Almost 4000 years ago, Babylonians had discovered the following approximation to. x 2 dy 2 =1, (5.0.2)

Transcription

1 Chater 5 Pell s Equation One of the earliest issues graled with in number theory is the fact that geometric quantities are often not rational. For instance, if we take a right triangle with two side lengths equal to, the hyootamus has length, which is irrational. But how can we do arithmetic with irrational numbers? Well, erhas the most basic thing is to work with rational aroximations. Almost 4000 years ago, Babylonians had discovered the following aroximation to : = = (5.0.) 600 In this chater we ll exlain how find the (integer) solutions to Pell s equation: x dy =, (5.0.) and how gives us good aroximations to d (see Proosition 5..3). Following Stigler s law of eonomy, Pell s equation was studied by the Indian mathematician and astronomer Brahmaguta in 68 (who discovered the comosition law Proosition 5..) and with a general method of solution by another Indian mathematician and astronomer, Bhaskara II, in 50. In Euroe, methods for solving Pell s equation were rediscovered hundreds of years later by Fermat and Lord Brouncker. Euler misattributed Lord Brouncker s solution after reading a discussion of Lord Brouncker s method written by the English mathematician John Pell (6 685). Throughout this chater d> is a ositive integer which is not a square. 5. Units and Pell s equation Recall that a unit u of Z[ d] was defined to be an element such that u has a multilicative inverse u Z[ d] (i.e., the real number u 6= 0and u Z[ d]). Further, by Lemma.., we know that x + y d Z[ d] (x, y, Z) isaunitifandonlyif N(x + y d)=(x + y d)(x y d)=x dy = ±. That no scientific discovery is named after it s first discoverer. The Pythagorean theorem is another famous examle. Of course there are many counterexamles to Stigler s law as well. Aroriately, Stigler s law itself is not. 30

2 Thus solutions to Pell s equation (5.0.) are in natural bijection with the units of Z[ d] with norm. On the other hand, we also know by Proosition that the set of units U = U d = Z[ d] of Z[ d] form an (abelian) grou. We also denote by U + = U + d the set of units in U = U d of norm, so we can think of the solutions to Pell s equation as the subgrou U + of U. Exercise 5... Check that U + is indeed a subgrou of U. This grou structure will hel us determine the set of solutions to Pell s equation. First, we have the following, which is similar to the comosition law for sums of two squares. Proosition 5... (Comosition law) If (x,y ) and (x,y ) are solutions to x dy = m, x dy = n. Then the comosition of these solutions defined by (x 3,y 3 )=(x,y ) (x,y ):=(x x + dy y,x y + y x ) (5..) is a solution of x 3 dy 3 = mn. Proof. We simly translate the above into a statement about norms. The hyothesis says N(x + y d)=m and N(x + y d)=n. Now observe that (x + y d)(x + y d)=x x + ny y +(x y + y x ) d = x 3 + y 3 d. Hence by the multilicative roerty of the norm, x 3 dy 3 = N(x 3 + y 3 d)=n(x + y d)n(x + y d)=mn. In articular, when m = n =, this says that we can comose two solutions to Pell s equation to get a third solutions. We can also comose two solutions to x dy = to get a solution to x dy =+. Both of these are summarized in this corollary. Corollary 5... Let u = x + y d and u = x + y d be units of Z[ d]. IfN(u )= N(u ),thenthecomosition(x,y ) (x,y ) defined in Eq. (5..) also a solution to Pell s equation (5.0.). The following should be obvious if you ve had an algebra class, but since we never covered isomorhisms, I m not entirely sure if this is obvious to you: 3

3 Exercise 5... Let G Z Z be the set of solutions to Pell s equation (5.0.). Show that the comosition law Eq. (5..) makes G into a grou. (Note the above corollary just says comosition is a binary oeration on G.) Now that we know the comosition law, the hoe is that if we can determine a few good solutions to Pell s equation, then maybe we can generate all solutions by comosing those we know. Moreover, ideally these good solutions should be the smallest nontrivial solutions to Pell s equation. By the trivial solutions to Pell s equation, we mean the obvious ones: (x, y) =(±, 0), which corresond to the elements of U + which lie in Z, i.e., ±. Examle 5... Consider d =. Note (, ) is a solution to x y =. The comosition (, ) (, ) = (3, ) is a nontrivial solution to Pell s equation x y =. Similarly, we comute (3, ) (3, ) = (7, ) and (3, ) (7, ) = (99, 70). Hence (3, ), (7, ) and (99, 70) are three nontrivial solutions to x y =. Exercise Find a nontrivial solution to x more (distinct) solutions to x 3y =. 3y =.Usecomositiontofindtwo We remark that in the case of d =3,unliked =, there are no units of norm fact, the following more general statement is true.. In Exercise Suose d 3 mod 4. Show Z[ d] has no units of norm, i.e., U d = U + d. The converse to the revious exercise does not hold, i.e., there may or may not be a unit of norm when d 6 3 mod 4. We ve seen there is such a unit when d =. The next exercise gives you an examle where there isn t a unit of norm but d 6 3 mod 4. Exercise Show that Z[ 6] has no units of norm. In general, it is an oen roblem to determine for what d there are units of norm in Z[ d]. It s not clear that there is a nice answer to this roblem, but there results about how often Z[ d] has units of norm. 5. Aroximation and existence of solutions At the end of the last section we saw that there are nontrivial solutions to Pell s equation when d =, 3. Next we will rove the existence of a non-trivial solution for all nonsquare d, which is originally due to Lagrange in 768. However, the roof we will give is due to Dirichlet (ca. 840). It uses the igeonhole rincile. You ve robably at least seen the finite version in your Discrete Math class. 3

4 Pigeonhole rincile (finite version) If m>kigeons go into k boxes, at least one must box must contain more than igeon. (infinite version) If infinitely many igeons go into k boxes, at least one box must contain infinitely many igeons). Proosition 5... (Dirichlet s aroximation theorem) For any nonsquare d> and integer B>, thereexista, b N such that b<b and a b d < B. This says that a b d < bb B, which is a recise way of saying a b is close to d.e.g.,it says we can find a rational aroximation a b for which is accurate within 00,000 (so a b and agree to 5 decimal laces, after rounding if necessary ) with denominator b<00, 000. Such an examle was exhibited at the beginning of this chater in (5.0.). Proof. Consider the B irrational numbers d, d,..., (B ) d. For each such number k d ( ale k ale B ), let a k N be such that 0 <a k k d<. Partition the interval [0, ] into B subintervals of length B. Then, of the B +numbers 0,a d, a d,..., ab (B ) d, in [0, ] two of them must be in the same subinterval of length B. Hence they are less than distance B aart, i.e., their difference satisfies a b d < B. Further their irrational arts must be distinct, so we have B<b<Bwith b 6= 0.Ifb>0 we are done; if b<0, simly multily a and b by. Clearly we need a>0 for a b d <. Theorem 5... Suose d N is nonsquare. Then x i.e., there is a unit in Z[ d] of norm other than ±. dy =has a nontrivial solution, Proof. Ste. Fix B >. Then by Dirichlet s aroximation theorem, there exist a,b N such that a b d < B < b. Let B >b such that B < a b d. Alying Dirichlet s aroximation again, we get a new air (a,b ) of integers such that a + b d < B < b. For instance, and are within of each other, but their first three digits are only equal 000 after rounding to the nearest 4 digits. 33

5 Reeating this we see there an infinite sequence of distinct integer airs (a j,b j ) such that a j b j d gets smaller and smaller, and for all j a j b j d < b j.. Then a j b j is an infinite sequence of increasingly good aroximations to d. Ste. Assume (a, b) satisfies a b d < b.notethat a + b d ale a b d + b d ale+b d ale 3b d. Then a db = a + b d a b d ale3b d b =3 d. Hence there are infinitely many a 3 d. b d Z[ d] whose norm, in absolute values, is at most Ste 3. By successive alications of the (infinite) igeonhole rincile, we have: (i) infinitely many a b d with the same norm n Z, where n ale3 d (and n 6= 0) (ii) infinitely many a b d with norm n and a a 0 mod n for some a 0. (iii) infinitely many a b d with norm n, a a 0 mod n, b b 0 mod n for some b 0. Hence, relabeling if necessary, we have a,b,a,b N such that N(a b d) = N(a b d)=n, a a mod n, b b mod n, anda b d 6= ±(a b d). Ste 4. Consider := a b d = (a b d)(a + b d) a b d a db Note and = a a db b n a a db b a a db b a db 0 mod n, a b b a a b b a 0 mod n. + a b b a d. n Thus the coefficients of are integers, i.e., = a + b d Z[ d] where a, b Z. Then hen since a db = N(a + b d)=n(a b d)n (a b d) = nn =, i.e., (a, b) is a solution of x dy =. Furthermore, it is a nontrivial solution since a b d 6= ±(a b d). Exercise 5... Exlain how to modify the above roof to conclude the existence of infinitely many solutions to x dy =. Conclude the real quadratic rings Z[ d] (d > nonsquare) have infinitely many units, in contrast to the case of imaginary quadratic rings Z[ d]. 34

6 The above aroximations suggest how solutions to Pell s equation are related to rational aroximations to d. Proosition Suose (x, y) is a nontrivial solution to x dy =.Assumex, y > 0. Then 0 < x d< y y( + d) < y. Proof. Let = x y d Z[ d]. Then N( ) = =.Note = x + y d + d since x, y. Thus ale + d,so x y d = y < y( + d). Since and N( ) are ositive, so is, andthus y, which finishes the asserted bounds. This roosition says ositive solutions (x, y) to Pell s equation, i.e., units of norm +, giverationalaroximationsto d, and solutions with larger values of y give better aroximations. Furthermore, we are always getting overestimates for d. One can similarly get underestimates with units of norm, i.e., solutions to x dy =, at least when they exist. When they don t, one could instead look for solutions to x dy = or x dy = 3 etc. We remark the aroximation in (5.0.) corresonds to the solution (30547, 600) to x y = 79, though I m not suggesting Babylonians came u with this aroximation by starting with the equation x y = 79! Exercise 5... Suose (x, y) is solution to x dy = with x, y > 0. Show x y < d and rove a (good) bound for d x y in terms of y. Exercise Suose (x, y) is solution to x dy = with x, y > 0. Show x y < d and rove a (good) bound for d x y in terms of y. Examle 5... Recall from Examle 5.., (3, ), (7, ) and (99, 70) are solutions to x y =. This gives the following aroximations, with the following error bounds y from the above roosition: x x y decimal error bound y 3.5 < < <

7 Exercise Recall the solutions in the revious examle came from the first three owers n (n =,, 3) oftheunit =3 U +.However,noneofthesearoximations are better than the (more comlicated) Babylonian one (5.0.). Using a calculator, comute a few successive aroximations to from higher owers n, along with the exact error (u to several decimal laces). What is the first aroximation you get in this way that is better (i.e., closer to )thantheonein(5.0.). Exercise Using 3 nontrivial solutions to x 3y =you found in Exercise 5..3, give 3 rational aroximations to 3 with error bounds. Using a calculator, comute the actual error in these aroximations (e.g., you can make a table as in the examle above). 5.3 Fundamental units Here we determine the structure of the grou of units U d, which will give us a method for generating all solutions to Pell s equation. Definition The fundamental unit " d of Z[ d] is the smallest unit x+y d Z[ d] such that x, y > 0. The fundamental +unit " + d of Z[ d] is the smallest unit x + y d Z[ d] such that x, y > 0 and N( ) =. 3 In real quadratic rings, smallest means with resect to the usual order on R, unlikethe case of imaginary quadratic rings where we (artially) ordered elements by their norm. Lemma For any (nonsquare) d>, thefundamentalunit" d and the fundamental +unit " + d exist and are uniquely defined. Proof. Since < defines a strict ordering of real numbers, the condition of smallest guarantees that " d and " + d will be unique if they exist, so it suffices to show existence. Recall we always have a nontrivial solution (x 0,y 0 ) to x dy =from Theorem 5... Moreover, we can assume x 0,y 0 > 0. Nownoteifx + y d<" 0 = x 0 + y 0 d with x, y > 0, we must have x<" 0 and y< " 0 d. Hence n " d =min x + y d :alex, o dy < " 0, x + dy =. Since the set on the right is finite, this minimum is well defined, hence " d exists. The case of " + d follows in the same way, simly using the equation x dy =instead of x dy =. 3 Note that most discussions you will find about fundamental units talk about fundamental units in the ring of integers O d of Q( d). Here, assuming d is squarefree, O d is just Z[ d] when d 6 mod 4 but is Z[ + d ] when d mod 4 (recall Definition.5.5). So be careful comaring what we say here and what is written other laces about fundamental units, as there may be a slight difference when d mod 4, though there is no serious difference in the theory. E.g., when d =5a fundamental unit in Z[ 5] is + 5 but in O 5 it is + 5. On the other hand, the fundamental unit in O 7 = Z[ + 7 ] is the same as the fundamental unit in Z[ 7], namely" 7 =4+ 7. Also, the term fundamental +unit is not standard as far as I know, there is no standard term for the hrase the smallest unit of norm. 36

8 Here is a naive algorithm for finding " d or " + d. First, ick some bound N. Then range over all ale x, y ale N to look for solutions to x dy =or x dy =. If we found any, then the smallest solution (x, y) gives us " d or " + d as x + y d.ifnot,weicka larger N and reeat. This rocess terminates at some oint by the existence of " d and " + d. Examle When d =, " =+ and " + units in Examle 5... = " =3+.Wesawbothofthese Examle When d =5,wecomute" 5 =+ 5 and " + 5 = " 5 = Examle Consider d =7. Recall from Exercise 5..4 that Z[ 7] has no units of norm. Wecomute" 7 = " + 7 = Exercise Comute " d and " + d for d =3, 6,. Exercise An alternative definition of fundamental unit (res. +unit) is the smallest "> in Z[ d] such that N(") =)(res.n( ) =). Prove that this is equivalent to the above definition as follows. (Suggestion: Show that " = x + y d> aunitimlies " <, which imlies x, y > 0.) Theorem For d> nonsquare, U d (res. U + d ) is the infinite abelian grou generated by " d (res. " + d )and. Exlicitly, and U d =...,±" d, ±" d, ±, ±" d, ±" d,... U + d =...,±("+ d ), ±(" + d ), ±, ±" + d, ±("+ d ),..., and all the elements listed in the sets on the right are distinct, i.e., " m d = ±"n d for m, n Z (res. (" + d )m = ±(" + d )n )ifandonlyifm = n and the lus/minus sign is +. Proof. We know U d and U + d are abelian grous by Proosition and Exercise 5... Thus U d and U + d must contain all the elements in the sets on the right. Write G denote U d or U + d, and let " denote " d or " + d, according to whether G = U d or G = U + d. Since ">, the sequence "n (n 0) is a strictly increasing sequence lying in [, ), and" n (n>0) is a strictly decreasing sequence lying in (0, ). From this one easily sees that all elements in the above sets on the right are distinct. Finally, we show any G Z[ d] is of the form ±" n for some n Z. Suose there is some which is not of this form. By taking the negative and/or inverse if necessary, we may assume >. Since " is the smallest element of G larger than (Exercise 5.3.) and " n!as n!, there must be some n>0 such that " m < <" m+. But then < " m <"and N( " m )=, contradicting the minimality of ". 37

9 Note that for any unit u U d, N(u) =uu = ± imlies u is either u or u, according to whether N(u) =or N(u) =. In articular, if (" + d )n = x n + y n d, then (" + d ) n = (" + d )n = x n y n d. Hence the above theorem says that once we find " + d, we can comute all elements of U + d by comuting (" + d )n = x n + y n d, n. Then the elements of U + d are n = {±}[ o ±x n ± y n d : n, (5.3.) U + d where we read the ± signs in ±x n ± y n indeendently. (A similar statement is also true for U d.) This immediately gives our desired descrition of solutions to (5.0.). Corollary For n, write(" + d )n = x n +y n d for n (with xn,y n Z). Then all solutions to Pell s equation x dy =are the trivial solutions (±, 0) and the nontrivial solutions (±x n, ±y n ) for n. Via Proosition 5..3, this gives us the following sequence of aroximations x n y n d of d. To rove that these aroximations are getting better (at least asymtotically), by this roosition we want to rove the y n s are increasing. Exercise With x n,y n as above, show the sequences (x n ) and (y n ) are strictly increasing sequences for n. Deducethatthesequence xn y n converges to d. Using the above theorem, we can also relate " d and " + d now. Exercise For d> anonsquare,show" + d = " d if Z[ d] has units of norm, and " + d = " d otherwise. Deduce in articular that " d ale " + d. Hence if we solve the roblem of finding the fundamental unit " d,wealsoknowthe fundamental +unit " + d. Since " d ale " + d, even if our goal is to comute "+ d, it may often be easier algorithmically to look for " d first, since the x and y aearing in the reresentation x + y d can be much smaller. Examle Consider d = 9. Then by the naive algorithm for finding fundamental units, we can check " 9 = , which has norm. Thus " + 9 = " 9 = , butthiswouldrequiremanymorecalculationstofindsolelybythenaive algorithm. Here s another consequence of the structure theorem for U d (or, if you refer, the revious exercise): if Z[ d] has no units of norm, we can rove this algorithmically by comuting " d and checking it has norm. For then ±" n d also has norm for all n, i.e., U d = U + d. 38

10 5.4 Continued fractions In the last section, we described how to find all solutions to Pell s equation in terms of the fundamental +unit " + d. Earlier, we also resented a naive algorithm to comute " d and " + d. The roblem is that as d gets even moderately large, the naive algorithm is not very efficient. This is already suggested by the case of d = 9 in Examle Here is a more imressive examle: Examle When d = 6, " + d = , i.e., the smallest ositive nontrivial solution to x 6y =is ( , ). The above examle was discovered by Bhaskara II in the th century in India, and indeendently (much later) rediscovered by Fermat in Euroe. How can one find such solutions, esecially without owerful comuting devices? The answer come from an alternative reresentation of numbers, not as decimals, but as continued fractions. First we exlain the continued fraction exansion with an examle. Examle Consider a b remainder: = a b = 3 5, which we write as a whole number lus a a b = 3 5 =+3 5. Now we can t exactly reeat this on the remainder, but we can on its recirocal: Thus we have a b := 5 3 =+ 3. a b =+ 3/ = Now reeat again with the recirocal of the remainder in a b : If we do this again, we get: a 3 b 3 := 3 =+. a 4 b 4 = =, a rational with no remainder, and so we sto. This leads to the following exression: a b = 3 5 = + + +, which we call the continued fraction exansion of 3 5.Notethatbecauseateachstagewe are taking recirocals, we ll see a sequence of s going down, and all that matters are the 39

11 numbers in the boxes. To simlify notation, we will also write this as [,,, ] = =+ + + Definition Let x R. Thecontinued fraction exansion of x is the exression [q,q,q 3,...]=q + q + q 3 + where q Z and q j Z 0 for j are defined as follows: q = bxc is the greatest integer ale x, so0aler < where r = x q ; for j, inductivelyset q j+ = so r j+ := r j q j satisfies 0 ale r j+ <. ( b r j c(the greatest integer ale r j ) r j 6=0 0 r j =0, If r j =0for all j > m, we also write the continued fraction exansion as the finite sequence [q,...,q m ] = q + q + q 3 + q m, in which case we call the continued fraction exansion finite. The q j and r j is used to make you think that these quantities are like quotients and remainders (which they are if x is rational). The rounding down function x 7! bxc (also often denoted by x 7! [x]) is called the greatest integer function or the floor function. Note for any x R, there is a unique continued fraction exansion [q,q,...]. Moreover, since at each ste 0 ale r j <, the recirocal will be at least if r j 6=0,andsoq j+ =0if and only if r j =0. Exercise Comute the continued fraction exansion of Exercise Let x R, and[q,q,...] be the continued fraction exansion. Let (x n ) denote the sequence of rational numbers by evaluation the artial continued fraction exansions: x n = q + q qn Show lim n! x n = x. Exercise For x R, showthecontinuedfractionexansionforx is finite if and only if x Q. 40

12 Examle Let s comute the continued fraction exansion [q,q,...] of 5. First set q = b 5c =, r = 5, so at the first stage our exansion looks like 5=q + r =+( 5 ) = + /( 5 ). The nice thing about quadratic numbers is we can rationalize the denominator in 5 by multilying by the conjugate (in Z[ 5]) ofthedenominator. NoteN(r )=r r = N( + 5) = 4 5=, so r = r =+ 5,i.e., So at the next stage we let = =+ 5. r 5 q = b+ 5c =4, r =+ 5 q = 5 =r. Thus at the next stage, we have the exansion 5= Since r = r,weseethatq 3 = q,sor 3 = r = r,andsoon. Sothesecomutations simly reeat, and we will have q j =4for all j, givingthecontinuedfractionexansion 5=[, 4, 4, 4,...]= Examle Now let s try finding the continued fraction exansion of 3. At the first stage we have q = b 3c =, r = 3. Then So Then So 3+ = = + 3. r 3 3+ q = b + 3 c =, r = = =+ 3. r =. q 3 = b+ 3c =, r 3 = 3 =r. 4

13 Since r 3 = r,wemustreeatafterthis,i.e.,q 4 = q, r 4 = r,andsoon,givingusthe continued fraction exansion 3=[,,,,,,,,,...]= Notice the above continued fraction exansions reeat. We make this notion recise as follows. Definition We say a continued fraction exansion [q,q,...] is eriodic if there exist s 0 and m N such that [q,q,...]=[q,...,q s,q s+,...,q s+m,q s+,...,q s+m,q s+,...,q s+m,...], i.e., if q j+m = q j for all j>s.inthiscase,wedenotethisexansionby [q,...,q s, q s+,...,q s+m ]. If the exansion is eriodic, the smallest such m for which the above condition holds (for some s) iscalledtheeriod of [q,q,...]. For instance, the examles above say 5=[, 4] is eriodic with eriod and 3= [,, ] is eriodic with eriod. Since any rational number has continued fraction exansion of the form [q,...,q s, 0], any rational number has a eriodic continued fraction exansion with eriod. Note eriodic continued fractions can be secified by a finite amount of data. Theorem (Lagrange). For any x R, thecontinuedfractionexansionofx is eriodic if and only if x Q( d) for some d. Inarticularthecontinuedfractionexansionof any element of Z[ d] is eriodic. Due to lack of time, we won t rove this. But the idea of the roof for the if direction is that at each stage in the continued fraction exansion, the quantities r j will be elements of Q( d) satisfying certain conditions, and then showing that there are only finitely many ossibilities, so for some j, m, wehaver j+m = r j by the igeonhole rincile. The only if direction is easier. For simlicity, we just illustrate the secial case s =0, so x =[q,...,q m ] (what is called urely eriodic). Consider =[q,...,q m ] Q. Then Then so taking recirocals shows i.e., x = + x = , + x = = x, =(x )x = x x, so x satisfies the quadratic equation x x =0with rational coefficients, from which it follows x Q( d) where d = +4. 4

14 Exercise Comute the continued fraction exansion of. Exercise Comute the continued fraction exansion of 7. Now we exlain (without roof) the connection with Pell s equation and fundamental units. Assume d> is squarefree, and write d =[q,...,q s, q s+,...,q s+m ] where s and m are chosen minimally so we can reresented this continued fraction eriodically. In articular m is the eriod. Consider the artial continued fraction exansions [q,...,q n ]. These are rational numbers, so we write them as x n y n =[q,...,q n ], x n,y n N, gcd(x n,y n )=. As they converge to d (Exercise 5.4.), we call (x n,y n ) the n-th convergent of the continued fraction. Theorem For d> squarefree, let (x n,y n ) be the n-th convergent in the continued fraction exansion of d. Then x m + y m d is the fundamental unit "d,wherem is the eriod of this continued fraction. More generally, x km + y km d is " k d for k. So in summary, we used units in real quadratic fields to determine all solutions to Pell s equation in terms of " d. Now we know how to comute " d in terms of continued fractions, and thus determine all solutions to Pell s equation. Moreover, this allows us to construct good rational aroximations to d. Of course, using continued fractions directly gives us rational aroximations to d, but in some sense the ones coming from solutions to Pell s equation (or x dy = ) are otimal in that they will have minimal remainder (see Proosition 5..3). (We also haven t roved that the continued fraction convergents give us good rational aroximations, though one can rove this.) Examle Recall 5=[, 4], which has eriod. So we look at the first convergent is given by x y =[q ] = [] = so the theorem says " 5 =+ 5, which matches with Examle Examle Recall =[,, ], which has eriod. Thus the second convergent is given by x y =[, ] = + =, so the theorem says " 3 =+ 3, which matches what you should have got in Exercise

15 Exercise Use the continued fraction exansion of 7 to comute " 7 (see also Examle 5.3.3). Then obtain a rational aroximation to 7 accurate to within 00. Exercise Use the continued fraction exansion of 9 to comute " 9. Use " 9 to obtain a rational aroximation to 9 accurate to within 000. (You may use a calculator.) Exercise Use continued fractions to obtain the exression for " + 6 asserted in Examle (Youmayuseacalculator.) 5.5 Aftermission: fundamental units and Fibonacci numbers We close this chater with an amusing connection with fundamental units and Fibonacci numbers, following ideas that we used to solve Pell s equation. The golden ratio = + 5 is the fundamental unit for the full ring of integers Z[ + 5 ]. For x, y Z, note N(x + y + 5 )=(x + y + 5 )(x + y 5 )=x + xy y. This exression is a binary quadratic form, which we also denote Q(x, y) =x + xy y. Recall the Fibonacci numbers F n are defined by F = F =,F n+ = F n+ + F n, n. Exercise Show the Fibonacci numbers satisfy F n+ +=F n+ F n+ + F n+. Put another way, the exercise says that (F n+,f n+ ) are solutions to Q(x, y) =, which is an analogous equation to Pell s equation. In other words F n+ + F n

16 is a unit of norm in Z[ + 5 ]. Note the golden ratio has norm, butitsquare" = = 3+ 5 has norm (i.e., is the fundamental +unit in Z[ + 5 ]. Then we can write " = 3+ 5 =+ + 5 Comuting a coule of owers of ", we see + 5 = F + F. " = = = F 3 + F 4, " 3 = This is art of a general rule. = = F 5 + F 6. Exercise Comute " 4 directly and then check that " 4 = F 7 + F Exercise Prove that " n = F n + F n for n. The above exression gives a way to comute Fibonacci numbers. While it s not exactly resented as a formula for F n, you robably noticed in the calculations above you immediately see F n as the coefficient b in the exression " n = a+b 5, and then a is just b +F n. One can rewrite these calculations into a well-known formula for F n : Exercise Prove that F n = n n for n. 45