Chakravala - a modern Indian method. B.Sury

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1 Chakravala - a modern Indian method BSury Indian Statistical Institute Bangalore, India sury@isibangacin IISER Pune, India Lecture on October 18,

2 Weil Unveiled What would have been Fermat s astonishment if some missionary, just back from India, had told him that his problem had been successfully tackled there by native mathematicians almost six centuries earlier? - André Weil Weil was talking about a 1657 challenge of Fermat to the English mathematicians and all others Writing to his friend Frenicle, he posed the problem of finding a solution of x 2 Ny 2 = 1 pour ne vous donner pas trop de peine like N = 61, 109 Cut back to 1150 AD when Bhaskara II gave the explicit solutions ( ) 2 = ( ) 2 = 1! Weil s comment was because - unbeknownst to Fermat - the ancient Indians had not merely given a solution but they had gone all the way! Brahmagupta, Jayadeva, Bhaskara Jayadeva (11th century) and Bhaskara had given the finishing touches to a wonderful algorithm called the Chakravala which finds all solutions to x 2 Ny 2 = ±1 for any positive integer N! Indeed, Brahmagupta ( ) had already solved this equation in 628 AD for several values like N = 83 and N = 92 Brahmagupta had remarked, a person who is able to solve these two cases within a year is truly a mathematician! Consider a natural number N which is not a perfect square We are basically interested in integer solutions of the equations x 2 Ny 2 = 1, x 2 Ny 2 = 1 More generally, suppose (p, q) is a solution of x 2 Ny 2 = m and (r, s) is a solution of x 2 Ny 2 = n Then is a solution of x 2 Ny 2 = mn (p, q) (r, s) := (pr + Nqs, ps + qr) This composition law or samasabhavana was discovered by Brahmagupta This was one of the first instances of a group-theoretic argument 2

3 Observe therefore that if (p, q), (r, s) are positive solutions of x 2 Ny 2 = 1, the Bhavana produces a larger solution In this manner, starting with one nontrivial (that is, (1, 0)) solution, one obtains infinitely many solutions of this equation We shall describe a method known as the chakravala method due to Jayadeva, Bhaskara and Narayana from the 11th and 12th centuries which solve this equation The amazing thing is that the chakravala method produces all solutions! This is what Weil was referring to As far as I know, the Indians did not state explicitly that the Chakravala method produces all solutions of the equations x 2 Ny 2 = 1 although they might have even been convinced in their minds that it does We shall find it convenient to use continued fractions to developed by Lagrange in the 1780 s to prove that the Chakravala does give all the solutions Moreover, it is an algorithm which can easily be implemented on a computer That is the reason we have called the Chakravala a modern method Some of the references to consult for more information are : (i) History of Algebra by VSVaradarajan in TRIM series, (ii) Number theory from Hammurapi to Legendre by AWeil, (iii) Theory of Numbers by Hardy & Wright Brahmagupta s shortcuts For finding a particular solution of x 2 Ny 2 = 1, Brahmagupta devised some shortcuts of the following kind: Write (u, v; n) to mean u 2 Nv 2 = n Then, (u, v; ±1) (2u 2 ± 1, 2uv; 1) (u, v; ±2) (u 2 ± 1, uv; 1) (2u, v; 4) (2u 2 1, uv; 1) (2u + 1, v; 4) ((2u + 1)(2u 2 + 2u 1), 2u(u + 1)v; 1) (2u, v; 4) (2u 2 + 1, v; 1) (u, v; 4) (p, q; 1) 3

4 with u odd, p = (u2 +2)((u 2 +1)(u 2 +3) 2) 2, q = uv(u2 +1)(u 2 +3) 2 Examples N = 13 Observing that (3) 2 = 4 Then, (11, 3; 4) and the 4th shortcut gives (649, 180; 1) N = 61 Observe that (5) 2 = 4 The last shortcut gives a solution x = 1523 (1522)(1524) 2 2 = and y = (39)(5)(1522)(1524) 2 = This happens to be the smallest solution! Chakravala algorithm Let us describe the Chakravala method roughly first The basic idea is to start with the initial values p 0 = [ N] and q 0 = 1 and look at them as solutions of the equation x 2 = m 0 y 2 = 1 where m 0 = p 2 0 N Taking an appropriate x 1 close to N, one has a solution (x 1, 1) of the second equation x 2 Ny 2 = x 2 1 N Then, one uses Samasabhavana to get a solution of the equation x 2 Ny 2 = m 0 (x 2 1 N) Now, the key point is to make the choice of x 1 in such a manner that the resulting solution (p 1, q 1 ) of x 2 Ny 2 = m 0 (x 2 1 N) as well as the modulus x2 1 N are all multiples of m 0 so that (p 1 /m 0, q 1 /m 0) would be a solution of x 2 Ny 2 = m 1 where m 1 = (x 2 1 N)/m 0 In this manner, recursively a suitable x i is so chosen that a solution of a new equation x 2 Ny 2 = m i is produced and the hope is that at some finite stage, the modulus m k = 1 and we stop! Let us describe this more precisely now Start with p 0 < N < p Take q 0 = 1 and m 0 = p 2 0 N (note m 0 < 0) Then, we have (p 0, q 0 ; m 0 ) Choose x 1 p 0 mod and x 1 < N < x 1 + Note that in forming the solution (p 0, q 0 ) (x 1, 1) = (p 0 x 1 + N, p 0 + x 1 ) of (p 0 x 1 + N, p 0 + x 1 ; m 0 (x 2 1 N)), the numbers p 0 x 1 + N, p 0 + x 1, x 2 1 N are all multiples of m 0 and x 2 1 N is as small as possible 4

5 Indeed, p 0 + x 1 0 mod m 0 and p 0 x 1 + N p N = m 0 0 mod m 0 Of course, we also have x 2 1 N p2 0 N = m 0 0 mod m 0 We have then (p 1, q 1 ; m 1 ) where p 1 = p 0x 1 +N, q 1 = p 0+x 1, m 1 = x2 1 N m 0 Also m 1 > 0 as x 2 1 N < 0 and m 0 < 0 Knowing p i, q i, m i, x i we shall describe (in that order) x i+1, m i+1, p i+1, q i+1 such that (p i+1, q i+1 ; m i+1 ) holds and stop when (and if!) we reach m k = 1 Recursive definition Suppose (p i, q i ; m i ) where we assume p i = p i 1x i + Nq i 1 m i 1, q i = p i 1 + x i q i 1, m i = x2 i N m i 1 m i 1 Define x i+1 x i mod m i with x i+1 < N < x i+1 + m i With this choice of x i+1, (p i+1, q i+1) := (p i, q i ) (x i+1, 1) = (p i x i+1 + Nq i, p i + q i x i+1 ) The key point to note is that the choice of the congruence defining x i+1 ensures that m i divides both p i+1 and q i+1 as well as x2 i+1 N: Indeed q i+1 = p i + q i x i+1 p i q i x i mod m i = p i 1x i + Nq i 1 p i 1 x i q i 1 x 2 i m i 1 = q i 1(N x 2 i ) m i 1 = ±q i 1 m i That is, m i divides q i+1 Also, we get p i+1 ±p i 1m i 0 mod m i The congruence x 2 i+1 N x2 i N mod m i ensures the divisibility of x 2 i+1 N by m i as m i m i 1 = x 2 i N Therefore, we can take p i+1 = p i+1 m i, q i+1 = q i+1 m i, m i+1 = x2 i+1 N m i The name Chakravala We shall show that each m i ( 2 N, 2 N) The m i s will repeat in cycles - hence called chakravala However, it is not obvious as yet that some m k = 1 but we will prove this later using continued fractions Moreover, it is not even clear that the p i, q i are integers Later, we show using continued fractions that these assertions are true 5

6 We now show simultaneously x i+1 < N, m i < 2 N = N p 2 0 < 2 N and x 1 < N < x 1 + So, x 1 > N > N 2 N > N Inductively, if m i < 2 N, then x i+1 < N < x i+1 + m i gives x i+1 < N as before, and m i+1 = x2 i+1 N m i = N + x i+1 N x i+1 m i < 2 N Formal algorithm summarized p 0 < N < p q 0 = 1 m 0 = p 2 0 N x 1 p 0 mod x 1 < N < x 1 + p 1 = p 0x 1 +N q 1 = p 0+x 1 m 1 = x2 1 N m 0 x i x i 1 mod m i 1 x i < N < x i + m i 1 p i = p i 1x i +Nq i 1 m i 1 q i = p i 1+x i q i 1 m i 1 m i = x2 i N m i 1 Example N = 13 Then p 0 = 3 < 13 < 4, q 0 = 1, m 0 = p = 4 Therefore (p 0, q 0 ; m 0 ) = (3, 1; 4) x 1 p 0 = 3(4) and x 1 < 13 < x gives x 1 = 1 p 1 = p 0x = 4, q 1 = p 0+x 1 = 1, m 1 = x m 0 = 3 Therefore, (p 1, q 1 ; m 1 ) = (4, 1; 3) and x 1 = 1 Now x 2 4 mod 3 and x 2 < 13 < x give x 2 = 2 x 2 = 2, p 1 = 4, q 1 = 1, m 1 = 3 So p 2 = p 1x 2 +13q 1 m 1 = 7, q 2 = p 1+q 1 x 2 m 1 = 2, m 2 = x m 1 = 3 Therefore, (p 2, q 2 ; m 2 ) = (7, 2; 3) and x 2 = 2 Now x 3 2 mod 3 and x 3 < 13 < x gives x 3 = 1 So p 3 = (7 + 26)/3 = 11, q 3 = (7 + 2)/3 = 3, m 3 = (1 2 13)/( 3) = 4 Therefore, (p 3, q 3 ; m 3 ) = (11, 3, 4) and x 3 = 1 x 4 1 mod 4 and x 4 < 13 < x gives x 4 = 3 So p 4 = ( )/4 = 18, q 4 = (11 + 9)/4 = 5, m 4 = (3 2 13)/(4) = 1 Therefore, (p 4, q 4 ; m 4 ) = (18, 5; 1) and x 4 = 3 x 5 3 mod 1 and x 5 < 13 < x gives x 5 = 3 6

7 So p 5 = ( )/1 = 119, q 5 = ( )/1 = 33, m 5 = (3 2 13)/( 1) = 4 Therefore, (p 5, q 5 ; m 5 ) = (119, 33; 4) and x 5 = 3 x 6 3 mod 4 and x 6 < 13 < x gives x 6 = 1 So p 6 = ( )/4 = 137, q 6 = (119+33)/4 = 38, m 6 = (1 2 13)/4 = 3 Therefore, (p 6, q 6 ; m 6 ) = (137, 38; 3) and x 6 = 1 x 7 1 mod 3 and x 7 < 13 < x gives x 7 = 2 So p 7 = ( )/3 = 256, q 7 = (137+76)/3 = 71, m 7 = (2 2 13)/( 3) = 3 Therefore, (p 7, q 7 ; m 7 ) = (256, 71; 3) and x 7 = 2 x 8 2 mod 3 and x 8 < 13 < x gives x 8 = 1 So p 8 = ( )/3 = 393, q 8 = ( )/3 = 109, m 8 = (1 2 13)/3 = 4 Therefore, (p 8, q 8 ; m 8 ) = (393, 109; 4) and x 8 = 1 x 9 1 mod 4 and x 9 < 13 < x gives x 9 = 3 So p 9 = ( )/4 = 649, q 9 = ( )/4 = 180, m 9 = (3 2 13)/( 4) = 1 Therefore, (p 9, q 9 ; m 9 ) = (649, 180; 1) and x 9 = 3 Voila! (ii) N = 67 (48842, 5967; 1) (iii) N = 61 ( , ; 1) (iv) N = 103 (227528, 22419; 1) (v) N = 97 ( , ; 1) Properties of the p i s,q i s (I) p i+1 q i q i+1 p i = ( 1) i (II) 0 < p 0 < p 1 <, 0 < q 1 < q 2 < (III) a i+1 = x i+1+x i m i are positive integers satisfying p i+1 = a i+1 p i + p i 1 q i+1 = a i+1 q i + q i 1 7

8 Proof of p i+1 q i q i+1 p i = ( 1) i As m i m i+1 = x 2 i+1 N < 0, m 0 < 0, ( xi+1 N 1 x i+1 ) ( ) pi q i m i m i = ( 1)i+1 = ( p i+1 q i+1 ) = ( ) pi+1 m i q i+1 m i Multiplying on the left by (q i, p i ), we get ( ) ( ) xi+1 N pi ( q i p i ) 1 x i+1 q i ( ) pi+1 m = ( q i p i ) i = (q q i+1 m i i p i+1 p i q i+1 ) m i The LHS q 2 i N p2 i = m i; so, p i+1 q i q i+1 p i = ( 1) i Note that (II) follows from (III) inductively; the beginning inequalities p 1 > p 0 > 0, q 1 > 0 are accomplished as follows Recall p 1 = p 0x 1 +N = p 0x 1 +p m 0 p = 1 + p 0 +x 1 0 = 1 + p 0 q 1 If we show that q 1 > 0, then it would follow that p 1 > p 0 As x 1 > N and 1 > N p 0, we have q 1 = p 0 + x 1 > p 0 + N = 1 N p0 1 > 0 Proof of (III) We leave the easy inductive proof of the fact that a i+1 = x i+1+x i m i are integers satisfying p i+1 = a i+1 p i + p i 1 q i+1 = a i+1 q i + q i 1 We prove that a i s are positive as follows As x i < N < x i + m i 1, we have m i 1 > N x i = N x2 i = m im i 1 = m im i 1 N + xi N + xi N + xi Hence N + x i > m i and so, a i+1 = x i + x i+1 m i > m i N + x i+1 m i > 0 Continued fractions - a crash course 8

9 A simple continued fraction is l = a a 1 + a 2 + := lim n (a a 1 + a One writes symbolically as l = [a 0 ; a 1, a 2, ] The successive quotients a n ) p 0 q 0 := a 0 1, p 1 q 1 = a 0a a 1 = a a 1, are called the convergents to the continued fraction Note p 0 q 0 < p 2 q 2 < < l < p 1 q 1 < p 3 q 3 < Inductively p n+1 = a n+1 p n + p n 1 q n+1 = a n+1 q n + q n 1 To see these, note that p n+2, q n+2 arise exactly like p n+1, q n+1 but with a n+1 replaced by a n a n+2 etc So which proves the claim p n+2 = (a n a n+2 )p n + p n 1 q n+2 (a n a n+2 )q n + q n 1 = a n+2(a n+1 p n + p n 1 ) + p n a n+2 (a n+1 q n + q n 1 ) + q n = a n+2p n+1 + p n a n+2 q n+1 + q n The ( above recursion ) ( is expressed ) ( better ) as : pn p n 1 an+1 1 pn+1 p = n 1 0 q n q n 1 q n+1 But, ( p 0 = a) 0,( q 0 = 1, p) 1 = ( a 0 a 1 + 1, ) q 1 = a 1 a0 1 a1 1 p1 p So = q 1 q 0 Thus, the recursion is expressed neatly as: ( a ) ( ) a q n ( ) an 1 = 1 0 Observe that immediately the determinants give p n q n 1 p n 1 q n = ( 1) n 1 ( ) pn p n 1 q n q n 1 9

10 So p n, q n are relatively prime and, from this, it can be shown that the limit defining the CF exists There is a simple way to recover l from any one of the so-called complete quotients l n = a n a n+1 + a n+2 + Since p n qn = a np n 1 +p n 2 a nq n 1 +q n 2 and l is obtained when a n above is replaced by l n, we have l = l np n 1 + p n 2 l n q n 1 + q n 2 CFs for rational numbers It is elementary to observe that finding continued fractions for rational numbers p/q is not only equivalent to the Euclidean algorithm of finding the GCD of p and q but also equivalent to (very efficiently) solving the linear Diophantine equation px qy = (p, q) Let us give an example Suppose we wish to solve 72x+5 7 y = 1 in integers x, y Look at the following division table: Then = [1085; 14, 2, 2] The penultimate convergent [1085; 14, 2] = /( ) = = provides us with a solution of 78125x 72y = 1; viz (x, y) = (29, 31467) So, it is not surprising that finding the continued fractions of certain numbers may lead to solutions of some Diophantine equations How does one find the CF for an irrational number like 7? = 2 + ( 7 2) = 2 + = ( 7 + 2)/ = 1 + ( 1) = = = ( 7 + 1)/2 = = = ( 7 2) 10

11 Thus, we have a repetition and So, 7 = [2; 1, 1, 1, 4] = Continued fraction of N Let N be a square-free positive integer How does the CF of N look? Now, N = a 1 + ( N a 1 ) with a 1 = [ N] Thus, N a 1 = N a2 1 N+a1 = r 1 N+a1, say, where r 1 = N = a 2 1 In other words, N = a 1 + r 1 1 N+a1 = a 1 + ( N+a 1 )/r 1 Write [ N+a1 r 1 ] = b 1 ; then N + a1 = b 1 r 1 + ( N a 2 ) where a 2 = b 1 r 1 a 1 So, N = a b 1 +( N a 2 )/r 1 = a The CF is N = b b b 2 + b 1 + b 0 = a 1 = [ N] r 1 = N a 2 1 b 1 = [ N+a1 r 1 ] 1 ( N+a 2 )/r 2 where r 1 r 2 = N a 2 2 a n = b n 1 r n 1 a n 1 r n 1 r n = N a 2 n b n = [ N+an r n ] The crucial point is that, at each stage if b n = [ N+an r n ] is replaced by l n = N+an r n, then we get the exact value N l n+1 p n + p n 1 N = = ( N+an+1 r n+1 )p n + p n 1 l n+1 q n + q n 1 ( N+an+1 r n+1 )q n + q n 1 Comparing the terms with and without N, a n+1 p n + r n+1 p n 1 = Nq n a n+1 q n + r n+1 q n 1 = p n Using p n 1 q n p n q n 1 ) = ( 1) n, we have a n+1 ( 1) n 1 = p n 1 p n Nq n 1 q n 11

12 r n+1 ( 1) n 1 = p 2 n Nqn 2 One may deduce from this a n+1, r n+1 > 0 Further, if we know some r k (say r n+1 ) equals 1, then ( 1) n 1 = p 2 n Nqn 2 Comparing Chakravala with CF Amazingy, the Chakravala method produces the same p k s and q k s as in the CF for N Recall that in Chakravala method, we have p i+1 = a i+1p i + p i 1 q i+1 = a i+1q i + q i 1 where a i+1 = x i+1+x i m i with a 0 = p 0 = [ N], x i < N < x i + m i 1 and m i m i 1 = x 2 i N We have written p i, q i, a i etc to distinguish from the p i, q i, a i in the CF In view of the recursions, one can show that the b i s in the CF of N are the same as the a i s here as follows Recalling the expression for complete quotient at any stage in the SCF of N, and recalling that Nqn 1 p n 1 we have l n+1 = This gives the expression p n Nq n N = l n+1 p n + p n 1 b n+1 = [l n+1 ] = [ Firstly, a 0 = p 0 = [ N] = b 0 Next, l n+1 q n + q n 1 Nqn 1 p n 1 p n Nq n ] a 1 = q 1 = p 0 + x 1 since x 1 < N < x 1 + Hence a 1 = [ b 0+ N ] = b N b 2 1 ; that is, 0 = b 0 + x 1 = [ b 0 + N ] a 1 = q 1 = b 1 12

13 Also, p 1 = 1 + p 0 q 1 = 1 + b 0b 1 = p 1 Thus, we have shown a 0 = b 0, p 0 = p 0, q 0 = q 0, p 1 = p 1, q 1 = q 1 We assume that b i = a i for all i n; then p i = p i, q i = q i for all i n We show b n+1 = a n+1 Recall that Nqn 1 p n 1 b n+1 = [ p n ] Nq n Hence, it suffices to show that Nq a n+1 = [ n 1 p n 1 p n ] ( ) Nq n Now, the Chakravala has which gives p i = p i 1 x i + Nq i 1, q i = p i 1 + x iq i 1 m i 1 m i 1 p n = p n 1 x n + Nq n 1 m n 1, q n = p n 1 + q n 1 x n m n 1 Using these, the right hand side of ( ) is Nq [ n 1 p n 1 p n m n 1 N + xn ] = [ ] = [ ] Nq n N xn m n where we have used m n m n 1 = x 2 n N Denote the integer [ N+xn m n ] by t Then, we have t < N + xn m n < t + 1 Hence t m n x n < N < t m n x n + m n As x n+1 is the unique integer satisfying x n+1 < N < x n+1 + m n as we defined in the Chakravala, we have x n+1 = t m n x n This means t = xn+x n+1 m n = a n+1 So b n+1 = a n+1 We have thus proved our assertion Recurrence of CF for N Let us start by proving the well-known fact that b n s recur Now 0 < r n 1 r n = N a 2 n implies that a 1 = [ N] a n n 13

14 So, each a n can only have the values 1, 2,, a 1 = [ N] From b n 1 r n 1 = a n + a n 1 2a 1, we have r n 1 2a 1 for all n This means that the complete quotients (these are N+an r n ) can take at the most 2a 2 1 values; so they recur So a n = a k, r n = r k, b n = b k for some n < k We shall show that a n 1 = a k 1, r n 1 = r k 1, b n 1 = b l 1 Now, r k 1 r k = N a 2 k = N a2 n = r n 1 r n gives r k 1 = r n 1 a k 1 a n 1 = (a k 1 + a k ) (a n 1 + a n ) = (b k 1 b n 1 )r n 1 We observe as follows that a 1 < a d + r d for all d r d 1 b d 1 r d 1 = a d 1 + a d < N + a d = N a2 d N ad = r d 1r d N ad So, it follows that r d > N a d > a 1 a d Thus, (b k 1 b n 1 )r n 1 = a k 1 a n 1 = (a 1 a n 1 ) (a 1 a k 1 ) < a 1 a n 1 < r n 1 ; we arrive at b k 1 = b n 1, a k 1 = a n 1 Proceeding in this manner, there is n such that a n+1 = a 1, r n+1 = r 1 So r n r 1 = r n r n+1 = N a 2 n+1 = N a 2 1 = r 1 which means r n = 1 So r n ( 1) n = ( 1) n = p 2 n 1 Nq2 n 1 Chakravala gives all The Chakravala method does give solutions of x 2 Ny 2 = 1 Indeed, we know how to get a solution of x 2 Ny 2 = 1 from one for x 2 Ny 2 = 1 The remarkable fact that it gives all the solutions is contained in the theorem below: Theorem Every solution of either of the equations x 2 Ny 2 = ±1 is of the form (p n, q n ) for some n Let us first observe that any solution of x 2 Ny 2 = ±1 satisfies x y N < 1 2y 2 We see this when x 2 Ny 2 = 1, for x y N = x y N 1 y = y(x+y < 1 N) 2y 2 since x + y N > 2y When x 2 Ny 2 = 1 and x y, the same argument works When x 2 Ny 2 = 1 with x < y, we have 1 = x 2 Ny 2 < y 2 Ny 2 = (1 N)y 2 which means (N 1)y 2 < 1, an impossibility 14

15 The rest of the proof then follows from the more general statement: If α is a real number which is irrational, and satisfies α r s < 1 where 2s 2 s > 0, then r/s is a convergent to the continued fraction of α We skip the rest of the proof of the theorem which is an easy consequence of the fact that we can find ζ such that α = ζp n + p n 1 ζq n + q n 1 and the key observation (which we prove) where the so-called modular group plays a role: Modular group comes in If α = ζp+r ζq+s where ζ > 1 and ps qr = ±1 and q > s > 0, then p/q, r/s are consecutive convergents to the CF of α For the proof, write p q = [a 0 ; a 1,, a n ] with p i /q i the convergents (so p n /q n = p/q) Choose n so that ps qr = ( 1) n 1 = p n q n 1 q n p n 1 Note that we have used the easy observation that a rational number has two simple continued fractions - one of even length and the other of odd length As p, q are relatively prime with q > 0 and p/q = p n /q n, it follows that p = p n, q = q n Thus, ( 1) n 1 = ps qr = p n s q n r = p n q n 1 q n p n 1 This gives p n (s q n 1 ) = q n (r p n 1 ) which means in view of coprimality of p n, q n that q n (s q n 1 ) As q n = q > s (hypothesis) and q n q n 1, we must have s = q n 1 (and hence r = p n 1 ) We are done Remarks It can be sown that if the continued fraction expansion of N looks like [b 0 ; b 1, b 2,, b r, 2b 0 ] and, the penultimate convergent [b 0 ; b 1, b 2,, b r ] gives a solution of x 2 Ny 2 = 1 or of x 2 Ny 2 = 1 according as to whether the period r + 1 above is odd or even Examples 7 = [2; 1, 1, 1, 4, ] gives the penultimate convergent before the period to be [2; 1, 1, 1] = 8/3 Then, (8, 3) is a solution of x 2 7y 2 = 1 (as the period is of even length) Note that x 2 7y 2 = 1 has no solution (look mod 4) 15

16 13 = [3; 1, 1, 1, 1, 6, ] gives the penultimate convergent before the period to be [3; 1, 1, 1, 1] = 18/5 Then (18, 5) is a solution of x 2 13y 2 = 1 (as the period is of odd length) From this a solution for x 2 13y 2 = 1 can be obtained as ( (5 2 ), 2(18)(5)) = (649, 180) 16