The three-hour car ride back was intended to be a winding down period. With my colleague Mark at the wheel, I had envisioned an event debrief between the two of us, interspersed with naps between the one of me. That dream died as my mind processed the remnants of a maths problem handed down by this seemingly crazed individual. Was there hidden depth to the old man’s ramblings?

My body was crying for rest, but my mind was buzzing. I set to work.

I knew right away that 49/20 could not possibly equal the square root of 6, and that 17/12 is no way the square root of 2. In each case, the square roots yield irrational numbers (those that can not be expressed as a fraction). This basic insight already gave some meaning to the man’s claims. He was chastising us mathematicians for our precision. In its place, he was inviting the alternative: approximation. So the question became:

In what sense is 49/20 a good approximation of √6 (and 17/12 of √2)?

The mathematician in me — largely dormant since 2011 — was slowly awakening. My first instinct was to bend the rules a little. Let’s suppose these values are in fact equal, even though we’ve confirmed otherwise. Let’s flirt with absurdity just to see where it gets us.

If 49/20 = √6, then squaring both sides gives: 2401/400 = 6

Interesting. 2400/400 is bang on 6. This simple exploration gives some credence to the old man’s approximation. It shows that, subject to a remainder of 1, 49/20 is a pretty good estimate of √6.

Is that what he was getting at? Was the old man’s point to just ignore the remainder 1? Since he had obliged with a second example, I could test-drive this idea further — let’s see what happens in the case of 17/12 and √2:

If 17/12 = √2, then squaring both sides: 289/144 = 2

Again, this statement is true up to that remainder of 1.

Having fun yet? I certainly was. The more I delved into the problem, the more I felt I was getting into the old guy’s head.

Sadly, the man had departed before I even had a chance to get his name. I would have no way of following up. Then again, I was making this problem my own. By now, I was asking sharper questions — questions such as:

How interesting is this property of numbers?

Can we find other such examples? How many?

It was here that I formalised the problem. Moving beyond the specifics of the two example, I asked:

For which positive integers a,b,c does a²/b² differ from c by a remainder of 1?

Some basic algebraic manipulation reduces this statement to the equation:

(a-1) * (a+1) = b² * c

This formulation lands the problem squarely in the domain of number theory. The most interesting question of all is just how interesting this relationship is: how often it occurs, and under what circumstances. Fermat mused over a vaguely similar problem almost 400 years ago; the rest is the stuff of mathematical legend. Our problem seems much narrower in scope; the approximation feels too crude to be of any real significance. Still, it was worth digging.

This is basically what got me into a tizzle this week

After a while (two hours, give or take) I retreated from the problem to focus on my two other commitments: the event debrief and the nap. Neither took place because, it transpired, I was not the only one fixated by the problem. My colleague Mark professed a long-term enjoyment of mathematics, curtailed only by the excessive focus of school exams. And now he was my collaborator. The interaction was telling; Mark was able to suggest creative and honest-to-god useful strategies, which combined nicely with my specific knowledge of prime decompositions to deliver a series of mini breakthroughs.

I will spare the reader all the details, except to say that one such mini breakthrough allowed us to generate an infinite supply of such examples. Great though this sounds, it need not mean we found them all. In fact, that procedure did not even recover the example of 17/12. The old man’s logic was not yet in our grasp.