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### Primitive Pythagorean Triples

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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/
```
Associated Topics:
High School Number Theory

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