Fermat s little theorem. RSA.

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2 Computing large numbers modulo n (a) In modulo arithmetic, you can always reduce a large number to its remainder a a rem n (mod n). (b) Addition, subtraction, and multiplication preserve congruence: if a b (mod n) and c d (mod n), then a + c b + d (mod n), a c b d (mod n), ac bd (mod n). and (c) Instead of a congruence relation, you can work with the equality of the remainders a b (mod n) if and only if a rem n = b rem n p. 2

3 Example Find x such that ( ) x (mod 21). In fact, we need to compute the remainder ( ) rem 21: (mod 21) 25 4 (25 2 ) (mod 21) 25 8 (25 4 ) (mod 21) (mod 21) (mod 21) Also we can say that ( ) rem 21 = 11. p. 3

4 Example Compute x = rem 137. p. 4

5 Example Compute x = rem (mod 137) 12 4 (12 2 ) (mod 137) 12 8 (12 4 ) (mod 137) (12 8 ) (mod 137) (12 16 ) (mod 137) (12 32 ) (mod 137) (12 64 ) (mod 137) (mod 137) p. 5

6 Lemma 1 Lemma. Let p be a prime. If k is not a multiple of p, then if ak bk (mod p) then a b (mod p) Proof. p (ak bk) p k(a b) By the lemma we proved in the previous lecture, p k or p (a b). Since p k, we conclude that p (a b), or equivalently, a b (mod p). p. 6

7 Observation x x rem When computing remainders of the division by p for a sequence of consecutive integers, we get all numbers from 0 to p 1. Can we come up with other ways for producing these numbers? p. 7

8 Observation For example, we take every second integer: 2, 4, 6, 8,... x x rem We generate the same numbers, even though they are permuted. Can we try another sequence: 3, 6, 9, 12, 15,... right? Yes, we can, but we can prove a more general result. p. 8

9 Lemma 2 Lemma. Let p be a prime. If k is not a multiple of p, the sequence k rem p, 2k rem p, 3k rem p,... (p 1)k rem p is a permuation of 1, 2, 3,... p 1. Proof. All remainders are in the interval between 1 and p 1. Let s prove by contradiction that all of them are different. Assume that exist two distinct integers 1 i j p 1 such that ik rem p = jk rem p ik jk (mod p) By Lemma 1, i j (mod p), and so i = j, because both are integers between 1 and p 1. Therefore, all remainders are different, and this proves the lemma. p. 9

10 Examples k = 2 and p = 5: 2 rem 5 = 2 4 rem 5 = 4 6 rem 5 = 1 8 rem 5 = 3 k = 12 and p = 7: 12 rem 7 = 5 24 rem 7 = 3 36 rem 7 = 1 48 rem 7 = 6 60 rem 7 = 4 72 rem 7 = 2 p. 10

11 Theorem. Let p be a prime. If k is not a multiple of p (i.e. p k) then k p 1 1 (mod p) Proof (p 1) = (k rem p )(2k rem p) (3k rem p) ((p 1)k rem p) }{{}}{{}}{{} k (mod p) 2k (mod p) (p 1) k 2k 3k (p 1)k (mod p) (p 1) k p 1 (mod p). 1 k p 1 (mod p). (by Lemma 1) p. 11

12 Theorem. Let p be a prime. If k is not a multiple of p then k p 1 1 (mod p) Using this, we can find an inverse, x, kx 1 (mod p). How can we do that? If k is not a multiple of p and p is a prime than k p 2 is an inverse! k k p 2 = k p 1 1 (mod p) Because we need an inverse modulo p, the remainder (k p 2 rem p) is an inverse too. Notice the difference with the Extended Euclid s algorithm that can be used for computing an inverse even when p is not a prime. p. 12

13 Compute an inverse of 12 modulo 137, that is find x such that 12x 1 (mod 137) 137 is a prime, so we apply the, (mod 137). The inverse is x = = Or, in fact, just the remainder rem 137 is also the inverse, and we have computed it already: rem 137 = 80. Let s check: (mod 137) Thus 80 is a multiplicative inverse, indeed. p. 13

14 Theorem. Let p be a prime. If k is not a multiple of p then k p 1 1 (mod p) Find x such that x (mod 137). By, (mod 137), so p. 14

15 Theorem. Let p be a prime. If k is not a multiple of p then k p 1 1 (mod p) Find x such that x (mod 137). By, (mod 137), so (mod 137) p. 15

16 Find x such that x (mod 2017). By, (mod 2017), so p. 16

17 Find x such that x (mod 2017). By, (mod 2017), so (mod 2017) p. 17

18 Theorem (Chinese remainder). Let n 1, n 2,..., n k be pairwise relatively prime positive integers greater than one and a 1, a 2,..., a n arbitrary integers. Then the system x a 1 (mod n 1 ), x a 2 (mod n 2 ),... x a k (mod n k ) has a unique solution modulo n = n 1 n 2 n k. (That is, there is a solution x with 0 x < n, and all other solutions are congruent modulo n to this solution.) p. 18

19 cryptosystem is a public key cryptosystem introduced by Ronald Rivest, Adi Shamir, and Leonard Adleman. Constructing a publuc key: 1. You pick two large primes p and q. (How large? In practice, if their decimal representations are more than 200 digits long, they are large enough). 2. Let n = pq. 3. You pick an integer e such that gcd e, (p 1)(q 1) = 1. The public key is a pair (n, e). p. 19

20 encryption You are given the public key (n, e), such that n = pq, and gcd e, (p 1)(q 1) = 1. Where p and q are unknown primes. 1. You break the message (a long string of digits) into blocks so that each is less than n Message = }{{} }{{} }{{} }{{} M 1 <n M 2 <n M 3 <n M 4 <n 2. Each block is encrypted separately: C i = M e i rem n 3. The encrypted blocks can be transmitted to the receiver. p. 20

21 decryption The decryption key is an integer d such that ed 1 (mod (p 1)(q 1)) So, d is the multiplicative inverse of e modulo (p 1)(q 1). If p and q are unknown, it s hard to compute d. To decrypt each encrypted block C i, you compute C d rem n. i Turns out that the original message M i = C d rem n. i Let s show that this is the case. p. 21

22 decryption The decryption key d is such that ed 1 (mod (p 1)(q 1)) ed = 1 + k(p 1)(q 1) C i M e i C d i (mod pq) (M e i )d M 1+k(p 1)(q 1) i (mod pq) M i M k(p 1)(q 1) i (mod pq) If a b (mod pq), there exists t s.t. a b = t pq, and so Therefore, a b (mod q) and a b (mod p). C d i C d i M i M k(p 1)(q 1) i (mod p) M i M k(p 1)(q 1) i (mod q) p. 22

23 decryption Take the first: C d i M i (M p 1 i ) k(q 1) (mod p) If p M i, then M i 0 (mod p). This means C d i Hence, C d 0 M i i (mod p) If p M i, then M p 1 i Therefore in any case, 0 (mod p). 1 (mod p) by, hence C d i M i 1 k(q 1) M i (mod p) C d i M i (mod p) Similarly, C d i M i (mod q) p. 23

24 decryption We have shown that C d M i i (mod p) C d M i i (mod q) Recall that we want to prove that C d i M i (mod pq) If this is the case, then we can decrypt C i by taking the remainder C d rem pq. i C d M i i is a multiple of p and a multiple of q, which are primes. But because the prime factorization is unique, the difference C d M i i should be also a multiple of pq. Thus C d i M i (mod pq) This is why decryption works! p. 24

25 decryption We have shown that C d M i i (mod p) C d M i i (mod q) Recall that we want to prove that C d i M i (mod pq) If this is the case, then we can decrypt C i by taking the remainder C d rem pq. i C d i M i is a multiple of p and a multiple of q, which are primes. But because the prime factorization is unique, the difference C d i should be also a multiple of pq. Thus M i C d i M i (mod pq) This is why decryption works! p. 25

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