HISTORY of MATHEMATICS 1 Pythagorean Triples

HISTORY OF MATHEMATICS Summer 2010 Pythagoras and Fermat

1 Pythagorean Triples

A Pythagorean triple is a triple of positive integers (a, b, c) such that a2 +b2 = c2. The simplest such triple is (3, 4, 5).

Here are some facts about these triples. Since this is a math course, proofs are provided. I’ll abbreviate Pythagorean triple to pt.

1. If (a, b, c) is a pt, then so is (ma, mb, mc) for any positive integer m (mul- tiples of pt’s are pt’s). In fact,

(ma)2 + (mb)2 = m2a2 + m2b2 = m2(a2 + b2) = m2c2 = (mc)2.

As an example, from the triple (3, 4, 5) we get the triples (6, 8, 10), (9, 12, 15), (12, 16, 20), etc.

2. If (a, b, c) is a pt and if the positive integer d divides a, b, and c, then (a/d, b/d, c/d) is a pt. In fact, ( ) ( ) ( ) a 2 b 2 a2 b2 a2 + b2 c62 c 2 + = + = = = . d d d2 d2 d2 d2 d

Because of these two ﬁrst properties, the basic pt’s are those in which a, b, c do not have a common divisor other than 1. We make the following deﬁnition: A pt is said to be primitive, or to be a primitive Pythagorean triple if the greatest common divisor of a, b, c is one. Every Pythagorean triple is either primitive or a multiple of a primitive one. Let’s continue with the properties 4. If (a, b, c) is a ppt, then c is odd and a, b have opposite parity; that is, one of a, b is odd, the other one even. In fact, suppose (a, b, c) is a ppt and (to get a contradiction) assume c is even. Then we can write c = 2k for some integer k > 0, hence c2 = 4k2 is divisible by 4. Because the triple is primitive, once c is even, a and b cannot both be even; otherwise 2 would be a common divisor of a, b, c. But because the square of an even number is even, of an odd number is odd and a2 + b2 is even, both a, b must be odd. (Just in case this isn’t crystal clear, if one were odd, the other one even, the sum of their squares would be odd). Then we can write a = 2m + 1, b = 2n + 1 for some non-negative integers m, n. Then

c2 = a2+b2 = (2m+1)2+(2n+1)2 = 4m2+4m+1+4n2+4n+1 = 4(m2+n2+m+n)+2

and this contradicts that c2 is a multiple of 4; the number on the far right leaves a remainder of 2 when dividing by 4. Having established that c, hence also c2, is odd, the equation a2 + b2 = even number cannot hold except if a, b have opposite parity.

Since one of a, b is even, the other one odd,in a ppt, we can agree to always write the odd one ﬁrst, the even one second. That is always assume that a is odd and b is even. We then have the following theorem that characterizes all ppt’s. In this theorem we will use a few basic number theory facts. Namely: 1. Every positive integer n can be written uniquely in the form

e1 ··· er n = p1 pr

where e1, . . . , er are nonnegative integers and p1 < p2 < ··· < pr are e1 primes. (Of course r = 1 is possible, in this case n = p1 is the power of a prime.) 2. If a, b are integers, we write a|b to mean a divides b; that is b = ka for some integer k. 3. If p is prime, then p|ab if and only if p|a or p|b. Theorem 1 The triple of positive integers (a, b, c) is a ppt if and only if there exist positive integers m, n such that m, n have opposite parity (one even, one odd) and no common divisor other than 1, m > n, and such that a = m2 − n2, b = 2mn. In this case, c = m2 + n2. Proof. The proof that if a = m2 − n2, b = 2mn then taking c = m2 + n2 one has a pt (a, b, c) will be left as an exercise. This is always true, even if m, n have the same parity, or have a common divisor. That a pt so determined is primitive if and only if m, n have opposite parities and no common divisor other than 1 is part of this proof. Assume (a, b, c) is a ppt; remember we assume that b is the even one, a the odd one. We can write b = 2k for some integer k > 0 and then we have 4k2 = b2 = c2 − a2 = (c − a)(c + a). Now comes the difficult part, which isn’t really so difficult. The greatest common divisor of c − a, c + a is 2. In fact, if d divides both, it also divides their sum, e1 ··· er which is 2c, and their difference 2a. Suppose p1 pr is the decomposition of d into powers of primes, p1 < ··· < pr. If in this list there is any prime p other than 2, then p divides both 2a and 2c; because they are odd, p has to divide a and c, a contradiction since a, c have no common divisor other than 1. For a similar reason, d cannot be divisible by a power of 2 higher than 1, since that would force a, c to be even. On the other hand, c a are even, and the conclusion is that the greatest common divisor of c − a, c + a is 2. We now can write c − a c + a k2 = · 2 2 − e1 ··· er and both (c a)/2, (c + a)/2 are positive integers. Suppose k = p1 pr is the 2e1 ··· 2er 2 2 prime power decomposition of k; then k = p1 pr is that of k . Since k is odd, p1 > 2. Let p be any of the primes in the decomposition of k and let e be its exponent. Then c − a c + a c − a c + a p2e| · , thus p| · . 2 2 2 2 c−a c+a If p divides both 2 , 2 it must divide their sum and difference, impossible since a, c have no common divisor other than 1. Thus p and hence also pe divides c−a c+a precisely one of 2 , 2 . This allows us to partition the prime factors of k into c−a c+a two groups, those that divide 2 , and those that divide 2 . If we denote by ek 2ek c+a m the product of all the prime powers pk such that pk divides 2 , and by n ek 2ek c−a the product of all the prime powers pk such that pk divides 2 , we get c − a c + a b = 2k = 2mn, = n2, = m2. 2 2

2 Then also c − a c + a c = + = m2 + n2 2 2 and ﬁnally a2 = c2−b2 = (m2+n2)2−4m2n2 = m4+n4+2m2n2−4m2n2 = m4+n4−2m2n2 = (m2−n2)2 implying a = m2 − n2 since m > n (why is m > n?). Finally, if m, n have a common factor other than 1, they have a prime common factro, such a prime common factor would also be a common factor of a, b, c. Because a, c are odd, m, n must have opposite parity. We now can easily generate all ppt’s. The simplest (smallest) choice for m, n is m = 2, n = 1. This gives the triple

(22 − 1, 2 · 1 · 2, 22 + 1) = (3, 4, 5).

The next smallest choice for m is m = 3. Then we must have n = 2 (opposite parity); we get the triple (5, 12, 13). With m = 4 we can have n = 1 or n = 3. This gives the triples (15, 8, 17) and (7, 24, 25); etc. Fermat’s proof that the equation x4 + y4 = z2 has no solutions in positive integers.

The idea is to proceed by contradiction, assuming there is a solution. Once on has one, one shows there is another one that is smaller in some sense, say with a smaller z. Since we are dealing with positive integers, this is a contradiction; once z is 1, it cannot get smaller. This method used to be called “inﬁnite descent.” Proof. Assume x, y, z are positive integers, satisfying x4 + y4 = z2. To continue.