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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
the steps that lead to your answer.  We would begin by assuming that
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 
this, please let me know!

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

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