Yesterday I attended a day of mathematical fun to celebrate the retirement of Jenny Piggott, who stepped down as NRICH director earlier this year. In the morning, we worked on problems which had been suggested by people who had worked with Jenny over the course of her career. These were varied – we talked about Pythagorean triples, methods of buttoning up one’s shirt, and what happens when you drop elephants into a lake! Some of these problems will no doubt be developed into tasks on the NRICH site in the next few months.

The Pythagorean triples task reminded me of something else I’ve tackled recently, as I ended up with the same set of equations to solve as I had done when working on another problem. Unfortunately I can’t find such a problem in my notes, and I can’t remember what it was I was working on – maybe it was all a very vivid dream! We were seeking triples (sets of three whole numbers that could be sides of a right-angled triangle, so satisfying Pythagoras’s theorem a^2+b^2=c^2) where two of the numbers differed by 1. Having worked on Pythagorean Triple problems before, I quickly found examples where the longer two sides differed by 1. It wasn’t immediately obvious whether it was possible for the perpendicular sides to differ by 1 though – I won’t post my further musings on this just yet in case people want to investigate it for themselves.

In the afternoon, we did some lovely mathematical origami. Alas, no camera with me today but my model ended up very like this. Read more about making them here. Then while tea was served, Bubblz entertained both young and old with mathematical bubbles which took me back to vague memories of an undergraduate course on calculus of variations that I attended ten years ago! It’s days like this that enthuse and energise us as maths educators to remember why we love our subject so much, and help us to pass on our passion to those we work with.

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July 15, 2010 at 05:26 |

Warning: spoilers!

Well, there’s certainly 3,4,5 and 20,21,29 at least, so it is possible.

There’s infinitely many like that, since if you generate them with the usual 2uv and u^2 – v^2 pattern, you have u=2, v=1 and then u=5, v=2, and the pattern continues with each v being the previous u and each u being twice the previous u, plus v. So we get u=12 v=5 next, which is 120 119 169 I think?

I’m pretty sure this family is the only ones that work but I don’t have a proof of that.

July 15, 2010 at 07:47 |

Some people working on the problem wondered whether 3,4,5 might be the only possibility – a good example of how starting with a false conjecture makes it harder to prove stuff 🙂

I found the 119, 120, 169 example just as the session came to an end and we had to stop working on the problem – managed to miss the much simpler 20, 21, 29 solution though!

July 15, 2010 at 20:52 |

I’d just like to confirm Alison’s conjecture that some of the problems we tackled on Tuesday will find their way onto NRICH. Indeed they will. We will be dedicating January’s website to problems inspired by the contributions for Jenny’s retirement event.