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More on Pythagorean Triples (1/29)

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 .

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More on Pythagorean Triples (1/29)

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  1. 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, (s2 – t2) / 2, (s2 + t2) / 2). • Every pair of numbers (s, t) with s > t > 0 give rise to a primitive Pythagorean triple (s t, (s2 – t2) / 2, (s2 + t2) / 2).

  2. And some questions on “modulo” • 23 and 17 are congruent modulo 11. • 39 and 17 are congruent modulo 11. • -5 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.

  3. 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).

  4. Outline of the Method • Starting with a Pythagorean triple a2 + b2 = c2, divide through by c2, 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 (x2 +y2 = 1). • Note that slope m of the line connecting that point to the point (-1, 0) is m = b2 / (a2 + c2) (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.

  5. Conclusion • We obtain, after some algebraic effort (see text), that(x, y) = ((1 – m2) / (1 + m2), 2m / (1 + m2)), 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) = ((u2 – v2) / (u2 + v2), 2uv / (u2 + v2)). • Multiplying through, we get (a, b, c) = (u2 – v2, 2uv, u2+ v2). • 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.

  6. 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.

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