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Pythagorean Triples


Date: 18 May 1995 19:31:20 -0400
From: Roger Gillies
Subject: equation for pythagorean triples

Dear Dr. Math:

Myself and another teacher have been tearing out our hair over this:

There is a relation between Pythagorean triples we have been working 
with. It is as follows:

let n, x and x+1 represent the numbers of a pythagorean triple where n
is always an odd number and x is always an even number.

The first triple is:

3, 4  and 5

The next is:

5, 12 and 13 

Followed by:

7, 24 and 25

and so on.........

For each triple, the first odd digit (n) is multiplied by its order in 
the sequence (7's order is 3rd, so 7x3), then the number 3 is added to 
the product, giving x.

The question is this:

How can this relationship be expressed as a formula?

Hope you can help, and thanks in advance for trying.

Sincerely,
Roger Gillies and John Reid 


Date: 20 May 1995 10:19:18 -0400
From: Dr. Ken
Subject: Re: equation for pythagorean triples

Hello there!

Let me first introduce (you may have seen this before) the formula for
Pythagorean triples: a = (m^2 - n^2), b = 2mn, c = (m^2 + n^2).  
You can verify that these three expressions will always form a P-triple 
for any positive m and n, and in fact I believe that all P-triples are 
expressible in this form.  This will be a handy tool.

In your case, you say that b is one less than c.  Then we have 

1 + 2mn = m^2 + n^2
1 = m^2 - 2mn + n^2
1 = (m - n)^2

Assuming that m is bigger than n (which it is by examining our first term
m^2 - n^2, which is positive), this gives us m = n+1.  Let's plug that 
back into our original formulae:

a = (n+1)^2 - n^2 = 2n + 1
b = 2(n+1)n       = 2n^2 + 2n
c = (n+1)^2 + n^2 = 2n^2 + 2n + 1

Now try plugging in n=1,2,3,... into these formulae.  This is really neat
stuff, thanks for writing us about it!!

-K
    
Associated Topics:
High School Geometry
High School Number Theory
High School Triangles and Other Polygons

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