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 2b 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 = 3b. For n = 3, again b can have any value, and a = 2b.

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).