13
91
28
3
y
12
8
40
42
32
64
18
84
60
96
4
z
13
17
41
58
68
80
82
85
109
100
5
Product
780
2040
14760
97440
130560
245760
118080
92820
595140
268800
60
Table 23 shows the product of x, y and z. If we look at all of the products of
x,y,z, notice that each answer ends with a 0. This means that each of the products is
divisible by 10. Let’s see if any other numbers divide into all of the products. The
numbers 2 and 5 have to go into all the products since they all end in a zero. The number
3 also goes into each of the products. We know this by adding the digits in each product
together. If that answer is divisible by 3 then the whole product is divisible by 3. The
numbers 4 and 6 also can be divided into each of the products. We know that the number
6 can be divided into each of the product since the numbers 3 and 2 can be divided into
each of the products. We know 4 can be divided into each of the products because if we
Amanda Magli
Page 32
look at the last two digits of each product, if that number is divisible by 4 then the whole
product is divisible by 4. If there are double zeros as the last two numbers of the product,
then look to the last 3 digits of the product and if 4 is divisible into that then the whole
product is divisible by 4. We can conjecture by knowing that all of these numbers can be
divided into each of the products, that 60 can also be divided into each of the products.
RECOMMENDATIONS FOR FURTHER RESEARCH
We generated 3 formulas to use to create Pythagorean Triples. One formula was called
the Euclidean method, another formula was created when the smallest leg was odd, and a
3rd formula was created when the smallest leg was even. There are some questions that
still have to be answered.
•
How many Pythagorean Triples are there?
•
Are these the only 3 sets of formulas known that generate Pythagorean Triples?
•
What patterns are there in the triples we find?
•
Is the product of the elements of any Pythagorean Triple divisible by 60?
Amanda Magli
Page 33
BIBLIOGRAPHY
Brown, S. A Surprsing Fact about Pythagorean Triples. Mathematics Teacher, Vol.78,
No.8, (October 1985) pp. 540- 541
Gerver, R. Writing Math Research Papers. 1997. Berkeley, Ca: Key Curriculum Press
Studyworks Mathematics Deluxe. CD-ROM. Boston: MathSoft, 2002.
Tirman, A. Pythagorean Triples. Mathematics Teacher, Vol. 79, No.9, (November 1986)
Pp. 652 – 655
http://dictionary.reference.com/
Amanda Magli
Page 34
Proving the Product of Any Pythagorean Triple is Divisible by 60
Now we are going to try to prove the pattern that all Pythagorean triples are divisible by
60. We are going to take the 3 formulas found by the Euclidean method and let
x = u2-v2
y = 2uv
z = u2 + v2
We then are going to multiply all these formulas together and see if it is divisible by 60.
(u2-v2)(2uv)( u2 + v2)
Since 2 is a factor, it must only be shown that
(u2-v2)uv( u2 + v2)
We are going to let
n = (v2 – u2)(uv)(u2 + v2)
This is therefore divisible by 30 since we cut the 60 in half after we factored out the 2. If
either u or v is divisible by 2, then n is divisible by 2. If u and v are odd, then so are u2
and v2. For example if
v=5
u=3
Then
v2 = 25
u2 = 9
Therefore v2- u2 is the difference of two odd numbers which gives us an even number.
Amanda Magli
Page 35
25 – 9 = 16
This proves that n is even and therefore is divisible by 2. This also must be divisible by 3
and 5 since the whole thing is divisible by 30.
Divisibility by 3
We now are going to show that every positive integer is divisible by 3 or leaves a
remainder of 1 or 2 after it is divided by 3.
Table 24: Integers Expressed as 3k, 3k+1, and 3k+2
Positive Integer
3k
3k+1
1
√
2
3
√
4
√
5
6
√
7
√
35
39
√
56
67
√
78
√
3k+2
√
√
√
√
As seen in Table 24, every integer can be expressed as 3k, 3k+1 or 3k+2.
In the next table, we show that every perfect square can be expressed as 3m or 3m+1.
Amanda Magli
Page 36
Table 25: Perfect Squares Expressed as 3m or 3m+1
Perfect Square
3m
3m + 1
4
√
16
√
25
√
36
√
49
√
64
√
81
√
100
√
121
√
144
√
Table 25 demonstrates the fact that every perfect square can be expressed as 3m or 3m+1.
To prove that any perfect square can be expressed as this, we began with the fact that an
integer can be written as 3k, 3k+1, or 3k+2.
Lets’ say
b = 3k
Then b2 would be
b2 = 9k2
If we factor a 3 out of this number we get
3(3k2)
This represents 3 times a number or 3m.
Now let
b = 3k+1
Square
b2 = 9k2 + 6k + 1
Factor out a 3
3(3k2 +2k) + 1
Amanda Magli
Page 37
This represents 3m +1.
Next we let
b = 3k+2
If we square this we get
b2 = 9k2 + 12k + 4
Change the 4 into a 3 + 1
Factor out a 3
3(3k2 + 4k + 1) + 1
This represents 3m + 1.
This proves that a perfect square can always be expressed as either 3m or 3m+1.
We now are going to introduce a new notation which will be used in the next few steps.
This new notation is the divisibility notation which is represented by a vertical line and
means “divides evenly into”. Here are some examples demonstrating the use of this new
notation.
6 |18
5 |10
10 | 7
10 | 5
These examples show that 6 divides evenly into 18, 5 divides evenly into 10, but 10 does
not divide evenly into 7 and 10 does not divide evenly into 5. We now are going to use
this notation in showing that the expression (v2-u2)uv( u2 + v2), is divisible by 3.
Amanda Magli
Page 38
If u2 or v2 can be written as 3m, then it is a multiple of 3 and so is its square root (u or v).
If
3 | u2
3|u
This shows that if 3 can divide evenly into u2 then 3 can divide evenly into u. An
example is if
3|9
Then
3|3
If 3 | uv, then 3 | n.
(v2-u2)uv( u2 + v2)
If 3 | u2 and 3 | v2 , let
v2 = 3k+1
u2 = 3b+1
We plug it into the formula v2 – u2
v2-u2 = (3k+1) – (3b+1)
Distribute the – sign
3k – 3b
Factor a 3 out
3(k-b)
Therefore since 3 | (v2-u2), 3 | n.
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From this we conclude that (u2-v2)uv( u2 + v2) must be divisible by 3.
Now since we found out that this expression is divisible by 2 and 3, we are going to
prove that this expression is also divisible by 5.
Divisibility by 5
Table 26 shows that every positive integer can be expressed as 5k, 5k+1, 5k+2, 5k+3 or
5k+4.
Table 26: Positive Integers Can Be Expressed as 5k, 5k+1, 5k+2, 5k+3 or 5k+4
Positive
Integer
1
2
3
4
5
6
7
23
37
42
55
68
5k
5k+1
5k+2
5k+3
5k+4
√
√
√
√
√
√
√
√
√
√
√
√
Table 26 expresses positive integers as 5k ,5k+1, 5k+2, 5k+3, or 5k+4. Every positive
integer can be expressed this way. For example, the positive integer 3 can be expressed
as
5k + 3
This is because 5 times 0 equals 0 and 0 + 3 is 3.
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Next we are going to show that every perfect square can be expressed as 5m, 5m+1, or
5m-1.
Table 27: Every Perfect Square Can Be Expressed as 5m, 5m+ 1 or 5m-1
Perfect Square
4
16
25
36
49
64
81
100
121
144
169
5m
5m+1
5m-1
√
√
√
√
√
√
√
√
√
√
√
As seen in Table 27, every perfect square can be expressed as 5m, 5m+1 or 5m-1. To
prove that this is true we begin with the fact that any integer can be written as 5k, 5k+1,
5k+2, 5k+3, or 5k+4.
Let
b = 5k
Square both sides
b2 = 25k2
Factor out a 5
5(5k2)
This can be represented as 5 times m where m represents any positive integer.
Now let
b = 5k+1
Square both sides
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b2 = 25k2 +10k +1
Factor out a 5
5(5k2 +2k) +1
This represents 5m +1.
Next we let
b = 5k+2
Square both sides
b2 = 25k2 +20k +4
Factor out a 5
5(5k2 +4k +1) -1
This represents 5m – 1.
Let
b = 5k+3
Square both sides
b2 = 25k2 +30k + 9
Express 9 as (10 -1)
b2 = 25k2 + 30k + (10-1)
Use the associative property
b2 = (25k2 + 30k + 10) - 1
Factor out a 5
5(5k2 +6k + 2) – 1
This represents 5m -1.
Finally we let
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b = 5k+4
We square both sides
b2 = 25k2 +40k +16
Factor out a 5
5(5k2 + 8k + 3) + 1
This represents 5m + 1.
This proves that a perfect square can be expressed as 5m, 5m+1 or 5m-1.
If u2 or v2 is a multiple of 5, then u or v is a multiple of 5 and so is the product of n.
Therefore 5 | n where
n = (u2-v2)uv( u2 + v2)
If neither u2or v2 is a multiple of 5 four other cases must be considered. One case is when
v2 = 5k + 1
u2 = 5b + 1
If we plug it into the v2 – u2 formula we get
v2 – u2 = (5k+1) – (5b + 1)
Distribute the – sign
5k – 5b
Factor out a 5
5(k-b)
If 5 | (v2-u2) then 5 divides evenly into the whole expression
(u2-v2)uv( u2 + v2)
The second case that must be considered is when
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v2 = 5k-1
u2 = 5b – 1
If we plug it into the formula v2 – u2 we get
v2-u2 = (5k-1) – (5b -1)
Distribute the – sign
5k – 5b
Factor out a 5
5 (k – b)
Again 5 is a factor of n. The 3rd case is when
u2 = 5b + 1
v2 = 5k-1
We plug it into the formula u2 + v2 this time
u2 + v2 = (5b+1) + (5k-1)
This equals
5b + 5k
Factor out a 5
5(b + k)
Again this shows that if 5 | (u2 + v2) then 5 | n. The last case is when
u2 = 5b – 1
v2 = 5k + 1
We plug it into the formula u2 + v2
u2 + v2 = (5b – 1) + (5k + 1)
This equals
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5b + 5k
Factor out a 5
5(b + k)
Again this shows that since 5 | u2+ v2, 5 | n. Thus n is divisible by 2, 3, and 5. Therefore
this shows that the product of a Pythagorean triple, a, b,c, is always divisible by 60.
60 | abc
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The next question that we will be answering is if there an infinite number of
Pythagorean Triples? We are going to prove this by using indirect reasoning. Indirect
reasoning is when we prove that something can not happen and therefore the other choice
has to be correct. We are going to use the formulas found by the Euclidean method to
complete this proof.
x = u2 – v2
y = 2uv
z = u2 + v2
There are an infinite number of positive integers to choose from for u and v. We are
trying to see by doing this proof, if from this infinite numbers that can be chosen for u
and v, if two different sets of numbers will produce the same triple. Therefore our first
step is that we are going to assume that there are a finite number of triples, which means
that two different sets of numbers produce the same triple. Therefore
u1 ≠‌ u2
v1 ≠ v2
and they form the same triple.
Let
x = u12 – v12
x = u22 – v22
Let
y = 2u1v1
y = 2u1v2
Let
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z = u12 + v12
z = u22 + v22
From the two expressions for x we can say
u12 – v12 = u22 – v22
From the two expressions for z we can say
u12 + v12 = u22 +v22
We can say this because the formulas above both equal z. If we add these two equations
together we get
2u12 = 2u22
Divide by 2
u12 = u22
This contradicts what we said first of all because we said that u1 and u2 did not equal each
other. Therefore from this, we conclude that our assumption, that there are a finite
numbers of triples, is wrong and therefore the only other possible answer to this proof is
that there are an infinite number of triples.
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RECOMMENDATIONS FOR FURTHER RESEARCH
We proved that the product of any Pythagorean Triple is divisible by 60. There is still
one question that is left that remains unanswered.
•
If prime number divides evenly into a perfect square, then a prime number divides
evenly into its square root.
Amanda Magli
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BIBLIOGRAPHY
Brown, S. A Surprsing Fact about Pythagorean Triples. Mathematics Teacher, Vol.78,
No.8, (October 1985) pp. 540- 541
Gerver, R. Writing Math Research Papers. 1997. Berkeley, Ca: Key Curriculum Press
Studyworks Mathematics Deluxe. CD-ROM. Boston: MathSoft, 2002.
Tirman, A. Pythagorean Triples. Mathematics Teacher, Vol. 79, No.9, (November 1986)
Pp. 652 – 655
http://dictionary.reference.com/
Amanda Magli
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