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 \((21 - 15)\).