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Given the Elliptic Curve E : Find all points of E y² = X u + x +1 over GF(5)

Given the Elliptic Curve E : Find all points of E y² = X u + x +1 over GF(5)

Elementary Geometry For College Students, 7e
Elementary Geometry For College Students, 7e
7th Edition
ISBN:9781337614085
Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.
ChapterP: Preliminary Concepts
SectionP.CT: Test
Problem 1CT
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**Given the Elliptic Curve \( E: y^2 = x^3 + x + 1 \) over \( GF(5) \)** Find all points of \( E \). --- To find all the points on the elliptic curve \( E \) over the finite field \( GF(5) \), we evaluate the equation for each \( x \) value in the field \( \{0, 1, 2, 3, 4\} \) and determine corresponding \( y \) values that satisfy the equation. 1. **Evaluate for \( x = 0 \):** - \( y^2 = 0^3 + 0 + 1 = 1 \) - Possible \( y \) values are solutions to \( y^2 \equiv 1 \mod 5 \): \( y = 1, 4 \) 2. **Evaluate for \( x = 1 \):** - \( y^2 = 1^3 + 1 + 1 = 3 \) - Possible \( y \) values are solutions to \( y^2 \equiv 3 \mod 5 \): None (no solutions, as 3 is not a quadratic residue mod 5) 3. **Evaluate for \( x = 2 \):** - \( y^2 = 2^3 + 2 + 1 = 11 \equiv 1 \mod 5 \) - Possible \( y \) values are \( y = 1, 4 \) 4. **Evaluate for \( x = 3 \):** - \( y^2 = 3^3 + 3 + 1 = 31 \equiv 1 \mod 5 \) - Possible \( y \) values are \( y = 1, 4 \) 5. **Evaluate for \( x = 4 \):** - \( y^2 = 4^3 + 4 + 1 = 69 \equiv 4 \mod 5 \) - Possible \( y \) values are solutions to \( y^2 \equiv 4 \mod 5 \): \( y = 2, 3 \) Thus, the points on the elliptic curve \( E \) over \( GF(5) \) include: - \( (0, 1
Transcribed Image Text:**Given the Elliptic Curve \( E: y^2 = x^3 + x + 1 \) over \( GF(5) \)** Find all points of \( E \). --- To find all the points on the elliptic curve \( E \) over the finite field \( GF(5) \), we evaluate the equation for each \( x \) value in the field \( \{0, 1, 2, 3, 4\} \) and determine corresponding \( y \) values that satisfy the equation. 1. **Evaluate for \( x = 0 \):** - \( y^2 = 0^3 + 0 + 1 = 1 \) - Possible \( y \) values are solutions to \( y^2 \equiv 1 \mod 5 \): \( y = 1, 4 \) 2. **Evaluate for \( x = 1 \):** - \( y^2 = 1^3 + 1 + 1 = 3 \) - Possible \( y \) values are solutions to \( y^2 \equiv 3 \mod 5 \): None (no solutions, as 3 is not a quadratic residue mod 5) 3. **Evaluate for \( x = 2 \):** - \( y^2 = 2^3 + 2 + 1 = 11 \equiv 1 \mod 5 \) - Possible \( y \) values are \( y = 1, 4 \) 4. **Evaluate for \( x = 3 \):** - \( y^2 = 3^3 + 3 + 1 = 31 \equiv 1 \mod 5 \) - Possible \( y \) values are \( y = 1, 4 \) 5. **Evaluate for \( x = 4 \):** - \( y^2 = 4^3 + 4 + 1 = 69 \equiv 4 \mod 5 \) - Possible \( y \) values are solutions to \( y^2 \equiv 4 \mod 5 \): \( y = 2, 3 \) Thus, the points on the elliptic curve \( E \) over \( GF(5) \) include: - \( (0, 1
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