Contents. 6 Continued Fractions and Diophantine Equations. 6.1 Linear Diophantine Equations

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1 Number Theory (part 6): Continued Fractions and Diophantine Equations (by Evan Dummit, 04, v 00) Contents 6 Continued Fractions and Diophantine Equations 6 Linear Diophantine Equations 6 The Frobenius Coin Problem 3 6 The Equation x + y = z : Pythagorean Triples 4 63 Continued Fractions 5 63 Convergents to Continued Fractions 6 63 Innite Continued Fractions Periodic Continued Fractions 0 64 The Farey Sequences 65 Rational Approximation 3 66 Pell's Equation 6 67 An Assortment of Diophantine Equations 8 68 Some Remarks on Fermat's Last Theorem 0 6 Continued Fractions and Diophantine Equations In this chapter, we discuss Diophantine equations, which are concerned with the problem of solving equations over the integers: one of the earliest nontrivial examples was posed by Diophantus, whence the general name of Diophantine equation for this class of problems One of the most famous Diophantine equations is the Fermat equation x n + y n = z n, which we will make a central focus of studying Another important equation is Pell's equation x Dy =, whose solution is related to another important topic in classical number theory: continued fractions and rational approximation We begin by analyzing a few simple Diophantine equations, then discuss the ideas behind rational approximation and develop the theory of continued fractions Finally, we apply these results, and others, to solve various Diophantine equations There is no general procedure for deciding whether a given Diophantine equation possesses any solutions, or (even if existence is known) for nding them all (This can in fact be formalized and proven using methods of mathematical logic) Thus, many of the methods for solving Diophantine equations are rather ad hoc, and so our goals in this chapter are primarily to provide a survey of elementary techniques One recurring theme, however, will be the structure of Z[ D] and closely related rings 6 Linear Diophantine Equations The simplest equations are linear equations in two variables (Solving a linear equation in one variable is, of course, trivial) The general form of such an equation is ax + by = c, for some xed integers a, b, and c: our goal is to determine when this equation has an integral solution (x, y), and then to characterize all the solutions Theorem: Let a, b, c be integers with ab 0, and set d = gcd(a, b) If d c, then the equation ax + by = c has no solutions in integers (x, y) If d c, then the equation has innitely many solutions, and if (x 0, y 0 ) is one solution, then all the others are (x 0 bt/d, y 0 + at/d), for some integer t

2 Remark: If a = b = 0, then the equation is either trivially true (if c = 0) or trivially false (if c 0), so we can assume that the gcd exists If one of a, b is zero, the equation is also trivial, so assuming that ab 0 loses nothing Proof: Observe that there is an integral solution to ax + by = c if and only if there is a solution to the congruence ax c (mod b), since then y = c ax b Now from our discussion of linear congruences, we know that ax c (mod b) has a solution only if d = gcd(a, b) divides c In this case, if we set a = a/d, b = b/d, and c = c/d, the set of all such x is given by the residue class x 0 modulo b, where x 0 c (a ) (mod b ) Now if (x, y) is any solution, then by the above, we see that x = x 0 bt/d for some integer t, and then y = y 0 + at/d Example: Find all solutions to 4x + 8y = in integers (x, y) First, we compute gcd(4, 8) =, and then divide through by the gcd to get 7x + 9y = 6 This is equivalent to solving 7x 6 (mod 9) We compute (via the Euclidean algorithm) that the inverse of 7 mod 9 is 4, so multiplying both sides by 4 yields x 4 6 (mod 9) Hence one solution is (x, y) = (6, 4) The set of all solutions is then (x, y) = (6 9t, 4 + 7t) for t Z We could also solve equations of this form using changes of variable: Example: Find all solutions to the equation 4x + 3y = 5 By the division algorithm, we have 3 = 3 4+, so we can write the system in the form 4x+(3 4+)y = 5, and rearrange this into the form 4(x + 3y) + y = 5 If we substitute u = x + 3y, this new system becomes 4u + y = 5, which we can easily solve to get y = 5 4u Substituting back yields x = u 3y = u 3(5 4u) = 5 + 3u Thus, we obtain the general solution (x, y) = ( 5 + 3u, 5 4u) This latter method, using changes of variable, is the proper way to solve systems of linear Diophantine equations The study of the solutions to a system of linear equations (with coecients in a eld, often Q, R, or C) is a basic problem of linear algebra The standard solution technique is to convert the system into matrix form, and then perform row and column operations on the matrix until it is in a suciently simple form that the solution to the original system is obvious The general procedure for solving a system of linear equations over Z is essentially the same, except for the added complication that all of the row and column operations need to be done over Z As with a system of equations over a eld, the end result will be either that the system has no solution, a unique solution, or an innite family of solutions with some number of free parameters We will not go into the technical details, since the procedure falls more properly into a course in linear algebra or abstract algebra Instead, we will just give an example Example: Find all solutions to 3x + 5y + 7z = in integers (x, y, z) Motivated by the division algorithm, we rewrite the equation as 3(x + y + z) + y + z =, and then substitute w = x + y + z The new equation is 3w + y + z =, which we can easily solve, obtaining z = y 3w Solving for x yields x = w y z = + 7w + 3y, so we obtain the general solution (x, y, z) = ( + 7w + 3y, y, y 3w), where w, y are arbitrary integers

3 6 The Frobenius Coin Problem Above, we characterized when there exists a solution to the equation ax + by = c in integers (x, y) In various settings, we are interested in knowing for which values of c this equation has a solution in nonnegative integers (x, y) If a and b are not relatively prime, clearly c must be divisible by their gcd, and (by dividing through by the gcd) we can reduce to the case where a and b are relatively prime One version of this problem uses postage stamps, which often cost irregular amounts: if, for example, there are postage stamps worth 5 cents and stamps worth 3 cents, is it possible to use them to put exactly 79 cents' worth of postage on an envelope? The most obvious method is simply to make a list of totals that are attainable: 0, 5, 0, 3, 5, 8, 0, 3, 5, 6, 8, 30, 3, 33, 35, 36, 38, 39, 40, 4, 43, 44, 45, 46, 48, 49, 50, 5, 5, 53, 54, 55, 56, 57, 58, 59, 60, Based on our list, it seems that every value above 47 is attainable (Indeed, it is easy to see that since 48, 49, 50, 5, and 5 are all attainable, we can obtain any larger number by adding additional 5-cent stamps) Thus, for example, we get exactly 79 cents of postage by using three 3-cent stamps and eight 5-cent stamps Another version occurs in sports: In American football, a team can score 3 points for a eld goal, or 7 points for a touchdown What possible scores can a team obtain? (Ignore safeties, missed extra points, and so forth) Like above, we can simply list the totals that are attainable: 0, 3, 6, 7, 9, 0,, 3, 4, 5, 6, 7, 8, 9, 0,,, 3, Based on our list, only a few values are unattainable:,, 4, 5, 8, and Indeed, it is easy to see that since, 3, and 4 are attainable, any larger number is also attainable by adding more eld goals The problem of describing the largest integer that cannot be written as a nonnegative linear combination of two integers (sometimes called the Frobenius coin problem) was rst solved by Sylvester: Theorem: If a and b are relatively prime integers, then there are exactly (a )(b ) integers that cannot be written in the form ax + by with x, y 0, and the largest such integer is ab a b Proof: For brevity, we say an integer is representable if it can be written in the form ax + by with x, y 0 Without loss of generality, assume a < b Arrange the nonnegative integers in an array in the following manner: 0 a a a + a + a a a + a + 3a ab a ab a + ab a + ab Now we use the array to mark all of the representable integers We rst circle all of the multiples of b: then an integer is representable precisely if it appears in the same column as some multiple of b, lower down For illustration, here is the array with a = 3 and b = 5:

4 Since a and b are relatively prime, the integers 0, b, b,, (a )b all lie in dierent columns Thus, the largest element that is left unmarked is the element one row above (a )b, which is ab a b, so this is the largest integer not expressible as ax + by with x, y 0 For the other part, we simply count the number of unmarked integers in the array: the number of integers a kb lying above kb is kb/a, so there are a total of unmarked integers in the array a k=0 We can interpret this as the number of lattice points lying under the line y = b x, with x a, a which is (equivalently) the number of lattice points lying below the diagonal of the rectangle with vertices (0, 0), (a, 0), (a, b), and (0, b) and not on any of the sides By symmetry, since there are no lattice points on the interior of the diagonal, this is exactly half of the lattice points lying strictly inside the a b rectangle, and there are (a )(b ) such lattice points Remark: An alternate method for obtaining this count is to prove that if 0 c ab a b, then c is representable if and only if ab a b c is not representable For a = 5 and b = 3, we see that the largest non-representable integer is = 47, which is indeed what we found above We could of course generalize this problem, to ask: for given integers a, a,, a k, what is the largest integer n that cannot be written as a nonnegative integer linear combination of the a i? It turns out that there is no general formula when k > In fact, for large k, there is not even a known algorithm for nding the answer that is appreciably faster than merely attempting to list the possibilities! 6 The Equation x + y = z : Pythagorean Triples Now that we have discussed solving linear equations in integers, we turn our attention to the simplest quadratic Diophantine equation: characterizing integer triples (x, y, z) such that x + y = z Such triples are naturally called Pythagorean triples because (by the Pythagorean theorem) they form the sides of a right triangle: a familiar example is the nearly-ubiquitous (3, 4, 5) triangle Since the dening equation is homogeneous (ie, all of the terms have the same degree), if we have one Pythagorean triple, we can create more by scaling x, y, and z: thus we obtain (6, 8, 0), (9,, 5), and so forth from (3, 4, 5) To exclude these essentially repetitious cases, we say a Pythagorean triple (x, y, z) is primitive if gcd(x, y, z) =, and would like to characterize the primitive triples First we note that if (x, y, z) is a primitive Pythagorean triple, x and y cannot both be even, since then z would also be even Also, x and y cannot both be odd, since then x + y (mod 4), but is not a square modulo 4 So we conclude that z must be odd, and that exactly one of x and y is also odd Theorem: Every primitive Pythagorean triple of the form (x, y, z) with x even is of the form (x, y, z) = (st, s t, s + t ), for some relatively prime integers s > t of opposite parity, and (conversely) any such triple is Pythagorean and primitive For the reverse direction, it is easy to see that (st) + (s t ) = (s + t ) simply by multiplying out Furthermore, it is easy to check that if s and t are relatively prime and have opposite parity, that gcd(s t, s + t ) =, so this triple is primitive We will give three proofs of the nontrivial direction: arithmetic of Z[i], and one using geometry one using the arithmetic of Z, one using the The central idea in the rst proof is to rearrange the equation and use the arithmetic of Z The central idea in the second proof is to exploit the arithmetic of Z[i], while the central idea in the third proof is to use the geometry of the dehomogenized curve x + y = to study the rational solutions Proof : Since y and z are both odd and x is even, we can rewrite the equation as z y 4 z + y ( x ) =

5 Now we claim that z y and z + y are relatively prime: their gcd divides their sum z and their dierence y, and since y and z are relatively prime, the gcd must be Since z y and z + y share no prime divisors and their product is a square, each of them must individually be a square, by the uniqueness of prime factorization Hence there exist integers s and t such that z y = t and z + y = s Then z = s + t and y = s t, and then we also obtain x = st, as claimed Furthermore, s and t are necessarily relatively prime and have opposite parity, since (x, y, z) is primitive Proof : In Z[i], we factor the equation as (x + iy)(x iy) = z Now we claim that x + iy and x iy are relatively prime: any gcd must divide x and y, hence divide However, x + iy is not divisible by the prime + i, since x and y are of opposite parity Hence, since x+iy and x iy are relatively prime and have product equal to a square, by the uniqueness of prime factorization in Z[i], there exists some s + it Z[i] and some unit u {, i,, i} such that x + iy = u(s + it) Multiplying out yields x + iy = u [ (s t ) + (st)i ] Since x is even and y is odd, we must have u = ±i: then writing out the various possibilities yields the given parametrization ( x ) ( y ) Proof 3: Dividing by z yields the equivalent equation + =, so it is sucient to describe z z all points (a, b) on the unit circle x + y = whose coordinates are both rational numbers To do this, consider all non-vertical lines passing through the point (, 0) Such a line will intersect the circle x + y = in exactly one other point If the coordinates of this point are rational, then the line will have rational slope Conversely, if the line has rational slope t s, its equation is y = t (x + ) so we can simply compute ( s s t ) the other intersection point to see that it is (x, y) = s + t, st s + t, which is rational ( s t ) Thus, the rational points on the unit circle are those of the form s + t, st s + t for some integers s and t Clearing the denominator yields the desired Pythagorean triples Remark: The third proof is closely related to the Weierstrass substitution u = tan(θ/), which transforms an integral of any rational function of sin(θ) and cos(θ) into an integral of a rational function of u, which can then be evaluated using partial fraction decomposition (With the notation above, u = s/t) Using the characterization above, we can easily generate a list of all the primitive Pythagorean triples: ordered by the length of the hypotenuse, the rst few are (3,4,5), (5,,3), (8,5,7), (7,4,5), (0,,9), (,35,37), (9,40,4), (8,45,53), 63 Continued Fractions Denition: A nite continued fraction is an expression of the form a 0 + a + a + a a k + a k where the a i are positive real numbers For brevity, we will denote this expression using the much more compact notation [a 0, a,, a k ] If the a i are integers, we term it a simple continued fraction, Example: [, 3, 4] = = 30 3 We note a few very basic properties: [a 0, a,, a k ] = a 0 + [a,, a k ] = [a 0, a,, a k + a k ] 5

6 Clearly, every simple continued fraction is a rational number Conversely, every rational number can be written as a simple continued fraction: if p/q is any positive rational number in lowest terms, then if we apply the Euclidean algorithm to write p = q q + r q = q r + r r = q 3 r + r 3 r k = q k r k + r k = q k+ where we know the last remainder will be since p/q is in lowest terms, then it is an easy induction to verify that p q = [q, q,, q k, q k+ ] Furthermore, by the uniqueness of the Euclidean algorithm, all of the quotients are unique, so the expression is unique, except for the fact that we can write [q, q,, q k ] = [q, q,, q k, ] If we exclude the case where the nal term is equal to, then every positive rational number can be written uniquely as a continued fraction Example: To convert 7 7 so that, by the above, 7 7 into a continued fraction, we rst compute = [,, 3] 7 = = = 3 Another way of doing this is just to work it out explicitly, by writing 7 7 = = + 7/3 = + Example: To convert 67 9 so that 67 = [3,,, 9] 9 into a continued fraction, we compute 67 = = = = Convergents to Continued Fractions Denition: If C = [a 0, a,, a k ] is given, then the continued fraction C n = [a 0, a,, a n ] for n < k is called the nth convergent to C Example: For 7 0 = [, 6, 3, 5], the successive convergents are [] =, [, 6] = 7, [, 6, 3] = 6 9, and [, 6, 3, 5] = 7 0 Observe that , while and Notice that the convergents are fairly close to the actual value of the continued fraction, and their accuracy improves as we take higher convergents We will return to this idea in a moment 6

7 There is a simple recursive relation that allows us to compute the convergents C i Proposition: Let C = [a 0, a,, a k ], and dene p = q = 0 p = a 0 q 0 = and for n k, set Then C n = p n / p n = a n p n + p n = a n Proof: For the rst statement, we use induction The cases n = and n = are trivial to verify, since [a 0 ] = a 0 / and [a 0, a ] = a 0 + /a = (a 0 a + )/a, as claimed For the inductive step, suppose we know that the result holds for n m By hypothesis, for any x, it is the case that [a 0, a,, a m, x] = p m x + p m q m x + q m Now observe that [a 0, a,, a m, a m, a m+ ] = [a 0, a,, a m, a m + a m+ ]: now apply the above result with x = a m + a m+ to obtain which is the desired result [a 0, a,, a m, a m + a m+ ] = ( a m + ) p m + p m a ( m+ a m + ) q m + q m a m+ = (a mp m + p m ) + p m /a m+ (a m q m + q m ) + q m /a m+ = p m + p m /a m+ q m + q m /a m+ = a m+p m + p m a m+ q m + q m = p m+ q m+, Using the expressions above, we can deduce a few simple corollaries about the p i and q i : Corollary: Let C = [a 0, a,, a k ], with the p i and q i as above p n p n = ( ) n a n Then p n p n = ( ) n and For the rst statement, by the recursion we can write p n p n = (a n p n + p n ) p n (a n ) = (p n p n ) so since p q 0 p 0 q =, by a trivial induction we see that p n p n = ( ) n The second statement follows in the same way (We skip the algebra) Using the result above, we can show that the convergents actually converge to the value of the original continued fraction C: Corollary: Let C = [a 0, a,, a k ], with the p i and q i as above For C i = p i /q i, we have C n C n = ( )n and C n C n = ( )n a n Furthermore, C > C 3 > C 5 > > C 6 > C 4 > C, and C C n < + q n 7

8 The rst two statements follow immediately by dividing the relations p n p n = ( ) n and p n p n = ( ) n a n by and respectively From the statement C n C n = ( )n a n, we see that C n < C n if n is odd, and C n > C n if n is even Hence, by a trivial induction, we see C > C 3 > C 5 > and > C 6 > C 4 > C Furthermore, since C n+ > C n for every n, we can combine the two chains of inequalities to obtain the third statement For the last statement, we simply observe that the inequalities above imply that C is between C n and C n+ for every n, hence the triangle inequality implies C C n C n+ C n = < + qn 63 Innite Continued Fractions We have discussed nite continued fraction expansions (of rational numbers) and shown that their convergents obey some nice relations, and we can compute the simple continued fraction expansion of any rational number using the Euclidean algorithm We would now like to extend our discussion to include irrational numbers: what, for example, does it mean to ask for the continued fraction expansion of? Or of π, or ln()? None of these is a rational number, so any such expansion cannot be nite To handle this, we simply extend our denition of continued fraction to an innite continued fraction by taking a limit Denition: Given a sequence a 0, a, of positive integers, we dene the innite continued fraction α = [a 0, a, ] to be the limit lim n [a 0, a,, a n ] of its nite continued fraction convergents It is worthwhile explaining why this limit exists From our work above, if C n = [a 0, a,, a n ], then C > C 3 > C 5 > > C 6 > C 4 > C So the sequence C, C 3, C 5, is monotone decreasing and bounded below (by C ), hence it has a limit Similarly, the sequence C, C 4, C 6, is monotone increasing and bounded above (by C ), hence it also has a limit These two limits must be equal because C n C n+ < qn, which tends to zero as n Alternatively, we could appeal to the fact that the intervals [C n, C n ] form a set of nested closed intervals of lengths tending to zero, so by standard properties of R, their intersection is a single point C equal to the limit of the sequence C i Proposition: If α = [a 0, a, ] is an innite continued fraction and C n = [a 0, a,, a n ], then α C n < + qn The proof follows identically from the nite case, since α lies between C n and C n+ Since rational numbers have a nite continued fraction expansion (unique up to replacing the last term in an essentially trivial way), we would expect that any innite continued fraction must be irrational We would also expect that distinct innite continued fractions have dierent values Proposition: If α = [a 0, a, a, ] is an innite continued fraction, then α is irrational Furthermore, any two dierent innite continued fractions converge to dierent values Proof: For the rst statement, suppose α = p/q were rational By the proposition above, we know that 0 < p q p n <, meaning that 0 < p p n q < q q n 8

9 However, q goes to zero as n, since q is xed but is a strictly increasing sequence This is impossible, since if > q the expression p p n q would be an integer between 0 and For the second statement, rst observe that C 0 < α < C, meaning that a 0 < α < a 0 + a, so we see that α = a 0 Next, observe that α = lim n [a 0, a,, a n ] = lim n ( a 0 + ) = a 0 + [a,, a n ] [a, a, ] Now suppose β = [b 0, b, ] and β = α By taking oors, we see that b 0 = a 0 Then [b, b, ] = = = [a, a, ] Taking oors again shows b = a β b 0 α a 0 Repeating the argument yields b i = a i for every i, so α and β are identical So far, we have discussed the ideas behind innite continued fractions, but we have not actually computed any! It is not hard, from the above, to work out the procedure for converting an irrational number α into an innite continued fraction [a 0, a, a, ] First, we must have a 0 = α, as we observed above Then, as we also observed, [a, a, ] =, so if we dene α = (which is greater than α a 0 α a 0 because 0 < α a 0 < by irrationality of α and the denition of the oor function), we must have a = α Now we repeat: we set α = and take a = α α a In general, we obtain the terms recursively, via the relations a 0 = α, α i =, and a i = α i α i a i As noted above, each of the a i will be a positive integer, because α i will always be greater than We should verify that the resulting continued fraction [a 0, a, a, ] actually converges to α To do this, we observe (essentially by the denition) that α = [a 0, a,, a n, α n+ ], and then compute α [a 0, a,, a n ] = p n α n+ + p n p n α n+ + = < ( α n+ + ) because α n+ is positive Since this tends to zero as n, we see that α is indeed equal to lim n [a 0, a,, a n ] Example: Find the continued fraction expansion of With α =, we nd, successively, n 0 α n + + a n α n a n and since each term after this will repeat, we see that = [,,,,, ] Example: Find the continued fraction expansion of 7 9

10 With α = 7, we nd, successively, n α n a n α n a n and since each term after this will repeat, we see that 7 = [,,,, 4,,,, 4, ] Example: Find the continued fraction expansion of π With α = π, we nd, numerically, that the rst ten terms are [3, 7, 5,, 9,,,,,, ] This is easy to do even with a hand calculator: simply subtract o the integer part, reciprocate, and repeat There is no apparent pattern, and the sequence does not seem to repeat, in the nice way that the two previous examples did 633 Periodic Continued Fractions Two of the continued fractions above both eventually repeat We will give this situation a special name: Denition: An innite continued fraction [a 0, a, a, ] is periodic if there is some integer n such that a r = a n+r for all suciently large r We employ the notation [a 0, a, a,, a k, a k+, a k+,, a k+n ] to indicate that the block of integers under the bar repeats indenitely This is the same notation used for repeating decimals (This is reasonable, since it is essentially the same situation, too) Example: Find the real number α = [] By the periodicity of the expansion, we know that α = + α = α + α This yields a quadratic equation for α, namely α = α +, whose solutions are α = ± 5 Since α >, we need the plus sign, so α = + 5 (This is the famous golden ratio) Remark: It is also worth computing the convergents to α: they are,, 3, 5 3, 8 5, 3 8, 3, 34, 55, and so 34 forth Notice that these are simply ratios of consecutive Fibonacci numbers, which (once noticed) is easy F n+ to prove by induction In fact, our results about convergence provide a proof that lim = + 5 n F n Example: Find the real number α = [, 5] By the periodicity of the expansion, we know that α = α = + α 5α + = α + 5α + This yields a quadratic equation for α, namely α(5α + ) = α +, whose solutions are α = 5 ± 35 5 Since α >, we need the plus sign, so α = So far, all of the periodic continued fractions we have seen have been solutions of a quadratic polynomial in Q[x] This is not an accident: 0

11 Theorem: If α is a periodic continued fraction, then α is an irrational root of a quadratic polynomial in Q[x] Conversely, if α > 0 is irrational and a root of a quadratic polynomial in Q[x], then α has a periodic continued fraction expansion Proof (forward): Let α = [a 0, a, a,, a k, a k+, a k+,, a k+n ], and γ = [a k+, a k+,, a k+n ] Then by the periodicity of the expansion, we have γ = [a k+,, a k+n, γ] Expanding this out yields γ = p n γ + p n γ +, which is a quadratic equation for γ Since γ is irrational (being an innite continued fraction), we conclude that γ = b + c for some integers d b, c, and d Then α = [a 0, a, a,, a k, γ] is also a rational function in γ (and irrational), so clearing the denominator shows that α = e + f for some integers e, f, and g, which is also a root of a quadratic polynomial g Remark: For the reverse direction, the idea is to show that for a given α = b + c, there are only a d nite number of possible α i that can appear when computing the terms in continued fraction expansion, and so eventually a repetition must occur There does not seem to be an especially brief argument doing this, so we will omit the details 64 The Farey Sequences If α is an irrational number, we have seen that the continued fraction convergents to α provide very good rational approximations of α Specically, if p/q is a convergent to α, then we showed earlier that α p q < q We would like to know how good of an approximation this gives, compared to choosing some other rational numbers One thing we might rst look at is the set of rational numbers of small denominator Since we want to understand distances between nearby numbers, we should arrange the rationals in increasing order Denition: The Farey sequence of order n is the set of rational numbers between 0 and whose denominators (in lowest terms) are n, arranged in increasing order Example: The Farey sequence of order 4 is 0, 4, 3,, 3, 3 4, Example: To obtain the Farey sequence of order 5, we simply insert the terms with denominator 5 in 0 the proper locations:, 5, 4, 3, 5,, 3 5, 3, 3 4, 4 5, There are several natural and immediate questions about this sequence: for example, how many terms does it have? Are consecutive terms related to each other? In going from the (n )st to the nth sequence, how many terms are added and where do they go? Several of these can be immediately answered: of all rational numbers of the form k, the ones in lowest n terms will be those with gcd(k, n) = Thus, the number of such terms added in going from the (n )st Farey sequence to the nth is simply ϕ(n) n By a trivial induction, we see that the length of the Farey sequence of order n is + ϕ(d) It is a nontrivial problem to estimate the rate of growth of this function, but it turns out to be approximately 3 π n A little numerical experimentation quickly leads to a discovery of the following result: d=

12 Proposition: If a b and c are consecutive terms in the Farey sequence of order n, then bc ad = d Proof: Suppose a b and c are consecutive terms in the Farey sequence d In the plane, draw the triangle whose vertices are (0,0), (b, a), and (d, c) By Pick's Theorem, the area of this lattice-point triangle is B + I, where B is the number of lattice points on the boundary and I is the number of points in the interior We claim B = 3 and I = 0 To see this, suppose there were a lattice point (x, y) in the interior, where (necessarily) y max(b, d) Then the slope of the line joining (0, 0) to (x, y) would lie strictly between a/b and c/d: but then y/x would be between a/b and c/d in the Farey sequence, which by hypothesis it is not Now suppose there were a lattice point on the boundary not equal to one of the vertices It cannot lie on the side joining (0,0) and (b, a), since a and b are relatively prime Similarly, it cannot lie on the side joining (0,0) and (d, c) If it were on the side joining (b, a) and (d, c), then by the same argument given above, there would be a term between a/b and c/d in the Farey sequence Thus, B = 3 and I = 0, so the triangle has area By basic geometry, the area of this triangle is ad bc, so since bc > ad we conclude immediately that bc ad = From this proposition, we obtain a number of very useful corollaries Corollary: If a b, e f, and c d are three consecutive terms in the Farey sequence, then e f = a + c b + d Notation: This last expression is sometimes called the mediant of a/b and c/d (It is also occasionally called the baseball average, since it is the expression, frequently computed in baseball, used when combining several hit percentages into a single statistic) Proof: By the result just proven, we know that be af = and that cf de = This is a system of two linear equations in the two variables e and f, so solving it yields e = (a + c)/(bc ad) and f = (b + d)/(bc ad): thus, e f = a + c, as claimed b + d One can check directly that a + c b + d appears between a b and c in the Farey sequence of order b + d d Corollary: If a b and c are rational numbers between 0 and with bc ad =, then these two terms are d consecutive entries in the Farey sequence of order max(b, d) The rst term that will appear between them (in a later sequence) is a + c, and this rst occurs in the Farey sequence of order b + d b + d Proof: Suppose e f is the term immediately following a in the sequence of order max(b, d) Then be af = b, so subtracting from bc ad = yields b(c e) a(d f) = 0, so b(c e) = a(d f) Since a and b are relatively prime, we conclude that b divides d f Since f max(b, d) < b + d, the only possibility is that f = d, and then e = c Alternatively, we could have observed that both (e, f) and (c, d) are solutions to the linear Diophantine equation bx ay =, and used the structure of the solutions to deduce this result For the second statement, we just showed that a/b and c/d are consecutive in the Farey sequence of order max(b, d) Now increase the order of the sequence in increments of : if e/f is the rst term to appear between a/b and c/d, then as shown above, it would necessarily be the case that e = (a+c)/(bc ad) = a+c and f = (b + d)/(bc ad) = b + d Corollary: The rational numbers a b and c d if bc ad = and b + d > n are consecutive terms in the Farey sequence of order n if and only Proof: If the rst condition fails, then there is a term between a/b and c/d whose denominator is at most max(b, d) If the second condition fails, then a + c is a term between a/b and c/d b + d

13 If both conditions hold, then the corollary above immediately implies that there are no terms between a/b and c/d in the Farey sequence of order n Corollary: If a/b and e/f are consecutive terms in the Farey sequence of order n, the term immediately n + b n + b following e/f is c/d, where c = e a and d = f b f f Proof: By the mediant property, we know that e f = a + c Thus, there must exist some integer k such b + d that a + c = ke and b + d = kf, so that c = ke a and d = kf b Since the closest term to e/f will n + b have k as large as possible, and since d n, the largest possible value of k is f Using the above results, we can construct the portion of any Farey sequence around any desired rational number, without needing to compute all of the terms in the sequence Example: Find the rst three terms after /0 in the Farey sequence of order 500 By the above results, if /0 and c/d are consecutive terms, then 0c d = Solving this Diophantine equation using the Euclidean algorithm produces the solutions (c, d) = (3 + k, k) for k Z The larger the value of k is, the smaller the value of c d 0 = will be The largest possible value 0d for k is k =, so the rst term is Now we can apply the two-term recursion to /0 and 5/457 to quickly nd the next terms: they are 4/55 and 7/308 Thus, the three terms are 5 457, 4 55, Similarly, we can ll in the portion of any desired Farey sequence between any two given rational numbers simply by taking mediants Example: Find all terms between 7/33 and 4/65 in the Farey sequence of order 00 First, we notice that these terms are not consecutive (in any Farey sequence), because = 7, not We start by nding terms between them: the mediant of these two terms is /98 = 3/4 Now 7/33 and 3/4 are consecutive in the Farey sequence of order 33, since = Also, 3/4 and 4/65 are consecutive in the Farey sequence of order 65, since = Now we just need to ll in the missing terms By the results above, these are all given by computing mediants of the terms already found We obtain the sequence 7 33, 7 80, 0 47, 3 6, 6 75, 9 89, 3 4, 0 93, 7 79, Rational Approximation Now that we have discussed the Farey sequences, we can use them to discuss the approximation of irrational numbers by rational numbers, and compare the results to the ones we obtained using continued fractions Proposition: If n is a positive integer and α is any real number, then there is a rational number p q 0 < q n and α p q q(n + ) such that Proof: By replacing α with α α as necessary, we can assume that α lies in [0, ] 3

14 Now consider the Farey sequence of order n, and let a b and c be two consecutive terms such that d a b x c By our earlier results, we know that bc ad = and b + d n + d [ a The number α either lies in the interval b, a + c ] [ a + c or in b + d b + d, c ] d In the rst case, α a a b b a + c ad bc b + d = b(b + d) b(n + ) In the second case, α c c d d a + c ad bc b + d = d(b + d) d(n + ) Thus, in either case, we obtain a rational number p q such that α p q q(n + ) Corollary: If α is any irrational real number, then there are innitely many distinct rational numbers p q such that α p q < q Proof: Apply the previous result to the Farey sequence of order n for each n: this yields a collection of rational numbers p n such that α p n < (n + ) < qn, and with n Since α is irrational, none of these dierences can be zero, and so there must be innitely many dierent terms p n, since the distances α p n become arbitrarily small, but remain nonzero Notice that we have already obtained this same result, but earlier we obtained it using continued fraction expansions One reasonable question, then, is: is there any relation between our two methods? It seems reasonable to guess that a suciently good rational approximation of an irrational number should be related to the continued fraction expansion somehow In fact, the terms of the continued fraction expansion give the best rational approximation: Proposition: Suppose α is any irrational real number and p/q is any rational number If p n / is the nth continued fraction convergent to α, and α p q < α p n, then q > In fact, if qα p < α p n, then q + Observe that the rst statement says that the best rational approximation to α, among all terms in the Farey sequence of order, is the convergent p n / Proof: Consider the Farey sequence of order : since p n p n =, we see that p n and p n are consecutive in this sequence Hence, there is no rational number with denominator less than that lies between them For the second statement, suppose that q < + By basic linear algebra, there exist integers x and y such that p = xp n + yp n+ and q = x + y+ (They are integers because the determinant is ±) Notice that since q < +, the second equation requires that one of x, y be positive and the other is negative Since α p n and α p ( n+ also have opposite signs, we conclude that x α p ) n and ( + y α p ) n+ have the same sign + Then we can write qα p = (x + y+ )α (xp n + yp n+ ) = x( α p n ) + y(+ α p n+ ) = x α p n + y + α p n+ α p n which establishes the contrapositive of the desired result 4

15 Proposition: If α is any irrational real number, then there are innitely many distinct rational numbers p q such that α p q < q Conversely, any rational number p satisfying this bound is a continued fraction q convergent to α Remark: The constant in the rst part is not sharp In fact, it is a result of Hurwitz that the can be replaced with 5, but with no larger constant Proof: For the rst statement, let p n be the nth continued fraction convergent to α We claim that at least one of p n and p n+ satises the desired inequality, so suppose that neither + does Then since α lies between p n and p n+, we have + p n p n+ + ( p n = α + p n+ α + > 4 p n α p n+ α + 4 = + q n q n+ where in the middle step we used the inequality (x + y) 4xy (which is equivalent to (x y) 0, and equality cannot hold in our case because α is irrational) Taking the square root gives p n p n+ + >, but this is false since these quantities are equal + For the second statement, suppose that p/q is not a convergent and that α p q < q Let n be such that q < + By the previous proposition, it must be the case that α p n < qα p = q α p q < q Thus, we conclude that α p n < qq n Now we get p n q p q q = p q p n p q α + α p n < + q qn But this implies q <, which is a contradiction We will remark at this point that the continued fraction expansion provides a good way to detect approximations of rational numbers using a decimal expansion: simply compute the continued fraction of the decimal, and then round o appropriately Example: Find a rational number of small denominator with decimal expansion We compute the continued fraction expansion of α = , which is easy to do with a calculator or computer We obtain the exact expression α = [0,, 5,, 57,, , 4,,, 6, 4] We truncate just before the huge term in the middle to get a guess of α = [0,, 5,, 57, ] = And indeed, we see that From our results above, we can see that any rational number that is a closer approximation will have denominator roughly on order of the next convergent [0,, 5,, 57,, ] = , so the number we found is clearly the simplest It is interesting to note that the period of the decimal expansion of 353 is 6, so in fact we have identied 765 the rational number before the expansion started repeating! ) 5

16 66 Pell's Equation The goal of this section is to study the integer solutions (x, y) to Pell's equation x Dy = r, for D a positive squarefree integer and r an arbitrary integer One reason that we should care about solving this equation is that x Dy = r is equivalent to N(x + y D) = r: thus, solving this equation in integers (x, y) will tell us which elements in Z[ D] have norm r In particular, if we can write down all solutions when r = ±, they will correspond to the units in Z[ D], since an element is a unit in Z[ D] i its norm is a unit in Z It is possible that there are no solutions to x Dy = : for example, if D 3 (mod 4), then modulo D this would say x (mod D), which we know to be false However, we will show that there are always solutions to x Dy = for any D Our rst result relates Pell's equation to our results with continued fractions: Proposition: Let D be a positive squarefree integer and r any integer with r + r < D If x and y are positive integers with x Dy = r, then x y is a continued fraction convergent to D Proof: We want to show that convergent to D Then we can write x y D < y By our previous results, this will show that x y is a x y D = = x Dy y x/y + D r y r/y + D + D r y r + r = y where we used the fact that r/y + D is at most r and D r, by the given information Since small values of x Dy all arise from continued fraction convergents, we should use the convergents to study the solutions to x Dy = r A rst step is to show that the Pell equation with r = always has a nontrivial solution: Proposition: If D is a positive squarefree integer, then the equation x Dy = always has a nontrivial solution in integers (x, y), and Z[ D] has a nontrivial unit x + y D Proof: If p q is a convergent to D, then p q + D < + D, so we see that p Dq + D < q q = + D Since D is irrational, there are an innite number of convergents but only a nite number of possible values for p dq, so by the pigeonhole principle there is some r such that p Dq = r has innitely many solutions Choose such an r: then there are again only nitely many possible pairs for the reduction of (p, q) modulo r, so there exist two distinct convergents x/y and s/t such that x Dy = s Dt = r, x s (mod r), and y t (mod r) Now we compute u = x + y D s + t xs Dyt xt + ys = + D, and observe that both xs Dyt D r r x Dy 0 (mod r) and xt+ys 0 (mod r), so the quotient u is of the form a+b D where a, b Z 6

17 But now N(u) = N(x + y ( D) xs Dyt N(s + t D) =, so u is a unit in Z[ D], and, r solution to Pell's equation ) xt + ys is a nontrivial r Now that we know there exists a nontrivial unit u in Z[ D], we can write down all of the units Denition: For a xed positive squarefree D, a fundamental solution (x, y) to Pell's equation is a pair (x, y) of positive integers such that x Dy = ± and x + y D is minimal The fundamental unit of Z[ D] is u = x + y D Note that this is well-dened: there will be a unique minimal positive value for x + y D over all pairs (x, y ) satisfying x Dy = ± Example: For D =, it is easy to see that (, ) is a fundamental solution to Pell's equation x y = ±, and the corresponding fundamental unit is u = + Proposition: Let u = x + y D be the fundamental unit in Z[ D] If w is an arbitrary unit in Z[ D], then w = ±u n for some integer n (possibly negative) Equivalently, each pair (x n, y n ) dened by x n + y n D = (x + y D) n is a solution to x Dy = ±, and these are all of the solutions up to changing the signs of x n or y n Proof: By scaling by (as necessary), assume w is positive Then there exists a unique integer n such that w [u n, u n+ ) since u is a real number greater than Now observe that w u n [, u), and w u n is also a unit in Z[ D] If this unit x + y D were not equal to, then (possibly after ipping signs on one of its terms) it would yield a positive solution (x, y) to Pell's equation x Dy = ± such that x + y D < u But this contradicts the minimality of u, so in fact we must have w u n =, whence w = u n The second statement is simply a rephrasing of the rst, after converting the statements about units to statements about solutions to Pell's equation (by taking norms) Remark (for those who like group theory): This result says that the group of units of Z[ D] is isomorphic to (Z/Z) Z (The rst term is from the ± sign and the second term is the power of the fundamental unit u) Example: Since the fundamental unit of Z[ ] is u = +, the set of units of Z[ ] are the elements ±( + ) n for n Z Taking the fth power (say) yields the element 4+9, and we can indeed compute that 4 9 = We have almost completely described the units of Z[ D]: all that is left is to explain how to nd the fundamental unit using the continued fraction expansion of D: Proposition: If D is a squarefree positive integer, the continued fraction expansion of D is periodic and of the form [a 0, a, a,, a r, a 0 ] with a 0 = D Then the fundamental unit u = x + y D satises x = p r and y = q r Furthermore, there exists a solution to x Dy = if and only if r is odd We will omit the details of this proof, since it is fairly lengthy We will content ourselves with verifying the result in one or two cases Example: Find the fundamental unit in Z[ 7] Earlier, we computed the expansion as 7 = [,,,, 4] This has the desired form with r = 4, so we conclude there is no solution to x 7y = Then the desired convergent is C 4 = [,,, ] = 8 3, and we can indeed verify that = We conclude that the fundamental unit of Z[ 7] is Example: Find the fundamental unit in Z[ 3] 7

18 We compute the continued fraction expansion of 3 = [3,,,,, 6] This has the desired form with r = 5, so we conclude there is a solution to x 3y = Then the desired convergent is C 5 = [3,,,, ] = 8 5, and we can indeed verify that = We conclude that the fundamental unit of Z[ 3] is If we want a solution to Pell's equation x 3y = instead, we simply square the fundamental unit: ( ) = : then the minimal nontrivial solution is given by (649, 80) 67 An Assortment of Diophantine Equations In this section we discuss a number of unrelated Diophantine equations Several will have the form y = x 3 + ax + bx + c for xed integers a, b, c Such equations are examples of elliptic curves, which are quite important in modern number theory Over a eld of characteristic zero, up to a change of variables, an elliptic curve has the general form y = x 3 + Ax + B Elliptic curves appear in a tremendous number of places, and are fundamentally important to modern number theory For example, they are the basis for elliptic curve factorization (currently one of the fastest integer factorization algorithms), elliptic curve cryptography (a frequently-used cryptosystem which is often used instead of RSA), and they are a key ingredient in the proof of Fermat's Last Theorem (see the next section) Proposition: There are no solutions to the Diophantine equation x + y + z = 4 a (8b + 7) The idea of this proof is to use modular arithmetic Proof: We prove the result by induction on a For the base case a = 0, consider the equation modulo 8 Each square is either 0,, or 4 mod 8, so it is not possible to obtain a sum of 7 mod 8 by adding three of them Now suppose there are no solutions for a k, and take a = k + Consider the equation x + y + z = 4 k+ (8b + 7) modulo 4 Each of the squares is 0 or, while the term 4 k+ (8b + 7) is 0 mod 4, so all of the squares must be 0 mod 4 ( x ) ( y ) ( z ) Then + + = 4 k (8b+7), but this is a contradiction since by induction, this equation has no solutions Remark: In fact, these are the only integers that cannot be written as a sum of three squares, as rst proven by Legendre (Gauss gave a formula for the number of such representations, similar to Fermat's formula for the number of ways of writing an integer as a sum of two squares) Proposition: The Diophantine equation y = x 4 + 4x 3 + x + x + has the solutions (x, y) = ( 4, ±3), (0, ±), (, ±3), (6, ±47), and no others The idea of this result is to attempt to complete the square of the x-terms, and then use some simple inequalities to bound how big x and y can be ( Proof: We complete the square of the x-terms and obtain x 4 +4x 3 +x +x+ = x + x 3 ) +8x 5 4 Thus, we should try comparing y to (x + x ) = x 4 + 4x 3 8x + 4 and (x + x ) = x 4 + x 3 + x 4x + We see that y (x +x ) = x +0x 3 is positive outside [ 03, 03], while (x +x ) y = x 6x is positive outside [0, 6] Hence, if x [ 0, 6], then we have the strict inequalities (x + x ) < y < (x + x ), which is impossible if x and y are both integers 8

19 Thus it must be the case that x [ 0, 6] Now it is (reasonably) easy to check the 7 cases to nd the solutions as listed above Remark: More generally, one can adapt this proof method to show that there are only nitely many solutions to any equation of the form y = x 4 + ax 3 + bx + cx + d for xed integers a, b, c, d Proposition: All solutions to the Diophantine equation x + y = z 3 with gcd(x, y) = are of the form (x, y, z) = (a 3 3ab, 3a b b 3, a + b ) for relatively prime integers a, b of opposite parity The idea of this proof is to work over the Gaussian integers Z[i] Proof: First observe that if x, y were both odd, then z 3 (mod 4), but is not a cube modulo 4 Since x, y are not both even since gcd(x, y) =, we conclude that one is even and the other is odd Now, over Z[i], factor the equation as (x + iy)(x iy) = z 3 We claim that x + iy and x iy are relatively prime: any common divisor would divide both x and y, hence divide But + i (the only Gaussian prime dividing ) does not divide x + iy, since x, y are of opposite parity Now, by the uniqueness of prime factorization in Z[i], we conclude that x + iy must be a unit times a cube But since each unit in Z[i] is actually a cube, we conclude that x+iy = (a+bi) 3 for some a+bi Z[i] Equating real and imaginary parts yields x = a 3 3ab, y = 3a b b 3, and then z = (a+bi)(a bi) = a + b, as claimed Remark: More generally, one can use an almost identical argument to write down the solutions to x + y = z d for any positive integer d Corollary: The only solution to the Diophantine equation y = x 3 is (x, y) = (, 0) Proof: Clearly, gcd(x, y) = Rearranging the equation into the form + y = x 3 and applying the previous result shows that = a 3 3ab for a, b Z Factoring this gives = a(a 3b ) Clearly, a = ±, and then the only solution is easily seen to be (a, b) = (, 0), yielding (x, y) = (, 0) We do not need to restrict to considering Diophantine equations where the terms involved are polynomials in the variables: we can also include variables in the exponents Proposition: The Diophantine equation 3 a b = has the two solutions (a, b) = (, ), (, 3), and no others The idea is to use congruence conditions Proof: Clearly a and b must be nonnegative, else the denominators of the rational numbers involved could not be equal Clearly b = 0 fails, and b = gives a = Now suppose b and consider the equation modulo 4: we obtain 3 a (mod 4), meaning that a is even, say, a = k Then we have b = 3 k = (3 k + )(3 k ), so 3 k + and 3 k must both be powers of But their dierence is, and so they must be 4 and respectively Thus, the only other solution is (a, b) = (, 3) Remark: This result is a special case of a result called Catalan's conjecture (recently proven in 00 by Mihailescu) that 8 and 9 are the only perfect powers that are consecutive: in other words, the only solutions to x a y b = in integers greater than is (a, b, x, y) = (, 3, 3, ) Proposition: The Diophantine equation y = x has no solutions The idea of this result is to rewrite the equation slightly, exploit congruence conditions, and then quadratic reciprocity to obtain a contradiction Proof: If x is even, then this equation yields y 3 (mod 4), which is not possible Then x (mod 4) meaning x (mod 4) Now we write the equation as y + = x = (x + )(x x + 4) 9

20 By our results about quadratic residues, any prime divisor of y + must be congruent to modulo 4, since y + 0 (mod p) is equivalent to being a quadratic residue modulo p Thus, any prime divisor (hence any divisor) of y + = x must be congruent to modulo 4 But x + is a divisor of x congruent to 3 modulo 4, so we have a contradiction Remark: This result is notable because there always is a solution to the equation y = x modulo p for every prime p (This is not trivial to prove) Theorem (Fermat): The Diophantine equation x 4 + y 4 = z has no solutions with xyz 0 This result, as noted, is originally due to Fermat Note in particular that it is a stronger result than the n = 4 case of Fermat's Last Theorem, which asks for solutions to x 4 + y 4 = z 4 We show the result using an innite descent technique: we consider the smallest nontrivial solution of the equation, and use it to construct a smaller solution Proof: If the equation has nontrivial solutions, let u be the smallest positive integer such that x 4 +y 4 = u has a solution Observe that gcd(x, y) =, otherwise we could replace x, y, u with x/d, y/d, u/d to obtain a smaller solution By reducing both sides modulo 4, we see that one of x, y is even and the other is odd: by interchanging, assume x is even Then (x, y, u) is a primitive Pythagorean triple, so from our parametrization we see that x = st, y = s t, and u = s + t for some integers s > t > 0 of opposite parity Since y = s t, it must be the case that s is odd and t is even: otherwise, y = s t would be congruent to modulo 4 ( x ) If we set t = k, then we see that = sk where gcd(s, k) =, so each of s and k is a perfect square Setting s = a and k = b yields the system y = s t = a 4 4b 4, so that y + (b ) = a 4 Then (y, b, t) is a primitive Pythagorean triple, so there exist relatively prime integers m and n such that b = mn, y = m n, and a = m + n The rst equation gives b = mn, so m and n are both squares: say, m = v and n = w Then, at last, we see that a = v 4 + w 4, meaning that we have a new solution (v, w, a) to the original equation Clearly a a = s < s + t = u, so this solution is smaller We have obtained a contradiction 68 Some Remarks on Fermat's Last Theorem One of the most famous Diophantine equations is Fermat's equation x n + y n = z n, for a xed integer n 3 Clearly, there are solutions if one of the variables is equal to 0: the question is whether this equation possesses any other solutions It is enough to prove that there are no solutions in the cases n = 4 and n = p where p is an odd prime, since any n > is divisible by 4 or an odd prime This result was famously conjectured by Fermat in 637, who wrote (in the margin of his book, in Latin) It is impossible to separate a cube into two cubes, or a fourth power into two fourth powers, or in general, any power higher than the second, into two like powers I have discovered a truly marvellous proof of this, which this margin is too narrow to contain It is now believed that Fermat probably did not have a correct proof of this result In the next section, we will give a proof that there are no solutions to the Fermat equation when n = 4 For the case n = 3, the idea is to factor the equation x 3 + y 3 = z 3 in the ring Z[ρ], where ρ = + 3 non-real cube root of unity is a 0