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It's As Easy As It’s As Easy As abc Andrew Granville and Thomas J. Tucker Introduction (2) xp−1x + yp−1y = zp−1z. Fermat’s Last Theorem We now have two linear equations (1) and (2) (think- In this age in which mathematicians are supposed ing of xp−1, yp−1, and zp−1 as our variables), which to bring their research into the classroom, even at suggests using linear algebra to eliminate a vari- the most elementary level, it is rare that we can turn able: Multiply (1) by y and (2) by y, and subtract the tables and use our elementary teaching to help to get in our research. However, in giving a proof of Fer- mat’s Last Theorem, it turns out that we can use xp−1(xy − yx)=zp−1(zy − yz). tools from calculus and linear algebra only. This p−1 p−1 − may strike some readers as unlikely, but bear with Therefore x divides z (zy yz ), but since us for a few moments as we give our proof. x and z have no common factors, this implies that Fermat claimed that there are no solutions to (3) xp−1 divides zy − yz. p p p (1) x + y = z This is a little surprising, for if zy − yz is nonzero, then a high power of x divides zy − yz, something for p ≥ 3, with x, y, and z all nonzero. If we assume that does not seem consistent with (1). that there are solutions to (1), then we can assume We want to be a little more precise. Since we dif- that x, y, and z have no common factor, else we ferentiated, we evidently never were working with can divide out by that factor. Our first step will be integers x, y, z, but rather with polynomials. Thus to differentiate (1) to get if zy − yz =0, then (y/z) =0, and so y is a con- pxp−1x + pyp−1y = pzp−1z, stant multiple of z, contradicting our statement that y and z have no common factor. Therefore (3) im- and after dividing out the common factor p, this plies that leaves us with (p − 1) degree(x) ≤ degree(zy − yz ) Andrew Granville is a Canadian Research Chair of math- ≤ degree(y) + degree(z) − 1, ematics at the Université de Montréal. His email address is [email protected]. since degree(y )= degree(y) − 1 and degree(z )= − Thomas J. Tucker is currently a visiting assistant profes- degree(z) 1. Adding degree(x) to both sides gives sor at the City University of New York Graduate Center. (4) p degree(x) < degree(x) + degree(y) + degree(z). His email address is [email protected]. This article is based on the first named author’s AMS- The right side of (4) is symmetric in x, y, and z. MAA invited lecture at the Orlando annual meeting in Jan- The left side is a function of x simply because of uary 1996. the order in which we chose to do things above. We 1224 NOTICES OF THE AMS VOLUME 49, NUMBER 10 could just as easily have derived the same state- Then if we add the first column to the second, we ment with y or z in place of x on the left side of get a(t) c(t) (4), so that ∆(t)= , a(t) c(t) p degree(y) < degree(x) + degree(y) + degree(z) and and similarly p degree(z) < degree(x) + degree(y) + degree(z). c(t) b(t) ∆(t)= c(t) b(t) Adding these last three equations together and then dividing out by degree(x) + degree(y) + degree(z) implies by adding the second column to the first, a beau- p<3, tiful symmetry. − and so Fermat’s Last Theorem is proved! We note that ∆(t) =0, else ab a b =0, so b is Well, not quite, but what we have proved (and a scalar multiple of a (with the same argument as so simply) is still of great interest: above), contradicting the hypothesis. To find the appropriate analogy to (3), we in- Proposition 1. There are no genuine polynomial so- terpret that as stating that the factors of x (as well ∈ C p p p lutions x(t),y(t),z(t) [t] to x(t) + y(t) = z(t) as of y and z) divide our determinant to a high ≥ with p 3. By “genuine” we mean that the triple power. So now suppose that α is a root of a(t) and (x(t),y(t),z(t)) is not a polynomial multiple of a that (t − α)e is the highest power of (t − α) which C solution of (1) in . divides a(t). Evidently (t − α)e−1 is the highest That Fermat’s Last Theorem is easy to prove for power of (t − α) which divides a (t), and thus it polynomials is an old result, going back certainly is the highest power of (t − α) which divides as far as Liouville (1851), although his proof, which ∆(t)=a(t)b (t) − a (t)b(t) (since α is not a root of goes through integration, is much more involved b(t)). Therefore (t − α)e divides ∆(t)(t − α). Multi- than that given here. The proof we have presented plying all such (t − α)e together, we obtain above is certainly some years old; for instance, a variant can be found in standard textbooks of fifty a(t) divides ∆(t) (t − α). years ago. After reading through it, one sees that a(α)=0 this argument is easily generalizable to other Dio- phantine problems, though it is not obvious what In fact, a(t) appears on the left side of this equation would be the ultimate generalization. only because we studied the linear factors of a; anal- ogous statements for b(t) and c(t) are also true, and Mason’s Generalization It takes a certain genius to generalize to some- since a(t), b(t), c(t) have no common roots, we can combine those statements to read thing far simpler than the original. But what could possibly be more simply stated, yet more general, (6) a(t)b(t)c(t) divides ∆(t) (t − α). than Fermat’s Last Theorem? It was Richard C. (abc)(α)=0 Mason (1983) who gave us that insight: The next step is to take the degrees of both sides Look for solutions to and see what that gives. Using the three different (5) a + b = c. representations of ∆ above, we have We will just follow through the proof above and − degree(a) + degree(b) 1, see where it leads: Start by assuming, with no loss ≤ − degree(∆) degree(a) + degree(c) 1, of generality, that a, b, and c are all nonzero poly- nomials without common factors (else all three degree(c) + degree(b) − 1. share the common factor and we can divide it out). Then we differentiate to get The degree of (t − α) is precisely the (abc)(α)=0 a + b = c . total number of distinct roots of a(t)b(t)c(t). In- serting all this into (6) we find that Next we need to do linear algebra. It is not quite so obvious how to proceed analogously, but what max{degree(a), degree(b), degree(c)} we do learn in a linear algebra course is to put our < #{α ∈ C :(abc)(α)=0}. coefficients in a matrix, and solutions follow if the determinant is nonzero. This suggests defining Put another way, this result can be read as: a(t) b(t) Proposition 2. If a(t),b(t),c(t) ∈ C[t] do not have ∆(t):= . a(t) b(t) any common roots and provide a genuine polyno- mial solution to a(t)+b(t)=c(t), then the maximum NOVEMBER 2002 NOTICES OF THE AMS 1225 of the degrees of a(t),b(t),c(t) is less than the num- their central importance. He showed, amongst ber of distinct roots of a(t)b(t)c(t)=0. other things, that for any finite subset S of Q there is a map π : C ∪ {∞} → C ∪ {∞} for which This is a “best possible” result in the sense that − π(S) ⊆{0, 1, ∞} , and π 1(z) = degree(π) for we can find infinitely many examples where there every z/∈{0, 1, ∞}. We can reinterpret this in is exactly one more zero of a(t)b(t)c(t)=0 than terms of polynomials as follows. the largest of the degrees: for example, the famil- iar identity Proposition 3. For any f (t) ∈ Z[t] there exist a(t),b(t),c(t) ∈ Z[t] which do not have any com- 2 2 − 2 2 2 (2t) +(t 1) =(t +1) mon roots and provide a genuine polynomial so- lution to a(t)+b(t)=c(t) for which f (t) divides or the rather less interesting a(t)b(t)c(t), and such that the maximum of the de- tn +1=(tn +1). grees of a(t),b(t),c(t) is exactly one less than the number of distinct roots of a(t)b(t)c(t)=0. Classifying such polynomial identities leads us naturally to the study of a special class of rational Thus we can use Belyı˘ maps to construct many functions, as we shall see next. “best possible examples” in Proposition 2. As we shall see later, this elegant construction is central Silverman’s Proof to several important results. Silverman provided a more sophisticated route to Proposition 2, via the theory of covering maps, an An Analogy for Integers? approach that will turn out to be very useful.
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