© Hindawi Publishing Corp. ON A FEW DIOPHANTINE EQUATIONS, IN PARTICULAR, FERMAT’S LAST THEOREM

This is a survey on Diophantine equations, with the purpose being to give the flavour of some known results on the subject and to describe a few open problems. We will come across Fermat’s last theorem and its proof by Andrew Wiles using the modularity of elliptic curves, and we will exhibit other Diophantine equations which were solved al aWiles. We will exhibit many families of Thue equations, for which Baker’s linear forms in logarithms and the knowledge of the unit groups of certain families of number fields prove useful for finding all the integral solutions. One of the most difficult conjecture in number theory, namely, the ABC conjecture, will also be described. We will conclude by explaining in elementary terms the notion of modularity of an elliptic curve.


Introduction. On
, at the Isaac Newton Institute of Cambridge (England), Professor Andrew Wiles (Princeton University) made a striking announcement. He had found a proof of Fermat's last theorem.
Fermat's last theorem. Let n be an integer greater than or equal to 3. Then there are no nonzero integers A, B, C such that A n + B n = C n . (1.1) The mathematical community became very excited and the news spread all over the world in a matter of days. For more than 350 years, many mathematicians and a greater number of amateurs had tried without success to obtain a proof of this conjecture. Already in 1918, Paul Wolfskehl poured some oil on the fire by promising a 100 000 German marks reward to whoever provides the first proof of this theorem. The mark devaluation almost annihilated the value of the prize, but did not decrease the interest of specialists in this question and similar problems.
In this survey article, we offer you an excursion over centuries into this fantastic world of Diophantine equations. Though we will not deal with the equations considered by Mordell and by Shorey and Tijdeman in their classical books [31,39], we plan to make you more familiar with other Diophantine equations. You will encounter some equations which have integral solutions, sometimes a finite number, sometimes an infinite number; you will also see some equations which have no solution at all, and you will come across some equations about which the only thing we know is that we know nothing about them. We will say a few words about the Fermat equation and in Section 9, you will get the flavour of Wiles' proof. We will exhibit other Diophantine equations which were solved à la Wiles, namely, by using some modular elliptic curves. You will also see that Baker's linear forms in logarithms and the knowledge of the unit groups of some families of number fields proved useful in solving some families of the so-called Thue equations. By the way, the delights of the ABC conjecture may make your mouth water: the assumption of this conjecture provides a short proof of Fermat's last theorem (for all n but a finite number). In Section 10, we will dare to open a parenthesis on the modularity notion of an elliptic curve, but we will rush to close it in order to avoid getting involved in technicalities.

Diophantus and Fermat. The
Greek mathematician Diophantus (born in 325) got interested in finding solutions of a given equation belonging to the set Q of rational numbers. However, under modern terminology, solving a Diophantine equation is looking for integral solutions, that is, for solutions belonging to the set Z of integers.
On the one hand, you may agree with the fact that the equation is easy to solve, but there are still open problems concerning this equation (as will be seen in Section 6). On the other hand, the situation often gets complicated if powers X n , where n is an integer greater than or equal to 2, come into play. Some solutions may be easily exhibited; for instance, a solution of the Diophantine equation is X = 2, Y = 1, and Z = 3, while another is X = 3, Y = 3, and Z = 6. Some Diophantine equations may happen to have no integral solution at all, like the equation The so-called Fermat-Pell equation has been around for many centuries, and the continued fraction expansion of √ D leads to its solution. This equation goes back to Archimedes (with the cattle problem) and was studied by the Indian mathematician Brahmagupta around 1630 and by the English mathematician William Brouncker around 1650. Pierre de Fermat (1601-1665) had a copy of the Latin translation (made by Bachet) of Diophantus book Arithmetica. Quite often, Fermat used to write personal notes in the margin of this book, and a Latin annotation of him (once translated) reads "It is impossible to write a cube as a sum of two cubes, a fourth power as a sum of two fourth powers, and in general, a power (except a square) as a sum of two powers with the same exponent. I possess a truly wonderful proof of this result, that this margin is too narrow to contain." This is equivalent to stating that the equations X 3 + Y 3 = Z 3 , X 4 + Y 4 = Z 4 , X 5 + Y 5 = Z 5 , and so on, have no solutions in positive integral integers.

Pythagoras.
Stating his theorem, Fermat assumed n ≥ 3, precisely because for n = 2, the Diophantine equation X 2 +Y 2 = Z 2 has integral solutions. As a matter of fact (see Figure 3.1), all the solutions of the equation are given by X = kU 2 − kV 2 , Y = 2kUV , and Z = kU 2 + kV 2 , in which one substitutes any integer for k, U, and V . Indeed, we come across a very old result.
Pythagoras' theorem. Suppose that in a given rectangle triangle, the length of the base is a, the height b, and the diagonal c. Then (3. 2) The proof appears as Proposition XLVII of the first book of Euclid's Elements [36, page 38]. A modern proof of this theorem is to let c = d in the statement of a result called the parallelogram law (easy to prove with some use of the scalar product). Parallelogram law. Let a and b be the two sides of a parallelogram, and let c and d be the two diagonals (as in Figure 3.2). Then (3. 3) The parallelogram law immediately leads to the median formula.
Median formula. Let a, b, and c be the sides of a triangle and let m be the median (as in Figure 3.3). Then Converse of Pythagoras' theorem. If a 2 +b 2 = c 2 , then the triangle of sides a, b, and c is rectangle with c as the hypotenuse.
Thanks to the median formula, one can supply a short proof. Let a 2 +b 2 = c 2 ; then 2m 2 +(1/2)c 2 = c 2 , that is, c = 2m. Hence, the two diagonals of Figure 3.4 are equal, and the parallelogram is a rectangle, that is, is made of two rectangle triangles. Babylonians applied the converse of Pythagoras' theorem to build an angle of 90 degrees. Indeed, they used ropes having knots at intervals of the same length and used them as in Figure 3.1. They were sure to obtain a right angle of 90 degrees between the horizontal line and the vertical line.
Nowadays, when it comes to fixing the wooden rectangle (wooden rail) on the foundations of a building to be built (see Figure 3.5), where the length between A and D is equal to the length between B and C, and where the length between A and B is equal to the length between C and D, home builders make sure that the length between A and C is equal to the length between B and D. Though they may not be aware of it, they "use" the parallelogram law (see Figure 3.2) and make sure that c 2 (= a 2 + b 2 ) = d 2 , and then use the converse of Pythagoras' theorem to conclude that the two glued triangles are rectangle triangles.
Thanks to Pythagoras (and to the converse of his theorem), we can solve a problem which became famous in San Francisco on July 28, 1993, during a public conference on Fermat.
Pizza problem. The owner of a restaurant advertizes a small pizza at $6, a medium size pizza at $9, and a large pizza at $15. Spending $15, do you get a better deal by buying a large pizza, or by buying a small pizza and a medium size one? You may use only a pizza knife to make your decision. Just cut the three pizzas in two equal parts and place the three half pizzas of different sizes as to form a triangle. Three cases can occur.
If the triangle is obtuse (Figure 3.8), your best deal is to take the large pizza since With Figures 3.9 and 3.10, one sees that   , it is better to order a small pizza and a medium size one since 4. Some Diophantine equations. Diophantine equations are often mysterious. Two very similar equations may have very different solution sets, and it may happen that one is difficult to deal with, and the other one is easy to study. We just saw that for n ≥ 3, the equation X n +Y n = Z n has no nontrivial solution. Nevertheless, for all n ≥ 1, the equation possesses the positive solution X = Y = Z = 1. Are there more for n ≥ 3? We will see later that the answer is no, though the proof is deep. We explain why, for n ≥ 2, the Diophantine equation has an infinite number of (nontrivial) solutions, and that it is easy to find all of them. For all n ≥ 0, let A n and B n be defined by where i = √ −1 and where a and b are variables running through Z. On the one hand, we have where z is the complex conjugate of z. On the other hand, (4.5) Hence A 2 n + B 2 n = (a 2 + b 2 ) n . Thus the equation X 2 + Y 2 = Z n has an infinite number of solutions given by X = A n , Y = B n , and Z = a 2 + b 2 , where the binomial expansion of (a + bi) n gives It turns out that all the solutions are of that form, and this is justified by the fact that the class number of the imaginary quadratic field Q(i) is 1. We write a few values of A n and B n in Table 4.1.
It is now easy to deduce that an infinite family of solutions of is given by According to Euler, the equation has an infinity of solutions; it suffices to consider or (4.11) Thanks to Elkies [12], we know that the Diophantine equation also possesses an infinite number of solutions (with W , X, Y , and Z different from 0), with the smallest, according to R. Frye, being W = 95800, X = 217519, Y = 414560, and Z = 422481. We first deal with Elkies [12] exhibited an infinite family of solutions of the latter Diophantine equation: (4.14) Using a judiciously chosen elliptic curve, he next showed that there exist an infinite number of integers a (effectively computable) for which Y is a perfect square, thus giving rise to an infinity of solutions of the equation The equation possesses, for instance, the solution V = 27, W = 84, X = 110, Y = 133, and Z = 144, though we still do not know whether it possesses an infinite number of solutions. Along the same lines, so far nobody could exhibit an integer n (with n ≥ 6) and n−1 nonzero integers such that the sum of the nth powers of these n−1 integers is an nth power. The case where n = 6 reads (4.16) and no solution with UV W XY Z ≠ 0 is known. We conclude this section with three other open problems. , that is, a rectangular box such that the lengths of the three sides and of the four diagonals are integers? In other words, we do not know whether or not there are positive integers a, b, c, d, e, f , and g such that In 1983, a major breakthrough was made by Faltings [13] when he proved that for a fixed n ≥ 4, the equation X n + Y n = Z n has only a finite number of solutions (with no common divisors). As a matter of fact, Faltings obtained in 1986 the Fields Medal for having proved the Mordell conjecture: every smooth algebraic curve of genus g ≥ 2 over any given algebraic number field K has a finite number of K-rational solutions. If one views an algebraic curve Ꮿ as a Riemann surface, the genus of Ꮿ is the number of holes. Since for n ≥ 4, the Fermat algebraic curve X n + Y n = Z n is of genus (n − 1)(n − 2)/2, the curve has only a finite number of positive integral solutions coprime to one another.
We know since 1993 that Fermat's last theorem is true for n ≤ 4000000; this was established with the help of computers. Moreover, a result of K. Inkeri implies that if there exist integers C ≥ B ≥ A ≥ 1 such that A n + B n = C n , then A > 4000000 11999996 . This last integer is so big that if we wanted to write it at full length, it would require more than 70 million digits, which would make it close to 100 kilometers long (with 6 digits per centimeter). However, Wiles wanted to prove Fermat's last theorem definitely without the help of a computer, and so he did! 6. A small detour: the ABC conjecture. As odd as this may look, there exists an open problem concerning the equality and it is called the ABC conjecture. This is one of the deepest conjectures in mathematics and it is far from being proved, though many experts think it is true. The ABC conjecture, formulated in 1985 by J. Oesterlé and D. W. Masser, provides an upper bound for |C| in terms of the product of the prime divisors of ABC. More precisely, first choose a real number ε > 0 (e.g., ε = 0.000001). Next suppose that A+B = C, where A and B have no divisor in common. Then the ABC conjecture asserts the existence of a constant M (depending only on ε) such that Assuming the ABC conjecture, one can give a short proof of the existence of a (noneffective) constant N such that Fermat's last theorem is true for all n ≥ N. Here is how it goes. Choose and fix ε with, for instance, 0 < ε < 1/10. Suppose next that there exist n ≥ 4 and some integers c > b > a > 0, coprime to one another, such that a n + b n = c n . Put A = a n , B = b n , and C = c n . Then the ABC conjecture guarantees the existence of a constant M (depending only on ε) such that a n < b n < c n < M from which we conclude that n is bounded.
Notice to amateurs: Let C > 0. Denote by E what was C, that is, A+B = E; use the letter C for what was R, and take ε = 1. Then the conjecture states that E ≤ MC 2 , and the relativity of this shaky conjecture will certainly scare physicists.
7. Some generalized Fermat equations. For a long time, it was conjectured that 8 and 9 are the only consecutive powers. Many mathematicians contributed numerous partial results till this so-called Catalan conjecture was officially proved by Mihȃilescu (see [30] or [4]), thanks to a clever use of the arithmetic of cyclotomic fields. The result can be stated in the following terms.
Using deep mathematics, namely, elliptic curves à la Wiles, Darmon and Merel [11] solved some variants of the Fermat equation. They proved that the Dénes conjecture is true: for n ≥ 3, the only positive solution of the Diophantine equation

2)
with XY Z = 0 and gcd(X,Y ,Z) = 1, is X = Y = Z = 1. They also proved the following: for n ≥ 4 and for q ∈ {2, 3}, the Diophantine equation has no integral solution with gcd(X,Y ,Z) = 1 and XY Z = 0. We justify their hypothesis gcd(X,Y ,Z) = 1. Assume that n is of the form n = 6m + 5 and that a n + b n = C; then Darmon and Granville [9] and Beukers [3] obtained great results on the Diophantine equation AX p + BY q = CZ r , (7.5) where A, B, and C are nonzero integers. Attach to the last equation the invariant Using a big result of Faltings, Darmon and Granville [9] proved that when w < 1, there are only finitely many integral solutions with gcd(X,Y ,Z) = 1.
Bennett also contributed major results on simultaneous Diophantine equations. In particular, he proved the following [2]: if a, b ∈ N\{0} with a = b, then the simultaneous Diophantine equations have at most three integral positive solutions with XY Z = 0. As a matter of fact, Bennett, supported by some of his results, conjectured that there are at most two solutions.

On certain families of Thue equations.
Consider an algebraic number field K = Q(ω), where ω is a solution of an irreducible polynomial f (X) = X m + a 1 X m−1 + a 2 X m−2 +···+a m−1 X + a m (8.1) of degree m with r real roots and 2s complex roots: m = r + 2s. Inside the ring ᏻ K of algebraic integers of K lives the unit group E K , which, by Dirichlet theorem [37], is isomorphic to a finite group of roots of unity times r + s − 1 copies of Z: It is classical to call {ε 1 ,ε 2 ,...,ε r +s−1 } a fundamental system of units of Q(ω).
For small values of m, some mathematicians exhibited a fundamental system of units {ε 1 ,ε 2 ,...,ε r +s−1 } of K (resp., a maximal independent system of units of K) and when the coefficient a m of f (X) is in {1, −1}, a natural problem is the following one: solve families of Thue equations naturally associated to f (X); namely, exhibit the integral solutions of the Thue equation with c ∈ {1, −1}. Most of the time, Baker's linear forms in logarithms are used and the knowledge of the unit group of K = Q(ω) proves useful.
(K) Heuberger [18] showed that, for |n| > 3.6 × 10 19 , the integral solutions of (L) In [14,15], Gaál and Lettl proved that, for all n ∈ Z, the two integral solutions of (M) Levesque and Mignotte [24] proved that, for all n ≥ 10 12 , the three solutions of the Diophantine equation (N) In [22,23], using hypergeometric methods and Baker's linear forms in logarithms, Lettl, Pethő, and Voutier [22] proved that, for all n ≥ 89, the six integer solutions of 9. Andrew Wiles. We come back to Fermat's last theorem and to the proof of Wiles. The main ingredient of the proof is the theory of elliptic curves. An elliptic curve E (see Figure 9.1) over the field Q of rational numbers can be characterized as the set of rational solutions (i.e., solutions in Q) of an equation of the form with a, b, c ∈ Z. We mention, by the way, that according to Mordell [31], the number of integral solutions (x, y) of E is finite. On the one hand, G. Frey showed in 1985 that if nonzero integers a, b, c ≥ 1 happen to verify a n + b n = c n , n ≥ 5, then the elliptic curve Y 2 = X X − a n X + b n (9.2) is semistable, that is, its conductor (to be defined in the next section) involves only prime integers raised to the power 1. On the other hand, K. A. Ribet proved a few years later that such an elliptic curve Y 2 = X(X −a n )(X +b n ) (built from a hypothetical solution a, b, and c of A n + B n = C n with a, b, c ≠ 0) cannot be modular , that is, cannot be written in terms of certain functions dubbed as modular functions. In a tour de force, Andrew Wiles next proved the following remarkable result: every semistable elliptic curve is modular. If you reread this paragraph, you will see that Fermat's last theorem is proved by contradiction, with the help of this striking result of Wiles.

Modular elliptic curves.
In this section, we give the flavour of the notions of conductor , semistability, and modularity of an elliptic curve E over Q which can be written in the affine plane as and whose discriminant is, by definition, ∆ = −16(4a 3 +27b 2 ) ≠ 0. If p ∆, we say that E has good reduction at p. If p|∆ and if the elliptic curve E, viewed as a curve over Z/pZ, has a double point with two different tangents (resp., with the same tangent), we say that E has multiplicative (resp., additive) reduction at p. The conductor N of E, a divisor of ∆, is by definition N = p|∆ p δp (10.2) with δ p = 1 if the reduction of E at p is multiplicative, δ p = 2 if the reduction of E at p for p ≥ 5 is additive, and δ p ≥ 2 if the reduction of E at p for p ∈ {2, 3} is additive. We say that E is semistable if the conductor N of E happens to be square-free. Denote by ᏺ p the number of solutions of the curve E modulo p, that is, ᏺ p is the number of pairs (x, y) ∈ Z/pZ × Z/pZ which are solutions of the equation of the curve, the equation being considered as a congruence modulo p. Since ∞ (which happens to be the identity element of the group of rational points of the curve) is also a solution, in practice, the number #E(Z/pZ) of points of the curve E over Z/pZ is ᏺ p +1. The pieces of information obtained for all primes p generate the numbers Two points z and w of Ᏼ will be considered as equivalent modulo Γ 0 (N) (in symbols w ∼ z) if there exists M ∈ Γ 0 (N) such that Mw = z. The quotient space Ᏼ/ ∼, classically written as Ᏼ/Γ 0 (N), can be compactified by adding a finite number of the so-called cusps to obtain a compact Riemann surface denoted by X 0 (N) = Ᏼ/Γ 0 (N) ∪ {cusps}. (10.8) A modular form f (z) of level N (and of weight 2) is a function f of a complex variable defined on Ᏼ with values in C, holomorphic on Ᏼ ∪ {cusps}, and verifying In particular, f (z +1) = f ((z + 1)/(0z + 1)) = f (z), which implies that f has a Fourier expansion f (z) = ∞ n=1 a n e 2πinz , a n ∈ C, (10.10) so one associates with f the Dirichlet series ∞ n=1 (a n /n s ). The study of an L-series associated with an elliptic curve E is easier if we know that the L-series associated with E is an L-series attached as above to a modular form f . This is exactly what the Shimura-Taniyama conjecture predicts: for every elliptic curve E over Q, there exists a modular form f whose L-series associated to f is the same as the L-series associated to E. As a matter of fact, the conjecture predicts more, and the reader is invited to look at [5,8,10,16,17,32,33,38].

Epilogue.
Starting from June 23, 1993, the members of the mathematical community intensively studied the proof of Wiles and realized that there was a gap. In a public e-mail dated December 6, 1994, Andrew Wiles himself confessed that the proof was not complete since an upper bound of the order of a so-called Selmer group was missing. With the help of his former Ph.D. student, Richard Taylor, Andrew Wiles overcame this difficulty by using other machinery.
One year later, the proof was complete. On October 11, 1994, a handful of mathematicians, including my colleague Henri Darmon (McGill University), received the long proof of Wiles together with a joint preprint of R. Taylor and A. Wiles. On October 25, 1994, about 20 mathematicians were officially sent this proof. Then fax machines and e-mails got into action, and we know the end of the story. Specialists agreed: this time, devil played no trick and Fermat may rest in peace. The proof appeared in [46]; the proof uses results of leaders in mathematics together with, for the final step, the results of a paper by Taylor and Wiles [40].
A few years later, Breuil et al. [6] proved along the lines of the programme of Wiles that indeed every elliptic curve over Q is modular . This result allowed mathematicians to unconditionally solve other Diophantine equations.