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1 Advanced Maths Lecture 3 Next generation cryptography and the discrete logarithm problem for elliptic curves Richard A. Hayden EC crypto p. 1

2 Public key cryptography Asymmetric cryptography two keys: Public key widely distributed Private key users keep secret Mathematically related, but private key not (thought to be!) practically computable from just public key EC crypto p. 2

3 What s wrong with RSA? There are sub-exponential attacks on RSA: ( (64 )) O exp( 9 b) 1/3 log(b) 2/3 GNFS None (discovered) for ECC. So ECC can achieve same security with smaller key sizes. Faster computations Lower power consumption Memory/bandwidth savings Think mobile devices, ubiquitous computing, the environment etc. EC crypto p. 3

4 Key sizes US National Institute of Standards and Technology says for exchanging AES symmetric keys: ECC/RSA key size relationship is not linear. ECC clearly much more future-proof. EC crypto p. 4

5 Trapdoor one-way functions To implement asymmetric crypto, We need a function (of the public key and plaintext), that is easy to compute (giving the ciphertext) EC crypto p. 5

6 Trapdoor one-way functions To implement asymmetric crypto, We need a function (of the public key and plaintext), that is easy to compute (giving the ciphertext) whose inverse is hard to compute, EC crypto p. 5

7 Trapdoor one-way functions To implement asymmetric crypto, We need a function (of the public key and plaintext), that is easy to compute (giving the ciphertext) whose inverse is hard to compute, unless given special information (the private key), in which case the inverse should be easy to compute EC crypto p. 5

8 The discrete logarithm problem Over R + : log(x) = 2 n=0 1 2n + 1 ( ) 2n+1 x 1 x + 1 Easy to compute to arbitrary accuracy, a x over R not one-way. EC crypto p. 6

9 The discrete logarithm problem Over R + : log(x) = 2 n=0 1 2n + 1 ( ) 2n+1 x 1 x + 1 Easy to compute to arbitrary accuracy, a x over R not one-way. For a discrete group (G, ), the discrete log log b (g) for b, g G is the least k Z 0 with: g = b k := b... b }{{} k times In the cyclic groups, b G g G k Z 0 [g = b k ] EC crypto p. 6

10 Harder: Z p Closure if 1 a, b p 1 by defn. 0 (ab mod p) p 1. If ab 0 mod p, p divides ab and thus p divides a or b, contradiction. EC crypto p. 7

11 Harder: Z p Closure if 1 a, b p 1 by defn. 0 (ab mod p) p 1. If ab 0 mod p, p divides ab and thus p divides a or b, contradiction. Associativity obvious because regular multiplication of integers is. EC crypto p. 7

12 Harder: Z p Closure if 1 a, b p 1 by defn. 0 (ab mod p) p 1. If ab 0 mod p, p divides ab and thus p divides a or b, contradiction. Associativity obvious because regular multiplication of integers is. Identity element is 1. EC crypto p. 7

13 Harder: Z p Closure if 1 a, b p 1 by defn. 0 (ab mod p) p 1. If ab 0 mod p, p divides ab and thus p divides a or b, contradiction. Associativity obvious because regular multiplication of integers is. Identity element is 1. Inverses If 1 < a p 1, and a 2 1 mod p, then a 2 a mod p or p divides a. Similarly, if a 3 1 mod p, then a 3 a 2 or a mod p etc. Eventually, we must find k, with a k 1 mod p as we will exhaust all (finite) other possibilities. EC crypto p. 7

14 E.g. p = DLP question: What is log 3 (5)? I.e. we want k such that 3 k = 5 EC crypto p. 8

15 E.g. p = DLP question: What is log 3 (5)? I.e. we want k such that 3 k = = = = = = = = = 5 so log 3 (5) = 5. Also, 3 generates the group, log 3 ( ) always defined. EC crypto p. 8

16 ElGamal key generation Generate an efficient description of some (large) cyclic group G of order q with generator g EC crypto p. 9

17 ElGamal key generation Generate an efficient description of some (large) cyclic group G of order q with generator g Choose a random 0 k q 1 EC crypto p. 9

18 ElGamal key generation Generate an efficient description of some (large) cyclic group G of order q with generator g Choose a random 0 k q 1 Compute h = g k EC crypto p. 9

19 ElGamal key generation Generate an efficient description of some (large) cyclic group G of order q with generator g Choose a random 0 k q 1 Compute h = g k Publish (G, q, g, h) as public key EC crypto p. 9

20 ElGamal key generation Generate an efficient description of some (large) cyclic group G of order q with generator g Choose a random 0 k q 1 Compute h = g k Publish (G, q, g, h) as public key k is private key EC crypto p. 9

21 ElGamal Encryption: Convert message m into an element of G Choose a random 0 y q 1, calculate c 1 = g y and c 2 = mh y Transmit ciphertext (c 1, c 2 ) EC crypto p. 10

22 ElGamal Encryption: Convert message m into an element of G Choose a random 0 y q 1, calculate c 1 = g y and c 2 = mh y Transmit ciphertext (c 1, c 2 ) Decryption: Compute c 2 c k 1 = mhy g ky = mgky g ky = m EC crypto p. 10

23 Breaking ElGamal Being able to solve the DLP problem for the group G lets you calculate k and thus the original message EC crypto p. 11

24 Breaking ElGamal Being able to solve the DLP problem for the group G lets you calculate k and thus the original message Still open question whether breaking ElGamal is as hard as solving DLP, has been shown in special cases (c.f. we don t know if breaking RSA is as hard as prime factorisation!) EC crypto p. 11

25 Breaking ElGamal Being able to solve the DLP problem for the group G lets you calculate k and thus the original message Still open question whether breaking ElGamal is as hard as solving DLP, has been shown in special cases (c.f. we don t know if breaking RSA is as hard as prime factorisation!) But no general sub-exponential DLP algorithm naive algorithm raising to higher and higher powers is exponential EC crypto p. 11

26 Breaking ElGamal Being able to solve the DLP problem for the group G lets you calculate k and thus the original message Still open question whether breaking ElGamal is as hard as solving DLP, has been shown in special cases (c.f. we don t know if breaking RSA is as hard as prime factorisation!) But no general sub-exponential DLP algorithm naive algorithm raising to higher and higher powers is exponential However, for DLP using Z p, the index calculus algorithm is sub-exponential, c.f. general number field sieve for RSA EC crypto p. 11

27 Elliptic curves over R A curve of the form: y 2 = f(x) = x 3 + ax + b where a and b R together with a point on the curve at infinity we call O. Also, the curve must be non-singular, this means the roots of f(x) must be distinct. EC crypto p. 12

28 Bezout s theorem Bezout s theorem says a line will intersect an elliptic curve in exactly three points as long as: We allow complex points EC crypto p. 13

29 Bezout s theorem Bezout s theorem says a line will intersect an elliptic curve in exactly three points as long as: We allow complex points We count intersection multiplicities EC crypto p. 13

30 Bezout s theorem Bezout s theorem says a line will intersect an elliptic curve in exactly three points as long as: We allow complex points We count intersection multiplicities We add the point at infinity (formally) to the curve, so points on curve is some A R 2 union {O} EC crypto p. 13

31 Intersection multiplicities What do we need to know about these? Not much, just that: If a line intersects an elliptic curve at P and is a tangent line at P, the multiplicity is greater than one EC crypto p. 14

32 Intersection multiplicities What do we need to know about these? Not much, just that: If a line intersects an elliptic curve at P and is a tangent line at P, the multiplicity is greater than one If the multiplicity of some intersection point P of a line and an elliptic curve is greater than one, then the line is a tangent at P EC crypto p. 14

33 Intersection multiplicities What do we need to know about these? Not much, just that: If a line intersects an elliptic curve at P and is a tangent line at P, the multiplicity is greater than one If the multiplicity of some intersection point P of a line and an elliptic curve is greater than one, then the line is a tangent at P Tangent lines at P are guaranteed unique by the non-singularity requirement (so we can replace a with the in the above), we will see later that this is important EC crypto p. 14

34 Real intersections (1) We define a line to include O iff it is vertical. So now we can talk of intersections between lines and elliptic curves as subsets of R 2 {O}. EC crypto p. 15

35 Real intersections (1) We define a line to include O iff it is vertical. So now we can talk of intersections between lines and elliptic curves as subsets of R 2 {O}. Assume an elliptic curve and a non-vertical line y = mx + c (m, c R) meet in at least 2 real points (counting multiplicities!). Sub into y 2 = x 3 + ax + b: (mx + c) 2 = x 3 + ax + b A cubic in x. Has at least one real root. EC crypto p. 15

36 Real intersections (2) Assume only one real root. Then they meet at one real point with multiplicity (at least) two. But cubic must then also have two complex conjugate roots, so also meets at two other complex points, so counting multiplicities, this is at least four points. Too many, contradicts Bezout. EC crypto p. 16

37 Real intersections (2) Assume only one real root. Then they meet at one real point with multiplicity (at least) two. But cubic must then also have two complex conjugate roots, so also meets at two other complex points, so counting multiplicities, this is at least four points. Too many, contradicts Bezout. Assume at least two real non-zero roots, so cubic is (x r 1 )(x r 2 )(x r 3 ) for r 1, r 2 R and since r 1 r 2 r 3 = b c 2 R, r 3 R, so cubic has three real roots (similarly if r 1 and/or r 2 is zero) EC crypto p. 16

38 Real intersections (2) Assume only one real root. Then they meet at one real point with multiplicity (at least) two. But cubic must then also have two complex conjugate roots, so also meets at two other complex points, so counting multiplicities, this is at least four points. Too many, contradicts Bezout. Assume at least two real non-zero roots, so cubic is (x r 1 )(x r 2 )(x r 3 ) for r 1, r 2 R and since r 1 r 2 r 3 = b c 2 R, r 3 R, so cubic has three real roots (similarly if r 1 and/or r 2 is zero) Key point: if we draw a non-vertical line through two real points on elliptic curve, will always meet at a third real point... we can forget about C, and only care about R 2 {O} EC crypto p. 16

39 Intersection patterns (1) Possible intersection patterns for non-vertical lines through two real points on an elliptic curve: 1: All multiplicities 1. Three distinct real points, all of multiplicity 1. 2: Line is tangent to Q multiplicity = 2, so meets at two real points, one of multiplicity 2, one of multiplicity 1. EC crypto p. 17

40 Real intersections (3) What about vertical lines, i.e. those through O which intersect in at least one real point? They have the form x = d for some d R so sub in: y 2 = d 3 + ad + b y = ± d 3 + ad + b EC crypto p. 18

41 Real intersections (3) What about vertical lines, i.e. those through O which intersect in at least one real point? They have the form x = d for some d R so sub in: y 2 = d 3 + ad + b y = ± d 3 + ad + b Since one of these is real, so is the other. Thus unless y = 0, such lines meet at two distinct real points as well as O, thus all of multiplicity 1. EC crypto p. 18

42 Real intersections (3) What about vertical lines, i.e. those through O which intersect in at least one real point? They have the form x = d for some d R so sub in: y 2 = d 3 + ad + b y = ± d 3 + ad + b Since one of these is real, so is the other. Thus unless y = 0, such lines meet at two distinct real points as well as O, thus all of multiplicity 1. If y = 0, dx dy = 0, so the line is tangent at the real point of intersection (d, 0) and thus it has multiplicity > 1, i.e. 2 and O has multiplicity 1. EC crypto p. 18

43 Intersection patterns (2) Possible intersection patterns for non-vertical lines through two real points on an elliptic curve: 3: P, Q and O all with multiplicity 1. 4: Tangent at P multiplicity 2, O with multiplicity 1. EC crypto p. 19

44 A binary operation (1) Write for an elliptic curve C, C(R) as the set of points on the curve contained in R 2 {O}, we are going to define a (commutative) binary operation on C(R), called : C(R) C(R) C(R) EC crypto p. 20

45 A binary operation (1) Write for an elliptic curve C, C(R) as the set of points on the curve contained in R 2 {O}, we are going to define a (commutative) binary operation on C(R), called : C(R) C(R) C(R) If P and Q R 2 are distinct, draw the unique line through them... if they both have multiplicity 1, they meet at a third point R with multiplicity 1, define P Q = R if one has multiplicity 2 say P, Q must have multiplicity 1, define P Q = P EC crypto p. 20

46 A binary operation (1) Write for an elliptic curve C, C(R) as the set of points on the curve contained in R 2 {O}, we are going to define a (commutative) binary operation on C(R), called : C(R) C(R) C(R) If P and Q R 2 are distinct, draw the unique line through them... if they both have multiplicity 1, they meet at a third point R with multiplicity 1, define P Q = R if one has multiplicity 2 say P, Q must have multiplicity 1, define P Q = P If P = Q R 2, draw the unique tangent to the curve at P = Q. At P must have multiplicity 2 or 3. If it is 2, meets elsewhere, say R and define P Q = R. Otherwise, P Q = P = Q (in fact, only the mult. 2 case occurs) EC crypto p. 20

47 A binary operation (1) Write for an elliptic curve C, C(R) as the set of points on the curve contained in R 2 {O}, we are going to define a (commutative) binary operation on C(R), called : C(R) C(R) C(R) If P and Q R 2 are distinct, draw the unique line through them... if they both have multiplicity 1, they meet at a third point R with multiplicity 1, define P Q = R if one has multiplicity 2 say P, Q must have multiplicity 1, define P Q = P If P = Q R 2, draw the unique tangent to the curve at P = Q. At P must have multiplicity 2 or 3. If it is 2, meets elsewhere, say R and define P Q = R. Otherwise, P Q = P = Q (in fact, only the mult. 2 case occurs) EC crypto p. 20

48 A binary operation (1) Write for an elliptic curve C, C(R) as the set of points on the curve contained in R 2 {O}, we are going to define a (commutative) binary operation on C(R), called : C(R) C(R) C(R) If P and Q R 2 are distinct, draw the unique line through them... if they both have multiplicity 1, they meet at a third point R with multiplicity 1, define P Q = R if one has multiplicity 2 say P, Q must have multiplicity 1, define P Q = P If P = Q R 2, draw the unique tangent to the curve at P = Q. At P must have multiplicity 2 or 3. If it is 2, meets elsewhere, say R and define P Q = R. Otherwise, P Q = P = Q (in fact, only the mult. 2 case occurs) EC crypto p. 20

49 A binary operation (2) What if we are given a real P and O? Draw the (unique) vertical line through P. Remember that either it meets P with multiplicity 2, in which case, we define P O = P or with multiplicity 1, in which case it meets also at some other point Q, in which case we define P O = Q. EC crypto p. 21

50 A binary operation (2) What if we are given a real P and O? Draw the (unique) vertical line through P. Remember that either it meets P with multiplicity 2, in which case, we define P O = P or with multiplicity 1, in which case it meets also at some other point Q, in which case we define P O = Q. Finally, define O O = O (makes sense if consider line at infinity meeting curve just at O with multiplicity 3, but can just take it formally). EC crypto p. 21

51 Some examples... EC crypto p. 22

52 A group law? You might be asking, is C(R) a group under? Unfortunately not, clear that there is no identity. EC crypto p. 23

53 A group law? You might be asking, is C(R) a group under? Unfortunately not, clear that there is no identity. Define another (commutative) binary operation + : C(R) C(R) C(R): first compute P Q, draw the line through P Q and O, then P + Q is defined to be the third intersection point of this line with the curve (counting multiplicities), i.e. P + Q := O (P Q) EC crypto p. 23

54 A group law? You might be asking, is C(R) a group under? Unfortunately not, clear that there is no identity. Define another (commutative) binary operation + : C(R) C(R) C(R): first compute P Q, draw the line through P Q and O, then P + Q is defined to be the third intersection point of this line with the curve (counting multiplicities), i.e. P + Q := O (P Q) EC crypto p. 23

55 Explicit formulae for + Before we prove that + does indeed make C(R) a group, we show how to derive explicit formulae for the binary operation +: Let λ = q q p p, then x = λ2 p q and y = p + λ(x p) (elliptic curves are symmetric about x-axis). If the two points are the same, we can similarly compute an explicit expression for 2P := P + P. EC crypto p. 24

56 Group law (1) Closure we ve already checked this EC crypto p. 25

57 Group law (1) Closure we ve already checked this Identity element is O EC crypto p. 25

58 Group law (1) Closure we ve already checked this Identity element is O Inverse element is of P = (x, y) is P := (x, y) EC crypto p. 25

59 Group law (2) Associativity hardest to check. Can check using the explicit formulae, but you have to make sure you consider all cases, i.e. those involving O, those where two/three of the points are the same etc. There is a more elegant proof, but it involves a bit more algebraic geometry, for which there is not time now so you will have to just trust me (or bash it out with the explicit formulae!) EC crypto p. 26

60 Fields Recall that a field was a set F together with two binary operations, + and, where (F, +) is an abelian group with identity we call 0 EC crypto p. 27

61 Fields Recall that a field was a set F together with two binary operations, + and, where (F, +) is an abelian group with identity we call 0 is associative and commutative EC crypto p. 27

62 Fields Recall that a field was a set F together with two binary operations, + and, where (F, +) is an abelian group with identity we call 0 is associative and commutative has an identity (we call 1) and inverses for all elements except 0 EC crypto p. 27

63 Fields Recall that a field was a set F together with two binary operations, + and, where (F, +) is an abelian group with identity we call 0 is associative and commutative has an identity (we call 1) and inverses for all elements except 0 distributes over +, i.e. a (b + c) = a b + a c EC crypto p. 27

64 Fields Recall that a field was a set F together with two binary operations, + and, where (F, +) is an abelian group with identity we call 0 is associative and commutative has an identity (we call 1) and inverses for all elements except 0 distributes over +, i.e. a (b + c) = a b + a c 1 0 EC crypto p. 27

65 Fields Recall that a field was a set F together with two binary operations, + and, where (F, +) is an abelian group with identity we call 0 is associative and commutative has an identity (we call 1) and inverses for all elements except 0 distributes over +, i.e. a (b + c) = a b + a c 1 0 Examples are Q, R, C under their usual addition and multiplication. Z is not a field though. EC crypto p. 27

66 Finite fields Recall the integers {1,..., p 1} (prime p) under multiplication modulo p. Knowing that this is a group, it is easy to see that {0, 1,..., p 1} is a field with addition modulo p. Call it F p. EC crypto p. 28

67 Finite fields Recall the integers {1,..., p 1} (prime p) under multiplication modulo p. Knowing that this is a group, it is easy to see that {0, 1,..., p 1} is a field with addition modulo p. Call it F p. None of what we did earlier, proving that C(R) was a group, was dependent on R, apart from that it is a field, the same thing works for any field, but we lose the geometric analogies and pretty pictures. EC crypto p. 28

68 Finite fields Recall the integers {1,..., p 1} (prime p) under multiplication modulo p. Knowing that this is a group, it is easy to see that {0, 1,..., p 1} is a field with addition modulo p. Call it F p. None of what we did earlier, proving that C(R) was a group, was dependent on R, apart from that it is a field, the same thing works for any field, but we lose the geometric analogies and pretty pictures. Bezout s theorem carries across, replacing R with F p and C with the algebraic closure of F p, so we get a group law on C(F p ) EC crypto p. 28

69 Example (1) over F 5. y 2 = x 3 + x + 1 EC crypto p. 29

70 Example (1) y 2 = x 3 + x + 1 over F 5. How can we find all the points on the curve? Plug all 5 possibilities for x in... gives 9 points: C(F 5 ) = {O, (0, ±1), (2, ±1), (3, ±1), (4, ±2)} EC crypto p. 29

71 Example (1) y 2 = x 3 + x + 1 over F 5. How can we find all the points on the curve? Plug all 5 possibilities for x in... gives 9 points: C(F 5 ) = {O, (0, ±1), (2, ±1), (3, ±1), (4, ±2)} So C(F 5 ) is an abelian group of order 9. In fact, it is easy to see that (0, 1) generates the group, so it is the cyclic group of order 9... EC crypto p. 29

72 Example (2) Let s try to compute (0, 1) 2 = (0, 1) + (0, 1) in C(F 5 ). We need the (formal) tangent to y 2 = x 3 + x + 1 at (0, 1). EC crypto p. 30

73 Example (2) Let s try to compute (0, 1) 2 = (0, 1) + (0, 1) in C(F 5 ). We need the (formal) tangent to y 2 = x 3 + x + 1 at (0, 1). Compute dy dx = 3x2 +1 2y = 1/2 = = 1 3 = 3, so tangent is y = 3x + 1. EC crypto p. 30

74 Example (2) Let s try to compute (0, 1) 2 = (0, 1) + (0, 1) in C(F 5 ). We need the (formal) tangent to y 2 = x 3 + x + 1 at (0, 1). Compute dy dx = 3x2 +1 2y = 1/2 = = 1 3 = 3, so tangent is y = 3x + 1. Subbing in, (3x + 1) 2 = x 3 + x + 1, roots x = 0 and x 2 9x 5 = x 2 4x 5 = x 2 + x = 0, x = 0 and x = 1 = 4. So (0, 1) (0, 1) = (4, 2). EC crypto p. 30

75 Example (2) Let s try to compute (0, 1) 2 = (0, 1) + (0, 1) in C(F 5 ). We need the (formal) tangent to y 2 = x 3 + x + 1 at (0, 1). Compute dy dx = 3x2 +1 2y = 1/2 = = 1 3 = 3, so tangent is y = 3x + 1. Subbing in, (3x + 1) 2 = x 3 + x + 1, roots x = 0 and x 2 9x 5 = x 2 4x 5 = x 2 + x = 0, x = 0 and x = 1 = 4. So (0, 1) (0, 1) = (4, 2). Then (0, 1) + (0, 1) = O (4, 2) = (4, 2) C(F 5 ). EC crypto p. 30

76 ECDLP So an elliptic curve over a (large) finite field gives us a (large) finite group, indeed a theorem of Hasse says: ( p 1) 2 C(F p ) ( p + 1) 2 for any elliptic curve C. We can do the discrete log problem in cyclic subgroups of such groups. EC crypto p. 31

77 ECDLP So an elliptic curve over a (large) finite field gives us a (large) finite group, indeed a theorem of Hasse says: ( p 1) 2 C(F p ) ( p + 1) 2 for any elliptic curve C. We can do the discrete log problem in cyclic subgroups of such groups. ECC is then the ElGamal algorithm in subgroups of the group of points on elliptic curves over finite fields EC crypto p. 31

78 ECDLP For an EC group of size p, roughly need same number of bits as for multiplication modulo p group (Hasse), in both cases your key is a constant number of log 2 (p)-bit numbers, i.e. proportional to p EC crypto p. 32

79 ECDLP For an EC group of size p, roughly need same number of bits as for multiplication modulo p group (Hasse), in both cases your key is a constant number of log 2 (p)-bit numbers, i.e. proportional to p Sub-exponential algorithm for Z p relied on a special result specific only to these groups EC crypto p. 32

80 ECDLP For an EC group of size p, roughly need same number of bits as for multiplication modulo p group (Hasse), in both cases your key is a constant number of log 2 (p)-bit numbers, i.e. proportional to p Sub-exponential algorithm for Z p relied on a special result specific only to these groups No such sub-exponential DLP algorithm has been found for elliptic curve groups, thus in terms of key sizes required for the same level of security, ECC beats both RSA and ElGamal over multiplication modulo p groups EC crypto p. 32

81 ECDLP Certain curves are vulnerable to special attacks and should be avoided, c.f. chinese remainder theorem attacks on RSA EC crypto p. 33

82 ECDLP Certain curves are vulnerable to special attacks and should be avoided, c.f. chinese remainder theorem attacks on RSA NIST recommends a small number of curves and fields, chosen for optimal security (i.e. not vulnerable to any known special case attacks) and implementation efficiency (some curves have properties making the implementation of ECC cheaper) EC crypto p. 33

83 ECDLP Certain curves are vulnerable to special attacks and should be avoided, c.f. chinese remainder theorem attacks on RSA NIST recommends a small number of curves and fields, chosen for optimal security (i.e. not vulnerable to any known special case attacks) and implementation efficiency (some curves have properties making the implementation of ECC cheaper) Hardest (publicly) broken ECC scheme to date had 109-bit key. 10,000 Pentium class PCs, > 540 days EC crypto p. 33

84 References A beautiful book: Rational Points on Elliptic Curves by Silverman and Tate, Chapter 4 addresses finite fields, very accessible EC crypto p. 34

85 References A beautiful book: Rational Points on Elliptic Curves by Silverman and Tate, Chapter 4 addresses finite fields, very accessible Elliptic Curves and their Applications to Cryptography by Andreas Enge, actually focussed on crypto expensive though! EC crypto p. 34

86 References A beautiful book: Rational Points on Elliptic Curves by Silverman and Tate, Chapter 4 addresses finite fields, very accessible Elliptic Curves and their Applications to Cryptography by Andreas Enge, actually focussed on crypto expensive though! Algebraic Aspects of Cryptography by Neal Koblitz, focusses on lots of interesting areas of algebra and number theory applicable to crypto EC crypto p. 34

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