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 $$n$$,

$$a + b = n(a - b)$$

Solving for $$a$$ we obtain,

$$a = {{nb + b} \over {n - 1}}$$

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:

$$\begin {eqnarray} a &=& {{(n - 1)b + 2b} \over {n - 1}} \\ &=& b + {2b \over n - 1} \end {eqnarray}$$

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 $$\frac 1 2 {(n - 1)}$$. 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 $$n \ne 1$$, 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 $$a = \frac 3 2 {b}$$. 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 two numbers that differ by 1, that is $$a - b = 1$$. Clearly, then, $$a + b = n$$. Combining these two equations, we obtain $$a = \frac 1 2 (n + 1)$$ and $$b = \frac 1 2 (n - 1)$$. If $$n$$ 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 $$a - b \ne 1$$, 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 $$n$$ is not always odd. In particular, if $$a$$ and $$b$$ differ by $$2$$, then $$a + b = 2n$$ and using similar reasoning, $$a = n + 1$$ and $$b = n - 1$$. In this case, when $$n$$ is even, then $$a$$ and $$b$$ are both odd. They do not have a common factor equal to their difference since their difference is even. Instead multiples of these equations yield cases where $$a$$ and $$b$$ have a common factor equal to half their difference. For example, $$21 + 15 = 6(21 - 15)$$ The common factor, $$3$$, is half of $$(21 - 15)$$.