Question

Q4. Consider the elliptic curve E11(1,6); that is, the curve is defined by y2 = x + x +6 with a modulus of p = 11. a) Determine all of the points in E11(1,6). Hint: 0 5x<p and 0 sy<p, x and y are integers. Coding may be the easier way. b) For P(2,4), calculate 13P. c) For P(2,4) and Q(2,7), calculate P+Q. show your steps.

Answer 1

The Sat E., (1,6) is the set of integere (ny) that satisfy y² = x3 + x + 6 (mod 11) y can see that (ny) = 6, 9) is in this set as (7²+7+6) (mod 11) 81 (mod 11) - 356 (mod 11) 4 = 4 9² (mod 11) the integer 00 into the these To find all the points intii (1,6), find all the possible values n3+h+6 (mod p) and then what values of y² will match. There 11 choices of {0, 1, . - 10} . values in turn subbing Cubic and reducing modulo I will 3 give the possible values of y? x=07 +h+6 = 16=5 = 103 +0+6 - 27+3+6 = 36=3 n=1 (13+1+6 12-4 h 446 8 -64th+ 1 = 2 (2+2 +6 748 8 +2+6

to calculats med Mg. = 153+5+6 = 125+5+6 = 136 = 4 q= 16+6+6 divide by 216+ 12 = 228 = 8 2.0/ 13 74/14 My = (7)? 7+6 -343+13 So on = 356= 4 no = (0+ 8 +6 = 526 = 9 ng (93+9+6 - 744 = 7 1 = (10)*((0) +6 - 1 016 34 can see the possible value of y2 are {3, 4, 5, 6, 7, 8, 93 nie - y Can not be 0, 1, 2 od 10. So that Next examine the lo possible values of y and identify which values of a they could be poired give a point on with to the curve

= (0,2 No points So y? y=1 = y² = No points Notes Mod Calculated Livided by 11 liki 16/11 5 (2 y = 2 = y2 (2) 2 2 = 5, 7, 10 25/11 = 3 1 = 4 y=3 = 1332 1 1 9 y=4 = (42 165=2 y=5= 152 - n=3 253 3 y=6= (6)2 36= 3 n-3 y 7 = 4955 N=2 y=8 = 18,² 64 = 9 = 8 y=9= (932 = 81 = 4 n = 5, 7, 10 Y=10 = 102 100= 1 No points 13 points in E,, (1,6) -(the 12 found above and oo So, there L

{(2,4), (2,7), (3,5), (3,6), (5,2),/5,9), (3, 2), (7,9), (8,3), (8,8), (16, 2), (10,9), 00} 10 S -(All points shown graph) (3) 008) (36)

solution & y² = x3+x+6 (mod 11) point (2,4) =(2, 3) lies the To compute 13 P. d = 3²+ a = (x3, ys) we have 29 = 3(23²+) 2 X 4 13 8 } ( (mod 1) 5 x3 = - *, -42 = (5) 22-2 = 25-4 21 (mod 11) - 10 y = 2(x-3) - Y) = 5(2-10) - 4 = - Yo-4 -yu mod!) E. 1 Find points finally P(2,4) = 110,0)

(6) For. P (2, 4) and a 12,7) Calculate (P+O) m= 7-4 3 릉 2-2 makes point addition an internal las Enure ham a neutral which is required for a group. That is So that for all point P on the curul PAN Some N - N+P =P how ot (2,7)= 2,7) 7 helds neutral o It holds because we define (2,7)+(2,4) as the neutral (12,21 +13,4)) +(2,7) = 0+ (2,7) =(2,7) An Alternative justification is commutativity and associatioity, combined with the addition law for normal curve points. (12,3)+(2,48) + (2,7) = {(2, 4)+(2,7)) + (2,7) = 12,4) + (12,7)+12,7) ) = (2,4) + (5,2) (2,7)