Consider two whole numbers such that their product is a multiple of their sum, i. e., $$pq = n(p + q)$$. Now let $$q = p + m$$, assuming $$q > p$$ and $$m > 0$$. Then $$p(p + m) = n(p + p + m)$$. Solving for $$p$$, we get:

$$p = {(2n - m) \pm \sqrt{m^2 + (2n)^2} \over 2}$$

For this equation to have a whole number solution, it is necessary that the expression $$m^2 + (2n)^2$$ be a perfect square. This expression is similar to the expression that makes a Pythagorean triple, $$a^2 + b^2 = c^2$$. In fact, it turns out that every Pythagorean triple yields a solution for $$p$$. This is true because, for every Pythagorean triple, at least one of $$a, b$$ must be even.

Suppose $$a$$ and $$b$$ are both odd, i. e., for some whole numbers $$j$$ and $$k$$, $$a = 2j - 1$$ and $$b = 2k - 1$$. Then $$a^2 + b^2$$ becomes $$(2j - 1)^2 + (2k - 1)^2$$. This can be resolved to the expression $$2[2(j^2 - j + k^2 - k) + 1]$$. The expression within the square brackets must be odd. Therefore, the entire expression cannot be a perfect square. It's square root must have an element of $$\sqrt{2}$$ in it. Thus there can be no Pythogorean triple where both $$a$$ and $$b$$ are odd.

In any Pythagorean triple, at least one of $$a$$ and $$b$$ must be even. The even term can be put in for $$2n$$ in the expression for $$p$$, or if both are even, either can be put in.

It remains then to show that the expression $$(2n - m) \pm \sqrt{m^2 + (2n)^2}$$ must always yield an even whole number.

It should be obvious that if $$m$$ is odd, both terms $$(2n - m)$$ and $$\sqrt{m^2 + (2n)^2}$$ are likewise odd, which makes their sum or difference even. If $$m$$ is even, both terms are even, and again their sum or difference is even. Therefore, as long as the $$\sqrt{m^2 + (2n)^2}$$ is a whole number, the entire expression for $$p$$ must be an integer. In addition, $$p > 0$$ if $$(2n - m) + \sqrt{m^2 + (2n)^2} > 0$$, which will always be so if $$m - 2n < \sqrt{m^2 + 4n^2}$$. This resolves to $$-4nm < 0$$ which will always be true by hypothesis since $$n$$ and $$m$$ are whole numbers. Therefore, every Pythagorean triple yields a solution.

This does not mean that every such solution will be unique. It is possible that different Pythagorean triples will yield the same solution to the original problem of finding whole numbers that satisfy the relation $$pq = n(p + q)$$.

We need not consider the case where $$p > q$$, i. e., $$m < 0$$, because we can always choose $$q$$ as the larger of $$p$$ and $$q$$ without altering the product or sum in the relation. Therefore, we need only still consider the case where $$q = p$$. In that case, our initial relation becomes $$p^2 = 2np$$. One solution is the trivial $$p = q = 0$$, where $$n$$ can be any whole number. If $$p \ne 0$$, then $$p = 2n$$, and a solution exists for every even whole number $$p$$. For example, if $$p = q = 26$$, then $$n = 13$$, and the relation $$26 \times 26 = 13(26 + 26)$$ is true.

Following are two examples where Pythagorean triples yield solutions to the initial problem of a product being a multiple of the sum.

Example 1: 3, 4, 5.

Given the Pythagorean triple $$3^2 + 4^2 = 5^2$$, we choose $$m = 3$$ and $$n = 2$$ to obtain $$p = {1 \pm 5 \over 2}$$. Ignoring the result where $$p < 0$$, we get $$p = 3$$, from which it immediately follows that $$3 \times 6 = 2(3 + 6)$$.

The only other solution where $$n = 2$$ is for $$p = q = 4$$, i. e., $$4 \times 4 = 2(4 + 4)$$. We know this because no other Pythagorean triples contain $$4$$ for $$a$$ or $$b$$.

Example 2: 7, 24, 25

Given the Pythagorean triple $$7^2 + 24^2 = 25^2$$, we choose $$m = 7$$ and $$n = 12$$ to obtain $$p = {17 \pm 25 \over 2}$$ Ignoring the result where $$p < 0$$, we get $$p = 21$$, from which it immediately follows that $$21 \times 28 = 12(21 + 28)$$.

There may, however, be other solutions than $$p = q = 24$$, i. e., $$24 \times 24 = 12(24 + 24)$$.

Let's apply the techniques on this page to see if there are other solutions where $$n = 12$$. Since $$2n$$ is even, we can ignore odd factors of $$(2n)^2$$. The even factors less than $$24$$ are $$2, 4, 6, 8, 12, 16, 18$$. These lead to Pythagorean triples and accompanying solutions listed below.

Pythagorean triple Calculated $$p$$ Solution
$$24, 143, 145$$ $$13$$ $$13 \times 156 = 12(13 + 156)$$
$$24, 70, 74$$ $$14$$ $$14 \times 84 = 12(14 + 84)$$
$$24, 45, 51$$ $$15$$ $$15 \times 60 = 12(15 + 60)$$
$$24, 32, 40$$ $$16$$ $$16 \times 48 = 12(16 + 48)$$
$$24, 18, 30$$ $$18$$ $$18 \times 36 = 12(18 + 36)$$
$$24, 10, 26$$ $$20$$ $$20 \times 30 = 12(20 + 30)$$
$$24, 7, 25$$ $$21$$ $$21 \times 28 = 12(21 + 28)$$
$$24, 0, 24$$ $$24$$ $$24 \times 24 = 12(24 + 24)$$

Taken all together, there are 8 solutions where $$n = 12$$. Similar methods will yield all the solutions for any $$n$$.