Linear Diophantine Equations

Date: 11/27/97 at 15:12:50
From: Ulrich Richers
Subject: Linear Diophantus Equations

Recently my son's homework has included several problems with a number 
of linear equations smaller than the number of unknown parameter. All 
solutions had to be integer and greater than zero.

I have a math book from East Germany with an example but without 
directions for how to get there.

   3x - 2y -5 = 0

there are several solution pairs:

   x = 2t + 1
   y = 3t - 1

How was the t-variable introduced, and what is the general method?

Thank you and apologies for my rusty math,

Ulrich Richers

Date: 11/27/97 at 19:24:08
From: Doctor Anthony
Subject: Re: Linear Diophantus Equations

Divide through 2, the smaller coefficient, and so

      3x/2 - y = 5/2

   x + x/2 - y = 2 + 1/2

 x-y + (x-1)/2 = 2 .........(1)

and so  (x-1)/2 must be an integer, say t, and we get

       (x-1)/2 = t

           x-1 = 2t

             x = 2t + 1

Substitute this for x in (1)

  2t+1 - y + t = 2

        3t - 1 = y

We therefore have  x = 2t+1,  y = 3t-1  where t is any integer.

-Doctor Anthony,  The Math Forum
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