INTRODUCTION TO DIOPHANTINE EQUATIONS

 Loren Spencer
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1 INTRODUCTION TO DIOPHANTINE EQUATIONS In the earl 20th centur, Thue made an important breakthrough in the stud of diophantine equations. His proof is one of the first eamples of the polnomial method. His proof influenced a lot of later work in number theor, including diophantine equations, transcendental number theor, and later eponential sums. In this lecture, we will introduce some basic questions and conjectures and eplain what Thue proved. 1. Naive guesses about diophantine equations The most famous diophantine equation is the Fermat equation d + d z d = 0. For d = 2 there are man integer solutions, and for d 3 there are no positive integer solutions. The proof of the second part is etremel deep and hard. But is there an simple reason to epect that this situation is likel? In this section, we eplore some naive guesses about diophantine equations. Suppose that P is a homogeneous polnomial of degree d in n variables, with integer coefficients. Let us consider the equation P() = A, for some integer A. Let s tr to guess how man solutions this equation is likel to have. Let s tr to guess the number of solutions of size 2 s. Guess the size of the set{ Z n P() = A, 2 s }. Notice that if 2 s, then P() 2 sd. It s hard to sa much else about P() based on the information so far, so we make a primitive probabilistic model. For each with 2 s, let P() be a random integer of norm 2 sd. The epected size of the set { Z n P() = A, 2 s } is 2 ns /2 ds = 2 (n d)s. If the polnomial P behaved randoml, then the set of solutions of size 2 s would be 2 (n d)s. This suggests the following naive conjectures. Naive conjecture 1. If degp < n, then the equation P() = A has infinitel man integer solutions, and the number of solutions of size 2 s is 2 (n d)s. Naive conjecture 2. If degp > n, then the equation P() = A has onl finitel man integer solutions. (The case degp = n is more delicate. Our heuristic gives that the number of solutions of size 2 s is 1, which would suggest infinitel man solutions. But having no solutions would not be such a large departure from the estimate in the heuristic...) 1
2 2 INTRODUCTION TO DIOPHANTINE EQUATIONS These conjectures are both false, but the are still useful. We give some countereamples. Consider the equation = 1. Our model predicts it should have man solutions, but it has none because the lefthand side is alwas even. Therefore, naive conjecture 1 is false. (To fi this particular countereample, we should also assume that the equation has lots of solutions modulo p for some small primes p.) Consider the equation ( z 2 ) 10 = 0. This equation has degree 20 in 3 variables, but ever Pthagorean triple is a solution. There are also eamples in two variables. The equation ( ) 10 = 1 has infinitel man solutions. The equation = 1 has infinitel man solutions (approimatel one for each scale 2 s, as predicted b the heuristic). Therefore, the equation ( ) 10 = 1 has infinitel man solutions. (We can rule out these particular countereamples b insisting that P is irreducible.) Although the conjectures are false, the give some useful intuition. If the degree d > n and there are infinitel man solutions, then that seems to be a big coincidence, and one ma hope that there is some structure that eplains what is happening. Two of the big achievements in diophantine equations from the earl 1900 s confirm this intuition. The circle method of HardLittlewood proves that equations have lots of solutions if the number of variables is much larger than the degree and if nothing bad happens modulo p for small primes p. Thue proved that Naive Conjecture 2 is actuall true in two variables, as long as the polnomial is irreducible. Theorem 1.1. (Thue) Suppose P Z[, ] is a homogeneous polnomial with degree 3 which is irreducible (over Z). If A is an integer, then the equation P() = A has onl finitel man integer solutions. 2. Diophantine approimation Thue actuall proved an even stronger theorem about rational approimations of algebraic numbers. To see the connection, let us consider the equation = 7. If (, ) Z 2 solves this equation, then we see that ( ) 3 2 = 7 3. Therefore, / is a good approimation of the cube root of 2, especiall if is large. A short calculation shows that 2 1/3 3. These are reall ver good rational approimations. For contet, consider the following.
3 INTRODUCTION TO DIOPHANTINE EQUATIONS 3 Proposition 2.1. For an ǫ > 0, for almost ever real number β, there are onl finitel man integer solutions to the inequalit β 2 ǫ. ( ) (The proof is a standard eercise in measure theor. Consider all the β in an interval so that ( ) has a solution with > Y. This set is a union of intervals of total length Y ǫ.) Although it s eas to prove this result for almost ever β R, it s hard to check it for an particular β, sa β = 2 1/3. Liouville gave the first estimates about diophantine approimation of algebraic numbers. Proposition 2.2. (Liouville, 1840 s?) If β is an irrational algebraic number and is a rational number, then β c(β) deg(β). Recall that an algebraic number is a solution to a polnomial with integer coefficients. The degree degβ is the minimal degree of such a polnomial. We will use a couple basic facts about algebraic numbers. There is actuall a unique minimal polnomial Q with Q(β) = 0. (Minimal here means that the degree and the size of the coefficients are minimal.) The polnomial Q will be irreducible over Z, and so it will have no rational roots. This polnomial also has Q (β) = 0. Proof. Notice that Q(/) is a nonzero rational number. The denominator can be taken to be deg(β). Therefore, Q(/) deg(β). If / is ver close to β, then Q(/) = Q(β) + Q (β)(β /) + lower order terms Q (β) β /. For eample, 2 1/3 c 3. This inequalit is not strong enough to sa anthing about the number of solutions of = 7. If we look back inside the proof of the Liouville inequalit, it boils down to saing that = 0 has no integer solutions and so But this does nothing to constrain the solutions of = 7. However, an inequalit even slightl stronger than Liouville s does constrain the solutions to diophantine equations. Theorem 2.3. (Thue) If β is an irrational algebraic number, and γ > deg(β)+2, then 2 there are onl finitel man integer solutions to the inequalit β γ.
4 4 INTRODUCTION TO DIOPHANTINE EQUATIONS The diophantine approimation theorem implies the diophantine equations theorem for the following reason Outline of Thue s proof In this section we outline Thue s proof, and we eplain how it is analogous to other arguments we have seen. We recall the main steps in the polnomial method b outlining the proof of the finite field Nikodm theorem. Outline of the proof of finite field Nikodm: Suppose that N is a small Nikodm set in F n. (1) Find a nonzero polnomial P with controlled degree that vanishes on N. (Use parameter counting.) (2) Because N is a Nikodm set, the polnomial P must also vanish at man other points. (Vanishing lemma.) (3) The polnomial P vanishes at too man points, so it must be zero. Contradiction. Here is the outline of Thue s proof. Suppose that the algebraic number β has two ver good rational approimations r 1 and r 2. (1) Find a nonzero polnomial P Z[, ] with controlled degree and coefficients that vanishes to high order at (β, β). (Use parameter counting.) (2) Because r 1 and r 2 are good approimations of β, the polnomial must also vanish to high order at (r 1, r 2 ). (3) The polnomial P vanishes too much at (r 1, r 2 ), and so it must be zero. Contradiction. The first step took Thue the longest to figure out. In the special case that β is the d th root of a rational number, he constructed the polnomial P b hand with some difficult. In this wa, we was able to prove his finiteness theorem onl for equations of the form A d + B d = C. After tring hard to construct the polnomial P for other values of β, Thue realized that he could find it b parameter counting. Another important point about Thue s proof is that it uses two good rational approimations r 1 and r 2. It might seem simpler to start with one rational approimation r and tr to get a contradiction. But it seems ver difficult to do this. We will come back to this point more later Final comment. Suppose that α and β are two algebraic numbers. Then α+β is an algebraic number. Wh? This is a bit in the same spirit as the polnomial method...
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