Number Theory and Cryptography

Transcription

1 Number Theory and Cryptography Paul Yiu Department of Mathematics Florida Atlantic University Fall 207 Chapters 9 October 23, 207

2 Contents 9 Sums of Two and Four Squares 95 9 Fermat s two-square theorem 95 9 Fermat s two-square theorem and Gaussian integers Representations of integers as sums of two squares Lagrange s four-square theorem Sums of three squares 99 0 Continued Fractions 0 0 Finite continued fractions 0 0 Euler s function F for finite continued fractions 0 02 The convergents of a continued fraction Cornacchia s algorithm for a prime as a sum of two squares Infinite continued fractions Approximations of irrational numbers Purely periodic continued fractions Lagrange s theorem 07 Continued fraction expansion of d 2 The Pell Equation 5 The equation x 2 dy 2 = 5 2 The equation x 2 dy 2 = 8 3 The equation x 2 dy 2 = N 8 4 Applications of Pell s equations 20 5 Sums of consecutive squares 22 5 Odd number of consecutive squares Even number of consecutive squares 23

3 Chapter 9 Sums of Two and Four Squares 9 Fermat s two-square theorem Theorem 9 Letpbe an odd prime p is a sum of two squares if and only ifp (mod 4) In this case, the expression is unique Proof (Euler) Let p (mod 4) be a prime number The following algorithm expresses p as a sum of two squares of integers () Let q be the smaller positive square root of (mod p), ie, q 2 (mod p) and 0 < q < p Then with (x 2 0,y 0 ) = (q,), we have x 2 0 +y0 2 = q 2 + = m 0 p for an integer m 0 > 0 ) 2 + p 2 4,m 0 p 4 Since q 2 + < ( p 2 (2) Now suppose we have a sequence of positive integers satisfying (x 0,y 0 ), (x,y ),, (x k, y k ) x 2 i +y 2 i = m i p for i = 0,,2,k, andm 0 > m > > m k If m k =, we are done: x 2 k +y2 k = p Otherwise, choose integers a k andb k such that { u k = x k m k a k, v k = y k m k b k satisfy u k, v k < m k 2 (u k and v k are the least absolute value residues of x k andy k modulo m k ) With this choice, u 2 k +v 2 k = (x 2 k +y 2 k ) 2m k (a k x k +b k y k )+m 2 k (a 2 k +b 2 k) = m k p 2m k (a k x k +b k y k )+m 2 k (a 2 k +b 2 k) = m k (p 2(a k x k +b k y k )+m k (a 2 k +b 2 k)) = m k m k

4 96 Sums of Two and Four Squares where m k = p 2(a k x k + b k y k ) + m k (a 2 k + b2 k ) Since u k, v k < m k m k < m k 2 From the 2-square identity, (x 2 k +y 2 k )(u 2 k +v 2 k) = (x k u k +y k v k ) 2 +(x k v k y k u k ) 2 ( ) Here, x k u k +y k v k and x k v k y k u k are both divisible by m k, ie, there are integers x k and y k such that From (*), m k x k = x k u k +y k v k, m k y k = x k v k y k u k m k p m k m k = m 2 k (x 2 k +y 2 k) = x 2 k +y 2 k = m k p (3) If m k =, then we are done: x 2 k +y2 k = p Otherwise, repeat step (2) (Uniqueness) Ifp = a 2 +b 2 = x 2 +y 2, wherea < b andx < y are all positive, then p 2 = (a 2 +b 2 )(x 2 +y 2 ) = (ax+by) 2 +(ay bx) 2 = (ax by) 2 +(ay +bx) 2 Note that (ax+by)(ay +bx) = ab(x 2 +y 2 )+(a 2 +b 2 )xy = p(ab+xy) This means that one ofax+by anday+bx is divisible byp Sinceax+by,ay+bx p, we must have ay bx = 0 or ax by = 0 In other words, x = a or b Indeed, y b a It follows that we must have x = a andy = b x y = a b We make the algorithm in the proof more explicit: () (x 0,y 0 ) = (q,) (2) m k = x2 k +y2 k p (3) u k+ andv k+ are the least absolute value residues of x k andy k modulo m k (4) (x k+,y k+ ) = ( xk u k+ +y k v k+ m k, x kv k+ y k u k+ m k ) (5) m k+ = x2 k+ +y2 k+ p (6) The process stops whenm k+ = Thenx 2 k+ +y2 k+ = p Otherwise, repeat (2) Example 9 Letp = 7933 The smaller positive root of (mod p) isq = 2950 From this, 7933 = k x k y k u k v k m k ,

5 9 Fermat s two-square theorem 97 9 Fermat s two-square theorem and Gaussian integers Since p (mod 4), is a quadratic residue This means that there exists an integer a p such thata 2 + is divisible by p Note thata 2 + < p 2 2 Regarded as Gaussian integers, a 2 + = (a + i)(a i) We claim that p does not divide a + i nor a i; otherwise, p 2 = N(p) N(a + i) = a 2 + < p 2, a contradiction This means that p is not a prime in Z[ ] and there is a factorization of p = αβ Z[ ], in which none of α, β is a unit, ie, N(α),N(β) > It follows from p 2 = N(p) = N(α)N(β) that N(α) = N(β) = p, and p is a sum of two squares of integers 92 Representations of integers as sums of two squares We say that a representation n = x 2 +y 2 is primitive if gcd(x,y) = Lemma 92 If n has a prime divisor q 3 (mod 4), then it does not have a primitive representation Proof Suppose to the contrary thatn = x 2 +y 2 is a primitive representation Since q divides n, it does not divide any of x and y In the field Z q, we write y = ax for somea This means that0 = x 2 +y 2 = x 2 (+a 2 ) Sincex 0, we havea 2 = in Z q, q 3 mod 4, a contradiction Theorem 93 n = 2 a i be the prime factorization of n in which the p s and q s are respectively primes of the form 4k + and 4k +3 The number n is expressible as a sum of two squares if and only if each of the exponents c j is even Proof (Sufficiency) Since 2 = 2 + 2, and every p i is a sum of two squares, if everyc j is even, by repeatedly using the composition formula p b i i j q c j j (a 2 +b 2 )(x 2 +y 2 ) = (ax+by) 2 +(ay bx) 2 we easily obtain n as a sum of two squares (Necessity) Let n be divisible by a prime q 3 (mod 4), with highest power q c, c odd Consider a representation n = x 2 +y 2, with gcd(x,y) = d > Let q c be the highest power of q dividing d (Possibly, c = 0) Write x = dx, y = dy Then gcd(x,y) = Let N = X 2 + Y 2 The highest power of q dividing N is q c 2c This is positive since c is odd, contradicting Lemma 92 above

6 98 Sums of Two and Four Squares 92 Lagrange s four-square theorem Theorem 94 Every positive integer can be represented as a sum of four squares of nonnegative integers Lemma 95 (4-square identity) where (x 2 +x 2 2 +x 2 3 +x 2 4)(y 2 +y 2 2 +y 2 3 +y 2 4) = z 2 +z 2 2 +z 2 3 +z 2 4, z = x y +x 2 y 2 +x 3 y 3 +x 4 y 4, z 2 = x y 2 x 2 y +x 3 y 4 x 4 y 3, z 3 = x y 3 x 2 y 4 x 3 y +x 4 y 2, z 4 = x y 4 +x 2 y 3 x 3 y 2 x 4 y Therefore it is enough to prove Lagrange s theorem for prime numbers Lemma 96 Let p be a prime number There are integers x and y such that x 2 + y (mod p) Proof The set S := {x 2 Z p : x Z} has exactly p+ 2 elements; so does the set T := { (x 2 +) Z p : x Z} Now, S T = S + T S T p+ 2 + p+ 2 p = Therefore, there are integers x and y satisfying x 2 (y 2 + ) (mod p), ie, x 2 +y (mod p) Proof of Theorem 94 Letpbe a prime number There are integersxandy such thatx 2 +y 2 + is divisible byp We write this in the formx 2 +x 2 2+x 2 3+x 2 4 = kp for some integerk Clearly, we may assume x, x 2, x 3, x 4 p < p (p) 2 This means kp < = p 2 and k < p If k, we shall show that x, x 2, x 3, x 4 can be replaced by another quadruple with a smaller k Then, by descent, we shall ultimately reachk = Suppose k is even We may assume x x 2 (mod 2) and x 3 x 4 (mod 2) Then ( ) 2 ( ) 2 ( ) 2 ( ) 2 x +x 2 x x 2 x3 +x 4 x3 x = x2 +x 2 2 +x 2 3 +x = k 2 p

7 93 Sums of three squares 99 with a smaller multiplier for p Suppose k is odd For i =,2,3,4, choose y i x i with y i < k Note that y 2 +y2 2 +y3 2 +y4 2 x 2 +x 2 2 +x 2 3 +x 2 4 (mod k) Write y 2 +y2 2 +y3 2 +y4 2 = kq for some q < k Note thatq must be nonzero Apply the four-square identity to the two quadruples x i and y i The left hand side is (kp)(kq) = k 2 pq On the right hand side, z 2, z 3, z 4 are clearly divisible by k; so is z because z = x y + x 2 y 2 + x 3 y 3 + x 4 y 4 x 2 + x x x (mod k) Writing z i = kw i for i =,2,3,4, we have, from the 4-square identity, wi 2 +w2 2 +w3 2 +w4 2 = qp for q < k 93 Sums of three squares Theorem 97 An integer of the form4 k (8m+7) cannot be a sum of 3 squares Proof Integer squares modulo 8 are congruent to 0,or4 x mod x 2 mod From these, a sum of 3 squares cannot be congruent to7(mod8) , 0+0+, , 0++ 2, , , ++ 3, ++4 6, +4+4, Note that x 2 + y 2 + z 2 4 (mod 8) only if x, y, z are all even In this case, x 2 +y 2 +z 2 = 4(x 2 +y 2 +z 2 ) for integers x, y, z It follows that if 4 k n is a sum of three squares, then so isn Since numbers of the form8m+7 are not sums of 3 squares, neither are 4 k (8m+7) Remark The converse of Theorem 97 is also true; but is a much more difficult theorem

8 00 Sums of Two and Four Squares

9 Chapter 0 Continued Fractions 0 Finite continued fractions 0 Euler s functionf for finite continued fractions Every rational number a b can be written as a finite continued fraction in the form a b = q 0 + q + q 2 + +q n + q n, where q 0, q, q 2,, q n are the quotients in the Euclidean algorithm sequence for (a,b): putting r 2 = a,r = b, we defineq k andr k for k = 0,,,n by Here,q k = r k 2 r k, and a = bq 0 +r 0, b = r 0 q +r, r 0 = r q 2 +r 2, r n 2 = r n q n +r n, r n = r n q n+ a > b > r 0 > r > r 2 > r 3 > r n > r n in which r n r n The number r n is the gcd of a and b If we assume the rational number given in its lowest terms, thenr n = We shall write the continued fraction above simply as [q 0,q,q 2,,q n ] Now, it is easy to compute the following

10 02 Continued Fractions [q 0 ] = q 0, [q 0,q ] = q 0q + q, [q 0,q,q 2 ] = q 0q q 2 +q 0 +q 2, q q 2 + [q 0,q,q 2,q 3 ] = q 0q q 2 q 3 +q 0 q +q 0 q 3 +q q 2 +, q q 2 q 3 +q +q 3 Euler has given a very elegant procedure of computing finite continued fractions: [q 0,q,q 2,,q n ] = F(q 0,q,q 2,,q n ) F(q,q 2,,q n ), where F is the function obtained in the following way: F(q 0,q,,q n ) is the sum of q 0 q q n and all products obtained by deleting pairs of consecutive factors, with the stipulation that if n+ is even, deleting all consecutive pairs leads to the empty product Lemma 0 (a)f is palindromic: F(q 0,q,,q n,q n ) = F(q n,q n,,q,q 0 ) (b) F(q 0,q,,q n,q n ) = F(q 0,q,,q n 2 )+F(q 0,q,q n ) q n (0) = q 0 F(q,,q n )+F(q 2,q n ) (02) Proof (a) is clear (b) The terms off(q 0,q,,q n,q n ) fall into two classes: (i) each of those containing q n is multiplied to a term in mathcalf(q 0,, n ), (ii) those not containing q b cannot contain q n and are all inf(q 0,,q n 2 This proves (0) The relation (02) follows from this and (a) Lemma 02 Letq 0,q,, q n be positive integers The numbers F(q 0,q,,q n ), F(q,q 2,,q n ),,F(q n,q n ), F(q n ) are pairwise relatively prime

11 0 Finite continued fractions 03 Proof From the above relations, gcd(f(q 0,q,,, F(q,q,,q n ) = gcd(f(q,,q n,q n ), F(q 2,,q n ) = gcd(f(q n,q n ),gcd(q n ) = gcd(q n q n +, q n ) = Proposition 03 Let a > b be relatively prime With r 2 = a and r = b, in the Euclidean algorithm sequence of(a,b), for k = 0,,n 2, and r n = r k = F(q k+2,,q n ), Proof Note thatr 2 = a = F(q 0,q,,q n ) and r = b = F(q,q 2,,q n ) r 0 F(q,q 2,,q n ) = r 0 r = a b q 0 = [0,q,q 2,,q n ] = [q,q 2,,q n ] = F(q 2,,q n ) F(q,q 2,,q n ) Therefore, r 0 = F(q 2,,q n ) The rest follows by induction 02 The convergents of a continued fraction Given a continued fraction [q 0,q,,q n ], by writing, for k = 0,,,n, P k = F(q 0,q,,q k ), Q k = F(q,,q k ), we obtain a sequence P k Q k, k = 0,,,n, in which the term Pn = a P b k and Q k can be computed recursively as follows P k = P k 2 +q k P k, P 2 = 0, P =, Q k = Q k 2 +q k Q k, Q 2 =, Q = 0

12 04 Continued Fractions Remark Comparison with the Euclidean algorithm table of (a,b) shows that for k = 0,,2,,n P k = y k = ( ) k y k, Q k = x k = ( ) k x k, Example 0 The continued fraction[, 2, 3, 4, 5, 6, 7, 8, 9, 0] are computed easily using these relations k q k P k Q k Lemma 04 Fork =,,n, P k Q k P k Q k = ( )k Q k Q k Proof Write P k Q k P k Q k = N k Q k Q k We have N k = P k Q k Q k P k = (P k 2 +q k P k )Q k (Q k 2 +q k Q k )P k = (P k Q k 2 Q k P k 2 ) = N k Since N =, we have by easy induction N k = ( ) k N = ( ) k, and the result follows Lemma 05 Fork = 2,,n, P k Q k P k 2 Q k 2 = ( )k q k Q k 2 Q k Proof P k Q k P k 2 Q k 2 = ( Pk P ) ( k 2 Pk + P ) k Q k Q k 2 Q k Q k = ( )k 2 Q k 2 Q k + ( )k Q k Q k = ( )k (Q k Q k 2 ) Q k 2 Q k Q k = ( )k q k Q k Q k 2 Q k Q k = ( )k q k Q k 2 Q k 02 Cornacchia s algorithm for a prime as a sum of two squares Theorem 06 (Cornacchia) Let p (mod 4) be a prime, and q the smaller positive square root of mod p If in the euclidean algorithm sequence of(p,q), x and y are the first two remainders smaller than p, thenp = x 2 +y 2

13 02 Cornacchia s algorithm for a prime as a sum of two squares 05 Proof Consider the Euclidean algorithm sequence r 2 = p, r = q, r 0, r,, r n =, r n+ = 0 for the pair(p,q), and its two associated sequences(x k ) and(y k ) fork = 2,,, n,n+ We complete the proof in the following steps n is even () Since = gcd(p,q) = px n +qy n qy n (mod p) andq 2 (mod p), we havey n q (mod p) It follows thaty n = q orp q The reversal of ( y k ) is a Euclidean algorithm sequence with the same length as that of(p,q) If y n = p q, the Euclidean algorithm sequence of (p, p q) would be p, p q, q, which is longer than that of(p,q), a contradiction Thus,( ) n y n = y n = q andnis even 2 The sequence ( y k ) is the reversal of (r k ); ie, y k = r n k for every 2 k n+ Furthermore, ( y k ) is the reversal of the Euclidean algorithm sequence of (p,q), which is exactly the sequence (r k ) 3 The sequence(q k ) is palindromic; ie,q n+ k = q k for everyk = 0,, 2,, n+ This is an immediate consequence of (2) 4 rk 2 +y2 k is divisible by p for everyk This follows from r k = px k +qy k qy k (mod p) Squaring, we have rk 2 q 2 yk 2 y2 k (mod p), andr2 k +y2 k 0 (mod p) 5 Let n = 2m In the sequence (r k ), r m and r m are the two terms smaller than p Write n = 2m Note that r m = F(q m+,q m+2,,q 2m ), and p = r 2 = F(q 0,,q 2m ) Now, since (q k ) is palindromic, r 2 m = r m r m = F(q m+,q m+2,,q 2m )F(q m+,q m+2,,q 2m ) = F(q 0,q,q m )F(q m+,q m+2,,q 2m ) It is clear that each term in the product is contained inf(q 0,q 2,,q 2m ) This shows that r 2 m < p On the other hand, r 2 m 2 = r m 2 r m 2 = F(q m,q m+,q m+2,,q 2m )F(q m,q m+,q m+2,,q 2m ) = F(q 0,q,, q m )F(q m,q m+,q m+2,,q 2m )

14 06 Continued Fractions This time, every term in F(q 0,q,q 2,,q 2m ) is contained in this product This shows that r 2 m 2 > p Therefore, r m and r m are the first two terms in the Euclidean algorithm less than p 6 p = r 2 m +r 2 m Now, since r m = r (n ) (m ) = y m, we have r 2 m +r 2 m = r 2 m +y 2 m 0 (mod p) Since r m < r m < p, the sum r 2 m +r 2 m being positive and smaller than 2p, must be p Example 02 p = is the00,000-th prime The square roots of mod p are ± mod p k q k r k x k y k = Infinite continued fractions Theorem 07 Let q 0,q,,q n, be an infinite sequence of positive integers The sequence of rational numbers defined by P n = [q 0,q,,q n ] is convergent ( ) P Proof By Lemma 05, (i) the subsequence of even indexed convergents 2n Q 2n is ( ) an increasing sequence, (ii) the subsequence of odd indexed convergents P2n+ Q 2n+

15 03 Infinite continued fractions 07 is a decreasing sequence By Lemma 04, ( P0, P ) ( P2, P ) ( 3 P2n, P ) 2n+ Q 0 Q Q 2 Q 3 Q 2n Q 2n+ By Lemma 04 again, the lengths of these intervals converges to0: lim n P 2n+ Q 2n+ P 2n Q 2n = lim n Q 2n+ Q 2n = 0 since Q 2n+ > Q 2n > 2n Therefore the intervals ( P n ) have a unique common point, which is the limit of the sequence Letζ be a real, irrational number, The continued fraction expansion ofζ can be found recursively as follows Then, ζ 0 = ζ, q 0 = [ζ 0 ]; ζ n+ = ζ = [q 0,q,q 2,,q n,] ζ n [ζ n ], q n+ = [ζ n+ ] Corollary 08 Every positive real number has a unique continued fraction expansion If the number is rational, the last quotient is greater than Proposition 09 Let ζ = [q 0,q,,q n,] with convergents Pn For every integer n, () +2 < ζ P n < +, (2) ζ P n < ζ P n, ζ (3) ζ Pn < P n, (4) ζ Pn < Q 2 n Proof We make use of the tails of the infinite continued fraction For each n, define ζ n+ := [q n+, q n+2,] This satisfies () ζ = [q 0,q,,q n,ζ n+ ] = P n +P n ζ n+ + ζ n+ Therefore, ζ P n = ζ n+p n +P n ζ n+ + P n ζ P n = = P n P n (ζ n+ + ) ζ n+ +

16 08 Continued Fractions Sinceq n+ < ζ n+ < +q n+, the denominator is betweenq n+ + = + and (+q n+ ) + = (The last equality holds only if q n+2 = ) Thus, +2 < ζ P n < + (2) Applying () twice, < ζ P n < < ζ P n < (3) This follows from (2) by dividing by, and the fact that > (4) Since ζ lies Pn and P n+ +, ζ P n < P n+ P n + = ( ) n + = < + Q 2 n since < + 04 Approximations of irrational numbers Lemma 00 Let0 < Q If P Q Pn, then ζ Q P > ζ P n Proof () Assume Q = Since P P n, P Pn Q On the other hand, ζ P n = P n +P n ζ n+ P n + ζ n+ = ( ) n ( + ζ n+ ) = ( +ζ n+ ) < < Q 2 n 2 From this, P Q P n ζ P Q + ζ P n ζ P Q + 2 Therefore, ζ P Q 2 ζ Pn Clearing denominators, we have ζq P ζ P n

17 05 Purely periodic continued fractions 09 (2) Assume ot equal to the denominator of any convergent of ζ It must be between two successive ones, say,q k < Q < Q k for k n Since P k Q k Q k P k = ±, there are unique integers a and b such that (P,Q) = a(p k,q k )+b(p k,q k ) Hence, ζq P = a(ζq k P k )+b(ζq k P k ) Note that a and b must have opposite signs, and so doζq k P k andζq k P k It follows thata(ζq k P k ) andb(ζq k P k ) have the same sign From this, ζq P > ζq k P k > ζq k P k > ζ P n Theorem 0 If ζ P Q <, then P is a convergent in the continued fraction 2Q 2 Q expansion of ζ Proof From ζ P Q <, we have Qζ P < 2Q 2 2Q Suppose P is not a convergent of ζ Then Q Q k < Q < Q k for some k, and by Lemma 00, ζq P > ζq k P k Therefore, From this, This contradicts ζ P k Q k = ζq k P k < ζq P < 2QQ k Q k Q k P Q P k < ζ P Q + ζ P k < QQ k Q k P Q P k Q k Q k = PQ k QP k QQ k QQ k 05 Purely periodic continued fractions Letζ be represented by a purely periodic continued fraction: ζ = [q 0,q,,q k ] This means ζ = [q 0,q,,q k,ζ] Let P k Q k and P k Q k be the last two convergents of the finite continued fraction [q 0,q,,q k ] Then, ζ = ζp k +P k ζq k +Q k

18 0 Continued Fractions From this, we see that ζ is a root of the quadratic equation f(x) := Q k x 2 (P k Q k )x P k = 0 Note that (i) f( ) = Q k +(P k Q k ) P k = (P k P k )+(Q k +Q k ) > 0, (ii) f(0) = P k < 0, (iii) f() = (P k Q k ) (P k Q k ) < 0 From these we conclude that ζ > is the only positive root, and the other root ζ = a b d lies between and0 We say that a quadratic irrationality ζ > is reduced if ζ lies between and 0 A purely periodic continued fraction is a reduced quadratic irrationality 2 An eventually periodic continued fraction is a quadratic irrationality Proof Letχ = [p 0,,p h,q,,q n ] If the last two convergents of[p 0,,p h ] are P h Q h and P h Q h, andζ = [q,,q n ], thenχ = ζ P h+p h ζ Q h +Q h is also a quadratic irrationality Lemma 02 If [q 0,q,,q k ] = ζ, then < ζ < 0 and [q k,,q,q 0 ] = ζ Proof From ζ = [q 0,q,,q k ], we have q 0 ζ = [ q 0, ] = ζ From this, q + [q,,q k,ζ] [q,,q k,ζ] = [q 0, [q,,q k,ζ] ζ ], [q 2,,ζ] = [q 0, ], ζ q + [q 0, ] = ζ [ q,q 0, ] = ζ [q 2,,ζ], [q 2,,ζ] Continuing, we obtain [ q k,q k,,q,q 0, ] = ζ ζ Therefore, ζ The positive root must be ζ is the negative root of the quadratic equation x = [q k,q k,,q 0,x]

19 06 Lagrange s theorem Corollary 03 If a+ d b = [q 0,,q k ], then[q k,,q 0 ] = a+ d b 06 Lagrange s theorem A quadratic irrationality can be expressed in the form ζ = a+ d, where a, b, and d b are integers such thatb = d a2 is an integer b Proof It is clear that a quadratic irrationality is one of the form a+ d for integers b a, b and d If d a2 = b for relatively prime integers b q b and q with q >, we replace a and b by qa and qb, and d by q 2 d In this case, ζ = a+ d = qa+ q 2 d, and q 2 d (qa) 2 qb = q d a2 b = b is an integer Theorem 04 The continued fraction expansion of a quadratic irrationality is eventually periodic Proof () Letζ = [q 0,,q n,ζ n ] We have ζ = ζnp n +P n 2 ζ n + 2 Consider the conjugate ζ = ζnp n +P n 2 ζ n + 2 From this, ζ n = ζ 2 P n 2 ζ P n = 2 ζ ζ P n b P n 2 2 Since the sequence Pn converges to ζ, and 2 <, we can choose N large enough so that for n > N, ζ n lies between and 0 In other words, ζ n is reduced for n > N (2) ζ n = an+ d b n for integers a n,b n satisfying b n d a 2 n Proof Clearly,q 0 = [ζ], and ζ = = ζ q 0 = a+ d b q 0 b a q 0 b+ d = b( a+q 0b+ d) d (a q 0 b) 2 = b( a+q 0b+ d) d a 2 +2q 0 ab q 2 0b 2 = a+q 0b+ d b +2q 0 a q 2 0b qb By putting ζ = a + d b d a 2 = b b with a = a+q 0 b, b = b +2q 0 a q 2 0b, we have Here,

20 2 Continued Fractions More generally, ifζ n = an+ d b n with d a 2 n = b n b n, thenζ n+ = a n++ d b n+ with These satisfyd a 2 n+ = b n+ b n+ a n+ = a n +q n b n, b n+ = b n +2q n a n q 2 nb n, b n+ = b n (3) If ζ = a+ d b is reduced, then 0 < a < d and0 < b < 2 d (4) Since ζ n is reduced for n > N, 0 < a n < d and 0 < b n < 2 d There are finitely many pairs of positive integers (a, b) satisfying these conditions We conclude that there are integers h < k satisfying (a h,b h ) = (a k,b k ) If we choose h and k = h+r to be the smallest possible integers for which these hold, then for every integer t 0 and0 s < r, then a h+tr+s = a h+s, b h+tr+s = b h+s In other words,ζ h+tr+s = ζ h+s, andq h+tr+s = q h+s This completes the proof of Langrange s theorem Corollary 05 The continued fraction expansion of a reduced quadratic irrationality is purely periodic Proof It is enough to show that if ζ = [q 0,q,,q r ] is reduced, then indeed, q 0 = q r (The general case follows by induction) Letθ = [q,,q r ] Sinceq 0 + θ is reduced, q 0 + θ > > 0 > q 0 + θ > From this, q 0 = [ θ ] However, θ has continued fraction expansion [q r,,q ] It follows that q r = q 0 07 Continued fraction expansion of d Theorem 06 Let d be a rational number which is not a square The continued fraction expansion ofdis of the form d = [q0,q,q 2,,q 2,q,2q 0 ], whereq 0 = [ d] Proof () Letq 0 = [ d] The number ζ = q 0 + d is reduced (2) The continued fraction expansion of q 0 + d is purely periodic: q 0 + d = [2q 0,q,,q k ]

21 07 Continued fraction expansion of d 3 for positive integers q,, q k (3) The continued fraction expansion of d q0 = [q k,,q,2q 0 ] (4) [ ] d = q 0, d q0 = [q 0,q k,,q,2q 0 ] (5) Also, d = [q 0,q,,q k,2q 0 ], and(q,,q k ) = (q k,,q ) Example 03 Continued fraction expansions of d, d < 50 Those with asterisks have periods of odd lengths 2 = [,2]; 27 = [5,5,0]; 3 = [,,2]; 28 = [5,3,2,3,0]; 5 = [2,4]; 29 = [5,2,,,2,0]; 6 = [2,2,4]; 30 = [5,2,0]; 7 = [2,,,,4]; 3 = [5,,,3,5,3,,,0]; 8 = [2,,4]; 32 = [5,,,,0]; 0 = [3,6]; 33 = [5,,2,,0]; = [3,3,6]; 34 = [5,,4,,0]; 2 = [3,2,6]; 35 = [5,,0]; 3 = [3,,,,,6]; 37 = [6,2]; 4 = [3,,2,,6]; 38 = [6,6,2]; 5 = [3,,6]; 39 = [6,4,2]; 7 = [4,8]; 40 = [6,3,2]; 8 = [4,4,8]; 4 = [6,2,2,2]; 9 = [4,2,,3,,2,8]; 42 = [6,2,2]; 20 = [4,2,8]; 43 = [6,,,3,,5,,3,,,2]; 2 = [4,,,2,,,8]; 44 = [6,,,,2,,,,2]; 22 = [4,,2,4,2,,8]; 45 = [6,,2,2,2,,2]; 23 = [4,,3,,8]; 46 = [6,,3,,,2,6,2,,,3,,2]; 24 = [4,,8]; 47 = [6,,5,,2]; 26 = [5,0]; 48 = [6,,2] 2 Some simple patterns: a2 + = [a,2a]; a2 = [a,,2a 2]; a2 +a = [a,2,2a]; a2 +2 = [a,a,2a]; a2 2 = [a,,a 2,,2a 2] Theorem 07 Let d = [q 0,q,,q k,2q 0 ] If ζ n = [q n,q n+,] = an+ d b n with b n d a 2 n, then P () n+ d d P n + d = a n++ b n (2) P 2 n d Q 2 n = ( ) n b n+ (3) b n = if and only if n is a multiple of the period length of d (4) P 2 n d Q 2 n = if and only ifn+ is a multiple of the period length of d Proof () By induction For n = 0, P 0 + Q 0 d = q0 + d, P + Q =, a 0 + b 0 d = d, and a + b d = d q0 = q 0+ d This means a = q 0 and b 0 = From these, P 0 + d P +Q d = a + d b 0 d q 2 0

22 4 Continued Fractions Now suppose b n (P n + d) = (an+ + d)(p n + d) (2) Therefore, P n+ ++ d = q n+ + P n + d P n + d P n + d b n = q n+ + a n+ + d = q n+ + b n( a n+ + d) d a 2 n+ = q n+ + a n+ + d b n+ = q n+b n+ a n+ + d b n+ = a n+2 + d b n+ P 2 n d Q 2 n P 2 n d Q 2 n = a2 n+ d b 2 n = b n+ b n Note that P 2 dq 2 = = b 0 From this, we have Pn 2 d Q 2 n = ( ) n b n+ for everyn (3) Let m be the period length of d = [q 0,q,,q m ] (with q m = 2q 0 ) Then ζ = q 0 + d = [q,,q m ] is purely periodic, with ζ jm = q 0 + d Corresponding, a jm = q 0 andb jm = Conversely, suppose b n = for some n Since ζ n is purely periodic, it is reduced This means < ζ n < 0 But ζ n = an d b n = a n d Therefore, 0 < d a n <, and a n = [ d] = q 0 and ζ n = q 0 + d Hence, n must be a multiple of m

23 Chapter The Pell Equation The equation x 2 dy 2 = Letd > 0 be a non-square integer Ifxandy are positive integers satisfyingx 2 dy 2 =, then x y of d Proof d x y = x y d = y x+y < d y 2y 2 Therefore, by Theorem 0, x is a convergent of d y is a convergent 2 If Pn is a convergent of d, then P 2 n dq 2 n < +2 d Proof By Theorem, d Pn < Q 2 n Therefore, Pn < d+ 2 < + d Pn dq 2 2 n = P n + d Qn d Pn < (2 d+) = +2 d 3 From P n = q n P n +P n 2 and = q n + 2, we have ( ) ( )( ) Pn P n Pn P = n 2 qn 2 0 If we write M n = ( ) qn for n = 0,,2,, then 0 ( ) ( ) Pn P n Pn P = n 2 M = n 2 ( ) P P = 2 M Q Q 0 M M n = M 0 M M n 2

24 6 The Pell Equation 4 If the period of d is l, then P 2 l d Q2 l = ( )l This is + or according aslis even or odd 5 The fundamental solution(a,b) ofx 2 dy 2 = is(p l,q l orp 2l,Q 2l ) according aslis even or odd ( ) a db 6 = M b a 0 M M l or (M 0 M M l ) 2 according as l is even or odd Theorem Letdbe a nonsquare, positive integer The totality of positive solutions of the Pell equation x 2 dy 2 = form an infinite sequence (x n,y n ) defined recursively by x n+ = ax n +dby n, y n+ = bx n +ay n ; x = a, y = b, where (x,y ) = (a,b) is the fundamental solution (with a, b smallest possible) obtained from the continued fraction expansion d = [q0,q,,q k ], as follows Let P k Q k the (k ) th convergent of d (a) If the length of the period is even, then(a,b) = (P k,q k ) is the smallest positive solution of the Pell equation x 2 dy 2 = (b) If the length of the period is odd, then the smallest positive solution of the equation x 2 dy 2 = is (a,b) = (P 2 k +dq2 k,2p k Q k ) Examples The fundamental solution of the Pell equation x 2 2y 2 = is (3,2) This generates an infinite sequence of nonnegative solutions (x n,y n ) defined by x n+ = 3x n +4y n, y n+ = 2x n +3y n ; x 0 =, y 0 = 0 The beginning terms are n x n y n Fundamental solution (a,b) of x 2 dy 2 = for d < 00:

25 The equation x 2 dy 2 = 7 d a b d a b d a b Pell s equations whose fundamental solutions are very large: d a b The equation x y 2 = arises from the famous Cattle problem of Archimedes, and has smallest positive solution x = , y =

26 8 The Pell Equation 2 The equation x 2 dy 2 = Indeed, if the length of the period of the continued fraction expansion of d is odd, then(p k,q k ) is the smallest positive solution of the equation x 2 dy 2 = Only when this period is odd does this equation have solutions Examples Smallest positive solution(a,b) ofx 2 dy 2 = for the first 24 values ofd: d a b d a b d a b If p (mod 4) is prime, then the equation x 2 py 2 = is solvable Proof Let (a,b) be the fundamental solution of x 2 py 2 = This means a 2 = pb 2 Note that a must be odd, for otherwise a 2 (mod 4), but pb 2 (mod 4), a contradiction Consequently, gcd(a+,a ) = 2, and we have (i) a+ = 2r 2,a = 2ps 2, or (ii) a+ = 2pr 2,a = 2s 2, for some nonnegative integers r ands In (i), we have r 2 ps 2 =, with r < a, a contradiction since (a,b) is the smallest positive solution of x 2 py 2 = It follows that (ii) holds, and we have s 2 pr 2 = 3 The equation x 2 dy 2 = N Let d > 0 be a nonsquare integer If d = [q 0,q,,q k,2q 0 ] with convergents P n, then Pn 2 dq 2 n < 2 d+ If N < d, every positive solution of x 2 dy 2 = N is a convergent of d (i) SupposeN > 0 If (x 0,y 0 ) is a positive solution ofx 2 dy 2 = N, (x 0 y 0 d)(x0 +y 0 d) = N = x0 > y 0 d Therefore, d 0 < x 0 y 0 d = N y 0 (x 0 +y 0 d) < N y 0 (y 0 d+y0 d < 2y 2 0 =, d 2y0 2

27 3 The equation x 2 dy 2 = N 9 and Px 0 y 0 is a convergent of d (ii) Suppose N < 0 We rewrite the equation as y 2 d x 2 = A similar N reasoning as above shows that y 0 x 0 is a convergent of d It follows that x 0 y 0 is a convergent of d Consider the equation x 2 dy 2 = N for an arbitrary integer N We say that a solution (x,y) of the equation is primitive ifgcd(x,y) = If(x 0,y 0 ) is a primitive solution ofx 2 dy 2 = N, then there exists an integer k such that (i) x 0 ky 0 (mod N) and (ii) k 2 d (mod N) Since gcd(x 0,y 0 ) =, gcd(y 0,N) = There exists an integer k such that x 0 ky 0 (mod N) x 2 0 dy 2 0 y 2 0(k 2 d) 0 (mod N) gcd(y 0,N) = = k 2 d 0 (mod N) We say that the primitive solution (x 0,y 0 ) of x 2 dy 2 = N belongs to the integer k Two solutions (x,y ) and (x 2,y 2 ) of x 2 dy 2 = N are in the same class if there is a solution (x 0,y 0 ) such that x +y d = (x0 +y 0 d)(s +t d), x 2 +y 2 d = (x0 +y 0 d)(s2 +t 2 d), for solutions of (s,t ) and (s 2,t 2 ) of s 2 dt 2 = Two solutions ofx 2 dy 2 = N are in the same class if and only if they belong to the same integer 2 Letk be an integer such thatk 2 d (mod N), andn = k2 d The equation N x 2 dy 2 = N has a positive solution belonging tok if and only if the equation t 2 du 2 = N has a solution (not necessarily positive) belonging tok such that ku 0 t 0 > 0 and du 0 kt 0 > 0 If this is the case, (x N N 0,y 0 ) = ( du 0 kt 0, ku 0 t 0 ) N N is a positive solution of x 2 dy 2 = N belonging to k Theorem 2 Letk 2 d (mod N) andn = k2 d The N equationx2 dy 2 = N has a positive solution belonging to ±k if and only if t 2 du 2 = N has a positive solution belong to ±k In this case, ( ) du 0 ±kt 0 (x 0,y 0 ) =, t 0 ±ku 0 N is a positive solution of x 2 dy 2 = N which belongs to ±k, the sign being chosen so that t 0 ±ku 0 0 (mod N ) N

28 20 The Pell Equation Example : Consider the equationx 2 3y 2 = 2 There are four possible classes: k 2 3 (mod )2 for k = ±,±5 For k = ±, N = 2 3 =, we use a 2 positive solution of(8,5) ofx 2 3y 2 =, and obtain (i) for k =, ( 3 5+8, 8+5 ) = (83,23), and (ii) for k =, ( 3 5 8, 8 5 ) = (47,3) For k = ±5, N = 52 3 =, we use the fundamental solution (649,80) of 2 x 2 3y 2 =, and obtain (iii) for k = +5, ( (iv) for k = 5 and ( , ) = (5585,549), and, ) = (905,25) Example 2: Consider x 2 +3y 2 = 5 Here,k 2 3 (mod 5) for k = ±8,±25 () Fork = ±8,N = 82 5 = 3 Using the fundamental solution (a,b) = (649,80) of x 2 3y 2 =, we obtain (i) for k = +8,(x 0,y 0 ) = ( ) , = (7532,2089), (ii) for k = 8,(x 0,y 0 ) = ( ) , = (2852,79) (2) For k = ±25, N = = 2 We make use of the four classes of solutions 5 of the equationx 2 3y 2 = 2 obtained above (i) from(83,23), fork = 25, we obtain ( 3 23±2583, 83±253 ) = (48,4), 2 2 (ii) from (47,3) for k = +25, we obtain (2,3), (iii) no integer solutions from (905,25) and (5585,549) 4 Applications of Pell s equations Example Which triangular numbers are squares? Suppose the k th triangular numbert k = 2 k(k+) is the square ofn n2 = 2 k(k+);4k2 +4k+ = 8n 2 +; (2k+) 2 8n 2 = The smallest positive solution of the Pell equationx 2 8y 2 = being(3,), we have the solutions (k i,n i ) of the equation given by This means 2k i+ + = 3(2k i +)+8n i, n i+ = (2k i +)+3n i, k 0 =, n 0 = k i+ = 3k i +4n i +, n i+ = 2k i +3n i +, k 0 =, n 0 = The beginning values ofk and n are as follows i k i n i Example 2 Find all integersnso that the mean and the standard deviation ofn consecutive integers are both integers

29 4 Applications of Pell s equations 2 If the mean of n consecutive integers is an integer, n must be odd We may therefore assume the numbers to be m, (m ),,, 0,,, m, m The standard deviation of these number is m(m+) For this to be an integer, we must have 3 m(m + ) = k2 for some integer k m 2 = m = 3k 2 ; n 2 = (2m + ) 2 = 2k 2 + The smallest positive solution of the Pell equation n 2 2k 2 = being (7,2), the solutions of this equations are given by (n i,k i ), where The beginning values of n andk are n i+ = 7n i +24k i, k i+ = 2n i +7k i, n 0 =, k 0 = 0 i n i k i Example 3 (Almost isosceles Pythagorean triangles) Find all Pythagorean triangles the lengths of whose two shorter sides differ by Let x and x + be the two shorter sides of a Pythagorean triangle, with hypotenusey Theny 2 = x 2 +(x+) 2 = 2x 2 +2x+ From this,2y 2 = (2x+) 2 + The equation With z = 2x +, this reduces to the Pell equation z 2 2y 2 =, which we know has solutions, with the of this equations are (z n,y n ) given recursively by smallest positive one (,), and the equation z 2 2y 2 = has smallest positive solution (3,2) It follows that the solutions are given recursively by z n+ = 3z n +4y n, y n+ = 2z n +3y n, z 0 =, y 0 = If we write z n = 2x n +, these become x n+ = 3x n +2y n +, y n+ = 4x n +3y n +2, x 0 = 0, y 0 = The beginning values of x n andy n are as follows n x n y n Example 4 Find eleven consecutive positive integers, the sum of whose squares is the square of an integer Answer: = 77 2,

30 22 The Pell Equation = 43 2, = 529 2, = , = , = , 5 Sums of consecutive squares 5 Odd number of consecutive squares Suppose the sum of the squares of 2k + consecutive positive integers is a square If the integers are b,b±,,b±k We require (2k +)b 2 + k(k +)(2k +) = a2 3 for an integer a From this we obtain the equation a 2 (2k +)b 2 = 3 k(k +)(2k +) (E k) Suppose 2k + is a square Show that (E k ) has solution only when k = 6m(m + ǫ) for some integers m >, and ǫ = ± In each case, the number of solutions is finite Number of solutions of (E k ) when2k + is a square 2k Find the unique sequence of 49 (respectively 2) consecutive positive integers whose squares sum to a square Answer: = ; = ; Remark: The two sequences of 69 consecutive squares whose sums are squares are = 62 2 ; = Suppose 2k + is not a square If k + is divisible 9 = 3 2 or by any prime of the form 4k +3 7, then the equation (E k ) has no solution 4 Show that for the following values of k < 50, the equation (E k ) has no solution: k = 6,8,0,3,7,8,20,2,22,26,27,30,32,

31 5 Sums of consecutive squares 23 34,35,37,40,4,42,44,45,46,48, 5 Suppose p = 2k + is a prime If the Legendre symbol ( 3 k(k+) p ) =, then the equation (E k ) has no solution 6 Show that for the following values of k < 50, the equation (E k ) has no solution:,2,3,8,9,4,5,20,2,26,33,39,44 We need only consider (E k ) for the following values of k: 5,7,,6,9,23,25,28,29,3,36,38,43,47,49 7 Check that among these, only fork = 5,,6,23,29 are the equations(e k ) solvable 8 From the data of Example 20, work out 5 sequences of 23 consecutive integers whose squares add up to a square in each case Answer: = 92 2 ; = ; = ; = ; 9 Consider the equation (E 36 ) : a 2 73b 2 = Check that this equation does in fact have solutions (u,v) = (4088,478),(23360,2734) 0 Make use of the fundamental solution of x 2 73y 2 =, namely, (a,b) = (228249, ), to obtain two sequences of solutions of (E 73 ): Answer: (4088, 478),( , ),( , ), (23360, 2734),( , ),( , ), This means, for example, the sum of the squares of the 73 numbers with center 478 (respectively 2734) is equal to the square of 4088 (respectively 23360) 52 Even number of consecutive squares Suppose the sum of the squares of the 2k consecutive numbers is equal toa 2 This means b k +,b k +2,,b,,b+k,b+k, (2a) 2 2k(2b+) 2 = 2k 3 (4k2 ) (E k )

32 24 The Pell Equation Note that the numbers 2k,4k 2 are relatively prime Show that the equation (E k ) has no solution if 2k is a square 2 Suppose 2k is not a square Show that if 2k + is divisible by 9, or by any prime of the form 4k +, then the equation (E k ) has no solution 3 Show that for k 50, the equation (E k ) has no solution for the following values of k: k = 3,4,5,9,,3,5,7,2,23,24,27,29,3,33, 35,38,39,40,4,45,47,49 4 Letk be a prime Show that the equation (E k ) can be written as (2b+) 2 2ky 2 = 4k2 3 By considering Legendre symbols, show that the equation (E k ) has no solution for the following values ofk 50: k = 5,7,7,9,29,3,4,43 5 By using Theorem 053, check that, excluding square values of 2k < 00, the equation (E k ) has solutions only for k =,2,37,44 The case2k = 2 has been dealt with in Example Show that (34, 0), (38, 3), (50, 7) are solutions of (E 2 ) Construct from them three infinite sequences of expressions of the sum of 24 consecutive squares as a square Answer: = 82 2 ; = ; = Show that (85, 2), (2257,26), and (2849, 330) are solutions of(e 37) Construct from them three infinite sequences of expressions of the sum of 74 consecutive squares as a square Answer: = ; = ; = Show that and (242, 4) and (2222,235) are solutions of(e 44) Construct from them two infinite sequences of expressions of the sum of 88 consecutive squares as a square Answer: = ;

33 5 Sums of consecutive squares = Remark: The equation (E 26) : x 2 52y 2 = does indeed have two infinite sequences of solutions generated by the particular solutions (338, 36), (2002,276), and the fundamental solution (649,90) of the Pell equationx 2 52y 2 = None of these, however, leads to a solution of (E 26) since all the y s are even