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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
challenging.

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

(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,
including

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).

back and show me what you have been able to do, and I will try to
offer further suggestions.

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

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