Theorem 3 If the product of two relatively prime positive integers is a square, then both integers are squares. == We will prove this later. As an example, 12 × 75 = 900 = 302. Even though neither 12 nor 75 is a square, this does not contradict the theorem because they are not relatively prime. What the theorem says is the only way to write 900 as the product of two rela- tively prime positive integers is to use squares. We can certainly do this in many ways: 900 22 x 152 = 52 x 62, for example. == Page 5 If we believe Theorem 3, then uv = y₁, with u and v relatively prime, so u = p² and v = q² for some integers p and q, which are relatively prime. Now x = uv = p²-q², z=u+v= p²+q², and y² = 4uv = 4p²q² so y = 2pq. Thus, we again have (x, y, z) = (p²-q², 2pq, p²+q²). This may seem much shorter, but it really isn't: With the geometric approach, we proved things about primitive triples that we just used here without reproving them. Moreover, this proof is incomplete because we have not yet proven Theorem 3. Pythagorean triples from an algebraic approach Everything above was based on the geometric idea that rational solutions of the equation x2 y21 can be found by considering lines of rational slope passing through one fixed rational point on the curve (we used (-1, 0) because it was the most convenient point to use). We now look at things from a purely algebraic point of view that goes back at least to Euclid. Again, we restrict our attention to primitive Pythagorean triples. Suppose that (x, y, z) is a primitive Pythagorean triple and that y is even. Then x² + y² = z² can be rearranged: y² = 2²x²= (z+x)(z - x) = 4 +x 2 2 Letting u = 2+x 2 v = 2-x 2 " y = 2y1, we have 4y² = 4uv, or y uv. We can show that u and v are relatively prime: if d is a divisor of both u and v, then d is a divisor of their sum, u+v and their difference, u - v. But u+v=z, u- v = x, so d would have to be a divisor of x and z. Since x and z are relatively prime, d = 1.
Theorem 3 If the product of two relatively prime positive integers is a square, then both integers are squares. == We will prove this later. As an example, 12 × 75 = 900 = 302. Even though neither 12 nor 75 is a square, this does not contradict the theorem because they are not relatively prime. What the theorem says is the only way to write 900 as the product of two rela- tively prime positive integers is to use squares. We can certainly do this in many ways: 900 22 x 152 = 52 x 62, for example. == Page 5 If we believe Theorem 3, then uv = y₁, with u and v relatively prime, so u = p² and v = q² for some integers p and q, which are relatively prime. Now x = uv = p²-q², z=u+v= p²+q², and y² = 4uv = 4p²q² so y = 2pq. Thus, we again have (x, y, z) = (p²-q², 2pq, p²+q²). This may seem much shorter, but it really isn't: With the geometric approach, we proved things about primitive triples that we just used here without reproving them. Moreover, this proof is incomplete because we have not yet proven Theorem 3. Pythagorean triples from an algebraic approach Everything above was based on the geometric idea that rational solutions of the equation x2 y21 can be found by considering lines of rational slope passing through one fixed rational point on the curve (we used (-1, 0) because it was the most convenient point to use). We now look at things from a purely algebraic point of view that goes back at least to Euclid. Again, we restrict our attention to primitive Pythagorean triples. Suppose that (x, y, z) is a primitive Pythagorean triple and that y is even. Then x² + y² = z² can be rearranged: y² = 2²x²= (z+x)(z - x) = 4 +x 2 2 Letting u = 2+x 2 v = 2-x 2 " y = 2y1, we have 4y² = 4uv, or y uv. We can show that u and v are relatively prime: if d is a divisor of both u and v, then d is a divisor of their sum, u+v and their difference, u - v. But u+v=z, u- v = x, so d would have to be a divisor of x and z. Since x and z are relatively prime, d = 1.
Algebra: Structure And Method, Book 1
(REV)00th Edition
ISBN:9780395977224
Author:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole
Publisher:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole
Chapter9: Systems Of Linear Equations
Section9.1: The Graphing Method
Problem 20WE
Related questions
Question
Please prove this method for Pythagorean Triples
Transcribed Image Text:Theorem 3 If the product of two relatively prime positive integers is a square, then both
integers are squares.
==
We will prove this later. As an example, 12 × 75 = 900 = 302. Even though neither
12 nor 75 is a square, this does not contradict the theorem because they are not relatively
prime. What the theorem says is the only way to write 900 as the product of two rela-
tively prime positive integers is to use squares. We can certainly do this in many ways:
900 22 x 152 = 52 x 62, for example.
==
Page 5
If we believe Theorem 3, then uv = y₁, with u and v relatively prime, so u = p² and v = q²
for some integers p and q, which are relatively prime. Now x = uv = p²-q², z=u+v=
p²+q², and y² = 4uv = 4p²q² so y = 2pq. Thus, we again have (x, y, z) = (p²-q², 2pq, p²+q²).
This may seem much shorter, but it really isn't: With the geometric approach, we proved
things about primitive triples that we just used here without reproving them. Moreover, this
proof is incomplete because we have not yet proven Theorem 3.
Transcribed Image Text:Pythagorean triples from an algebraic approach
Everything above was based on the geometric idea that rational solutions of the equation
x2 y21 can be found by considering lines of rational slope passing through one fixed
rational point on the curve (we used (-1, 0) because it was the most convenient point to
use). We now look at things from a purely algebraic point of view that goes back at least
to Euclid. Again, we restrict our attention to primitive Pythagorean triples. Suppose that
(x, y, z) is a primitive Pythagorean triple and that y is even. Then x² + y² = z² can be
rearranged:
y² = 2²x²= (z+x)(z - x) = 4
+x
2
2
Letting u =
2+x
2
v =
2-x
2
"
y = 2y1, we have 4y² = 4uv, or y
uv. We can show that u
and v are relatively prime: if d is a divisor of both u and v, then d is a divisor of their sum,
u+v and their difference, u
-
v. But u+v=z, u- v = x, so d would have to be a divisor
of x and z. Since x and z are relatively prime, d = 1.
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