number theory powerpoint presentation free download

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NUMBER THEORY - PowerPoint PPT Presentation

NUMBER THEORY

Number theory ts.nguy n vi t ng * 4.quadratic residues we now present a criterion for deciding whether an integer is a quadratic residue of prime. – powerpoint ppt presentation.

• TS.Nguy?n Vi?t Ðông
• 2.Primer Factorization
• 3.Congruence
• 4.Quadratic Residues
• Theorem 1.1. Division Algorithm. Let n and d ? 1 be integers. There exist uniquely determined integers q and r such that n qd r and 0 ? r lt d.
• Proof. Let X n - tdt ? Z, n - td ? 0. Then X is nonempty (if n?0,then n ? X if n lt 0, then n(1 - d) ? X). Hence let r be the smallest member of X . Then r n - qd for some q ? Z, and it remains to show that r lt d. But if r ?d, then 0 ? r - d n - (q 1)d, so r - d is in X contrary to the minimality of r.
• As to uniqueness, suppose that n qd r, where 0? rlt d. We may assume that r ? r(a similar argument works if r ? r). Then
• 0 ? r - r (q - q)d, so (q - q)d is a nonnegative multiple of d that is less than d (because r - r ? r lt d). The only possibility is
• (q - q)d 0, so q q, and hence r r.
• Given n and d ? 1, the integers q and r in Theorem 1.1 are called, respectively, the quotient and remainder when n is divided by d.
• For example, if we divide n -29 by d 7, we find that -29 (-5) 7 6, so the quotient is -5 and remainder is 6.
• The usual process of long division is a procedure for finding the quotient and remainder for a given n and d ? 1. However, they can easily be found with a calculator. For example, if n 3196 and d 271 then n/d 11,79 approximately, so q 11. Then r n - qd 215, so 3196 11 271 215, as desired.
• If d and n are integers, we say that d divides n, or that d is a divisor of n, if n qd for some integer q. We write dn when this is the case. Thus, a positive integer p gt1is prime if and only if p has no positive divisors except 1 and p. The following properties of the divisibility relation are easily verified
• (i) nn for every n.
• (ii) If dm and mn, then dn.
• (iii) If dn and nd, then d n.
• (iv) If dn and dm, then d(xm yn) for all integers x and y.
• Given positive integers m and n, an integer d is called a common
• divisor of m and n if dm and dn.
• If m and n are integers, not both zero, we say that d is the
• greatest common divisor of m and n, and write d gcd(m, n),
• if the following three conditions are satisfied
• (i) d ? 1. (ii) dm and dn.
• (iii) If km and kn, then kd.
• Theorem 1.2. Let m and n be integers, not both zero. Then d gcd(m, n) exists,and d xm yn for some integers x and y.
• Proof. Let X sm tn s, t ? Z sm tn ?1. Then X is not empty since m2 n2 is in X, so let d be the smallest member of X. Since d ? X we have d ? 1 and
• d xm yn for integers x and y, proving conditions (i) and (iii) in the definition of the gcd.
• Hence it remains to show that dm and dn.We show that dn the other is similar. By the division algorithm
• write n qd r, where 0 ? r lt d. Then
• r n - q(xm yn) (-qx)m (1 - qy)n. Hence, if r ? 1, then r ? X, contrary to the minimality of d. So r 0 and we have dn.
• When gcd(m, n) xm yn where x and y are integers, we say that gcd(m, n) is a linear combination of m and n. There is an efficient way of computing x and y using the division algorithm.
• The following example illustrates the method.
• Example . Find gcd(37, 8) and express it as a linear combination of 37 and 8.
• Proof. It is clear that gcd(37, 8) 1 because 37 is a prime however, no linear combination is apparent. Dividing 37 by 8, and then dividing each successive divisor by the preceding remainder, gives the first set of equations.
• 37 4 8 5 1 3 - 1 2 3 - 1(5 - 1 3)
• 8 1 5 3 2 3 - 5 2(8 - 1 5) - 5
• 5 1 3 2 2 8 - 3 5 2 8 - 3(37 - 4 8)
• 3 1 2 1 14 8 - 3 37
• The last nonzero remainder is 1, the greatest common divisor, and this turns out always to be the case. Eliminating remainders from the bottom up (as in the second set of equations) gives 1 14 8 - 3 37.
• Theorem 1.3. Euclidean Algorithm. Given integers m and n ? 1, use the division algorithm repeatedly
• m q1n r1 0 ? r1 lt n
• n q2r1 r2 0 ? r2 lt r1
• r1 q3r2 r3 0 ? r3 lt r2
• r k-2 qkrk-1 rk 0 ? rk lt rk-1
• where in each equation the divisor at the preceding stage is divided by the remainder. These remainders decrease
• r1 gt r2 gt ? 0
• so the process eventually stops when the remainder becomes zero. If r1 0, then gcd(m, n) n. Otherwise, rk gcd(m, n), where rk is the last nonzero remainder and can be expressed as a linear combination of m and n by eliminating remainders.
• Proof. Express rk as a linear combination of m and n by eliminating remainders in the equations from the second last equation up. Hence every common
• divisor of m and n divides rk. But rk is itself a common divisor of m and n (it divides every riwork up through the equations). Hence rk gcd(m, n).
• Two integers m and n are called relatively prime if gcd(m, n) 1.
• Hence 12 and 35 are relatively prime, but this is not true for 12 and 15
• Because gcd(12, 15) 3. Note that 1 is relatively prime to every
• integer m. The following theorem collects three basic properties of
• relatively prime integers.
• Theorem 1.4. If m and n are integers, not both zero
• (i) m and n are relatively prime if and only if 1 xm yn for some integers x and y.
• (ii) If d gcd(m, n), then m/d and n/d are relatively prime.
• (iii) Suppose that m and n are relatively prime.
• (a) If mk and nk, where k ? Z, then mnk.
• (b) If mkn for some k ? Z, then mk
• Proof. (i) If 1 xm yn with x, y ? Z, then every divisor of both m and n divides 1, so must be 1 or -1. It follows that gcd(m, n) 1. The converse is by the euclidean algorithm.
• (ii). By Theorem 1.2, write d xm yn,
• where x, y ? Z. Then
• 1 x(m/d)y(n/d) and (ii) follows from (i).
• (iii). Write 1 xm yn, where x, y ? Z. If k am and k bn, a, b ? Z then k kxm kyn (xb ya)mn, and (a) follows. As to (b), suppose that
• kn qm, q ? Z. Then k kxm kyn (kx qn)m, so mk.
• Recall that an integer p is called a prime if
• (ii) The only positive divisors of p are 1 and p.
• The reason for not regarding 1 as a prime is that
• we want the factorization of every integer into
• primes to be unique. The following result is needed.
• Theorem 2. 1. Euclids Lemma. Let p denote a prime.
• (i) If pmn where m, n ? Z, then either pm or pn.
• (ii) If pm1m2 mr where each mi ? Z, then pmi for some i.
• Proof. (i) Write d gcd(m, p). Then dp, so as p is a prime, either d p or d 1.
• If d p, then pm if d 1, then since pmn, we have pn by Theorem 1.4 .
• (ii) This follows from (i) using induction on r.
• Theorem 2.2. Every integer n gt1 is a product of primes.
• Proof. Let pn denote the statement of the theorem. Then p2 is clearly true.
• If p2, p3, . . . , pk are all true, consider the integer k 1. If k 1 is a prime, there is nothing to prove. Otherwise,
• k 1 ab, where 2 ? a, b ? k. But then each of a and b are products of primes because pa and pb are both true by the
• (strong) induction assumption. Hence ab k 1 is also a product of primes, as required.
• Theorem 2.3. Prime Factorization Theorem. Every integer n ? 2 can be written as a product of (one or more) primes. Moreover, this factorization is unique except for the order of the factors. That is,
• if n p1p2 pr and n q1q2 qs ,
• where the pi and qj are primes, then r s and the qj can be relabeled so that pi qi for each i.
• Proof. The existence of such a factorization was shown in Theorem 2.2. To prove uniqueness, we induction the minimum of r and s. If this is 1, then n is a prime and the uniqueness follows from Euclids lemma. Otherwise, r ? 2 and s ? 2. Since
• p1n q1q2 qs Euclids lemma shows that p1 divides some qj , say p1q1 (after possible relabeling of the qj ). But then p1 q1 because q1 is a prime. Hence n/p1 p2p3 pr q2q3 qs , so, by induction,
• r - 1 s - 1 and q2, q3, . . . , qs can be relabeled such that pi qi for all i 2, 3, . . . , r. The theorem follows.
• It follows that every integer n ? 2 can be written in the form n p1n1 p2n2 prnr ,where p1, p2, . . . , pr are distinct primes, ni ? 1 for each i, and the pi and ni are determined uniquely by n. If every ni 1, we say that n is square-free, while if n has only one prime divisor, we call n a prime power.If the prime factorization
• n p1n1 p2n2 prnr of an integer n is given, and if d
• is a positive divisor of n, then these pi are the only possible prime divisors of d (by Euclids lemma). It follows that
• Definition 3.1.. If m ? 0 is fixed, then integers a and b are congruent modulo m,denoted by a ? b (mod m)
• if m ? (a b ). Usually, one assumes that the modulus
• m gt1 because the cases m 0 and m 1 are not very interesting if a and b are integers, then a ? b (mod 0) if and only if 0 ? (a b), that is, a b, and so congruence mod 0 is ordinary equality.
• The congruence a ? b (mod 1) is true for every pair of integers a and b because 1 ? (a b) always. Hence, every two integers are congruent mod 1.
• If a and b are positive integers, then a ? b (mod 10) if and only if they have the same last digit more generally, a ? b (mod 10n )if and only if they have same last n digits. For example, 526 ? 1926 (mod 100).
• London time is 6 hours later than Chicago time. What time is it in London if it is 1000 A.M. in Chicago? Since clocks are set up with 12 hour cycles, this is
• really a problem about congruence mod 12. To solve it, note that 10 6 16 ? 4(mod 12) and so it is 400 P.M. in London.
• Proposition 3.1. If m gt 0 is a fixed integer, then for all integers a, b, c,
• (i) a ? a (mod m)
• (ii) if a ? b (mod m), then b ? a (mod m)
• (iii) if a ? b (mod m) and b ? c (mod m), then a ? c (mod m).
• Proposition 3.2. Let m gt 0 be a fixed integer.
• (i) If a qm r , then a ? r (mod m).
• (ii) If 0 ? r lt r lt m, then r and r are not congruent mod m in symbols, r r (mod m).
• (iii) a ? b (mod m) if and only if a and b leave the same remainder after dividing by m.
• Proposition 3.3. Let mgt 0 be a fixed integer.
• (i) If ai ? ai (mod m) for i 1 2 n, then
• a1 ... an ? a1 ... an (mod m)
• In particular, if a ? a (mod m) and b ? b (mod m), then
• a b ? a b (mod m)
• (ii) If ai ? ai (mod m) for i 1 2 n, then
• a1 ... an ? a1 ... a'n (mod m)
• In particular, if a ? a (mod m) and b ? b (mod m), then ab ? ab (mod m)
• (iii) If a ? b (mod m), then an ? bn (mod m) for all n gt0.
• Theorem 3.5. If (am) 1, then, for every integer b, the congruence ax ? b (mod m) can be solved for x in fact,
• x sb, where sa ? 1 (mod m). Moreover, any two solutions are congruent mod m.
• Proof. Since (am) 1, there is an integer s with
• as ? 1( mod m) (because there is a linear combination
• 1 sa tm). It follows that b sab tmb and
• asb ? b (mod m), so that x sb is a solution. (Note that Proposition 3.2(i) allows us to take s with 1? s lt m.)
• If y is another solution, then ax ? ay mod m, and so
• m ? a(x - y). Since (am) 1, Theorem 1.4 gives m ?(x y) that is, x ? y (mod m).
• Definition 4.1. If m is a positive integer ,we say that the integer a is a quadratic residue of m if (a,m) 1 and the congruence x2 ? a (mod m) has a solution.
• If the congruence x2 ? a (mod m) has a no solution, we say a is quadratic nonresidue of m.
• Example. To detemine which integer are quadratic residues of 11, we compute the squares of the integer 1, 2, 3, , 10.We find that 12 ? 102 ? 1(mod 11), 22 ? 92 ? 4 (mod 11), 32 ? 82 ? 9 (mod 11), 42 ? 72 ? 5 (mod 11), and 52 ? 62 ? 3(mod11).
• Hence , the quadratic residues of 11 are 1, 3, 4, 5, 9 the integer 2,6,7,8,10 are quadratic nonresidues of 11.
• Lemma 4.1. Let p be odd prime and a an integer not divisible by p. Then the congruence x2 ? a (mod p) has either no solutions or exactly two incongruent solutions modulo p.
• Proof. If x2 ? a (mod p ) has a solution, say x x0, then we can easily demonstrate that x - x0 is second incongruent solution. Since (-x0)2 x02 ? a (mod p ) we see that x0 is solution. We note that x0 x0 (mod p), for if x0 ? - x0 (mod p), then we have 2x0 ? 0 (mod p). This is impossible since p is odd and p x0 (since x02 ? a (mod p ) and p a ).
• To show that there are no more than two incogruent solutions,
• assume that x ? x0 and x ? x1 are both solutions of x2 ? a (mod p). Then we have x02 ? x12 ? a (mod p) , so that
• x02- x12 (x0 x1)(x0- x1) ? 0 (mod p).
• Hence , p (x0 x1) or p (x0- x1), so that x1 ? - x0 (mod p) or
• x0 ? x1 (mod p). Therefore if there is a solution of x2 ? a (mod p), there are exactly two incongruent solution.
• Theorem 4.2. If p is an odd prime , then there are exactly
• (p-1 )/2 quadratic residues of p and ( p 1 )/2 quadratic nonresidues of p among the integer 1, 2, , p 1 .
• Proof. To find all the quadratic residues of p among the integers 1, 2, , p 1 we compute the least positive residues modulo p of the squares of the integers 1, 2, p 1 .
• Since there are p 1 squares to consider and since each congruence x2 ? a (mod p) has either zero or two solotions , there must be exactly ( p 1 )/2 quadratic residues of p among the integer 1, 2, , p 1 . The remaining p 1 ( p 1 )/2
• ( p 1 )/2 positive integers less than p 1 are quadratic nonresidues of p . ?
• The special notation associaed with quadratic residues is described in the following definition.
• Definition 4.2. Let p be an odd prime and a an integer not divisible by p . The Legendre symbol is defined by
• The symbol iz named after the French mathematician Andrien Marie Legendre who introduced the use of this notation
• Example . The previous example shows that the Legendre symbol
• have the followings values
• We now present a criterion for deciding whether an integer is a quadratic residue of prime. This criterion is useful in demonstraing propeties of the Legendre symbol.
• Theorem 4.3. Eulers Criterion.
• Let p be an odd prime and let a be positive integer not divisible by p. Then
• Firt ,assume that .Then , the congruence x2 ? a (mod p)
• has a solution, say x x0. Using Fermats little theorem , we see
• Hence,if ,we know that
• Now cosider the case where .Then , the congruence
• x2 ? a (mod p) has no solutions. For each integer i such that
• 1?i?p-1, there is a unique integer j with 1?j?p-1, such that
• ij ? a (mod p ). Furthermore , since the congruence x2 ? a
• (mod p) has no solutions, we know that i?j. Thus, we can
• group the integer 1, 2,, p - 1 into (p 1 )/2 pairs each with
• product a . Multiplying these pairs together, we find that
• (p 1 )! ? a (p 1 )/2 (mod p).Since Wilsons theorem tell us that
• (p 1 )!? - 1 (mod p), we see that 1 ? a (p -1 )/2 (mod p) .In this
• case, we also have
• Theorem 4.4. Let p be an odd prime and a and b integers not
• divisible by p.Then
• If a ? b (mod p ) then x2 ? a (mod p) has s solution if an
• only if x2 ? b (mod p) has solution.Hence .
• By Eulers criterion we know that
• Since the only posible values of a Lagendre symbol are ?1, we
• conclude that
• iii) Since , from part (ii) it folows that

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NUMBER THEORY

Apr 02, 2019

760 likes | 967 Views

NUMBER THEORY. Chapter 5: Cryptology. Era of Electronic. Electronic communication Electronic Banking. Terminology. Cryptology: Discipline Cryptography: Design & Implementation Cryptoanalysis : Breaking the system. Terminology. Plaintext: secret message

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• secret message
• number theory
• complete system
• cryptoanalysis
• electronic banking

Presentation Transcript

NUMBER THEORY Chapter 5: Cryptology

Era of Electronic • Electronic communication • Electronic Banking

Terminology • Cryptology: Discipline • Cryptography: Design & Implementation • Cryptoanalysis: Breaking the system

Terminology • Plaintext: secret message • Cipher: method to alter the plaintext • Cybertext: message has been altered

Terminology • encryption/enciphering: process • decryption/deciphering: process

Letter ↔ Integer

Character Cipher Caesar cipher

How to decipher?

Character Cipher Affine transformation

Complete system of residue!

Secure? • Character cipher is not safe

Block Cipher Vigenere cipher

Exercise • Use the key “SECRET”

Block Cipher Digraphic cipher

Numerical equivalent

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