Pythagorean Triples Pythagorean triple similar primitive


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1 Pythagorean Triples One of the most farreaching problems to appear in Diophantus Arithmetica was his Problem II8: To divide a given square into two squares. Namely, find integers x, y, z, so that x 2 + y 2 = z 2. Clearly, this problem can be restated in the form: Find a right triangle with integer side lengths, we call such a triple of (positive) numbers {x, y, z} a Pythagorean triple. Some easy observations will focus our attention here: note that if {x, y, z} is a Pythagorean triple, then so is {kx, ky, kz} for any integer k; so in fact, we can group Pythagorean triples into classes by saying that two sets of triples are similar if one is a rational multiple of the other. For instance, {6,8,10} and {9,12,15} are similar since the second is 3/2 times the first. In each similarity class of triples, then, there is a unique primitive one, namely, a triple {x, y, z} for which (x, y,z) = 1. Notice also, that the primitive triple in each similarity class of triples has the smallest sized values (why?). Indeed, not only are the three members of a primitive Pythagorean triple {x, y, z} relatively prime as a group, but they are relatively prime in pairs, for if, say, (x, y) = d > 1, then since x 2 + y 2 = z 2, d 2 z 2, which implies that d z. (Similar arguments
2 work for the other pairings within {x, y, z}.) It follows, for instance, that only one of the three numbers can be even. But we can say more, for if x and y were both odd, then they would both have to be congruent to either ±1 (mod 4). This implies that z 2 x 2 + y (mod 4), but this is not possible, for since z is even, 4 z 2, so z 2 0 (mod4). It follows that z must be odd, and one of x or y is even and the other is odd. We will assume thenceforth that x is odd and y = 2w is even. (Our textbook reverses this standard convention!) We can now write 4w 2 = y 2 = z 2 x 2 = (z + x)(z x) (*) w 2 = z + x z x 2 2 where, since x and z are odd, the factors on the right are known to be integers. Put d = z + x 2, z x. Then d z and d x, so d (z, x). 2 But this means d = 1. We then have from (*) that
3 there exist positive integers a and b so that z + x 2 = a2, z x 2 = b2 ; also, a and b must be relatively prime, and on adding the two expressions above, z = a 2 +b 2, x = a 2 b 2. Further, (*) implies that w 2 = a 2 b 2 y = 2w = 2ab, and since z is odd, a and b must have opposite parity. These arguments can be extended by the Theorem If x, y are relatively prime positive integers with x odd, then there exists a positive integer z for which x 2 + y 2 = z 2 if and only if there exist relatively prime integers a, b with a odd, for which x = a 2 b 2, y = 2ab, z = a 2 +b 2. Proof ( ) Given above. ( ) If {x, y, z} are given in terms of a, b as in the formulas, it is easy to check that x 2 + y 2 = z 2, and that x is odd and y even. We leave it as an exercise to check that (x, y) = 1. //
4 In the early 1600s, Fermat studied the work of Diophantus, and used the result of the previous theorem in formulating an extension of the Pythagorean triples problem to what would be his most famous unsupported claim: Fermat s Last Theorem (FLT) There do not exist positive integers x, y, z that satisfy for any exponent n > 2. // x n + y n = z n The theorem was finally proven true in 1994 by Andrew Wiles, who employed extremely sophisticated tools from algebraic geometry to accomplish it. Still, Fermat was successful in proving a special case of the theorem, for n = 4, and in the process he developed a technique called the method of descent, whereby, on assuming the existence of a single solution in positive integers, he derives the existence of a smaller solution in positive integers. The ability to generate smaller solutions contradicts the Well Ordering Principle, showing that there could not have existed a solution to begin with!...
5 Theorem There do not exist positive integers x, y, z that satisfy x 4 + y 4 = z 2. Proof An analysis similar to the one we used for Pythagorean triples shows that any solution {x, y, z} can be reduced to a primitive solution, that is, one for which (x, y) = 1. We assume as before that x and z are odd and y is even. In fact, we may assume by the WellOrdering Principle that {x, y, z} is a solution for which z is as small as possible. Therefore, { x 2, y 2,z} is a primitive Pythagorean triple, so we may conclude that there exist relatively prime integers a, b (with a odd), for which x 2 = a 2 b 2, y 2 = 2ab, z = a 2 +b 2. Then a 2 = b 2 + x 2, and (b, x) divides b and x, hence also a; but (a,b) = 1 implies that (b, x) = 1. That is, {x, b, a} is a primitive Pythagorean triple, so there exist relatively prime integers u, v (with u odd), for which x = u 2 v 2, b = 2uv, a = u 2 +v 2. Note that 1 2 y ( ) 2 = a 1 2 b is the product of the relatively prime integers a and 1 b; so there also 2
6 exist relatively prime integers s and t so that a = s 2 and 1 2 b = t 2. But then t 2 = uv and since (u,v) = 1, we must have relatively prime integers m and n such that u = m 2 and v = n 2. It follows that m 4 +n 4 = u 2 +v 2 = a = s 2 ; in other words, {m, n, s} is a second solution our the original equation. However, s a < z, so {m, n, s} is smaller in size than the solution {x, y, z}, contradicting the minimality of the choice of {x, y, z}. This contradiction shows that our original assumption, that there existed a solution in the first place, must have been false. // Corollary (FLT4) There do not exist positive integers x, y, z that satisfy x 4 + y 4 = z 4. Proof If {x, y, z} were a solution to x 4 + y 4 = z 4, then {x, y, z 2 } would give a solution to x 4 + y 4 = z 2. //
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