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### Equations with Rational Expressions in Two Variables

```Date: 12/07/2002 at 10:36:27
From: Tanis Glass
Subject: Solving equations with rational expressions in two variables

Determine all positive integers a and b that satisfy the equation:
1/a + a/b + 1/ab = 1.

I have determined (by trial and error) two possible (a,b) solutions:
(a=2, b=5)  and (a =3, b=5)

I also realize that certain restrictions must be placed on the
variables, such as: a, b cannot equal 0; a cannot equal 1; a cannot
equal b; and b must be greater than a.

However, I have no clue as to how to coordinate this information to
come up with a solution that would allow me to figure out more
positive integers except by trial and error. This is a grade 11 math
course in functions so my background knowledge is limited.

```

```
Date: 12/07/2002 at 12:10:36
From: Doctor Ian
Subject: Re: Solving equations with rational expressions in two
variables

Hi Tanis,

Note that the restriction that a and b be positive integers rules out
a=0 and b=0.

The first thing that occurs to me would be to use something like
Mathematica to make a three-dimensional plot of the surface described
by the equation. That is, you'd have values of a along one axis,
values of b along the other, and the z axis would contain the values
of the function

f(a,b) = 1/a + a/b + 1/(ab) - 1

The points where the surface intersects the xy plane would be
solutions of the equation. If you plot it only for integer values of
a and b, you won't have to worry about non-integer solutions.

This also suggests that you could write a small computer program to
start checking values of a and b.

But what else can we do? One thing we can do is solve for one variable
in terms of the other.

1/a + a/b + 1/ab = 1

b/(ab) + a^2/(ab) + 1/(ab) = 1

b + a^2 + 1 = ab

a^2 + 1 = ab - b

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

(a^2 + 1)/(a - 1) = b

Now, you still have to use trial and error. But this form makes it a
little easier to do. You can start choosing values of a, and see what
values of b are forced:

a     (a^2 + 1)/(a - 1)    b
---    -----------------  -----
1     undefined
2     5/1                  5
3     10/2                 5
4     17/3
5     26/4
6     37/5

On the other hand, unless there is a bug in the computer program I
just wrote a few minutes ago, trial and error isn't going to find you
any more solutions for values of a up to a million. Which doesn't
prove, but strongly suggests, that there are no other solutions.

So why would that be the case?  Let's try doing the division, and see
what happens:

a +  1
_____________
a - 1 ) a^2 + 0a + 1
a^2 -  a
--------
a + 1
a - 1
-----
2
So

(a^2 + 1)/(a - 1) = (a + 1) + 2/(a - 1)

and now our table looks like this:

a     (a + 1) + 2/(a - 1)    b
---    -------------------  -----
1     undefined
2     3 + 2/1                5
3     4 + 2/2                5
4     5 + 2/3
5     6 + 2/4

Does this make it easier to see how many other solutions there will
be?

- Doctor Ian, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 12/07/2002 at 12:33:26
From: Tanis Glass
Subject: Thank you (Solving equations with rational expressions in two
variables)

Thanks very much, Doctor Ian. The use of a graph in 3-D was a bit too
complex (for me, anyway) because we have not studied that in school;
however, the rest of your help was just great and I understand your
explanation. You were a big help.

I have consulted the Dr. Math archives in the past for help with other
topics and have had success in locating similar problems; but this one
```
Associated Topics:
High School Analysis
High School Number Theory

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