Pell's Equation. Luke Kanczes

Size: px

Start display at page:

Download "Pell's Equation. Luke Kanczes"

Katrina Marshall
5 years ago
Views:

1 Pell's Equation Luke Kanczes

2 Introduction. Pell's Equation Pells equation is any Diophantine equation which takes the form [] x Dy = () for positive integers x and y, where D is a xed positive integer that is not a perfect square. Note that trivially x = and y = 0 always solves equation (). We refer to () as a Diophantine equation since only integer solutions are allowed. [4] The name of this equation arose from Euler mistakenly attributing its study to John Pell, when it had in fact been found by William Brouncker in response to a challenge set by Fermat. [] The rst recorded appearance of Pell's equation was in the Cattle problem of Archimedes. For the purposes of this report we will use illustrative examples with D = 4.. The Cattle problem of Archimedes This problem asks the reader to consider the following situation [] Compute the number of cattle of the Sun, who once upon a time grazed on the elds of the Thrinacian isle of Sicily, divided into four herds of dierent colours, one milk white, another a glossy black, a third yellow and the last dappled. Subject to a series of conditions: White bulls = ( + 3) black bulls + yellow bulls Black bulls = ( 4 + 5) dappled bulls + yellow bulls Dappled bulls = ( 6 + 7) white bulls + yellow bulls White cows = ( 3 + 4) black herd Black cows = ( 4 + 5) dappled herd Dappled cows = ( 5 + 6) yellow herd Yellow cows = ( 6 + 7) white herd With the further restriction that When the white bulls mingled their number with the black, they stood rm, equal in depth and breadth and the restriction that when the yellow and the dappled bulls were gathered into one herd they stood in such a manner that their number, beginning from one, grew slowly greater till it completed a triangular gure. We can denote these as follows: White bulls + black bulls = square number Dappled bulls + yellow bulls = triangular number The rst part of the problem can be solved as a system of seven equations with eight unknowns. We can denote the rst three equations as x = ( + ( 3) y + t, y = 4 + ( 5) z + t and z = 6 + 7) x + t. The general solution to these three equations is given by [3] (x, y, z, t) = m (6, 60, 580, 89) Where m is a positive integer. The next set of equations can be denoted as x = ( )(y + y ), y = ( )(z + z ), z = ( )(t + t ) and t = ( )(x + x ). The general solution to these four equations is solvable if and only if m is divisible by 4657; with m = 4597 k we have the solution [3] (x, y, z, t ) = k (706360, , 35580, 54393)

3 Now we are required to choose k such that x + y = k is a square and z + t = k is a triangular number. Taking the prime factorisation = we can re-write the rst condition as k = al where a = and l is an integer. Since z + t is a triangular number if and only if 8(x + t) + is a square, we obtain the equation h = 8(z + t) + = al + which is Pell's equation: for h dl = d = ( 4657) = The rst to solve the problem was A. Amthor in 880, who made the observation that d = can be re-written as ( 4657) d where d = is squarefree. Hence if (x, y) solves the Pell equation for d then (x, 4657 y) solves the Pell equation for d. Reducing the problem to the easier problem of solving the Pell equation for d. We now ask how do we go about nding a solution to equation (). Pell's Equation Theorem Given Pell's equation as denoted in () we rst note the condition that D cannot be a perfect square. If it were then we could denote D as D = c for cɛz and the equation could be factorised as (x cy)(x + cy) =. We now observe that we can re-write () as follows (x + y D)(x y D) = () [ D ] So that nding a solution comes down to nding a nontrivial unit of the ring Z of norm ; here the norm [ D ] [ D ] Z Z = {±} between unit groups multiplies each unit by its conjugate, and the units ± of Z are considered trivial. This reformulation implies that once one knows a solution to Pell's equation, one can nd innitely many. [3] We can demonstrate this as follows. Given we know (x, y ) is a solution, if we square both sides of equation () we obtain a new solution = = (x + y D) (x y D) = ((x + y D) + x y D)((x + y D) x y D) = (x + y D) (x y ) D Therefore, (x + y D, x y ) is our new solution. Taking the third power, then the fourth power, and so on, are able to obtain an innite number of additional solutions. We formalise this in the following Theorem. []. Pell's Equation Theorem Let D be a positive integer that is not a perfect square. Then Pell's equation x Dy = always has solutions in positive integers. If (x, y ) is the solution with smallest x, which we shall refer to as the fundamental solution, then every solution (x k, y k ) can be obtained by taking powers x k + y k D = (x + y D) k for k =,, 3,... (3) This however doesn't provide a method to nd our fundamental solution (x, y ). To make matters even more complex, sometimes we nd the fundamental solution is relatively small, other times, however, the fundamental solution is quite large. For example, Pell's equation with D = 60 has fundamental solution x = 3, y = 4. However, Pell's equation with D = 6 has fundamental solution x = , y = [] 3

4 .. Example for D=4 For the case of Pell's equation with D = 4, we might observe via brute force that the fundamental solution is (x, y ) = (5, 4). From this we are able to nd all the solutions by taking powers The second and third smallest solutions are ( ) = ( ) 3 = ( )( ) = Hence (x, y ) = (449, 0) and (x 3, y 3 ) = (3455, 3596).. Reversing the problem If we assume a fundamental solution to Pell's equation exists in positive integers (x, y) then we can divide () by y to obtain: ( ) x D = y y We can factorise the left-hand side and re-arrange to obtain ( x y ) D = ( x y y + ) (4) D We now observe that the right-hand side of expression (4) is small, in particular for large y and therefore the left-hand side of (4) implies that x y is close to D. Hence a solution (x, y) to Pell's equation provides a good approximation to D. We now reverse the question. Rather than asking how do we nd (x, y) let's instead consider the rational approximations to D. 3 Diophantine Approximation We now turn our attention to the problem of how small we can make follows ( x y ) D. We can reformulate expression (4) as x y D = x + y D And instead focus on how small we can make x y D. We rst observe that trivially for any positive integer y, if we take x to be the closest integer to y D, then x y D since any real number lies between two integers and therefore the distance to the nearest integer is at most. We can improve on this using the following theorem rst obtained by Dirichlet in 84. [5] 3. Dirichlet's Diophantine Approximation Theorem Suppose that D is a positive integer that is not a perfect square. Then there are innitely many pairs of positive integers (x, y) such that x y D < (6) y (5) 3. Proof of Dirichlet's Diophantine Approximation Theorem We will consider the more general case and prove that there exists innitely many pairs of positive integers (x, y) such that x yσ < (7) y for σ an irrational. Clearly if (7) holds then (6) holds for σ = D, where D isn't a perfect square. The result can be derived using The Pigeonhole Principle, which states that 4

5 If you have more pigeons than pigeonholes, then at least one of the pigeonholes contains more than one pigeon. If we start by considering the Y + numbers 0, σ, σ,..., Y σ. We can write each of these Y + numbers in the form i σ = N i + F i for i = 0,..., Y, N i an integer and 0 F i <. If we divide the numbers F 0,..., F Y into Y disjoint subintervals, each of length Y then it follows that two of the Y + numbers must lie in one of the Y sub-intervals. The dierence between these two numbers, denoted as F m and F n for n > m, must satisfy the equality F m F n < Y. We can re-write this as Re-arranging (mσ N m ) (nσ N n ) < Y (N n N m ) (n m)σ < Y And we can now denote x = N n N m and y = n m, which are both positive integers. Since we have that both n > m and these two numbers lie between 0 and Y, we have 0 < m < n Y. It therefore follows, for y = n m, that 0 < y Y, which implies that Y y. Hence x yσ < y We nally note that there are an innite number of solutions (x, y). If we were to suppose there are a nite set of solutions, which we'll denote as (x i, y i ) for i =,..., k then we can order these solutions as x y σ x y σ... x k y k σ. However, we can dene our set Y to be arbitrarily large, therefore taking Y > x k y k σ we obtain x k y k σ > Y. Contradiction. Hence there are an innite number of solutions (x, y). 3.3 Application of Dirichlet's Diophantine Approximation Theorem We can use expression (7) to provide an approximation of an irrational number, for example, we can approximate 4 = The following solution approximations were found using a simple brute force algorithm programmed in R [Appendix]. However there exists a far more elegant method to nd x and y using continued fractions and considering whether x y D y <. The continued fraction algorithm is outlined in Section Diophantine Approximation of 4 Table lists all the (x, y)'s with x < 00 and y < 00, such that y x x y 4 < y x y 4 y x y x 4y Table : Diophantine Approximation of 4 We can see that 6 3 provides a fairly good approximation of 4. 5

6 4 Finding the fundamental solution We now return to Pell's equation as stated in (). Dirichlet's Diophantine Approximation Theorem tells us that there exists innitely many pairs of positive integers (x, y) that satisfy the inequality x y D < y (8) We aim to show that every solution in positive integers (x, y) to (8) also satises x Dy < 3 D (9) Suppose that (x, y) satises expression (8). We note x Dy = x y D x + y D. Further, from (8), we see that x is bounded by x < y D + y. Therefore, x + y D < (y D + y ) + y D < 3y D and hence yielding expression (9). x Dy = x y D x + y D < x y D 3y D < y (3y D) = 3 D We now aim to apply the pigeonhole principle. Dene T = 3 D. We consider the nite set T, T +, T +,...,, 0,,..., T, T, T as our pigeonholes and take the innite set of positive solutions (x, y) to expression (8) as our pigeons. Given that x, y and D are all positive integers, we know x Dy is an integer. Further since x Dy < 3 D and x Dy is an integer, then x Dy must lie between T and T. Therefore, we can assign each solution (x, y) to the pigeonhole numbered x Dy = k for kɛ{ T, T +,..., T, T }. Discussion This eectively means we have assigned an innite set of pigeons to a nite set of pigeonholes. One might wish to draw a similarity between such an act of faith, applied here with the axiom of choice. This states that given any set of mutually disjoint nonempty sets, there exists as least one set that contains exactly one element in common with each of the nonempty sets. [6] For example, this states that if we have an innite number of pigeonholes, each containing at least one pigeon, then it is always possible to choose one pigeon from each hole. One simply is required to either accept or reject the axiom of choice. In a similar vain, the reader is encouraged to either accept or reject that assigning an innite set of pigeons to a nite set of pigeonholes is possible. The author retains the view that such an act is possible, in particular if we assert that there must be some pigeonholes which contain innitely many pigeons. For example, one might choose to take the innite set N as our pigeons and split this into the two set's even and odd, which we could consider to be our pigeonholes. Providing one example where such a technique might be possible, however, may provide little encouragement to accept such a technique for all cases. Assuming, however, that we do choose to accept, we can continue accordingly. Proceeding... If we assume that pigeonhole, corresponding to x Dy =, contains innitely many solutions (x, y), then we can list these solutions as (X, Y ), (X, Y ),... We now look for two solutions (X j, Y j ) and (X k, Y k ) that satisfy X j X k mod and Y j Y k mod. We now consider the following pairs as pigeonholes... - (, ) (, ) (, -) (, ) (, ) (, ) (, -) (, ). - (-, ) (-, ) (-, -) (-, ) (, ) (, ) (, -) (, ) and consider our pigeons as the innite set of solutions (X, Y ), (X, Y ),... We assign the pigeon (X i, Y i ) to pigeonhole (A, B) by choosing A and B to satisfy X i A mod and Y i B mod, 0 A, B <. We have therefore assigned innitely many pigeons to a nite number of pigeonholes. Therefore, we again must have at least one pigeonhole which contains innitely many pigeons. We can therefore nd two pairs of positive integers (X j, Y j ) and (X k, Y k ) such that: 6

7 We now claim that x = X jx k Y j Y k D are solutions to x Dy =. We rst check that (x, y) satisfy Pell's equation. X j X k mod Xj DY j = Y j Y k mod X k DY k = and y = X jy k X k Y j ( ) ( ) x Dy XjX = k Y jy k D XjY D k X k Y j = (X j DY j )(X k DY j ) = We can now verify that x and y are integers. If we consider the numerators of x and y X j X k Y j Y k D X j Y j D = 0 mod X j Y k X k Y j X j Y j X j Y j 0 mod Hence the numerators are divisible by and therefore x and y are integers. Finally, we note that clearly x. We now just need to show that y 0. Assume y = 0, then X j Y k = X k Y j, so that Yk = Y k (X j DY j ) = (X j Y k ) D(Y j Y k ) = (X k Y j ) D(Y j Y k ) = Yj (X k DY k ) = Y However, we chose Y j and Y k to be positive and unequal. Contradiction. Hence y Example for D=4 j To nd our fundamental solution, we consider the Pell-like equation x Dy = for an a positive integer. We then seek two positive pairs of integers (X, Y ) and (X, Y ) such that X X mod X DY = Y Y mod X DY = If we can nd such integers we are able to solve Pell's equation x Dy = by dening x = X X Y Y D and y = X Y X Y For the case when D = 4, we refer to Table and we note that the pairs of integers (4, ) and (6, 3) satisfy the above criteria for = mod 4 4 = 6 4 = Therefore, 3 mod = = x = = 5 and y = Hence our fundamental solution is (x, y ) = (5, 4) = 4 7

8 5 Continued Fractions Another method we can use to nd our fundamental solution is via continued fractions. 5. Continued fraction algorithm Continued fractions allow us to denote an irrational number as an innite set of integers. The algorithm proceeds as follows [5] :. Let σ be some irrational number. Dene a 0 = [σ] and write σ = a 0 + σ. Where [x] denotes the integer part of x.. Now write σ = a + σ, where a = [σ ]. 3. Repeat Step for σ i = a i + σ i+ where a i = σ i for i =, 3,... We often for ease of notation denote σ as σ = [a 0, a, a,...]. For an irrational number of the form σ = D, the partial fraction has the form D = [a 0, a, a,..., a m ]. The number m of repeated partial quotients is called the period of D. It's also of interest to note that a m = a 0. We now denote p n and q n as p n = a n p n + p n and q n = a n q n + q n with p 0 = a 0, q 0 = and p = a 0 a +, q = a. Then the fundamental solution to Pell's equation is [7] (x, y ) = { (p m, q m ) m even (p m, q m ) m odd The proof of this is beyond the requirements of this paper, however, it is of signicant academic interest. One further interesting observation is that one could view the solvability of Pell's equation as a special case of Dirichlet's unit theorem, from algebraic[ number theory, which describes the structure of the group of units of a general ring of algebraic integers; D ] for the ring Z, it is the product of {±} and an innite cyclic group. [3] 5. Finding the fundamental solution using continued fractions for D = 4 If we consider Pell's equation with D = 4, we obtain 4 = We can therefore write 4 = [3,,,, 6]. Therefore the period length is m = 4 and hence the fundamental solution is (x, y ) = (p 3, q 3 ). We have p 3 = a 3 p + p = a 3 (a p + p 0 ) + a 0 a + = a 3 (a (a 0 a + ) + a 0 ) + a 0 a + = ( (3 + ) + 3) = 5 q 3 = a 3 q + q = a 3 (a q + q 0 ) + a = a 3 (a (a ) + ) + a = ( () + ) + = 4 Therefore, (x, y ) = (5, 4), as before. It's also an interesting note that for D = as in the cattle problem of archimedes, one is able to write D in continued fractions, with period length 0354 [3] and hence nd the fundamental solution in a similar manner. 6 Negative Pell's Equation One nal interesting equation is the negative Pell's equation x Dy = (0) We can solve this in a similar manner as Pell's equation using continued fractions. In this case there is no solution when the period m of D is even. The fundamental solution to the negative Pell's equation are given as [5] (x, y ) = { (p m, q m ) m odd No Solution m even 8

9 Therefore, for the case when D = 4, we have 4 = [3,,,, 6] and since the period length is even, there are no solutions. 7 References. J. Silverman. A friendly introduction to number theory. Prentice Hall. Chapters GREEK ATHEATICAL WORKS Translated by Ivor Thomas. The Loeb Classical Library, Harvard University Press, Cambridge, A, 94 Volume II, Pages [Online. Retrieved 4//0] 4. [Online. Retrieved 4//0] 5. Alan Baker. A concise introduction to the theory of numbers. Chapter [Online. Retrieved 4//0] 7. [Online. Retrieved 4//0] 8 Appendix The following R code was used to provide diophantine approximations to 4 vec <- NULL vec <- NULL for (y in :00) { for (x in :00) { if((abs(x-y*sqrt(4))*y)<) vec[y]<-y else print("") } } for (y in :00) { for (x in :00) { if(abs(x-y*sqrt(4))*y<) vec[y]<-x else print("") } } cbind(vec,vec) 9

Solving Pell s Equation with Fibonacci s Rabbits Teacher s Circle, October 2009 Aaron Bertram, University of Utah

Solving Pell s Equation with Fibonacci s Rabbits Teacher s Circle, October 2009 Aaron Bertram, University of Utah Pell s Equation was a recurring theme in the development of number theory, predating the