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Theorem 5.1. Every Pythagorean Triple has the form: (n(p2-q²), 2npq, n(p² + q²)) for some integers p and q with p>q> 0 and some positive integer n. Proof: Suppose (a, b, c) is a Pythagorean Triple. Suppose n = ged(a, b). Then n❘a and n|b, so a=nx, b=ny; x, y € Z+, gcd(x, y) = 1. ⇒ a² + b² = (nx)² + (ny)² = n²(x² + y²) = c², so n²|c² 8 If n² then n|c, so cnz; z Є Z By assumption that n = gcd(a,b), we know that gcd(x, y) = 1. Since a²+b² 2 = ⇒ (nx)² + (ny)² = (nz)² ⇒n=0 or r² + y² = 2² Deduce (x, y, z) is a primitive Pythagorean Triple and (a, b, c) = n(x, y, z). Thus, every Pythagorean Triple has the form (n(p² - q²), n(2pq), n(p² + q²)); n = Z+, 0 > 9>0. Therefore, (a, b, c) = n(x, y, z).

Theorem 5.1. Every Pythagorean Triple has the form: (n(p2-q²), 2npq, n(p² + q²)) for some integers p and q with p>q> 0 and some positive integer n. Proof: Suppose (a, b, c) is a Pythagorean Triple. Suppose n = ged(a, b). Then n❘a and n|b, so a=nx, b=ny; x, y € Z+, gcd(x, y) = 1. ⇒ a² + b² = (nx)² + (ny)² = n²(x² + y²) = c², so n²|c² 8 If n² then n|c, so cnz; z Є Z By assumption that n = gcd(a,b), we know that gcd(x, y) = 1. Since a²+b² 2 = ⇒ (nx)² + (ny)² = (nz)² ⇒n=0 or r² + y² = 2² Deduce (x, y, z) is a primitive Pythagorean Triple and (a, b, c) = n(x, y, z). Thus, every Pythagorean Triple has the form (n(p² - q²), n(2pq), n(p² + q²)); n = Z+, 0 > 9>0. Therefore, (a, b, c) = n(x, y, z).

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter1: Fundamental Concepts Of Algebra
Section1.1: Real Numbers
Problem 39E
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Please make corrections to this proof 

Theorem 5.1. Every Pythagorean Triple has the form: (n(p2-q²), 2npq, n(p² + q²)) for some integers p and q with p>q> 0 and some positive integer n. Proof: Suppose (a, b, c) is a Pythagorean Triple. Suppose n = ged(a, b). Then n❘a and n|b, so a=nx, b=ny; x, y € Z+, gcd(x, y) = 1. ⇒ a² + b² = (nx)² + (ny)² = n²(x² + y²) = c², so n²|c² 8 If n² then n|c, so cnz; z Є Z By assumption that n = gcd(a,b), we know that gcd(x, y) = 1. Since a²+b² 2 = ⇒ (nx)² + (ny)² = (nz)² ⇒n=0 or r² + y² = 2² Deduce (x, y, z) is a primitive Pythagorean Triple and (a, b, c) = n(x, y, z). Thus, every Pythagorean Triple has the form (n(p² - q²), n(2pq), n(p² + q²)); n = Z+, 0 > 9>0. Therefore, (a, b, c) = n(x, y, z).
Transcribed Image Text:Theorem 5.1. Every Pythagorean Triple has the form: (n(p2-q²), 2npq, n(p² + q²)) for some integers p and q with p>q> 0 and some positive integer n. Proof: Suppose (a, b, c) is a Pythagorean Triple. Suppose n = ged(a, b). Then n❘a and n|b, so a=nx, b=ny; x, y € Z+, gcd(x, y) = 1. ⇒ a² + b² = (nx)² + (ny)² = n²(x² + y²) = c², so n²|c² 8 If n² then n|c, so cnz; z Є Z By assumption that n = gcd(a,b), we know that gcd(x, y) = 1. Since a²+b² 2 = ⇒ (nx)² + (ny)² = (nz)² ⇒n=0 or r² + y² = 2² Deduce (x, y, z) is a primitive Pythagorean Triple and (a, b, c) = n(x, y, z). Thus, every Pythagorean Triple has the form (n(p² - q²), n(2pq), n(p² + q²)); n = Z+, 0 > 9>0. Therefore, (a, b, c) = n(x, y, z).
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