More on Pythagorean Triples (1/29) First, some True-False Questions (please click A for True and B for False): (9, 14, 17) is a Pythagorean triple. (9, 12, 15) is a Pythagorean triple. (9, 12, 15) is a primitive Pythagorean triple. (9, 40, 41) is a primitive Pythagorean triple. Every pair of numbers (s, t) with s > t > 0 give rise to a Pythagorean triple (s t, (s 2 – t 2 ) / 2, (s 2 + t 2 ) / 2). Every pair of numbers (s, t) with s > t > 0 give rise to a primitive Pythagorean triple (s t, (s 2 – t 2 ) / 2, (s 2 + t 2 ) / 2).

And some questions on “modulo” 23 and 17 are congruent modulo and 17 are congruent modulo and 17 are congruent modulo 11. Recall, by “a mod m” we mean the remainder upon division of a by m. (We always assume m > 0.) Note that this must be a number r for which 0  r < m. 63 mod 11 is 8 34 mod 7 is -1. If a is any integer, then there are exactly m possible values for a mod m.

A “Non-elementary” Approach to Pythagorean Triples An “elementary” solution to a number theory problem involves staying in the realm of integers (+ and –). When we venture outside that set, the solution becomes “non- elementary”. Many of the hardest problems (e.g., Fermat’s Last Theorem) seem to require non-elementary tools. In Chapter 3, we look at an easy but non-elementary approach to finding Pythagorean triples by going into the real plane (as in pre-Calc, Calc I and Calc II).

Outline of the Method Starting with a Pythagorean triple a 2 + b 2 = c 2, divide through by c 2, obtaining (a/c) 2 + (b/c) 2 = 1. Note that a/c and b/c are positive proper fractions, i.e., positive proper rational numbers, and they represent a point in the first quadrant on the unit circle in the plane (x 2 +y 2 = 1). Note that slope m of the line connecting that point to the point (-1, 0) is m = b 2 / (a 2 + c 2 ) (work this out!).In particular, m is itself positive, proper and rational. On the other hand, if we take any positive proper rational number m, we can solve the simultaneous equations of the circle and of the line with slope m connecting (-1, 0) to (x, y) (i.e., y = m(x + 1)) to get a rational point on the circle.

Conclusion We obtain, after some algebraic effort (see text), that (x, y) = ((1 – m 2 ) / (1 + m 2 ), 2m / (1 + m 2 )), which must be rational since m is rational. Specifically, if m = v / u (where v and u are positive whole numbers with v < u), then we get (again after algebra) (x, y) = ((u 2 – v 2 ) / (u 2 + v 2 ), 2uv / (u 2 + v 2 )). Multiplying through, we get (a, b, c) = (u 2 – v 2, 2uv, u 2 + v 2 ). For example, v = 1 and u = 2, give us the triple (3, 4, 5). Finally, connecting back to Chapter 2 (algebra again!), s = u + v and t = u – v. Hence we get primitive PT’s if u and v relatively prime and one is even, one odd.

Assignment for Friday Read Chapter 3 carefully. Do Exercises 3.1 a and b and 3.2. Had enough of Pythagoras for now? Me too. We will (after checking this material) move on.