The other day I came across a meme posing a math puzzle. It showed two equations.

$$ab + c = 85$$ $$a + bc = 86$$

I did not pause to see what solution might be offered, but the problem stayed with me, and I decided to investigate it.

I began by making some assumptions. First, I assumed the solution would consist of three whole numbers. If any of the numbers were not whole numbers, then there would not be a unique solution. Likewise, if any were 0, there would be noninteger solutions or trivial solutions. For example:

$$a = 0 \text{ yields } c = 85, b = {86 \over 85}$$

and

$$b = 0 \text{ yields } c = 85, a = 86$$

To make the problem more malleable, I replaced $$c$$ with $$a + \epsilon$$, where $$\epsilon$$ is some positive integer. (This assumes that $$a < c$$, but if it were not, I could just swap $$a$$ and $$c$$.) This yields two new equations:

$$ab + (a + \epsilon) = 85$$ $$a + ab + b\epsilon = 86$$

Subtracting the first from the second gives:

$$b\epsilon - \epsilon = 1$$

It immediately follows that $$\epsilon = {1 \over b - 1}$$. Since $$\epsilon$$ must be a whole number, $$b$$ can only be $$2$$, and $$\epsilon = 1$$. Substituting back into one of the equations above gives:

$$a \times 2 + (a + 1) = 85$$ $$a = 28$$

It follows, then, that the only unique solution is $$a = 28$$, $$b = 2$$, and $$c = 29$$.

What about a more generalized solution? Consider the follwoing:

$$ab + c = m$$ $$a + bc = n$$

Applying the same reasoning gets us to:

$$\epsilon = {n - m \over b - 1}$$

This equation will yield a whole number solution for $$\epsilon$$ whenever $$b - 1$$ divides $$n - m$$. For example, if $$n - m = 6$$, then $$\epsilon = 6, 3, 2, 1$$ for $$b - 1 = 1, 2, 3$$, and $$6$$, so there could be four solutions. However, these four values for $$\epsilon$$ yield four cases:

$$2a + a + 6 = m, a = {m - 6 \over 3}$$ $$3a + a + 3 = m, a = {m - 3 \over 4}$$ $$4a + a + 2 = m, a = {m - 2 \over 5}$$ $$7a + a + 1 = m, a = {m - 1 \over 8}$$

So $$m$$ must be a whole number that satisfies at least one of the conditions:

$$m \equiv 0 \mod(3)$$ $$m \equiv 3 \mod(4)$$ $$m \equiv 2 \mod(5)$$ $$m \equiv 1 \mod(8)$$

If $$m$$ satisfies exactly one of these conditions, then the solution is unique. If it satisfies none of them, then there is no solution for that $$m$$. No number can satisfy all four conditions. In particular, any $$m$$ such that $$m \equiv 3 \mod(4)$$ must also satisfy either $$m \equiv 3 \mod(8)$$ or $$m \equiv 7 \mod(8)$$. It cannot also satisfy $$m \equiv 1 \mod(8)$$. The smallest number $$m$$ that satisfies the first 3 conditions is 27. It yields the following three solutions:

$$a = 7, b = 2, c = 13, m = 27, n = 33$$ $$a = 6, b = 3, c = 9, m = 27, n = 33$$ $$a = 5, b = 4, c = 7, m = 27, n = 33$$

It's easy to see that any initial conditions where $$n - m = 1$$ will yield a unique solution where $$b = 2$$ and $$\epsilon = 1$$. In such cases, $$m = 3a + 1$$, meaning that solutions exist only when $$m \equiv 1 \mod(3)$$. Otherwise, $$a$$ cannot be a whole number, since $$a = {m - 1 \over 3}$$. For every $$n - m > 2$$ there are at least 2 conditions that lead to solutions, and if $$n - m$$ is a prime number, they are the only possible conditions, either of which will lead to solutions. They are $$b = 2$$ and $$b = n - m$$. In the first case $$\epsilon = n - m$$, and in the second case $$\epsilon = 1$$. This leads to two families of solutions:

$$a = {2m - n \over 3}, b = 2, c = {2n - m \over 3}$$ $$a = {n \over n - m + 1} - 1, b = n - m, c = {n \over n - m + 1}$$ Clearly, it would be easy to make other similar Internet memes with unique whole number solutions.