Pythagorean triples 427







has a square root, it must be the case that a+

ib equals (p+ iq)2for some complex number

p+ iq. We have a+ ib = (p+ iq)2= p2– q2+

2pqi, from which it follows that a= p2– q2

and b= 2pq. Also, c2= a2+ b2= (p2– q2)2+

(2pq)2= (p2+ q2)2yielding c= p2+ q2. One

now checks that the numbers pand qmust be

integers with the stated properties.

(To complete the final step, write the integers (p2+ q2)

+ (p2– q2) = 2p2and (p2+ q2) – (p2– q2) = 2q2each as

a product of prime numbers. Use this to show that if p

and qare not each themselves integers, then the triple

obtained is not primitive.)

Every Pythagorean triple (a,b,c) is a solution to the

equation . Thus, for example,

is a point on the

UNIT CIRCLE

x2+ y2= 1 with rational

coordinates, as are and . That

there are infinitely many Pythagorean triples shows

that the unit circle passes though infinitely many points

in the plane with rational x- and y-coordinates.

A study of

CONTINUED FRACTIONS

gives an alterna-

tive method for generating Pythagorean triples. If

(a,b,c) is a (primitive) triple with “legs” aand bdiffer-

ing by k, then (2a+ b+ 2c, a + 2b+ 2c, 2a+ 2b+ 3c) is

a new (primitive) triple with legs that also differ by k.

For example, from the triple (3,4,5), whose first two

terms differ by one, we obtain the new triples

(3,4,5) →(21,20,29) →(119,120,169) →…

From (5,12,13), whose first two terms differ by seven,

we obtain the new triples

(5,12,13) →(55,48,73) →(297,304,425) →…

This procedure will produce all the (primitive) triples

with legs that differ by a given amount k.

Although there are infinitely many integer solu-

tions to the equation a2+ b2= c2, Fermat’s last theorem

shows that there are no nonzero solutions to the com-

panion equations an+ bn= cnfor n> 2. The planar

curves given by xn+ yn= 1 thus never pass through

points with rational coordinates.