I’ve just finished reading The Girl Who Played with Fire, the second book in the series that begins with The Girl with the Dragon Tattoo. I’m not giving away any significant plot point when I tell you that there’s a character who works on Fermat’s Last Theorem as a hobby, or that the book was clearly written (or perhaps translated) by somebody with no clue how mathematics works or what Fermat’s Last Theorem is about. I particularly liked the reference to Andrew Wiles using the “world’s most complicated computer program” to solve the problem. It’s my understanding that Andrew barely even uses email. And certainly if you understood anything about the nature of the problem and/or the solution, you’d recognize the absurdity of trying to tackle it with a complicated computer program.

Be that as it may, I finished the novel with a few hours left to spare, so of course I was inspired to work on Fermat’s Last Theorem, or at least on the simplest cases. The problem, if you’ll recall, is to show that there are no integer solutions to any of the equations x³+y³=z³ , x⁴+y⁴=z⁴ and so on, except for the so-called trivial solutions in which one or more variables take the value zero.

This is relatively easy to prove in the n=4 case (that is, for the equation x⁴+y⁴=z⁴), and in fact I was able to reconstruct two separate proofs, one using elementary algebra and the other using a little geometry. (“Reconstruct” means that there was a time in my life when I knew these proofs well—and even taught them at a graduate level—but that was long long ago.) And I was able to reconstruct Lamé‘s flawed proof, which, when supplemented with some more work, can be converted to a correct proof for a large class of exponents (beginning with n=5). The attempt to understand when Lame´’s argument can (or can’t) be patched up inspired a century of progress in algebraic number theory. Alas, that work reveals that there are plenty of exponents for which the proof is irreparable, beginning with n=37. The only known proof, associated in the popular imagination with the great Andrew Wiles, but more properly attributed to Frey, Serre and Ribet, is nothing like Lamé’s (and about one octillion times more difficult).

But what really surprised me was that I didn’t have a clue how to solve the case n=3. And even now, I have no memory of ever having known how to solve the case n=3. I was aware that it took Euler to solve it in the first place, and that I am not as smart as Euler (by a factor of about one octillion), but I was also aware that I know a lot of fancy techniques that Euler didn’t have. So, like the character in the novel, I thought I’d give it a go.

My first idea was to use Fermat’s favorite technique: Pretend you’ve got a solution, and show that from that solution, you can construct a smaller solution. Keep repeating and your solutions get smaller forever, which is quite impossible with integers. (If your first solution involved x=100 and x gets smaller each time, you’re going to get stuck after 100 iterations—x can’t go below zero). This means you never had a solution in the first place. (Fermat called this the “Method of Infinite Descent”.)

So I pretended I had a solution—that is, a set of numbers x, y, z that satisfy x³+y³=z³—and used a little geometry to construct a new solution. I did this using what is, for a geometer, the obvious idea. Namely:

• Set X=x/z and Y=y/z, and observe that X³+Y³=1
• Observe that (0,1) and (X,Y) are both points on the curve defined by the equation x³+y³=1
• Draw the line connecting these two points. Because the curve is defined by a third degree equation, that line will hit the curve three times. We already know it hits at (0,1) and (X,Y). Compute the third point. Because everything else in sight is a rational number, that third point will have rational coordinates.
• Write the coordinates of that point as (a/c,b/c), where a, b and c are integers. (You can always make the two denominators equal by choosing a common denominator). Then because this point sits on the curve, it satisfies the equation (a/c)³+(b/c)³=1. This in turn implies that a³+b³=c³.

So starting with one solution (x,y,z), we get a new solution (a,b,c). If (a,b,c) is in any reasonable sense smaller than (x,y,z), we can keep repeating till we get a contradiction.

When I did this, I got a = x(1+y³), b = -y(1+x³) and c = x³-y³. (You can check by hand that if x,y,z solve the Fermat equation then so do a,b,c.) Sadly, this doesn’t help because the new solution is not smaller than the old solution in any reasonable sense that I can think of. (I’d expected as much, because if something this simple had any chance of working, it wouldn’t have taken Euler to solve the problem.)

So I futzed around with a few other ideas that didn’t work (e.g. instead of drawing the line that connects two points, you could draw the tangent line at the point (X,Y)) and finally looked up Euler’s proof, which I must say, rang absolutely no bells with me, meaning either that I must have been curiously uncurious about this when I was younger or that my memory is failing even more precipitously than I realized. On a side note, I also learned (for the first time, as far as I can recall) that Euler’s first published attempt was incorrect.

Well, at least I got a blog post out of this, and more importantly it was fun. Sometimes it pays to have a short memory. Every now and then (especially when I’m stuck in a boring meeting) I compute the sum of the infinite series 1 + 1/2ⁿ + 1/3ⁿ + 1/4ⁿ + … for various values of n, which is another problem that Euler got to before I did. The main idea stays with me, but the details are new every time.

Edited to add: For those who are playing along at home—I copied incorrectly from my notes. The a, b and c announced above come not from the line that connects (0,1) to (X,Y), but from the tangent line at (X,Y). If you use the line connecting (0,1) to (X,Y), you get a=-x, b= z, c=y, which is even less useful.