Suppose we want to find pairs of whole numbers such that their sum is a multiple of their difference. That is,
for whole numbers *a*, *b*, and *n*,

Solving for *a* we obtain,

But how do we guarantee that *a* is a whole number? Clearly, *nb* + *b* must be a multiple of
*n* − 1. Let's make an adjustment:

So for *a* to be a whole number, we require only that 2*b* be a multiple of (*n* − 1).
When *n* is odd, *b*
must be a multiple of ½(*n* − 1). When *n* is even, *b* must be a multiple of
(*n* − 1). Choosing *n*
tells us which values we can choose for *b*, which in turn tells us the values of *a*. Since *n* ≠ 1,
the smallest value
*n* can have is 2. In that case, *b* can have any value, and *a* = 3*b*.
For *n* = 3, again *b* can have any
value, and *a* = 2*b*.

For a sum of *a* and *b* to be 5 times their difference, *b* must be even, and *a* must be 1½ times *b*.
For example, choose *b* = 18,
then *a* = 27, and (27 + 18) = 5 × (27 − 18).

Another way to think of this problem is to start with numbers that differ by 1,
that is *a* − *b* = 1. Clearly, then,
*a* + *b* = *n*.
Combining these two equations, we obtain *a* = ½(*n* + 1) and
*b* = ½(*n* − 1). Since *n* is always odd,
*a* and *b* are guaranteed to be whole numbers. Now we can generate more solutions by
multiplying the original equation by another factor, say, *m*. Thus,
(*ma* + *mb*) = *n*(*ma* − *mb*).
In a similar manner, we can show that whenever *a* − *b* ≠ 1,
then both *a* and *b* have a common factor
which is equal to *a* − *b*. In the example above, the common factor is
(27 − 18) = 9.

Careful readers will have noticed that sometimes *n* is not odd. In particular, if *a* and *b* differ by 2,
then using similar reasoning, *a* = *n* + 1 and *b* = *n* − 1.
In this case, if *n* is even, then *a* and *b* are both odd. They do not have a common factor equal
to their difference. Instead multiples of these equations yield cases where *a* and *b* have a common factor equal to
half their difference. For example, 21 + 15 = 6 × (21 − 15).