pythagorean triples

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Pythagorean Triples: Avenues for Exploration Author(s): Lawrence O. Cannon Source: The Mathematics Teacher, Vol. 105, No. 4 (November 2011), pp. 311-315Published by: National Council of Teachers of MathematicsStable URL: http://www.jstor.org/stable/10.5951/mathteacher.105.4.0311Accessed: 15-11-2015 14:10 UTC

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Vol. 105, No. 4 • November 2011 | MATHEMATICS TEACHER 311

DELVINGdeeperLawrence O. Cannon

Pythagorean Triples: Avenues for Exploration

nates (a/c, b/c) that lies on the unit circle satisfies the equation x2 + y2 = 1. That is, we can derive a2 + b2 = c2 from (a/c)2 + (b/c)2 = 1. Therefore, we can state this as follows:

Every point on the unit circle with rational coor-dinates names a Pythagorean triple, and every Pythagorean triple names a point on the unit circle with rational coordinates.

At this point, students may wonder how many rational points may be on the unit circle. From the symmetry of the circle, we know that there are eight points that we associate with any one given point with rational coordinates: We can interchange coor-dinates or change the sign of either one. Thus, the points associated with (3/5, 4/5) are (±3/5, ±4/5) and (±4/5, ±3/5), which correspond, respectively, to the equations (±3)2 + 42 = 52, 32 + (±4)2 = 52, (±4)2 + 32 = 52, and 42 + (±3)2 = 52. In terms of the circle, these eight points come from the Pythago-rean triple (3, 4, 5). How do we generate new sets of rational points that produce other Pythagorean triples?

Teachers can introduce notation that indicates both a point with rational coordinates and a Pythag-orean triple. To that end, we use (a, b)/c for both purposes, so that (3, 4)/5 denotes both the point (3/5, 4/5) on the unit circle and the Pythagorean triple {3, 4, 5}. Since 6/8 = 3/4, we can use our nota-tion to make the identification (6, 8)/10 = (3, 4)/5.

Some students will already be familiar with addi-tional Pythagorean triples, such as (5, 12)/13 and (8, 15)/17, but we are interested in finding a sys-tematic procedure to generate more (or all) of them. The following technique is different from tradi-tional methods of generating Pythagorean triples in that we use lines and circles. Even if students have seen formulas for generating Pythagorean triples, they may not have used simple equations for lines and circles to derive their own formulas.

Delving Deeper offers a forum for classroom teachers to share the mathematics from their own work with the journal’s readership; it appears in every issue of Mathematics Teacher. Manuscripts for the department should be submitted via http://mt.msubmit.net. For more background information on the department and guidelines for submitting a manuscript, visit http://www.nctm.org/publications/content.aspx?id=10440#delving.

Edited by J. Kevin Colligan, [email protected] Center of SRA International, Columbia, MD

Maria Fung, [email protected] State College, Worcester, MA

Dan Kalman, [email protected] University, Washington, DC

Jeffrey Wanko, [email protected] University, Oxford, OH

Extending the study of Pythagorean triples beyond right-triangle geometry or simple num-ber theory can provide many avenues of explo-

ration for students. These explorations can make use of algebra, geometry, trigono metry, and even com-plex numbers and group theory. Here are a few ways in which I use Pythagorean triples to encourage stu-dents to investigate important and useful ideas. I will show how students can be encouraged to extend Pythagorean triples—defined here as any set of three integers {a, b, c| a2 + b2 = c2, c > a, c > b}—beyond traditional applications to triangles.

CIRCLES, LINES, AND RATIONAL POINTSWhen divided through by 25, the equation 32 + 42 = 52 produces

3

5

4

51

35

45

12

2

2

2

2 2

+ =+ =

+

2 2

2 2

2 2

2 2

=, .1, .1, .5

, .5

, .5

, .5

, .

, ., .

, .

, ., .

+, .+

, .

, ., .

, .

, ., .

or , . or , .

This means that the point with rational coordinates (3/5, 4/5) satisfies the equation x2 + y2 = 1 and lies on the unit circle. Any point with rational coordi-

Copyright © 2011 The National Council of Teachers of Mathematics, Inc. www.nctm.org. All rights reserved.This material may not be copied or distributed electronically or in any other format without written permission from NCTM.



312 MATHEMATICS TEACHER | Vol. 105, No. 4 • November 2011

,ac

bc

:

a = 2uv, b = u2 – v2, c = u2 + v2 (1)

Finally, have students calculate the slope of the line that passes through the point (0, –1) and any other rational point Q on the unit circle. They will find that the slope is a rational number. A little experimentation allows students to observe that they get a primitive Pythagorean triple when the slope is in reduced form and the numerator and denomina-tor have opposite parity (one odd and one even).

Curious students may want to play with rela-tionships between slopes of the line y = mx – 1 and the associated Pythagorean triples. For example, the coordinates of point Pm are both positive when Pm

is in quadrant I, or when m > 1. The y-coordinate is smaller than the x-coordinate (meaning that a > b in equation (1)) when 1 < m < 1 + 12, but the relationship is reversed for values of m greater than 1 + 12; that is, the relative sizes of a and bswitch when the line y = mx – 1 passes through the point (1/12, 1/12). By choosing values of m for which the point P is near (1, 0) or (0, 1), we can get Pythagorean triples with one of the values of aand b much smaller than the other. Thus, the slope m = 1.1 = 11/10 generates the Pythagorean triple (220, 21)/221. By choosing m to be near 1 + 12, we can find Pythagorean triples where a and b are rela-tively close together. For example, 1 + 12 ≈ 2.414, and taking m = 24/10 = 12/5, we get the Pythago-rean triple (120, 119)/169.

Students who follow the process this far can derive the standard formulas for Pythagorean triples. They also can observe their association with rational points on the unit circle and their corre-sponding rational slopes. This can be summed up in the following theorem:

There are infinitely many rational points on the unit circle (and, hence, infinitely many Pythago-rean triples {a, b, c}), given by the following equations—

a = 2uv, b = u2 – v2, c = u2 + v2 (1) —for each pair of integers, u, v where the frac-tion u/v is reduced and either u is even and v is odd or u is odd and v is even.

Now students are ready for an additional obser-vation. Not only are there infinitely many rational points on the unit circle, but also they are densely, thickly spread around the circle in the sense that every segment of the circle, no matter how small, contains rational points. To locate one, all we need to do is to find an appropriate rational slope. For

We begin with the unit circle with equation x2 + y2 = 1 and use any nonhorizontal line, y = mx – 1 (m ≠ 0), that meets the unit circle at the point (0, –1) (see fig. 1). We are interested in the coordinates of the other intersection point, Pm, which depends, obviously, on the slope m.

Have students substitute y = mx – 1 into the equation x2 + y2 = 1 and solve for x. Most students have no difficulty in finding that the other point of intersection of the line and circle can be repre-sented by

2

12=

+P

m

mm, .,, .,m

m

2

2, .

2, .

1

1, .

1, .

−+

, .

, ., ., .

, .

, ., .

, .

For any integer m, the coordinates of Pm are obviously fractions, so we have a means for gen-erating a whole family of rational points and their corresponding Pythagorean triples. Setting m = 1, we get the trivial (but very important) point

,22

02

( ), or 1, 0 .

Using our Pythagorean triple notation, we have (2, 0)/2 = (1, 0)/1. If we continue this process by taking the first few integer values for m, we get the triples (4, 3)/5, (6, 8)/10 = (3, 4)/5, (8, 15)/17, (10, 24)/26 = (5, 12)/13, …. Ask students to observe what happens when m is even (the triples are in reduced form—that is, they are primitive Pythagorean triples).

What happens if we allow other rational values for m? Another useful algebraic exercise is for students to show that if u and v are integers and m = u/v, then the coordinates of Pm are rational numbers

Fig. 1 Students can fi nd the coordinates of the other

intersection point, Pm.




or b is a multiple of 3, then the other must be a multiple of 3.) If a number is not a multiple of 3, then when the number is divided by 3, there is a remainder of either 1 or 2; so each of a and b must have a remainder of 1 or 2 when divided by 3. Give students the following possible case and let them investigate other cases: Suppose that a = 3n + 1 and b = 3m + 2. Then a2 + b2 = (3n + 1)2 + (3m + 2)2 = 9n2 + 9m2 + 6n + 12m + 1 + 4, which is not a multiple of 3 and contradicts the assumption that a2 + b2 = 3c2.

3. What values of k result in circles x2 + y2 = k that contain infinitely many rational points? We know that the circle x2 + y2 = 2 contains infinitely many rational points and that the circle x2 + y2

= 3 contains no rational points, so what about other circles? It is true that there are rational points on the circle x2 + y2 = 4 and that if a circle contains a single rational point, then it contains infinitely many rational points. We leave open the question of determining which circles x2 + y2

= k contain infinitely many rational points.

4. Show that the ellipse

+ =+ =16 9

12 2x y

+ =x y

+ =2 2x y2 2

contains infinitely many rational points; find a formula for integer solutions of the equation 9x2 + 16y2 = 144.

A NEW OPERATION: COMBINING TWO PYTHAGOREAN TRIPLES TO GET A NEW ONEEuler established a procedure that used infinite improper integrals to prove that the set of rational points on certain kinds of curves forms a group. Students can experience a new, less complicated operation on their group of Pythagorean triples if we assign the following definition to the operation. Given Pythagorean triples (a, b)/c and (d, e)/f, we define

(a, b)/c ⊗ (d, e)/f = (ad – be, ae + bd)/cf. (2)

To make sense of this definition, we obvi-ously must show that the operation always gives a Pythagorean triple, that (ad – be)2 + (ae + bd)2 = (cf)2. This is a straightforward algebraic manipula-tion for students, but I like to have my students play with the definition first to get a feeling for what the operation does. What do we get when we use (2) for (3, 4)/5 ⊗ (5, 12)/13 or (5, 12)/13 ⊗(3, 4)/5? Is the operation commutative? How about the product (3, 4)/5 ⊗ (1, 0)/1 or (5, –12)/13 ⊗(1, 0)/1? What conclusion do these results suggest? Can you prove your conjecture? Compute

example, consider the familiar, nonrational point A

−

12

32

, ,

, ,, ,

, ,2

, ,2

one-third of the way around the unit circle from the point (1, 0). The slope of the line through A and (0, –1) is the irrational number –(2 + 13) ≈ –3.732. We can choose rational slopes as near to this number as we wish and thus locate rational points as close to A as desired. For example, if we take m = –37/10, we get the Pythagorean triple (–740, 1269)/1469 (see fig. 2). We could get still closer by taking m = –373/100, giving the Pythago-rean triple (–74,600, 129,129)/149,129. Clearly, there is no limit to how close we can get to A.

At this point, students may want to explore other related questions. Are there rational points on similar circles? How about x2 + y2 = 2 or x2 + y2 = 3? The answers may be surprising.

1. The circle x2 + y2 = 2 contains infinitely many rational points, and each point of the form (a/c, b/c) yields integer solutions to the equation a2 + b2 = 2c2. Challenge students to write an equation for an arbitrary line with nonzero slope m that passes through the point (–1, –1) and intersects the circle x2 + y2 = 2 at a different point. Express the coordinates of the other point of intersection in terms of the slope m. Let m be a reduced frac-tion and find integer solutions to a2 + b2 = 2c2.

2. On the other hand, there are no rational points anywhere on the circle x2 + y2 = 3. If there were a rational point (a/c, b/c) on the circle x2 + y2 = 3, then we would have integers a, b, and c satisfying a2 + b2 = 3c2, where a, b, and c are pairwise relatively prime. (Note that if one of a

Fig. 2 The rational point P is very near the irrational point A.



314 MATHEMATICS TEACHER | Vol. 105, No. 4 • November 2011

(3, 4)/5 ⊗ (3, –4)/5. Can you find a triple such that (5, –12)/13 ⊗ (a, b)/c = (1, 0)/1?

Experimentation suggests that operation (2) behaves something like the multiplication of real numbers. The triple (1, 0)/1, corresponding to the special point (1, 0) on the unit circle, serves as the identity element, and for every triple (a, b)/c there is another triple that acts like a reciprocal (see fig. 3). To verify the last equality, we need to use the fact that in (a, b)/c, we have a2 + b2 = c2: (a, b)/c ⊗(a, –b)/c = (a2 + b2, 0)/(a2 + b2) = (c2, 0)/c2 = (1, 0)/1.

All that remains to show that ⊗ forms a group over the Pythagorean triples is to verify that the associative property holds:

[a, b)/c ⊗ (d, e)/f ] ⊗ (g, h)/i = (a, b)/c ⊗ [(d, e)/f ⊗ (g, h)/i]

The justification is tedious but not difficult. All these verifications can be done without mention-ing the word (or the idea) group, but students can verify a result that is probably not widely known:

The set of Pythagorean triples forms an infinite commutative group under the operation ⊗ as defined in (2).

PYTHAGOREAN TRIPLES, TRIGONOMETRY, AND COMPLEX NUMBERSEvery point (a, b) on the unit circle can be named either by trigonometric functions or by a complex number. If q is the central reference angle from the positive x-axis, then the coordinates of (a, b) are (cosq, sinq). The same point can also be named as the complex number a + bi (see fig. 4a).

These names for rational points relate directly to the operation we have defined for Pythagorean triples. In terms of the central angles, suppose that the point (a, b)/c has reference angle q and that point (d, e)/f has reference angle b. Then the refer-ence angle for product triple (a, b)/c ⊗ (d, e)/f is q + b. That is, when we take the product of two triples, we add their corresponding reference angles (see fig. 4b).

Students who are familiar with complex num-bers may be interested to know that the way we defined our opera tion for Pythagorean triples also relates directly to the complex number naming of points. Using the product of complex numbers, we have the following : (a + bi)(c + di) = (ac – bd) +(ad – bc)i.

Using the idea of complex numbers, if we make the identification of i with (0, 1)/1, then we have as its powers, as we should expect, i 2 = (0, 1)/1 ⊗(0, 1)/1 = (–1, 0)/1 = –1, i 3 = (0, –1)/1 = –i, and i 4 = (1, 0)/1 = 1. But these four triples, which cor-respond to the points where the coordinate axes meet the unit circle (and which also correspond to the four distinct powers of the complex number i: i, –1, –i, 1), are the only elements of finite order in the group. This means that given any Pythagorean triple A = (a, b)/c, there is no power n such that An = (1, 0)/1 = 1.

Fig. 4 The point in the complex plane can be represented in several different ways (a), and the product of two triples can

be obtained as shown (b).

(a) (b)

Fig. 3 The existence of an identity element and “reciprocals” suggests that ⊗ might

form a group with respect to Pythagorean triples.

Real Numbers, Multiplication Pythagorean Triples, ⊗

1 × a = a × 1 = a (1, 0)/1 ⊗ (a, b)/c = (a, b)/c ⊗ (1, 0)/1 = (a, b)/c

a × (1/a) = (1/a) × a = 1 (a, b)/c ⊗ (a, –b)/c = (a, –b)/c ⊗ (a, b)/c = (1, 0)/1




This assertion requires justification that is beyond the scope of this paper; however, it is a consequence of the fact that for rational points on the unit circle, other than the points where the coordinate axes meet the circle, all the reference angles for rational points are profoundly irratio-nal. In other words, except for the angles 0°, 90°, 180°, 270°, and so forth, either the sine and cosine (or both) of any angle that is a whole number of degrees is irrational and, hence, cannot be the refer-ence angle for a rational point on the unit circle or for any Pythagorean triple (Niven 1961, p. 68).

To illustrate, consider successive powers of the triple A = (3, 4)/5, with reference angle q approxi-mately (but not exactly) equal to 53.13°. Then A2 = (–7, 24)/25, with reference angle 2q ≈ 106.26°; A3 = (–117, 44)/125, with reference angle 3q ≈ 159.39°; A4 = (–527, –336)/625, with reference angle 4q ≈212.52°; and so on. By the seventh power, we have moved more than once around the unit circle. How-ever, no matter how many times we go around the circle, we can never end up at (1, 0)/1 because doing so would imply that some multiple of q is a whole number of degrees—which cannot happen.

It is not necessary to go as far as we have gone to benefit from this kind of study, nor do we need to stop here; there are other directions we could

explore. But from this beginning, it should be clear that Pythagorean triples can lead curious students into many delightful and rewarding byways.

BIBLIOGRAPHYHahn, Karl. 2011. “Rational Unit Circle Points/

Pythagorean Triples.” KCT: Karl’s Calc Tutor. www.karlscalculus.org/pythtrip.html.

Niven, Ivan. 1961. Numbers: Rational and Irrational (see section 5.2). Washington, DC: Mathematical Association of America.

Silverman, Joseph H. 2006. A Friendly Introduction to Number Theory. 3rd ed. Englewood Cliffs, NJ: Prentice-Hall.

LAWRENCE O. CANNON, [email protected], teaches mathematics at Utah State University in Logan and is a co-creator of the National Library

of Virtual Manipulatives (www.nlvm.usu.edu), a collection of teaching and learning tools for pre-K through college students and teachers. He has long been interested in relationships between geometry and number theory.

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