Introduction Proof by unique factorization in Z Proof with Gaussian integers Proof by geometry Applications. Pythagorean Triples

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1 Pythagorean Triples Keith Conrad University of Connecticut August 4, 008

2 Introduction We seek positive integers a, b, and c such that a + b = c. Plimpton 3 Babylonian table of Pythagorean triples (1800 BC). Eleventh row is (3, 4, 5).

3 Reduction Step a + b = c a b c Examples of Pythagorean Triples If d a and d b then d c, so d c. Similarly, if d a and d c then d b, and if d b and d c then d a. Therefore (a, b) = (a, b, c). Writing a = da, b = db, and c = dc, a + b = c = a + b = c. From now on we focus on primitive triples: (a, b) = 1.

4 Classification a + b = c, (a, b) = 1. Certainly a and b are not both even. Also they are not both odd: otherwise, c = a + b mod 4, which is impossible. So one of a or b is even and the other is odd. Then c = a + b is odd, so c is odd. Our convention: take b even. Theorem The primitive Pythagorean triples (a, b, c) where b is even are given by a = u v, b = uv, c = u + v, where u > v > 0, (u, v) = 1, and u v mod. For u and v in Z +, need u > v so that a > 0. The conditions (u, v) = 1 and u v mod are forced by primitivity.

5 Classification a = u v, b = uv, c = u + v, u > v > 0, (u, v) = 1, u v mod u v a b c Which u and v give the triple (a, b, c) = (19081, 7830, 05769)?

6 Classification Can solve for u and v : a = u v, b = uv, c = u + v, u > v > 0, (u, v) = 1, u v mod u = a + c, v = c a. For (a, b, c) = (19081, 7830, 05769), a + c So u = 445 and v = 88. = = 445, c a = 7744 = 88.

7 Primitive Triples of Nonzero Integers Theorem Triples (a, b, c) of nonzero integers where a + b = c, (a, b) = 1, b is even, and c > 0, are given by a = u v, b = uv, c = u + v, where u, v Z {0}, (u, v) = 1, and u v mod. Why? Suppose a > 0, b > 0, and c > 0, so the classification says a = u v, b = uv, c = u + v, u > v > 0, (u, v) = 1, u v mod. In terms of this u and v, how do the parametric formulas apply to (a, b, c)? To ( a, b, c)? To ( a, b, c)? To (a, b, c)?

8 Outline Classify primitive Pythagorean triples by unique factorization in Z. Classify primitive Pythagorean triples by unique factorization in Z[i]. Classify primitive Pythagorean triples by analytic geometry. See additional use of each method of proof.

9 First proof: unique factorization in Z, I a + b = c = b = c a = (c + a)(c a). Both c + a and c a are positive and even. What s their gcd? If d (c + a) and d (c a) then d c and d a, so d because (a, c) = 1. Since c + a and c a are even, (c + a, c a) =. So with factors relatively prime. Theorem ( ) b = c + a c a, If xy = in Z + and (x, y) = 1 then x = and y =. So (c + a)/ = u and (c a)/ = v with u, v Z +. Solving, c = u + v and a = u v ; b = (c + a)(c a) = (uv).

10 First proof: unique factorization in Z, II Let s try a different subtraction: a + b = c = a = c b = (c + b)(c b). Both c + b and c b are positive and odd. What s their gcd? If d (c + b) and d (c b) then d c and d b, so d. Since c + b and c b are odd, (c + b, c b) = 1. Then c + b = k and c b = l where k and l are in Z + and odd. Must have (k, l) = 1. Adding and subtracting, c = k + l, b = k l. Then a = (c + b)(c b) = k l, so a = kl. We expect that a = u v, and so on. Since u v = (u + v)(u v), try to get k = u + v and l = u v. Define u = k + l, v = k l.

11 Another parametrization Theorem The primitive Pythagorean triples (a, b, c) where b is even are given by a = kl, b = k l, c = k + l, where k > l > 0, (k, l) = 1, and k and l are both odd. k l a b c

12 Second proof: unique factorization in Z[i] Write c = a + b = (a + bi)(a bi). Suppose δ (a + bi) and δ (a bi) in Z[i]. Taking the norm, N(δ) divides a + b = c, which is odd, so N(δ) is odd. Also δ divides a and bi in Z[i], so N(δ) 4a and N(δ) 4b in Z. Since (a, b) = 1, N(δ) 4, so N(δ) = 1. Thus δ = ±1 or ±i. Theorem If αβ = in Z[i] and (α, β) = 1 then α and β are squares up to unit multiple. Either a + bi = (u + vi) or a + bi = i(u + vi). First case: a + bi = u v + uvi a = u v and b = uv. Second case: a + bi = uv + (u v )i a = uv, but a odd! Choose sign on u so u > 0. Then v > 0 and c = a + b = N((u + vi) ) = N(u + vi) = (u + v ).

13 Pythagorean triples from Gaussian integers Pythagorean triples arise from squaring Gaussian integers: (u + vi) = a + bi Norm = (u + v ) = a + b. α α Triple 1 + i 3 + 4i (3, 4, 5) + 3i 5 + 1i (5, 1, 13) 7 + 4i i (33, 56, 65) 7 + 5i i (4, 70, 74) i i (91, 60, 109) From 7 + 5i get nonprimitive (4, 70, 74) = (1, 35, 37). In Z[i], and 7 + 5i 1 + i = (7 + 5i)(1 i) (1 + i)(1 i) = 1 i (6 i) = 35 1i, which gives the primitive triple (35, 1, 37). = 6 i

14 Third proof: analytic geometry a + b = c = r + s = 1, r = a c, s = b c. The line through ( 1, 0) and (r, s) is y = m(x + 1), where m = s. If r, s Q then m Q. r + 1

15 Third proof: analytic geometry Conversely, for m Q where does y = m(x + 1) meet the circle? so 1 = x + y = x + (m(x + 1)) = (m + 1)x + m x + m, ( ) 0 = x + m m + 1 x + m 1 m = (x + 1) x + m m. + 1

16 Third proof: analytic geometry ( ) 0 = (x + 1) x + m 1 m + 1 The second point of intersection is at (r, s), where and We have a correspondence r = m 1 m + 1 = 1 m 1 + m s = m(r + 1) = m 1 + m. {rational points (r, s) ( 1, 0) on x + y = 1} m Q given by (r, s) m = s r + 1 ; 1 m m r = 1 + m, s = m 1 + m. Slope m gives point in first quadrant when 0 < m < 1.

17 From triples to slopes ( a (a, b, c) c, b ) m = b/c c a/c + 1 = ( ) 1 m m 1 + m, m 1 + m. b a + c, a b c m 1/ /3 3/4 1/4 4/5 9/14 It looks like m = v/u in our earlier notation: (u, v) (,1) (3,) (4,3) (4,1) (5,4) (14,9) a b c

18 From triples to slopes a = u v, b = uv, c = u + v, Earlier, we said we can solve for u and v : u = a + c, v = c a. For (a, b, c) = (19081, 7830, 05769), earlier we found a + c = = 445, c a = 7744 = 88, so u = 445 and v = 88. Now geometry makes us notice that so b a + c = uv u = v u, b a + c = = = v u.

19 From slopes to triples m 1/ 1/3 /3 1/4 3/4 (1 m )/(1 + m ) 3/5 4/5 5/13 15/17 7/5 m/(1 + m ) 4/5 3/5 1/13 8/17 4/5 m 1/5 /5 1/17 19/101 (1 m )/(1 + m ) 1/13 1/9 145/ /581 m/(1 + m ) 5/13 0/9 408/ /581 If m (x, y) then 1 m (y, x). 1 + m

20 Application 1: Polynomial Pythagorean triples Consider polynomials f (x), g(x), h(x) satisfying f (x) + g(x) = h(x) and all nonzero. Call the triple primitive if (f (x), g(x)) = 1. Theorem The primitive Pythagorean triples in R[x] are given by f (x) = c(u(x) v(x) ), g(x) = cu(x)v(x), h(x) = c(u(x) + v(x) ), where c R {0} and (u(x), v(x)) = 1. There is a proof by unique factorization in R[x], as in Z. Even/odd considerations drop out since is a unit as a polynomial.

21 Application : a + b = c Suppose a + b = c in Z + and (a, b) = 1. Then a is odd: if a is even then b is odd so c mod 4: NO. From a odd, also c odd, so b = c a mod 8, so b is even. Theorem The solutions (a, b, c) to a + b = c in Z + with (a, b) = 1 are given by a = u v, b = uv, c = u + v, where u, v > 0, (u, v) = 1, and u is odd. u v a b c

22 Application 3: a + b = c 3 In Z[i], (u + vi) 3 = u 3 + 3u (vi) + 3u(vi) + (vi) 3 Take norms of both sides: = (u 3 3uv ) + (3u v v 3 )i. (u + v ) 3 = (u 3 3uv ) + (3u v v 3 ). u v a = u 3 3uv b = 3u v v c = u + v Exercise: All integral solutions to a + b = c 3 with (a, b) = 1 arise in this way with (u, v) = 1 and u v mod.

23 Application 4: Rational parametrizations of other conics x + y =

24 Rational parametrizations of other conics Theorem The rational solutions to x + y = have the form for m Q, and (1, 1). x = m m 1 1 m m 1 + m, y = 1 + m m 1 3/ 5/7 8/5 1 x 1 7/13 3/37 41/89 119/145 y 1 17/13 47/37 119/89 167/ = 13, = 37, = 89.

25 Rational parametrizations of other conics x dy = 1

26 Rational parametrizations of other conics Theorem The rational solutions to x dy = 1 have the form with m Q, and ( 1, 0). x = 1 + dm 1 dm, y = m 1 dm m 1/ 1/3 /3 8/9 0 x 3 11/ /47 801/799 y 6/ /47 40/799 Solutions to x y = 1 There s no simple formula for integral solutions to x dy = 1!

27 Application 5: Factoring quadratics In Z[x], x + 4x + 3 = (x + 1)(x + 3), x + 4x 3 irreducible. but x + 5x + 6 = (x + )(x + 3), x + 5x 6 = (x 1)(x + 6). Question: When do x + mx + n and x + mx n factor in Z[x]? Here m and n are nonzero. If x + mx + n = (x r 1 )(x r ) then x mx + n = (x + r 1 )(x + r ). So we may assume m > 0. May take n > 0 too.

28 Factoring quadratics Roots of x + mx ± n are m ± m ± 4n, which are integers exactly when m ± 4n =, since m ± 4n m mod. So we can factor x + mx + n and x + mx n if and only if m 4n = d, m + 4n = e, d and e Z. Then d + e = m, so d e mod. Solving, m = d + e ( ) e + d ( ) e d = +. Thus we have a Pythagorean triple (without specified even term) ( e d, e + d ) e d, m, < e + d < m. Exercise: This triple is primitive if and only if (m, n) = 1.

29 Factoring quadratics Theorem (J. L. Poet, D. L. Vestal, 005) There is a one-to-one correspondence between Pythag. triples (a, b, c) with a < b < c and reducible pairs x + mx ± n with m, n > 0, given by (a, b, c) x + cx ± ab ( e d, x + mx ± n, e + d ), m, with m 4n = d and m + 4n = e. a b c m n x + mx + n x + mx n (x + )(x + 3) (x 1)(x + 6) (x + 3)(x + 10) (x )(x + 15) (x + 5)(x + 1) (x 3)(x + 0) Exercise. Factor x + (u + v )x ± uv(u v ) in Z[x].

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