SOME MATHEMATICAL FOUNDATIONS OF CRYPTOGRAPHY

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1 SOME MATHEMATICAL FOUNDATIONS OF CRYPTOGRAPHY PARKER HAVIZA Abstract. Ths paper ntroduces three mathematcal methods of sharng encrypted nformaton: the Pohlg-Hellman Key Exchange and the ElGamal and RSA cryptosystems. These systems are based on the hardness of nteger factorzaton and the Dscrete Logarthm Problem. Defntons of these problems and attacks aganst these cryptosystems are dscussed. Ths paper assumes no pror experence wth cryptography or modular arthmetc and bulds up from basc defntons. Contents 1. Groundwork General Concepts Essental Formulas and Algorthms 2 2. Dscrete Logarthm Problem Collson Algorthms Pohlg-Hellman Algorthm 6 3. Cryptosystems Dffe-Hellman Key Exchange ElGamal RSA 8 4. Fndng Prme Numbers Prmalty Testng Factorzaton 10 Acknowledgements 11 References Groundwork Before explanng the fundamental problems of cryptography or explorng the mathematcs of ndvdual cryptosystems, t wll be helpful to cover some basc concepts whch form a framework for later algorthms General Concepts. Defnton 1.1. A fnte commutatve rng R s a fnte set wth two operatons, denoted + (addton) and (multplcaton). Both operatons are commutatve and assocatve, and are lnked by the dstrbuton property a (b + c) = a b + a c for a, b, c R. The set contans dentty elements for both operatons. The set also contans addtve nverses. 1

2 2 PARKER HAVIZA The bass for all calculatons n ths paper s modular arthmetc, also known nformally as clock arthmetc for the way t descrbes the addton of tmes of day. Defnton 1.2. Let a, b, m be ntegers wth m postve. We call m the modulus. We say a and b are congruent modulo m f a = b + mk for some k 0. We express ths relaton wth the notaton a b (mod m). Proposton 1.3. If a 1 a 2 (mod m) and b 1 b 2 (mod m) then a 1 ± b 1 a 2 ± b 2 (mod m) and a 1 b 1 a 2 b 2 (mod m). Proof. By defnton, a 1 = a 2 + mk and b 1 = b 2 + mj for some m, j 0. Then we have a 1 ± b 1 a 2 ± b 2 + m(k ± j) a 2 ± b 2 (mod m) and a 1 b 1 a 2 b 2 + m(b 2 k + a 2 j + mkj) a 2 b 2 (mod m). Defnton 1.4. The rng of ntegers modulo m s the set {0, 1,..., m 1} wth addton and multplcaton operatons mod m. It s denoted Z/mZ. Defnton 1.5. A fnte feld s a fnte rng wth a multplcatve nverse for every element n the set except zero. Defnton 1.6. The nverse of a non-zero nteger a (mod p) s the unque nteger denoted a 1 that satsfes a a 1 1 (mod p). To dvde both sdes of an equaton by a s to multply both sdes by a 1. Defnton 1.7. A unt s an element of a rng that has a multplcatve nverse. The group of unts modulo m s denoted (Z/mZ) Essental Formulas and Algorthms. The followng formulas and algorthms are essental for later proofs. Algorthm 1. (Eucldean Algorthm) Assume wthout loss of generalty that for ntegers a and b, a > b. Let r 0 = a and r 1 = b. Repeat the followng untl r +1 = 0: (1) Startng wth = 1, dvde r 1 by r and denote the remander as r +1 (2) Increment Proposton 1.8. The algorthm termnates when r = gcd(a, b). Proof. For some ntegers q and the remander after dvson r we can wrte the equatons a = bq 1 + r 2, b = r 2 q 2 + r 3,..., r t 2 = r t 1 q t 1 + r t, r t 1 = r t q t denotng the greatest as t. Note that r > r +1, so eventually we reach an r +1 such that r +1 = 0 = r t+1. We can wrte the teraton step r +1 = r 1 r q

3 SOME MATHEMATICAL FOUNDATIONS OF CRYPTOGRAPHY 3 and see that a common dvsor of r 1 and r s also a dvsor of r +1. Smlarly, a common dvsor of r and r +1 s a dvsor of r 1. It follows that (1.9) gcd(r 1, r ) = gcd(r, r +1 ) for all = 1, 2, 3... Snce r t 1 = r t q t + 0 we have Usng = 1 and (1.9) we have gcd(r t 1, r t ) = gcd(r t q t, r t ) = r t. gcd(r 0, r 1 ) = gcd(a, b) = r t. Proposton (Extended Eucldean Algorthm) For postve ntegers a and b, there are always ntegers u and v such that au + bv = gcd(a, b). Proof. Note that r 1 = b and r 2 = 1 a q 1 b are ntegers. The teraton step tells us r +1 = r 1 r q. By nducton we see that r 1 and r are lnear nteger combnatons of a and b. In other words, for some ntegers x 1, x 2, y 1, and y 2 we can wrte r = ax 1 + by 1 and r 1 = ax 2 + by 2. Substtutng these equatons nto the teraton step we have r +1 = r 1 r q = ax 2 + by 2 (ax 1 + by 1 )q = a(x 2 y 1 q ) + b(x 2 y 1 q ). Ths gves r +1 as a lnear nteger combnaton of a and b. Denote u +1 = x 2 y 1 q and v +1 = x 2 y 1 q. Ths means that r t = gcd(a, b) s also a lnear nteger combnaton of a and b wth ntegers u t and v t. Proposton Let p be prme. Then the rng of ntegers modulo p s a feld (every non-zero nteger has a multplcatve nverse). Proof. If a Z/pZ then gcd(a, p) = 1. The Extended Eucldean Algorthm tells us that there are ntegers u, v such that 1 = au + pv. Ths can be rewrtten n the modular form au = 1 pv 1 (mod p) So there exsts an nverse u for every non-zero a n the rng. A rng of ntegers modulo a prme number s called a feld of prme order and s denoted F p. Perhaps the most mportant theorem underlyng modern cryptography s Fermat s Lttle Theorem, whch s the bass for many proofs that wll follow n ths paper. It states the followng: Theorem (Fermat s Lttle Theorem) For any prme number p and nteger a, { (1.13) a p 1 1 (mod p) p a 0 (mod p) p a

4 4 PARKER HAVIZA Proof. If p a then a 0 (mod p) so we wll look at the case where p a. Then the lst a, 2a, 3a,..., (p 1)a (mod p) wll contan p 1 dstnct elements. There are p 1 dstnct numbers between 1 and p 1, so ths lst must contan the same elements as 1, 2,..., p 1 (mod p). These lsts may not be n the same order, but the products of ther elements are the same. We equate these products to see that After factorng we see that a 2a... (p 1)a (p 1) a p 1 (p 1)! (p 1)! (mod p). (mod p). And notng that (p 1)! s non-zero and hence nvertble mod p by Proposton 1.11 we obtan a p 1 1 (mod p). We can easly calculate nverses mod p usng Theorem Note that a a p 2 a p 1 1 (mod p). Thus the nverse can be found by computng a p 2 (mod p), so we can easly perform dvson n a feld of fnte prme order. If we are workng modulo a composte number n the rng Z/(pq)Z where p, q are dstnct prmes, we can stll make use of Fermat s Lttle Theorem wth Euler s Formula. Theorem (Euler s Formula) Gven two dstnct prmes p and q, let g = gcd((p 1)(q 1)). Then a (p 1)(q 1)/g 1 (mod pq). Proof. Consderng the equaton mod p we have a (p 1)(q 1)/g = (a p 1 ) (q 1)/g 1 (q 1)/g 1 (mod p) Repeatng ths process mod q shows that a (p 1)(q 1)/g 1 (mod q). Then because a (p 1)(q 1)/g 1 s dvsble by both p and q, t s dvsble by pq. Smlar formulas can be derved when the modulus can be factored nto more than two prmes. Ths s not necessary for cryptographc purposes, snce the modulus s usually chosen as the product of two large prmes n order to make factorzaton computatonally dffcult. The Chnese Remander Theorem s a way of solvng a system of equatons n modular arthmetc. Algorthm 2. Let the nput of ths algorthm be ntegers a 1...a t and m 1...m t where gcd(m, m j ) = 1 for j. Let x 1 = a 1. Repeat the followng steps for 2 t: (1) Let nteger y satsfy (1.15) y (m 1 m 2 m 1 ) 1 (a x 1 ) (mod m ).

5 SOME MATHEMATICAL FOUNDATIONS OF CRYPTOGRAPHY 5 (2) Let (1.16) x = x 1 + y (m 1 m 2 m 1 ). Complete the algorthm by outputtng x t. Proposton Denote x as the output of Algorthm 2. Then x a (mod m ) for all 1 t and s unquely determned mod m 1 m 2...m t. Proof. Pluggng (1.15) nto (1.16) for some nteger q we have x = x 1 + (a x 1 ) + m q (m 1 m 2 m 1 ) = a + m q (m 1 m 2 m 1 ). Ths smplfes to x a (mod m ). Note then that (1.16) can be rewrtten as x x 1 a 1 (mod m 1 ). Hence, by nducton on, x satsfes x a j (mod m j ) for all j <. 2. Dscrete Logarthm Problem Cryptography reles on operatons known as trapdoor functons whch are easy to compute but take prohbtvely long to nvert usng modern computng power. One example of a potental trapdoor functon s multplyng two large prmes and then attemptng to recover the product s orgnal factors. Many modern cryptosystems rely on the Dscrete Logarthm Problem, or DLP, whch s a trapdoor functon. Defnton 2.1. A prmtve root g s a number such that every element of F p can be wrtten as g rased to some power. Thus, snce g p 1 1 (mod p) by Fermat s Lttle Theorem, the lst g, g 2, g 3,..., g p 1 s a lst of every element n F p, the set of unts of F p. The Dscrete Logarthm Problem s the queston of fndng an exponent x such that (2.2) g x h (mod N) for ntegers g and N. For cryptographc purposes, we are nterested n the Dscrete Logarthm Problem where g s a prmtve root and N s ether prme or the product of prmes Collson Algorthms. A collson algorthm s a brute-force method of solvng an equaton. The general strategy n collson algorthms s to create two lsts and fnd a match between them. We can apply collson algorthms to the DLP. Algorthm 3. (Shank s Babystep-Gantstep Algorthm) Choose an nteger n such that N < n < N where N s the modulus of the partcular Dscrete Logarthm Problem. Generate the lsts and e, g, g 2, g 3,..., g n babystep h, h g n, h g 2n,..., h g n2 gantstep. Fnd a match between these lsts of the form g = h g jn.

6 6 PARKER HAVIZA Proposton 2.3. Gven a match, x = + jn solves the dscrete logarthm g x h (mod N). Proof. For ntegers q, r let x = nq + r where 0 n < r. Then because 1 x < N and n > N we have q = x r < N n n < n2 n < n. Rewrtng the Dscrete Logarthm Problem we have So, g x = g r nq = h g r = h g nq where g r s n the babystep lst and g qn s n the gantstep lst. Thus ths algorthm wll always return a match and have a soluton of the form x = +jn. Another collson algorthm uses random numbers to create a hgh probablty of fndng a match n relatvely few steps. Proposton 2.4. Wrte the Dscrete Logarthm Problem lettng x = y z so that g y h g z (mod p) for prme p and prmtve root g. Choose random values of y and z to form the lsts (2.5) g y1, g y2,..., g yn 1 < y < p 1 and (2.6) h g z1, h g z2,..., h g zn. Then for randomly generated y and z j of a matchng par of numbers, x = y z has a 1 e n2 /(p 1) chance of solvng the dscrete logarthm problem g x h (mod p). Proof. The lst (2.5) s a sze-n subset of the sze-(p 1) set {g 1, g 2,..., g p 1 } because g s a prmtve root. The probablty of gettng a match between (2.5) and at least one member of (2.6) s 1-Pr(No match)=1-( p 1 n ) n = 1 (1 n n p 1 )n. After notng that e x 1 x ths gves us the lower bound 1 e n2 /(p 1) Pohlg-Hellman Algorthm. The ablty to factor p 1 s crucal n the Pohlg-Hellman Algorthm for solvng the DLP. Algorthm 4. Gven the dscrete logarthm problem g x h (mod p) for prme p and prmtve root g, suppose L = p 1 can be factored n the form L = q k1 1, qk2 2 For 1 t let g and h be defned by,..., qkt t. g = g L/qk and h = h L/qk. Solve the t dscrete logarthms of the form g y h (mod p)

7 SOME MATHEMATICAL FOUNDATIONS OF CRYPTOGRAPHY 7 usng Shank s Algorthm or by usng the Pohlg-Hellman algorthm agan. Use the Chnese Remander Theorem to fnd x by solvng x y 1 (mod q k1 1 ), x y 2 (mod q k2 2 ),..., x y 1 (mod q kt t ). Proposton 2.7. Ths x solves the dscrete logarthm g x h (mod p). Proof. Let x y (mod q k Wrtng (g x ) L/qk (g x ) L/qk h L/qk (2.8) (L/q k ). Ths means x = y + q k j for some j. Then we have (mod p) (g L/qk ) y g Lj (mod p) (g ) y 1j (mod p) h (mod p) h L/qk (mod p). (mod p) n logarthmc form we have ) x (L/qk ) log g(h) (mod p). Snce the greatest common dvsor of all L/q k s 1, the Extended Eucldean Algorthm tells us there are ntegers c 1...c t whch satsfy (2.9) L/q k1 1 c 1 + L/q k2 2 c L/q kt t c t = 1. If we multply both sdes of 2.8 by c and sum over 1 t we get t t L/q k c x L/q k c log g (h) (mod p). =1 Usng (2.9) we have =1 x log g (h) (mod p). 3. Cryptosystems Cryptography s the study of prvately transmttng a message across potentally open channels. Encrypton s the process of obscurng a message from everyone except the ntended recpent, who decrypts the message. Together encrypton and decrypton algorthms form a cryptosystem. The followng cryptosystems are based on the dffculty of ether factorng a product of large prmes or solvng the dscrete logarthm problem Dffe-Hellman Key Exchange. Defnton 3.1. A key s a secret number used n a cryptosystem when encryptng and decryptng messages. Wthout t, messages are hard to recover from the encrypted text. One major queston n cryptography deals wth how to establsh a secret key between two people who can only communcate across open channels. One method was proposed by Whtfeld Dffe and Martn Hellman n 1978 [3]. Algorthm 5. (Dffe-Hellman Key Exchange) Let prme p and prmtve root g be known. Person A: choose a secret nteger α and send a g α (mod p). Calculate b α (mod p).

8 8 PARKER HAVIZA Person B choose a secret nteger β and send b g β (mod p). Calculate a β (mod p). Proposton 3.2. Persons A and B now share the same key. Proof. It must be shown that b α (mod p) a β (mod p). Ths s the shared key. b α (mod p) (g β ) α (mod p) (g α ) β (mod p) a β (mod p) To uncover the key, a thrd person nterceptng the publcly sent nformaton would have to be able to fnd α from g α (mod p) or β from g β (mod p). Ths s the Dscrete Logarthm Problem. Numbers n cryptosystems are chosen suffcently large that t would take decades to fnd these values wth current algorthms. An attacker could also fnd a way of fndng g αβ from ether g α or g β, but no effcent method for ths s known ElGamal. The ElGamal cryptosystem s a means of encryptng and decryptng a secret message m. Algorthm 6. Encrypton Scheme: Let prme p and prmtve root g be known. Person A: Select a random prvate key α and dstrbute a publc key a g α (mod p). Person B: choose a random prvate key k and prvate message m and send c 1 g k (mod p) and c 2 ma k (mod p). Proposton 3.3. The message can be decrypted by computng m (c α 1 ) 1 c 2 (mod p). Proof. Person A knows α, so they can calculate (c α 1 ) 1 c 2 (g ak ) 1 (ma k ) (mod p) (g ak ) 1 (m(g a ) k ) (mod p) m (mod p) Agan, the strategy n breakng ths encrypton les n the Dscrete Logarthm Problem of dscoverng α from g α RSA. Another cryptosystem s RSA, ntroduced by Rvest, Shamr, and Adelman n ther 1978 paper [2]. The securty of ths system s based on the factorng of the product of large prmes. Algorthm 7. Person A: dstrbute a product of prmes N = pq and an encrypton exponent e such that gcd(e, (p 1)(q 1)) = 1. (We denote the nverse of e as d.) Person B: choose a secret message m and send c m e (mod N). Person A: compute d whch by defnton satsfes ed 1 (mod (p 1)(q 1)).

9 SOME MATHEMATICAL FOUNDATIONS OF CRYPTOGRAPHY 9 Proposton 3.4. Then the message can be decoded wth the formula m c d (mod N). Proof. By the Eucldean Algorthm there exsts ntegers d and k such that de = 1 (p 1)(q 1) k, so we know that there exsts an nverse for every e. c d m de (mod N) m 1 (p 1)(q 1) k (mod N) m (m p 1 m q 1 ) k (mod N) m (1 1) k (mod N) m (mod N) It takes a lot of computng power to fnd d for a large p, q unless the factorzaton of N s known, makng the sent message transparent for the publsher of N but not for other recpents of the encoded text. 4. Fndng Prme Numbers 4.1. Prmalty Testng. If large prmes are necessary for these cryptosystems, how do we fnd them? It s necessary to determne whether a randomly chosen large number s prme. It s only necessary to look at large odd numbers because the only even prme s 2. Defnton 4.1. A wtness a for the composteness of a number n s an nteger such that a n a (mod n). Indeed, multplyng both sdes of (1.13) by a gves a n a (mod n) for prme n. However, for some ntegers a Fermat s Lttle Theorem s stll satsfed for composte n. Ths brngs us to the Mller-Rabn Wtness Test. Algorthm 8. (Mller-Rabn Wtness Test) Gven an n wrtten as n 1 = 2 k q for odd nteger q, choose an nteger a such that a n, and check the followng condtons: (1) a q 1 (mod n) (2) a 2jq 1 (mod n) for some j < k Stop test f a number volatng one of these condtons s found. Proposton 4.2. If n s prme, both condtons are true. Proof. Consder the lst a q, a 2q, a 4q,..., a 2kq. Note that f n s prme a 2kq a n 1 1 (mod n) by Fermat s Lttle Theorem. Each number s the square of the prevous number wth the last term beng 1. Thus ether the frst term n the lst s 1, or some number n the lst s 1 when squared. In other words, f a q 1 (mod n) there s an nteger b such that b 1 (mod n) and b 2 1 (mod n), hence b = 1. So for some j < k, a 2jq 1 (mod n).

10 10 PARKER HAVIZA To test whether a number s prme, the Mller-Rabn test chooses several values for a n the hopes of dscoverng a wtness for the composteness of n (a number that volates one of the condtons). The chance of n beng prme mproves wth every non-wtness. Ths test does not conclusvely prove prmalty, but can show that n s lkely to be prme Factorzaton. Modern cryptography reles on numbers that are hard to factor. If the modulus used n a cryptosystem can be factored, the correspondng cryptosystem can be cracked. The smplest way to factor a large number s dvde t by every nteger startng wth 1, hopng to get an nteger. The numbers that are used n cryptosystems are suffcently large to make ths approach mpossble wth modern computng power. There are other methods to consder. Proposton 4.3. (Pollard s p-1 Method) Gven N = pq for p and q prme, choose L such that p 1 L and q 1 L and choose a such that a k 1 (mod q) where k L (mod q 1). Then p can be computed by the formula p = gcd(a L 1, N). Proof. By the defnton of dvsor, L = (p 1) = j(q 1) + k for nonzero ntegers, j, k where 1 k < q 1. Usng Fermat s Lttle Theorem shows that and a L = a (p 1) = (a p 1 ) 1 1 (mod p) a L = a j(q 1)+k = (a q 1 ) j a k 1 j a k a k (mod q). From the equatons above, we see that p a L 1 and q a L 1. Thus p =gcd(a L 1, N). If k s small and q s large, there s a good probablty that a k 1 (mod q) does not hold for a randomly chosen a. In fact, f p 1 s a product of small prmes t suffces to check gcd (a n! 1, N) for n = 1, 2, 3... because f p 1 can be wrtten as a product of small prmes t wll dvde n! for a small value of n. Thus we see that Pollard s Method s a fast factorng method n the case that p 1 factors nto small prmes. Another method of factorng reles on a very smple formula, X 2 Y 2 = (X Y )(X + Y ), whch we can adapt to factor N. Algorthm 9. Gven N = pq where p, q are prmes, calculate c = N + b 2 for b = 1, 2, 3... check each c to see f t can be wrtten as z 2 (so c s a perfect square). Stop when a perfect square s found. Proposton 4.4. If such a c s found, N can be factored as p = z b and q = z +b. Proof. N = pq = z 2 b 2 = (z b)(z + b), gvng p = z b and q = z + b. As b ncreases, each calculaton takes more computng tme. We can try usng multples of N to mprove the effcency of the algorthm by only checkng small b values. If we look at equatons of the form kn = (z b)(z + b) then some factor of N may dvde both (z b) and (z + b). If so, we can check values of the form c = kn + b 2 for b = 1, 2, 3...

11 SOME MATHEMATICAL FOUNDATIONS OF CRYPTOGRAPHY 11 to fnd (z b) and (z +b). One of these terms wll be dvsble by k because N = pq, so we fnd p = gcd(n, z b) and q = gcd(n, z + b) to complete a factorzaton of N. Acknowledgements It s my pleasure to thank my mentor John Wlmes for hs wsdom, gudance, and patence. Ths paper was made possble by hs strong desre to see me truly understand the underlyng concepts. I would also lke to thank Professor Peter May for organzng an amazng REU and Professor Laszlo Baba for hs stmulatng lectures on lnear algebra. References [1] Jeffrey Hoffsten, Jll Pper and Joseph H. Slverman. An Introducton to Mathematcal Cryptography. Sprnger, [2] R. L. Rvest, A. Shamr, and L. Adleman. A method for obtanng dgtal sgnatures and publc-key cryptosystems. Comm. ACM, 21(2): , [3] W. Dffe and M. E. Hellman. New drectons n cryptography. IEEE Trans. Informaton Theory, IT-22(6): , 1976.

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