## Fermat – putting the Last Theorem first

### March 29, 2009

While my main concern is how journalists use numerical skills in everyday reporting, I’ m also interested in how mathematical and science-based stories are treated. These present themselves less frequently to the average reporter, but nevertheless still deserve to be reported accurately and at a level appropriate to the audience.

I am developing some coursework to help students with this, and in order to make it concrete am thinking of using Fermat’s Last Theorem as a case study, since its solution was very widely reported in the early 1990s (the story appears in Amir Aczel’s Fermat’s Last Theorem and Simon Singh’s book of the same name), and the basic problem is easy enough to understand without any mathematical training.

My lesson plan is to explain to students what Fermat’s Last Theorem means so that they are then able to write about it themselves, as if they had been reporting on Andrew Wiles’s original discovery and their editor had asked for a 150 word sidebar on what the theorem says.

I am going to motivate discussion by looking at squares with various edge lengths and asking, when is it possible to take two squares and add them together to make a new square? (ie, the case when n=2 so that x^{2}+y^{2}=z^{2}; (the Pythagorean triplets, 3^{2}+4^{2}=5^{2}, for instance).

We’ll do this by taking 9 small squares of card arranged into a larger 3×3 square, and 16 small squares arranged into a 4×4 square, then jumbling all 25 small squares together and see if they can be arranged into a larger square (and, hey presto! we find we can make a 5×5 square).

So we have proved Fermat’s theorem for squares, when n=2. We’ll also see there are infinitely many solutions, before turning our attention to the case of cubes (when n=3). Can we find any two cubes, made up of smaller cube blocks, such that we can take the smaller blocks and make them into a larger cube? Of course, we know now that we can’t … and that there is no (whole number) answer for **any** value of n apart from n=2.

So by starting with the case where we’re dealing with squares – one that can be easily visualised – we can get the students to think about the situation involving cubes, and eventually to higher numbers that have no intuitive physical counterpart. And because we can present the squares and the cubes geometrically, we don’t have to use equations until the end, when the notation becomes a succinct way of expressing the physical fact that squares can be re-ordered into larger squares but cubes cannot.

It is this principle I’d want to get across, rather than any of the technical detail of Wiles’s solution.