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a² + b? = (b+ 1)² (2) %3D 1. Suppose that a and b satisfy equation (2). Expand the right hand side of equation (2) and express a? in terms of b. Conclude that a is an odd integer. 2. Now let a = 2k + 1 be any odd integer. Show it's the smallest side of a right triangle satisfying (2), i.e. a² + b? = (b+ 1)² %3D by expressing b in terms of k. 3. Check that your expression for b in terms of k is relatively prime to a = 2k + 1.

a² + b? = (b+ 1)² (2) %3D 1. Suppose that a and b satisfy equation (2). Expand the right hand side of equation (2) and express a? in terms of b. Conclude that a is an odd integer. 2. Now let a = 2k + 1 be any odd integer. Show it's the smallest side of a right triangle satisfying (2), i.e. a² + b? = (b+ 1)² %3D by expressing b in terms of k. 3. Check that your expression for b in terms of k is relatively prime to a = 2k + 1.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Some Pythagorean Triples. Recall that a Pythagorean triple, or simply a triple, is an integral solution to the equation a² + b² = c? (1) For example, (3, 4, 5), (5, 12, 13), and (8, 15, 17) are triples. A triple is called primitive if (a, b) = 1. Notice that two of the triples above have the property that the hypotenuse is one greater than the second largest side. We'll find all primitive solutions to equation (1) for triangles of this form, i.e. all primitive solutions to: a² + b? = (b+ 1)² (2) 1. Suppose that a and b satisfy equation (2). Expand the right hand side of equation (2) and express a² in terms of b. Conclude that a is an odd integer. 2. Now let a = 2k + 1 be any odd integer. Show it's the smallest side of a right triangle satisfying (2), i.e. a² + b? = (b+ 1)² by expressing b in terms of k. 3. Check that your expression for b in terms of k is relatively prime to a = 2k + 1. 4. It follows that all primitive solutions to equation (2) are of the form (2k + 1, expression for b, expression for b+1) k = 1, 2, 3, 4, ... Write down the triples for k = 1, 2, 3, 4. 5. Write down all Pythagorean triples (not necessarily primitive) solving (2) with hypotenuse smaller than 50. All solutions to the special triples were known to Pythagoras. The general solution to (1) is in the Elements. Read what's posted on the website about the Plimpton Library tablet.
Transcribed Image Text:Some Pythagorean Triples. Recall that a Pythagorean triple, or simply a triple, is an integral solution to the equation a² + b² = c? (1) For example, (3, 4, 5), (5, 12, 13), and (8, 15, 17) are triples. A triple is called primitive if (a, b) = 1. Notice that two of the triples above have the property that the hypotenuse is one greater than the second largest side. We'll find all primitive solutions to equation (1) for triangles of this form, i.e. all primitive solutions to: a² + b? = (b+ 1)² (2) 1. Suppose that a and b satisfy equation (2). Expand the right hand side of equation (2) and express a² in terms of b. Conclude that a is an odd integer. 2. Now let a = 2k + 1 be any odd integer. Show it's the smallest side of a right triangle satisfying (2), i.e. a² + b? = (b+ 1)² by expressing b in terms of k. 3. Check that your expression for b in terms of k is relatively prime to a = 2k + 1. 4. It follows that all primitive solutions to equation (2) are of the form (2k + 1, expression for b, expression for b+1) k = 1, 2, 3, 4, ... Write down the triples for k = 1, 2, 3, 4. 5. Write down all Pythagorean triples (not necessarily primitive) solving (2) with hypotenuse smaller than 50. All solutions to the special triples were known to Pythagoras. The general solution to (1) is in the Elements. Read what's posted on the website about the Plimpton Library tablet.
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