Download Pythagorean Triples Solution Commentary:

Pythagorean Triples
Solution Commentary:
Solution of Main Problems:
1. If m2 = (2n + 1), n = [m2-1]/2 and n+1 = [m2+1]/2. Then, by substitution, the
2
2
 m2 − 1 
 m 2 +1 
 + m 2 = 
 . Now,
equation n + (2n + 1) = (n+1) becomes 
 2 
 2 
because an even number squared is even and the expression (2n+1) is odd,
we know that both m2 and m are odd. Thus, for a short sequence of odd
numbers, we produce these Pythagorean triples (increasing side length):
m
a
b
c
3
3
4
5
5
5
12
13
7
7
24
25
9
9
40
41
11
11
60
61
13
13
84
85
Though this process generates many Pythagorean triples (seemingly all
primitive?), but not all Pythagorean triples. For example, the triple (8, 15, 17)
will never be generated for two reasons. First, the lowest term is 8, not an odd
number. And second, the difference between the two larger numbers in the
triple is not 1. Thus, perhaps at best, this process generates “all” Pythagorean
triples where the hypotenuse differs from one side by 1.
2
2
2. Let m be an odd number and one side of the right triangle. Then, by
Pythagoras’ method, the other side is (m2-1)/2 numbers. To get the
hypotenuse, add to get [(m2-1)/2] + 1 = (m2+1)/2. This is exactly the same
2
2
 m2 − 1 
 m 2 +1 
 + m 2 = 
 .
method, since 
 2 
 2 
3. Let p be an even number and one side of the right triangle. Then, by Plato’s
method, the hypotenuse is [p/2]2+1 and the other side is [p/2]2-1. Thus, for a
short sequence of even numbers, we produce these Pythagorean triples
(increasing side length):
p
a
b
c
4
3
4
5
6
6
8
10
8
8
15
17
10
10
24
26
12
12
35
37
14
14
48
50
 m2 − 1 
 m2 +1 
 and 
 will
4. In Pythagorean’s method, the two larger sides 
 2 
 2 
always differ by 1, and thus have no common factor. However, in Plato’s
method which starts with an even side, the two sides [p/2]2-1 and [p/2]2+1 not
only differ by 2 but will also both be even whenever [p/2] is odd.
Extension 1: With a = p2 – q2, b = 2pq, and c = p2 + q2, we have a2 + b2 = (p2 – q2)2 +
(2pq)2 = p4 – 2p2q2 + q4 + 4p2q2 = p4 + 2p2q2 + q4 = (p2 + q2)2 = c2. If p and q had a
common factor d > 1, then d|p and d|q implies that d|(p2-q2), d|(2pq), and d|(p2+q2),
thereby producing a non-primitive triple.
p
q
a
b
c
2
1
3
4
5
3
1
8
6
10
3
2
5
12
13
4
1
15
8
17
4
3
7
24
25
5
1
24
10
26
5
2
21
20
29
5
3
16
30
34
5
4
9
40
41
6
1
35
12
37
6
5
11
60
61
7
1
48
14
50
7
2
45
28
53
7
3
40
42
58
7
4
33
56
65
7
5
24
70
74
7
6
13
84
85
Many non-primitive triples are still generated, such as (6,8,10), (10,16,24),
(16,30,34), (14,48,50), (40,42,58), and (24,70,74). By comparison, this table
encompasses all of the triples listed in the two previous tables. Nonetheless, some
“new” Pythagorean triples are generated, such as (20,21,29), (28,45,53), and
(33,56,65). NOTE: Sierpinski (1962) proves that this set of expressions will generate
all primitive Pythagorean triples.
Extension 2: First, if x= 9, let p = 1 and q = 9. Then, a = 9 = 3, b = (9-1)/2 = 4, and c
= (9+1)/2 = 5, producing the triple (3,4,5). And, if x= 25, let p = 1 and q = 25. Then, a
= 25 = 5, b = (25-1)/2 = 12, and c = (25+1)/2 = 13, producing the triple (5,12,13).
To prove a2 + b2 = c2 starting with a = pq, we can prove that a2 = b2 – c2 = (b+c)(b-c).
 p + q p − q  p + q p − q   2 p  2q 
2
Thus, substituting, (c − b)(c − a ) = 
+
−

=
  = pq = a .
2
2
2
2
2
2


 
 
Note:
Extension 3: Using the conditions in Problem #1 and the right triangle’s diagram, let
 m2 −1 
 m2 +1 
 , BC = c = 
 where a2 + b2 = c2. The area of the
AB = a = m, AC = b = 
 2 
 2 
triangle can be visualized in two ways, where A =
ab
ar br cr
=
+ + . Solving for r and
2
2 2 2
ab
. Substituting in Pythagoras’ expressions for the triple (a,b,c)
a+b+c
 m2 −1 

m
2 
m −1

and simplifying algebraically, r =
=
. Expanding the table
2
2
2
 m −1   m +1 
 + 

m + 
 2   2 
generated in Problem #1, we obtain an interesting relationship, showing that the
respective radii are the set of whole numbers:
m
a
b
c
r
3
3
4
5
1
5
5
12
13
2
7
7
24
25
3
9
9
40
41
4
11
11
60
61
5
13
13
84
85
6
simplifying, r =
Open-Ended Exploration: This exploration can be quite productive, with many
perhaps unknown discoveries yet to be made. Freitag (1992), Bertucci (1991), and
Horadam (1961) are some useful resources.
Teacher Commentary:
When exploring Pythagorean triples and trying to connect them back to the visual
interpretation of proofs of the Pythagorean Theorem, students may want to explore
the connection to the Greek idea of “applying areas.” As a first step, the square of
side length c can be visualized as containing a square of side length a. Once the
latter square is removed, is it possible to transform the remaining “area” into a square
of side length b? You might want to have students explore Schwaller’s (1979)
interesting approach to this question, which ends up generating Pythagorean triples.
In Problem #1, Pythagoras’ method produced some Pythagorean triples, but it
seemed to miss some generated by Plato’s method, such as (8,15,17). But, if m were
not restricted to odd integers, other triples can be produced. For example, using m=2
in the expressions, generates the non-integer Pythagorean triple (4,7½,8½), which
can be multiplied by 2 to get the integer Pythagorean triple (8,15,17). If fact, if m is
allowed to be a rational number, it can be shown that Pythagoras’ method would
generate all of the Pythagorean triples. And, if m = p/q is substituted into Pythagoras’
expressions for generating triples, the more common set of expressions 2pq, p2 - q2
and p2 + q2 appears as explored in Extension Problem #1.
Extension Problem #2 explored Euclid’s method for producing Pythagorean triples.
This method has been analyzed and discussed often as to whether or not Euclid’s
method produces all of the triples. Laubenbacher & Pengelley (1999 pp. 172-178)
document thoroughly that this is the case, though such may be deceptive.
Boulger (1993) was the source for the solution shown in Extension Problem #3.
However, the problem has been discussed in many references, and even solved in
very different ways, such as in Maor (2007).
As a good source of writing projects, students can explore any of the following ideas
relative to Pythagorean triples:
• Extension Problem #1 focused on the standard set of expressions for
generating Pythagorean triples. Investigate and compare different approaches
to deriving these expressions, such as the number theoretic approach (Ore,
1988), the geometric approach (Kleiman, 1972), the trigonometric approach
(Maiorano, 1995), or even the complex number approach (Kerins et al, 2003).
• Though Pythagorean triples usually are viewed as a topic in number theory,
they can be viewed visually. Investigate the visual aspects, using sources
such as the visual patterns in Nelson (2007), the geoboard approach in
Bonsangue (1997), the 3-D graphs in Shabanian (1995), and the distribution
patterns in Oliver (1971).
• Investigate other unusual properties of Pythagorean triples, such as:
1. The product of the members of every Pythagorean triple (a,b,c) is a
multiple of 60. (Brown, 1985; Rudolph, 1987)
2. The connection between Pythagorean triples and special triangles
where all three sides and the area are integers, plus the area is an
integer multiple of the perimeter. (Markov, 2006)
3. For Pythagoras’ method, explore how “his” triangle’s semiperimeters
and areas are connected to triangular numbers and the sums of
consecutive square numbers. (Maiorano, 1994)
4. Based on an interpretation of Plimpton-322 clay tablet (ca. 1500 B.C.),
the Babylonians basically formed an infinite chain of Pythagorean
triples (a,b,c) by using a “rectangular rearrangement technique” where
a square with integral side length a visually becomes a rectangle with
integral side lengths b and c, such that a2 = c2 - b2 = (c + b)(c - b).
(Brudner, 1983)
Additional References:
Bertucci, J. (1991). “From Fibonacci numbers to Pythagorean triples.” Mathematics
Teacher. May, p. 356.
Bonsangue, M. (1997). A geometrical representation of primitive Pythagorean
triples.” Mathematics Teacher. Vol. 90#5. pp. 350-354.
Boulger, W. (1993). “Triple your fun again.” Mathematics Teacher. November, p. 704.
Brown, S. (1985). “A surprising fact about Pythagorean triples.” Mathematics
Teacher. October, pp. 540-541.
Brudner, H. (1983). Babylonian doubles and Pythagorean triples.” Educational
Technology. March, pp. 34-36 and April, pp. 33-36.
Freitag, H. (1992). “Generating triples.” Mathematics Teacher. January, p. 10.
Heath, T. (1956). The Thirteen Books of Euclid’s Elements. Vol 1. Dover
Publications.
Horadam, A. (1961). “Fibonacci number triples.” American Mathematical Monthly.
Vol. 68, pp. 751-753.
Kerins, B. et al. (2003). “Gauss, Pythagoras, and Heron.” Mathematics Teacher. Vol.
96#5. pp. 350-357.
Kleiman, S. (1972). “Pythagorean triples by geometry.” Two-Year College
Mathematics Journal. Spring, pp. 39-41.
Laubenbacher, R. and Pengelley, D. (1999). Mathematical Expeditions: Chronicles
by the Explorers. Springer.
Maiorana, P. (1994). “More triples fun.” Mathematics Teacher. November, p. 662.
Maiorano, P. (1995). “Triples by way of trigonometry.” Mathematics Teacher. March,
pAGE****.
Maor, E. (2007). The Pythagorean Theorem: A 4000 Year History. Princeton
University Press.
Markov, L. (2006). “Pythagorean triples and the problem A = mP for triangles.”
Mathematics Magazine. Vol. 79#2, pp. 114-121.
Nelson, R. (2007). “Visual gems of number theory.” Math Horizons, November, 7-9,
31.
Oliver, B. (1971). “The pattern of Pythagorean numbers.” Mathematics Teacher. May,
pp. 449-454.
Ore, O. (1988). Number Theory and Its History. Dover Publications.
Rudolph, M. (1987). “Triple’s surprise.” Mathematics Teacher. May, pp. 343-344.
Schwaller, R. (1979). “A method of generating Pythagorean triples.” Int. J. Math.
Educ. Sci. Tech. Vol. 10#1, pp. 75-77.
Shabanian, S. (1995). “Computer graphical modeling of Pythagorean triplets in 3-D
space.” Scientific Computing & Automation. June, pp. 45-48.
Sierpinski, W. (1962). Pythagorean Triangles. New York: Yeshiva University. pp. 5-7.