MPRI Cours I. Motivations. Lecture VI: continued fractions and applications. II. Continued fractions

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1 F Morain École olytechnique MPRI cours /25 F Morain École olytechnique MPRI cours /25 MPRI Cours 222 I Motivations F Morain Aroximate π by rationals Solve 009 = u 2 + v 2 Lecture VI: continued fractions and alications 200/0/2 Break some RSA arameters Factor integers (CFRAC, see EThomé s art) The slides are available on htt://wwwlixolytechniquefr/labo/francoismorain/mpri/200 I Motivations II Continued fractions III Continued fractions with integer coefficients IV Aroximating x by n / V Alications F Morain École olytechnique MPRI cours /25 II Continued fractions Good reading: Hardy & Wright, A M Rockett and P Szüsz, etc Rem Pre-dates LLL, and has still some interest F Morain École olytechnique MPRI cours /25 Aroximating reals numbers by rationals (/2) Lemma (Dirichlet) Let θ R, Q N: (, q) Z N st q θ qq q 2 Proof: sread the {nθ} for 0 n Q in [0, /Q[,, [(Q )/Q, ] Pigeon hole rincile (rincie des tiroirs de Dirichlet): i, h, k < h st {hθ}, {kθ} [i/q, (i + )/Q[ Let q = h k q Q and {qθ} /Q qθ = qθ + {qθ} = + {qθ} qθ /Q q θ qq Coro If θ Q, an infinity of /q st q θ q 2 Proof: If θ = a/b Q: (b > 0) a b q q 2 (q > 0 and a/b /q) imlies bq q 2 q b

2 F Morain École olytechnique MPRI cours /25 F Morain École olytechnique MPRI cours /25 Aroximating reals numbers by rationals (2/2) If θ Q: let /q, 2 /q 2,, s /q s all the rational numbers st i, θ i q i q 2 i ε = min θ i > 0 i s Let Q be an integer > /ε Dirichlet, q st θ q qq q 2 qq < ε q ε q i Finite continued fractions [a 0,, a N ] = a 0 + a + + a i i-th artial quotient (i-ième quotient artiel) a N + a N M 0, [a 0, a,, a M, [a M+,, a N ]] = [a 0, a,, a N ] 2 = 0, =, q = 0, q 2 = n = a n n + n 2, = a n + 2 Pro n N, [a 0,, a n ] = n n / n-th convergent (n-ième convergent ou réduite) of the continued fraction F Morain École olytechnique MPRI cours /25 Proerties of convergents F Morain École olytechnique MPRI cours /25 III Continued fractions with integer coefficients Pro n, n n = ( ) n Coro If a i > 0, i, q i 0 and strictly increasing st n, [a 0, a,, a n ] = n = a 0 + q q 0 q 2 q + + ( )n Pro For all n : n 2 n 2 = ( ) n a n Coro 2n /q 2n is increasing, 2n+ /q 2n+ is decreasing Moreover, for all n 2n < 2n+ q 2n q 2n+ Pro For all n, n and for all n 2, + Pro n and are rime together Pro If x Q +, there are two continued fractions reresenting x Proof: let x = [a 0, a,, a N ] If a N 2, then (with N + terms) If a N = : (with N terms) x = [a 0, a,, a N, ] x = [a 0, a,, a N + ]

3 F Morain École olytechnique MPRI cours /25 F Morain École olytechnique MPRI cours /25 Comlete quotients Unicity of the exansion (/2) Def a n = [a n, a n+,, a N ], n-th comlete quotient (n-ième quotient comlet) Pro [a 0,, a N ] = [a 0,, a n, a n] Pro Consider [a 0, a,, a N ] For all 0 n N, we have a n = a n excet when a N =, in which case a N = a N Proof: Let a n = [a n,, a N ] If n = N, a n = a N If n = N : a n = a n + a N If a N =, we have a N = a n and if a N 2, we have a n = a n If n N 2: a n = [a n, a n+] = a n + a n+ Thm Let [a 0, a,, a N ] = [b 0, b,, b M ] = x with a i, b j > 0, a N 2, b M 2 Then N = M and a i = b i for all i Proof: assume N M; we use induction and rove a i = b i for all i For all n, a n = a n, b n = b n (since a N 2, b M 2) In articular x = a 0 = b 0 a 0 = b 0 Hy a i = b i our i n < N Write x = a n n + n 2 a n + 2 = b n n + n 2 b n + 2 with n, n 2,, 2 only deending on (a i = b i ) for 0 i n with a n+ >, hence a n = a n F Morain École olytechnique MPRI cours /25 Unicity of the exansion (2/2) F Morain École olytechnique MPRI cours /25 Exansion of a rational number Reorganizing things or (a n n + n 2 )(b n + 2 ) = (b n n + n 2 )(a n + 2 ) a n( n 2 n 2 ) b n( n 2 n 2 ) = 0, or a n = b n, or a n = b n If M > N: hence N q N = N q N x = N q N = b N+ N + N b N+ q N + q N Thm Let x = h/k Q +, (h, k) = ; we can exand x as [a 0,, a N ] where (a i ) is given by the Euclidean algorithm alied on (h, k) Proof: h = a 0 k + k, 0 k < k, k = a k + k 2, k i = a i k i + k i+, 0 k i+ < k i, k N 2 = a N k N + k N, k N = a N k N, and k N+ = 0 Since (h, k) =, k N = and k N 2, which imlies a N 2 i, k i k i Rem h/k = N /q N ; hq N k N = = a i = [a i,, a N ]

4 F Morain École olytechnique MPRI cours /25 F Morain École olytechnique MPRI cours /25 Continued fraction exansion of a real number (/3) Continued fraction exansion of a real number (2/3) If x i 0, then x i+ is defined by a 0 = x, x = a 0 + x, 0 x <, x = a = a + x 2, 0 x 2 < x i = a i = a i + x i+, 0 x i+ < When x Q, the algorithm terminates since x i = k i+ /k i and (k i ) decreases If x Q, x i Q and the algorithm does not terminate Define ( n ) and ( ) as usual and introduce q n = a n + 2 Pro For all n x n = ( )n q n+ Coro Let x R Q, x > 0 For all n 0 2n q 2n x < 2n+ q 2n+ Rem The two sequences ( 2n /q 2n ) et ( 2n+ /q 2n+ ) are adjacent, since 2n+ 2n = q 2n+ q 2n q 2n q 2n+ 2n(2n + ) These two sequences define x F Morain École olytechnique MPRI cours /25 Continued fraction exansion of a real number (3/3) F Morain École olytechnique MPRI cours /25 IV Aroximating x by n / π = [3, 7, 5,, 292,,,, 2,, 3,, 4, 2,,, 2, ] π , π e = [2,, 2,,, 4,,, 6,,, 8,,, 0,,, 2, ] Thm For all n x n + Thm The robability that a n = a is aroximately (a) = log( + /a) log( + /(a + )) log 2 a (a) Thm (Best aroximation) Let n / be a convergent of x > 0 Then, for all integer and all q <, or equivalently qx > x n x q > x n

5 F Morain École olytechnique MPRI cours /25 F Morain École olytechnique MPRI cours /25 Aroximating (2/3) Thm Let n / and n+ /+ be two consecutive convergents of x > 0 One of them satisfies q x 2q 2 Proof: remark that n / x and n+ /+ x are of oosite sign: n+ n + = n+ x + + x n = + If then n x > 2q 2 n and > + 2q 2 + n 2q 2 n+ n+ x + > 2q 2 n+ ( ) 2 < 0 +, Aroximating (3/3) Thm One of n /, n+ /+, n+2 /+2 satisfies q x 5q 2 Thm (Tong) Let τ R + One of 2n /q 2n, 2n /q 2n, 2n+ /q 2n+ satisfies τ a 2 2n+ + 4τ q 2 < x q < a 2 2n+ + 4τ q 2 F Morain École olytechnique MPRI cours /25 Converse theorems F Morain École olytechnique MPRI cours /25 Examle: solving the Pell-Fermat equation Pb For given d >, find all integer solutions of Thm Let /q st q x 2q 2 Then /q is a convergent of the continued fraction exansion of x Thm If q x < q 2 then /q is a convergent or a an intermediary one: sitting in between n / and n / are ( n ± n )/( + ) Since x > d y, we get or x 2 dy 2 = x y d < 2 d y x y d < 2y 2 x y is a convergent of d Thm All solutions are obtainable via (x + y d) n

6 F Morain École olytechnique MPRI cours /25 F Morain École olytechnique MPRI cours /25 Periodic continued fractions (/2) Periodic continued fractions (2/2) [a 0, a,, a n0, a n0 +,, a n0 +T ] Thm The exansion of x is eriodic iff x is a root of some ax 2 + bx + c = 0 with integer coefficients, b 2 4ac 0, a 0 Thm We have d = [(a n )] = [a 0, a, a 2,, a n, 2a 0 ] with u 0 = 0, v 0 =, α n = a n = [a n, ] = u n + d v n, a n = α n, Rem Direct alication to Pell-Fermat (and/or finding units in real quadratic fields) Ex a 2 + = [a, 2a] Conj n d log log d for d mod 8 and d log log(4d) otherwise u n+ = a n v n u n, v n+ = d u2 n+ v n, n = [a 0,, a n ], 2 n dq 2 n = ( ) n v n, v n 2 d, 2 = 0, = F Morain École olytechnique MPRI cours /25 V Alications F Morain École olytechnique MPRI cours /25 Cont d A) Solving = x 2 + y 2 Thm Let be rime Then = x 2 + y 2 iff = 2 or mod 4 Alication: comute easily #E : Y 2 = X 3 + X over F, which is + ± 2x or + ± 2y (this can be given exactly) Necessary condition: (x/y) mod has a solution iff mod 4 Algorithm: solve a mod ; x/y will be some convergent of a/ Rem Can be generalized to x 2 + dy 2 = (Cornacchia s algorithm) Pro Let n 2, a and n a 2 + : (s, t) st n = s 2 + t 2 Proof: write a/n = [a 0, a,, a N ] Choose s = q k st q k n < q k+ If q = a = and q 2 > n, ut k = and s = q = a/n is in between of k /q k and k+ /q k+ : a n k k+ k q k+ = < q k q k+ nqk q k Let t = aq k n k : t 2 < n, s 2 n s 2 + t 2 2n But s 2 + t 2 = q 2 k + (aq k n k ) 2 q 2 k ( + a2 ) 0 mod n The only ossible multile is n and s 2 + t 2 = n q k

7 B) RSA with small exonent (Wiener) N = q, ed mod ϕ(n) Rem Given ϕ(n) = ( )(q ), we can get + q, from which we get and q using a degree 2 equation Pro for all N, N 3 N < ϕ(n) < N Thm If < q < 2, e < N et d < 4 N/3, we can recover the factorization of N Proof: k, ed kϕ(n) = ; one finds k d e N < 3k d N < 2d 2 Rem DeWeger02: use ϕ(n) N + 2 N Rem Many imrovements by Boneh, Durfee, etc F Morain École olytechnique MPRI cours /25