Generating Pythagorean Triples

Date: 03/17/2001 at 10:57:37
From: Samantha
Subject: Pythagorean Triplets

Dear Dr. Math,

I am doing a math extra credit project and I need to figure out the 
sixteen primitive Pythagorean triples. I have 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; . 

The second part of the question is to figure out how many Pythagorean 
triples there are such that the numbers of the triple lie between 
1 and 100. 

Please help.
Sam

Date: 03/17/2001 at 11:42:20
From: Doctor Paul
Subject: Re: Pythagorean Triplets

Let's talk about how to generate primitive Pythagorean triples.

We'll start with some of the ones you gave:

a   3   5   7   9  11  13
b   4  12  24  40  60   ?
c   5  13  25  41  61   ?

Do you see a pattern?  The first row is just the odd numbers: 3,5,7,9, 
etc...

Is there a relation between the second row and the third row?  Looks 
like the third row is just the second row plus one.

If there is a way to generate the number 'b' when given 'a', we will 
be done because once we know 'b' we can get 'c' just by adding one.

Perhaps it will help if we add another row:

a^2  9  25  49  81  121  169
a    3   5   7   9   11   13
b    4  12  24  40   60    ?
c    5  13  25  41   61    ?

Now do you see a relation between 'a' and 'b'?

It looks as if a^2 = 2b+1 or, solving for b, 

   b = (a^2 - 1)/2

So if a = 13, b = (169 - 1)/2 = 168/2 = 84.  It follows then that 
c = 84+1 = 85.

So the claim is that 13,84,85 is a primitive Pythagorean triple.  This 
is in fact the case.

Using the same method, we get 15,112,113 and we can go on forever.

Why does this work?

We want to show that a^2 + b^2 = c^2

Just make the substitution: 
   b = (a^2 - 1)/2 and 
   c = b + 1 = ((a^2 - 1)/2) + 1

Multiply everything out and you'll see that the two sides are equal.

So that works when a is odd.  What about when a is even?

The triples aren't going to all be primitive in this case, but let's 
look at them anyway for pattern analysis.

a   4   6   8  10  12  14
b   3   8  15  24  35   ?
c   5  10  17  26  37   ?

Again we ask the question - given a value for 'a', can we generate 'b' 
and 'c'?  Well it looks like if we know what the value of 'b' is, we 
can add two to it to get 'c'.  Let's again add the a^2 row to help us 
out:

a^2  16  36  64  100  144
a     4   6   8   10   12   14
b     3   8  15   24   35    ?
c     5  10  17   26   37    ?

Do you see that a^2 = 2(b+c)  ?

Substitute c = b+2 to obtain a^2 = 2(2b+2).  Solving for b yields:

   b = (a^2 - 4)/4

So given a = 14, b = (196-4)/4 = 192/4 = 48 and c = 48 + 2 = 50.

A similar procedure shows that 16,63,65 and 18,80,82 are also 
Pythagorean triples.

It appears to be the case that whenever 'a' is a multiple of four, the 
triple is primitive. When it isn't, the triple is not primitive. Can 
you figure out why this is the case?

Again, we wonder why this works.

You can show that a^2 + b^2 = c^2 by making the substitution:

   b = (a^2 - 4)/4 and c = b + 2 = b = ((a^2 - 4)/4) + 2

and simplifying the terms.


This should help you generate all of the Pythagorean triples you need 
to complete your project.

- Doctor Paul, The Math Forum
  http://mathforum.org/dr.math/