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.