Primitive Pythagorean Triples
Date: 02/23/98 at 12:56:13 From: Cyndy Subject: Primitive Pythagorean Triples A primitive Pythagorean triple is a triple of numbers (a, b, c) such that a, b, and c have no common factors and satisfy a^2+b^2 = c^2. By examining the list of primitive Pythagorean triples, make a guess about when a, b, or c is a multiple of 5. Try to show that your guess is right. Here are some primitive Pythagorean triples I found: (3,4,5), (5,12,13), (8,15,17), (7,24,25), (20,21,29), (9,40,41), (12,35,37), (11,60,61), (28,45,53), (33,56,65), (16,63,65), .... From the list I know one has to be a multiple of 5. But I'm stuck here because I can't find a pattern for when a, b, or c is a multiple of 5. Another fact about Pythagorean triples is that one of a and b is odd and the other even, and that c is always odd. I don't know if the fact has anything to do with the problem.
Date: 02/23/98 at 13:06:02 From: Doctor Rob Subject: Re: Primitive Pythagorean Triples What do you know about squares and their remainders when divided by 5? Are you aware that the remainders are always 0, 1, or 4? The remainder of a^2 + b^2 will have to be the sum of two of these numbers. The remainder of c^2 will also have to be one of these numbers. How can that happen? Then recall that remainder 0 means divisibility by 5, and if x^2 is divisible by 5, so is x. The odd-even property you observed has only this to do with this problem. Squares divided by 4 must leave remainder 0 or 1. The remainder of a^2 + b^2 will have to be the sum of two of these numbers. The remainder of c^2 will also have to be one of these numbers. How can that happen? Then recall that remainder 0 means divisibility by 4, and if x^2 is divisible by 4, x is even. You can use the same method to solve both problems. -Doctor Rob, The Math Forum Check out our web site http://mathforum.org/dr.math/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.