HOMEWORK 8 SOLUTIONS MATH 4753

Transcription

1 HOMEWORK 8 SOLUTIONS MATH 4753 In this homework we will practice taking square roots of elements in F p in F p 2, and study the encoding scheme suggested by Koblitz for use in elliptic curve cryptosystems. We will start with some basic examples. Question 1. Solve for a in the equation: 5 = a 3 in F49. (Hint: Your answer should be an element of F 7.) Solution. We square to get an equation modulo 7: 5 a 2 3 mod 7 This yields a 2 4 mod 7, so a ±2 mod 7 works. Question 2. Compute the square root of 6 in F 49. Proof. Really this question is asking us to express this root in a basis we chose like F 49 = F 7 [ 3], so we repeat the computation of the previous question: 6 = a 3, squaring both sides yields 6 3a 2 mod 7, and we find that a ±4 mod 7, so 6 = ±4 3. Define the field F p 2 = F p [ 3] where p are defined as in the hw8-data.txt file in Sage. These numbers satisfy that p 3 mod 4 and 3 is not a square modulo p. Remember, we do this as in lecture: R.<x> = PolynomialRing (GF( p ) ) F.<t> = GF( pˆ2, modulus = xˆ2 3) Here the modulus option tells Sage what the defining polynomial of t is. Thus our t will actually be 3. Check that you defined this correctly by computing t^2 and seeing if you get 3. Let E : y 2 = x 3 + ax + b be an elliptic curve defined over F p. As we mentioned in lecture, Koblitz suggested the following encoding function for elliptic curve cryptosystems: (1) Encode the message in, say, ASCII, as a number 0 m < p. Date: May 6,

2 2 HOMEWORK 8 SOLUTIONS MATH 4753 (2) Encoding the message as a point on an elliptic curve by sending m P m = (m, m 3 + am + b) E(F p 2), where the choice of square root is arbitrary. Question 3. For the given p in the data file, how many ASCII characters can you encode per message? Question 4. Use the GF(p).random element() member function as we did in lecture to select random coefficients a,b for an elliptic curve. Create the elliptic curve E associated to these parameters (but over the field F!) in Sage, and be sure that your parameters do not result in a non-zero discriminant. Implement Koblitz s encoding function for your elliptic curve. Be sure to include a check to make sure the message does not have too many characters. You may find it helpful to use the encoding function from HW7 s data file as a model. Question 5. Use random element() to find a random x and find a point P = (x, y) E(F p 2) to use as P in your public key for the El Gamal cryptosystem. Use randint to find a secret k and then generate Q = k*p to complete your public key. Question 6. My public key is given as the points P2, Q2 and the curve E2 in the data file (I used the same field F p 2). Create a new copy of your encoding function using my curve E2. A certain mathematics professor decides to give his students a surprise exam sometime during the week, either Monday, Wednesday, or Friday in lecture. But he tells the students, If you can guess when the exam will be, I will not give you an exam. A student (let s call him Vizzini) answers: But it s so simple! All I have to do is divine from what I know of you: are you the sort of professor who would give the exam early in the week, or later? Now, a clever professor would give the exam later in the week, because he would know that only a great fool would fail to study for an exam on Monday. I am not a great fool, so I can clearly I cannot say the exam would be on Friday. But you must have known I was not a great fool you would have counted on it! so I can clearly not expect the exam earlier in the week. The professor asks if the student has made his decision, but the student replies: Not remotely. Because our professor studied at Harvard, and as everyone knows, Harvard is entirely peopled with people who love giving exams as early and often as possible, so I can clearly not choose a time later in the week!

3 HOMEWORK 8 SOLUTIONS MATH The professor remarks that the student truly has a dizzying intellect, and the student continues, Wait till I get going! Where was I? Harvard! And you must have suspected I d have known your academic background you d have counted on it! so I clearly cannot choose a time earlier in the week! Now the professor interrupts Vizzini and says that not only are you just stalling you have missed the entire logic of the puzzle. Can you answer the question better than Vizzini did? Question 7. Create an encrypted response to me (using my public key) explaining why the exam cannot be on Friday. (Hint: Is the exam still a surprise if given on Friday?) Conclude that it is logically impossible for me to give you a pop quiz. (You may wish to break up your message into several fragments. Be sure you included a length check as in HW7 s ECencode function to avoid your messages wrapping around modulo p and being corrupted.) Question 8. your encrypted response from the previous question, as well as your curve s public key to me (i.e., the a, b, P, Q), and decrypt my subsequent response. Question 9. Let p 2 mod 3 throughout this problem. (1) Prove that the cubing map x x 3 : F p F p is a bijection. Proof. We compute the inverse of cubing map. In exponent space, we want an exponent k such that 3k 1 mod p 1. Since p 2 mod 3, we know that p 1 1 mod 3, so in particular, 3 p 1, so gcd(3, p 1) = 1. That means that 3 is invertible modulo p 1, so an inverse k does exist mod p 1. Then x x k is the inverse map to the cubing map, because (x 3 ) k x 3k x 1+l(p 1) x mod p for some l Z, where the p 1 power of x goes away by Euler s theorem when x 0 mod p, and the equation is trivially true when x 0 mod p. So essentially the k power is a cube root, but note, it is an integer power, so it still makes sense in modular arithmetic. (2) For some b F p, let E be the elliptic curve over F p given by Prove that #E(F p ) = p + 1. E : y 2 = x 3 + b. Proof. Since f(x) = x 3 is a bijection, so is f(x) = x 3 + b. So the possible values of x 3 + b are just all of the different numbers modulo p, in some permuted order. Since (p 1)/2 of those possible x coordinates

4 4 HOMEWORK 8 SOLUTIONS MATH 4753 are nonzero squares, those each yield 2 distinct y-values so 2 points in the group, the value 0 yields 1 point, and the point O adds one more, yielding 2 (p 1)/ = p + 1 points, as claimed. Question 10. Let E : y 2 = x over p = 5. Compute #E(F p 2), #E(F p 3), and #E(F p 4) using the previous question and the formula from lecture. Solution. Since p = 5 2 mod 3, we know that there are #E(F 5 ) = p+1 = 6 points on the elliptic curve over F p. As usual, set t = p + 1 #E(F p ) = 0, and then we solve for z in the characteristic equation of Frobenius: z 2 tz + p = 0 = z = ± p Call these roots α, β. Then our formula is that: Here, and #E(F p k) = p k + 1 α k β k. #E(F p 2) = p (+ p) 2 ( p) 2 = p p, #E(F p 3) = p (+ p) 3 ( p) 3 = p Bonus Question 1 (25 pts). Let p 2 mod 3 be an odd prime. Recall from the previous question that the elliptic curve where 0 b F p, satisfies E : y 2 = x 3 + b, #E(F p ) = p + 1. Such a curve is called supersingular. 1 Recall from lecture that if t p = p + 1 #E(F p ) so that #E(F p ) = p + 1 t p, then τ 2 p (P ) t p τ p (P ) + pp = O for all P E(F p ), where τ p (x, y) = (x p, y p ) is the Frobenius map on E and τp 2 (P ) = τ p (τ p (P )), that is, τp 2 (x, y) = (x p2, y p2 ), and F p = k 1 F pk, identified with the natural inclusions. 1 This definition applies when p 5, otherwise the notion is defined slightly differently. Here the word singular does not mean, as it usually does, having a singularity on the curve like a cusp, where the tangent line is undefined. But rather, the word singular is used in the original sense of special. These supersingular curves were special amongst the original class of special curves, and thus supersingular is used here in the sense of very special.

5 HOMEWORK 8 SOLUTIONS MATH (1) Prove that the p-torsion E[p] = {P E(F p ) : pp = O} of E(F p ) satisfies E[p] = {O}. (2) Compute E[p n ] = {P E(F p ) : p n P = O} for all n N. (3) Prove that E[p+1] = E(F p 2). Compute #E[p+1] by using the formula from lecture to compute #E(F p n) when n = 2. (4) Suppose b is not a square modulo p. Consider the point P = (0, b) E(F p 2). Compute p n P for all n N. (5) Now suppose to the contrary that b = β 2 mod p for some β F p, that is, b is a square. Let P = (0, β) E(F p ) and compute p n P for all n N. Oklahoma State University, Spring 2017