Rational Points On Elliptic Curves - Solutions. (i) Throughout, we ve been looking at elliptic curves in the general form. y 2 = x 3 + Ax + B

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1 Rational Points On Elliptic Curves - Solutions (Send corrections to cbruni@uwaterloo.ca) (i) Throughout, we ve been looking at elliptic curves in the general form y 2 = x 3 + Ax + B However we did claim that an elliptic curve has equation of the form y 2 equals a cubic (with nonzero discriminant). Show that if we have an elliptic curve of the form y 2 = x 3 + Rx 2 + Sx + T Then we can shift coordinates so that we get an equation that eliminates the x 2 term. Hint: Send (x, y) on the elliptic curve y 2 = x 3 + Rx 2 + Sx + T to the value (x R/3, y) then simplify. Note: In the form y 2 = x 3 +Ax+B, the discriminant is given by = 16(4A 3 27B 2 ). Also note that this trick doesn t always work in finite fields (and the definition of an elliptic curve needs to be a bit more general to work for Z 2 and Z 3 but I ll leave this for interest reading). Solution: Proceeding as the hint suggests, we have (x R/3) 3 + R(x R/3) 2 + S(x R/3) + T = x 3 Rx 2 + R 2 x/3 R 3 /27 + Rx 2 2R 2 x/3 + R 3 /9 + Sx RS/3 + T = x 3 + (S R 2 /3)x R 3 /27 + R 3 /9 RS/3 + T (ii) What is the discriminant of an elliptic curve associated to the congruent number problem y 2 = x 3 N 2 x? Solution: = 16(4( N 2 ) 3 27(0) 2 ) = 64N 6 (iii) Compute the following values in their respective fields. Reduce to the least nonnegative integer in Z in Z in Z in Z 5 5. (1)(2)(3)(4) in Z 5 6. (1)(2)(3)(4)(5)(6) in Z 7 1

2 7. (1)(2)(3)(4)(5)(6)(7)(8)(9)(10) in Z 11 (Do you notice a pattern? Can you prove it?) Solution: (mod 7) (mod 7) (mod 11) in Z 5 (Can you prove in general if a c (mod p) then a k c k (mod p)?) 5. (1)(2)(3)(4) 1 (mod 5) in Z 5 6. (1)(2)(3)(4)(5)(6) 1 (mod 7) in Z 7 7. (1)(2)(3)(4)(5)(6)(7)(8)(9)(10) 1 (mod 11) in Z 11 (The pattern in the last three questions is known as Wilson s Theorem) (iv) Find the number of points on the elliptic curve y 2 = x 3 + x + 1 over the finite fields Z 3, Z 7, Z 11 and Z 13. Hint: Sometimes it s easier to compute a value using an equivalent negative value. For examine in Z 11 the number 10 1 (mod 11). Solution: For both this and the next problem, it will help us to compute the squares in Z p. The table below helps us achieve this y y 2 (mod 3) (mod 5) (mod 7) (mod 11) (mod 13) (There is lots of symmetry in the above table!) Notice that there are many unnecessary entries above (for example, there are only three numbers in Z 3 so we needed to only compute the squares of 0, 1 and 2). 2

3 In Z 3, we have that for x = 0, 1, 2, y 2 = (0) 3 + (0) + 1 = 1 1 (mod 3) y 2 = (1) 3 + (1) + 1 = 3 0 (mod 3) y 2 = (2) 3 + (2) + 1 = 11 2 (mod 3) Hence we have that (0, 1), (0, 2) [since 1 and 2 squared give 1 in Z 3 ] and (1, 0) are the three solutions (plus the point at infinity). In Z 7, y 2 = (0) 3 + (0) + 1 = 1 1 (mod 7) y 2 = (1) 3 + (1) + 1 = 3 3 (mod 7) y 2 = (2) 3 + (2) + 1 = 11 4 (mod 7) y 2 = (3) 3 + (3) (mod 7) y 2 = (4) 3 + (4) (mod 7) y 2 = (5) 3 + (5) (mod 7) y 2 = (6) 3 + (6) (mod 7) Only 0, 1, 2 and 4 are squares in Z 7 hence the points on this elliptic curve over Z 7 are (0, 1), (0, 6), (1, 0), (2, 2), (2, 5), (4, 3), (4, 4) and the point at infinity. In Z 11, y 2 = (0) 3 + (0) + 1 = 1 1 (mod 11) y 2 = (1) 3 + (1) + 1 = 3 3 (mod 11) y 2 = (2) 3 + (2) + 1 = 11 0 (mod 11) y 2 = (3) 3 + (3) (mod 11) y 2 = (4) 3 + (4) (mod 11) y 2 = (5) 3 + (5) (mod 11) y 2 = (6) 3 + (6) (mod 11) y 2 = (7) 3 + (7) (mod 11) y 2 = (8) 3 + (8) (mod 11) y 2 = (9) 3 + (9) (mod 11) y 2 = (10) 3 + (10) (mod 11) Only 0, 1, 3, 4, 5 and 9 are squares in Z 11 hence the points on this elliptic curve over Z 11 are (0, 1), (0, 10), (1, 5), (1, 6), (2, 0), (3, 3), (3, 8), (4, 5), (4, 6), (6, 5), (6, 6), (8, 2), (8, 9) and the point at infinity. 3

4 In Z 13, y 2 = (0) 3 + (0) + 1 = 1 1 (mod 13) y 2 = (1) 3 + (1) + 1 = 3 3 (mod 13) y 2 = (2) 3 + (2) + 1 = (mod 13) y 2 = (3) 3 + (3) (mod 13) y 2 = (4) 3 + (4) (mod 13) y 2 = (5) 3 + (5) (mod 13) y 2 = (6) 3 + (6) (mod 13) y 2 = (7) 3 + (7) (mod 13) y 2 = (8) 3 + (8) (mod 13) y 2 = (9) 3 + (9) (mod 13) y 2 = (10) 3 + (10) (mod 13) y 2 = (11) 3 + (11) (mod 13) y 2 = (12) 3 + (12) (mod 13) Only 0, 1, 3, 4, 9, 10 and 12 are squares in Z 13 hence the points on this elliptic curve over Z 13 are (0, 1), (0, 12), (1, 4), (1, 9), (4, 2), (4, 11), (5, 1), (5, 12), (7, 0), (8, 1), (8, 12), (10, 6), (10, 7), (11, 2), (11, 11), (12, 5), (12, 8) and the point at infinity. (v) Find the number of points on the elliptic curve y 2 = x 3 4x over the finite fields Z 3, Z 5 and Z 7. Solution: In Z 3, we have that for x = 0, 1, 2, y 2 = (0) 3 4(0) (mod 3) y 2 = (1) 3 4(1) (mod 3) y 2 = (2) 3 4(2) (mod 3) Hence we have that (0, 1), (0, 2), (1, 1), (1, 2) and the point at infinity are the solutions over Z 3. In Z 5, y 2 = (0) 3 4(0) (mod 5) y 2 = (1) 3 4(1) (mod 5) y 2 = (2) 3 4(2) (mod 5) y 2 = (3) 3 4(3) (mod 5) y 2 = (4) 3 4(4) (mod 5) Only 0, 1 and 4 are squares in Z 5 hence the points on this elliptic curve over Z 5 are (0, 1), (0, 4), (4, 1), (4, 4) and the point at infinity. 4

5 In Z 7, y 2 = (0) 3 4(0) (mod 7) y 2 = (1) 3 4(1) (mod 7) y 2 = (2) 3 4(2) (mod 7) y 2 = (3) 3 4(3) (mod 7) y 2 = (4) 3 4(4) (mod 7) y 2 = (5) 3 4(5) (mod 7) y 2 = (6) 3 4(6) (mod 7) Only 0, 1, 2 and 4 are squares in Z 7 hence the points on this elliptic curve over Z 7 are (0, 3), (0, 4), (2, 1), (2, 6), (3, 0), (4, 3), (4, 4), (6, 2), (6, 5) and the point at infinity. (vi) There are many algorithms available for computing rational points. Sage Math Cloud has some of these available. Using these for the curves y 2 = x 3 N 2 x with N = 13, 14 and 15, we find that y 2 = x x contains the point P = ( 36/25, 1938/125) y 2 = x x contains the point Q = (18, 48) y 2 = x x contains the point R = ( 9, 36) Doubling each points gives 2P = ( / , / ) 2Q = (4225/144, /1728) 2R = (289/16, 2737/64) See if you can find the associated triangles in the case of N = 14 and N = 15. The case even when N = 13 is far too large to want to do by hand! Hint: You may want to use a calculator. Solution: For N = 14, we see that x = 4225/144 and y = /1728. Let u = x = 65/12, d = 12 and v = y/u = 3713/144. If we follow the steps from the notes, we can see that a = d/2( x + N + x N) = 63 and b = d/2( x + N x N) = 16. Thus, the triangle should have side lengths 2a/d = b/d = 8 3 2u = For N = 15, we see that x = 289/16 and y = 2737/64. Let u = x = 17/4, d = 4 and v = y/u = 161/16. If we follow the steps from the notes, we can see that a = d/2( x + N + x N) = 63 and b = d/2( x + N x N) = 16. Thus, the triangle should have side lengths 2a/d = 15 2b/d = 4 2u = (For N = 13, the a and b values are and respectively and the u value is / Yikes!) 5