Only **TWO** questions have to be completed, we chose #3 and #2. # Question 2 ## Listing All Points on E and Finding |E| We now search for all solutions (x, y) ∈ 𝔽₇ × 𝔽₇ to y² ≡ x³ + 2x + 4 (mod 7). Because our field 𝔽₇ is small, we can check x = 0, 1, 2, 3, 4, 5, 6, and for each x compute the right-hand side and see if it is a quadratic residue in 𝔽₇. We remember the squares (quadratic residues) mod 7 are {0, 1, 2, 4}. 1. **x = 0** - RHS = 0³ + 2·0 + 4 = 4. - We need y² ≡ 4. Then y ≡ ±2 = 2, 5 (mod 7). - Points: (0, 2), (0, 5). 2. **x = 1** - RHS = 1³ + 2 + 4 = 7 ≡ 0. - We need y² ≡ 0 → y = 0. - Point: (1, 0). 3. **x = 2** - RHS = 2³ + 2·2 + 4 = 8 + 4 + 4 = 16 ≡ 2. - We need y² ≡ 2, so y ≡ ±3 = 3, 4 (mod 7). - Points: (2, 3), (2, 4). 4. **x = 3** - RHS = 3³ + 6 + 4 = 37 ≡ 37 − 35 = 2 (mod 7). - Again y² ≡ 2 → y = 3, 4. - Points: (3, 3), (3, 4). 5. **x = 4** - RHS = 4³ + 8 + 4 = 76 ≡ 76 − 70 = 6 (mod 7). - We need y² ≡ 6, but 6 is not in {0,1,2,4}. - No solutions. 6. **x = 5** - 5³ = 125 ≡ 125 − 119 = 6 (mod 7). - RHS = 6 + 2·5 + 4 = 6 + 10 + 4 = 20 ≡ 6 (mod 7). - We need y² ≡ 6, which fails again. - No solutions. 7. **x = 6** - 6³ = 216 ≡ 6 (mod 7). - RHS = 6 + 2·6 + 4 = 6 + 12 + 4 = 22 ≡ 1 (mod 7). - We need y² ≡ 1, so y ≡ ±1 = 1, 6 (mod 7). - Points: (6, 1), (6, 6). Therefore, the affine points on E are: (0, 2), (0, 5), (1, 0), (2, 3), (2, 4), (3, 3), (3, 4), (6, 1), (6, 6). In total, there are **9** affine points. But in elliptic-curve notation, we always include the point at infinity (the identity) as well. Thus: |E| = 9 + 1 = 10. Therefore, **the total number of points on E is 10**.