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Pythagorean Triples (x,c,y) with Fixed c

Date: 05/09/2003 at 11:42:14
From: Shanon
Subject: Polynomials

Is there a shortcut to finding the integer solutions to equations of 
the type x^2 + c = y^2, where c is a constant of known value?  
Obviously, (1/2)(c-1) = x and (1/2)(c+1) = y are solutions for odd 
values of c.  But what about other whole number solutions?

It seems that there should be some relation between the solutions and 
if we know some of them it should be possible to deduce or at least 
zero in on the others. I am most interested in finding the smallest 
integer solution.


Date: 05/09/2003 at 14:46:33
From: Doctor Luis
Subject: Re: Polynomials

Hi Shanon,

Yes, there is a way. Numbers a,b,c that satisfy the equation

  a^2 + b^2 = c^2

are called Pythagorean Triples.

To read more about them, I suggest the following links from our
site:

   Formula for Pythagorean Triples - Dr. Math archives
   http://mathforum.org/library/drmath/view/55811.html 

   Pythagorean Triples - Dr. Math FAQ
   http://mathforum.org/dr.math/faq/faq.pythag.triples.html 

Now, the way you generate triples (a,b,c) is by specifying any two 
integers m,n and then using the following formulas

   a = m^2 - n^2 
   b = 2 m n
   c = m^2 + n^2

If you do quick check on these formulas, you can verify that
a^2 + b^2 = c^2 holds true for all values of m and n.

So, what you're asking, then, is to generate Pythagorean Triples 
(x,c,y) such that c is fixed. Let's try plugging these in the formulas 
above:

  x = m^2 - n^2
  c = 2 m n
  y = m^2 + n^2

Therefore, x^2 + c^2 = y^2 definitely have integer solutions if c is 
even. You pick m and n from the factors in c/2.

If c is odd, you might want to switch the formulas above

  c = m^2 - n^2
  x = 2 m n
  y = m^2 + n^2

In which case, you'll have to express the odd number c as a difference
of squares. A quick way to do this is by noticing that

  c = (m + n)(m - n)

Therefore, if you write c as the product of two numbers, say p and q,
you can solve the two simultaneous equations m+n=p, m-n=q for m,n.
Once you have m,n, then you can find x and y.

Does this make sense?

I hope this helped answer your question. Let us know if you have any 
others.

- Doctor Luis, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
College Number Theory

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