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Diophantine Equation Solved by Reasoning

Date: 04/05/2006 at 12:12:45
From: Arka
Subject: Solving an equation.

Find all integral solutions of (1/m) + (1/n) - (1/mn^2) = 3/4.

I am completely at a loss.  I am sure that m and n cannot be zero. 
But after that I'm stuck.

Date: 04/05/2006 at 22:39:08
From: Doctor Vogler
Subject: Re: Solving an equation.

Hi Arka,

Thanks for writing to Dr. Math.  That's a good question.  A polynomial
equation where you want integer solutions is called a "Diophantine
equation."  I love solving Diophantine equations.  They can be quite

One way to solve your equation is to solve it for m.  It's a quadratic
equation in n, so solving for n would get ugly, but you can solve for
m, since your equation is

  (1/m)(1 - 1/n^2) = 3/4 - 1/n

and therefore

       4n^2 - 4
  m = -----------
       3n^2 - 4n

Then you just need to find all integers n where

  4n^2 - 4

is divisible by

  3n^2 - 4n.

When n is large (or a large negative), then the ratio (m) is close to
4/3, so that means that n must be small for m to be an integer.  In
particular, you can prove that for most values of n (all except
certain ones which you can list), m will be strictly between 0 and 4/3.

Another way to solve your equation is to consider that if m and n are
both large, then all three fractions on the left will be very small,
so they won't add up to 3/4.  So you can prove something like:  If n
has absolute value at least 2, then

  3/4 = abs(3/4) = abs(1/n + (1/m)(1 - 1/n^2))
                 < abs(1/n) + abs(1/m)*1,

which is impossible if abs(m) >= 4, for example, or if abs(n) >= 3 and
abs(m) >= 3.  That only leaves a handful of remaining cases to check,

  n = 0, which you already showed was impossible,
  n = -1 or +1, which you can easily show is impossible,
  n = -2 or +2, which you can solve for m,
  m = 0, which you already showed was impossible,
  m = -1, +1, -2, or +2, each of which you can solve for n
      by the quadratic formula (if the discriminant is not
      a perfect square, then there are no integer roots).

If you have any questions about this or need more help, please write
back and show me what you have been able to do, and I will try to
offer further suggestions.

- Doctor Vogler, The Math Forum 
Associated Topics:
College Number Theory

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