Associated Topics || Dr. Math Home || Search Dr. Math

### Finding Pythagorean Triplets

```Date: 02/02/2006 at 17:07:24
From: Faizan
Subject: Pythagorean triplets

Without using the standard a = n^2 - m^2, b = 2nm, c = n^2 + m^2, how
can you work out primitive Pythagorean triplets?

For example, you can work out an equation for cases when c and b
differ by 1 using (2n+1)^2 + b^2 = (b+1)^2.

I cannot work it out when c and b differ by 2. How would you do it?

```

```
Date: 02/04/2006 at 14:21:28
From: Doctor Ricky
Subject: Re: Pythagorean triplets

Hey Faizan,

Thanks for writing Dr. Math!

We should be able to find a Pythagorean triplet where the hypotenuse
and a leg differ by 2.  Think of the most basic triplet - 3, 4, 5 -
and note that 5 - 3 = 2.

Assume we are trying to find a triplet where the hypotenuse and a
leg differ by 1.

You had the right idea with your work, but let's look more closely at
one leg and the hypotenuse differ by 1, or in other words:

b (leg) = n
c (hypotenuse) = n + 1

So, using the Pythagorean theorem and plugging in these assumptions,
we get:

a^2 + (n^2) = (n+1)^2.  Simplifying, we get:

a^2 + n^2 = n^2 + 2n + 1, and again we simplify to get:

a^2 = 2n + 1

So in other words,

a = sqrt(2n + 1)

Therefore, for any integer value of n that makes sqrt(2n + 1) an
integer, we have a Pythagorean triplet where:

a = sqrt(2n + 1)
b = n
c = n + 1

For example, if n = 4, we get the triplet 3, 4, 5.

If n = 12, we get the triplet 5, 12, 13.

We can keep going with this for all integer values of a.

So now let's attack the situation of when a leg and the hypotenuse
differ by 3 and hopefully you will be able to see how to prove it for
a difference of 2.

So we'll assume:

b (leg) = n
c (hypotenuse) = n + 3

Using the Pythagorean theorem, we get:

a^2 + (n^2) = (n+3)^2

Simplifying, we get:

a^2 + n^2 = n^2 + 6n + 9

Simplifying again, we get:

a^2 = 6n + 9

and

a = sqrt(6n + 9)

Therefore, for any integer value of n that makes sqrt(6n + 9) an
integer, we have a Pythagorean triplet where:

a = sqrt(6n + 9)

b = n

c = n + 3

For example, if we choose n = 12, we get the triplet: 9, 12, 15.

If n = 36, we get the triplet: 15, 36, 39.

Finding values for n that make this work won't be very easy as we
continue to increase the difference between a leg and the hypotenuse,
which is why the general formulas you mentioned in your questions are
great guides for creating Pythagorean triplets.

If you have any more questions or if you were confused about any of

- Doctor Ricky, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Number Theory

Search the Dr. Math Library:

 Find items containing (put spaces between keywords):   Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words

Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search