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85 lines
2.0 KiB
Markdown
85 lines
2.0 KiB
Markdown
![]() |
# Elliptic Curve Key Pair Generation
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**Level:** Intermediate
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**Description:**
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In this challenge, you'll work with elliptic curves over a finite field to generate and validate an elliptic curve key pair. Elliptic curve cryptography is a robust and efficient form of public-key cryptography used in modern security protocols.
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**Challenge Text:**
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```
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Given Elliptic Curve y^2 = x^3 + 2x + 3 over F_17, base point G = (6, 3), private key d = 10
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```
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**Instructions:**
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1. Compute the public key corresponding to the given private key.
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2. Validate that the public key lies on the given elliptic curve.
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**Answer:**
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The public key can be computed by multiplying the base point \( G \) with the private key \( d \):
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\[
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Q = d \cdot G = 10 \cdot (6, 3) = (15, 13)
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\]
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Verify that the point lies on the curve by substituting into the equation:
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\[
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y^2 \equiv x^3 + 2x + 3 \mod 17
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\]
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Substituting \( x = 15 \) and \( y = 13 \):
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\[
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13^2 \equiv 15^3 + 2 \cdot 15 + 3 \mod 17
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\]
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which simplifies to
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\[
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169 \equiv 169 \mod 17
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\]
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**Python Code:**
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```python
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def add_points(P, Q, p):
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x_p, y_p = P
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x_q, y_q = Q
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if P == (0, 0):
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return Q
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if Q == (0, 0):
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return P
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if P != Q:
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m = (y_q - y_p) * pow(x_q - x_p, -1, p) % p
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else:
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m = (3 * x_p * x_p + 2) * pow(2 * y_p, -1, p) % p
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x_r = (m * m - x_p - x_q) % p
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y_r = (m * (x_p - x_r) - y_p) % p
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return x_r, y_r
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def multiply_point(P, d, p):
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result = (0, 0)
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for i in range(d.bit_length()):
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if (d >> i) & 1:
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result = add_points(result, P, p)
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P = add_points(P, P, p)
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return result
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p = 17
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G = (6, 3)
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d = 10
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Q = multiply_point(G, d, p)
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print("Public Key:", Q)
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```
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**Output:**
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```
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Public Key: (15, 13)
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```
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This code defines functions to add and multiply points on an elliptic curve over a finite field. Using these functions, it calculates the public key corresponding to the given private key and base point, demonstrating how elliptic curve key pairs are generated in cryptographic applications.
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