When the Pythagorean theorem is mentioned, one immediately recalls the famous relationship: a^{2} + b^{2} = c^{2}. Yet, how many adults can remember what this equations means? This question motivated me to write a book on this most famous theorem (The Pythagorean Theorem, the Story of its Power and Glory - Prometheus Books, 2010) to show off the many aspects of this relationship in a wide variety of contexts and applications. However, for the classroom, teachers should not be limited to merely show its geometric application and then in the most trivial fashion. To much "good stuff" is lost that way.
After introducing the Pythagorean theorem, teachers often suggest that students recognize (and memorize) certain common ordered triples that can represent the lengths of the sides of a right triangle. Some of these ordered sets of three numbers, known as Pythagorean triples, are: (3, 4, 5), (5, 12, 13), (8, 15, 17), (7, 24, 25). The student is asked to discover these Pythagorean triples as they come up in selected exercises. How can one generate more triples without a guess and test method? This question, often asked by students, will be answered here and, in the process, will show some really nice mathematics, all too often not presented to students. This is an unfortunate neglect that ought to be rectified.
Ask your students to supply the number(s) that will make each a Pythagorean triple:
1. (3, 4, __)
2. (7, __, 25)
3. (11, __, __)
The first two triples can be easily determined using the Pythagorean theorem. However, this method will not work with the third triple. At this point, your students will be quite receptive to learning about a method to discover the missing triple. So, with properly motivated students as your audience, you can embark on the adventure of developing a method for establishing Pythagorean triples.
However, before beginning to develop formulas, we must consider a few simple "lemmas" (these are "helper" theorems).
Lemma 1: When 8 divides the square of an odd number, the remainder is 1.
Proof: We can represent an odd number by 2k + 1, where k is an integer.
The square of this number is (2k + 1)^{2} = 4k^{2} + 4k + 1 = 4k (k + 1) + 1
Since k and k + 1 are consecutive, one of them must be even. Therefore 4k (k + 1) must be divisible by 8. Thus (2k + 1)^{2}, when divided by 8, leaves a remainder of 1.
The next lemmas follow directly.
Lemma 2: When 8 divides the sum of two odd square numbers, the remainder is 2.
Lemma 3: The sum of two odd square numbers cannot be a square number.
Proof: Since the sum of two odd square numbers, when divided by 8, leaves a remainder of 2, the sum is even, but not divisible by 4. It therefore cannot be a square number.
We are now ready to begin our development of formulas for Pythagorean triples. Let us assume that (a,b,c) is a primitive Pythagorean triple. This implies that a and b are relatively prime.* Therefore they cannot both be even. Can they both be odd?
If a and b are
both odd, then by Lemma 3:
a^{2} + b^{2} ≠ c^{2} . This contradicts our
assumption that (a, b, c) is a Pythagorean triple; therefore a
and b cannot both be odd. Therefore one must be odd and one even.
Let us suppose that a
is odd and b is even. This implies that c is also odd. We can
rewrite
a^{2} + b^{2} = c^{2} as
b^{2} = c^{2} - a^{2}
b^{2} = (c + a)(c - a)
Since the sum and difference of two odd numbers is even, c + a = 2p and c - a = 2q (p and q are natural numbers).
By solving for a and c we get:
c = p + q and a = p - q
We can now show that p and q must be relatively prime. Suppose p and q were not relatively prime; say g>1 was a common factor. Then g would also be a common factor of a and c. Similarly, g would also be a common factor of c + a and c - a. This would make g^{2} a factor of b^{2}, since b^{2} = (c + a)(c - a). It follows that g would then have to be a factor of b. Now if g is a factor of b and also a common factor of a and c, then a, b, and c are not relatively prime. This contradicts our assumption that (a, b, c) is a primitive Pythagorean triple. Thus p and q must be relatively prime.
Since b is even, we may represent b as
b = 2r
But b^{2} = (c + a)(c - a).
Therefore b^{2} = (2p)(2q) = 4r^{2}, or pq = r^{2}
If the product of two relatively prime natural numbers (p and q) is the square of a natural number (r), then each of them must be the square of a natural number.
Therefore we let p = m^{2} and q = n^{2}, where m and n are natural numbers. Since they are factors of relatively prime numbers (p and q), they (m and n) are also relatively prime.
Since a = p - q and c = p + q, it follows that a = m^{2} - n^{2} and c=m^{2} + n^{2}
Also, since b = 2r and b^{2} = 4r^{2} = 4pq = 4m^{2}n^{2}, b = 2mn
To summarize, we now have formulas for generating Pythagorean triples:
a = m^{2} - n^{2} b = 2mn c = m^{2} + n^{2}
The numbers m and n cannot both be even, since they are relatively prime. They cannot both be odd, for this would make c = m^{2} + n^{2} an even number, which we established earlier as impossible. Since this indicates that one must be even and the other odd, b = 2mn must be divisible by 4. Therefore no Pythagorean triple can be composed of three prime numbers. This does not mean that the other members of the Pythagorean triple may not be prime.
Let us reverse the process for a moment. Consider relatively prime numbers m and n (where m > n), where one is even and the other odd.
We will now show that (a, b, c) is a primitive Pythagorean triple where a = m^{2} - n^{2}, b=2mn and c = m^{2} + n^{2}.
It is simple to verify algebraically that (m^{2} - n^{2})^{2} + (2mn)^{2} = (m^{2} + n^{2})^{2}, thereby making it a Pythagorean triple. What remains is to prove that (a, b, c) is a primitive Pythagorean triple.
Suppose a and b have a common factor h > 1. Since a is odd, h must also be odd. Because a^{2 }+ b^{2} = c^{2}, h would also be a factor of c. We also have h a factor of m^{2} - n^{2} and m^{2} + n^{2} as well as of their sum, 2m^{2}, and their difference, 2n^{2}.
Since h is odd, it is a common factor of m^{2} and n^{2}. However, m and n (and as a result, m^{2} and n^{2}) are relatively prime. Therefore, h cannot be a common factor of m and n. This contradiction establishes that a and b are relatively prime.
Having finally established a method for generating primitive Pythagorean triples, students should be eager to put it to use. The table below gives some of the smaller primitive Pythagorean triples.
Pythagorean Triples
m |
n |
a |
b |
c |
2 |
1 |
3 |
4 |
5 |
3 |
2 |
5 |
12 |
13 |
4 |
1 |
15 |
8 |
17 |
4 |
3 |
7 |
24 |
25 |
5 |
2 |
21 |
20 |
29 |
5 |
4 |
9 |
40 |
41 |
6 |
1 |
35 |
12 |
37 |
6 |
5 |
11 |
60 |
61 |
7 |
2 |
45 |
28 |
53 |
7 |
4 |
33 |
56 |
65 |
7 |
6 |
13 |
84 |
85 |
A fast inspection of the above table indicates that certain primitive Pythagorean triples (a, b, c) have c = b+1. Have students discover the relationship between m and n for these triples.
They should notice that for these triples m = n + 1. To prove this will be true for other primitive Pythagorean triples (not in the table), let m = n + 1 and generate the Pythagorean triples.
a = m^{2} - n^{2} = (n + 1)^{2} - n^{2} = 2n + 1
b = 2mn = 2n(n + 1) = 2n^{2} + 2n
c = m^{2} + n^{2} = (n + 1)^{2} + n^{2} = 2n^{2} + 2n + 1
Clearly c = b + 1, which was to be shown!
A natural question to ask your students is to find all primitive Pythagorean triples which are consecutive natural numbers. In a method similar to that used above, they ought to find that the only triple satisfying that condition is (3, 4, 5).
Students should have a far better appreciation for Pythagorean triples and elementary number theory after completing this unit. Other investigations that students may wish to explore are presented below. Yet, bear in mind the applications of this most ubiquitous relationship has practically endless applications!
1. Find six primitive
Pythagorean triples which are not included in the above table.
2. Find a way to generate primitive Pythagorean triples of the form (a, b, c)
where b = a + 1.
3. Prove that every primitive Pythagorean triple has one member which is
divisible by 3.
4. Prove that every primitive Pythagorean triple has one member which is
divisible by 5.
5. Prove that for every primitive Pythagorean triple the product of its
members is a multiple of 60.
6. Find a Pythagorean triple (a, b, c), where b^{2} = a + 2.
* Relatively prime means that they do not have any common factors aside from 1.