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Show that if b0 = 0, then (g^xm)≡ 1 (mod p). Q2: Show that if b0 = 1, then (g^xm) ≡ p − 1 (mod p).

Show that if b0 = 0, then (g^xm)≡ 1 (mod p). Q2: Show that if b0 = 1, then (g^xm) ≡ p − 1 (mod p).

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Q1:Show that if b0 = 0, then (g^xm)≡ 1 (mod p).
Q2: Show that if b0 = 1, then (g^xm) ≡ p − 1 (mod p). 

The aim of this question is to show that there are some groups in which the discrete logarithm problem (DLP) is easy. In this example, we will consider the multiplicative group G whose elements are exactly the set Z where p is a prime and the multiplication operation is multiplication modulo p. In particular, p = 2t + 1 for some positive integer t > 2. The number of elements in Z, i.e., the order of the group, is 2t. Recall that under DLP, we are given g and h such that g = h (mod p) for some unknown x, and we need to find z. We will assume that g is a generator of this group. As an example, you may consider p = 28 +1 = 257. Then g = 3 is a generator of this group. (Hint: It might
Transcribed Image Text:The aim of this question is to show that there are some groups in which the discrete logarithm problem (DLP) is easy. In this example, we will consider the multiplicative group G whose elements are exactly the set Z where p is a prime and the multiplication operation is multiplication modulo p. In particular, p = 2t + 1 for some positive integer t > 2. The number of elements in Z, i.e., the order of the group, is 2t. Recall that under DLP, we are given g and h such that g = h (mod p) for some unknown x, and we need to find z. We will assume that g is a generator of this group. As an example, you may consider p = 28 +1 = 257. Then g = 3 is a generator of this group. (Hint: It might
Since z is a number in the set {0, 1,..., 2}, we can write z in binary as: x=bo 20 + b₁-2¹ + b₂ 2²+... + b₂ -2¹, where b; are bits. If bo= 0, then x=b₁-2¹ + b₂-2² + + b₁-2¹ = 2y, for some integer y, i.e., z is an even number. On the other hand, if bo = 1, then z = 1+ b₁-2¹ + b₂ · 2²+.... + b 2 = 2y + 1, = 2t-1. for some integer y, i.e., z is an odd number. Let m =
Transcribed Image Text:Since z is a number in the set {0, 1,..., 2}, we can write z in binary as: x=bo 20 + b₁-2¹ + b₂ 2²+... + b₂ -2¹, where b; are bits. If bo= 0, then x=b₁-2¹ + b₂-2² + + b₁-2¹ = 2y, for some integer y, i.e., z is an even number. On the other hand, if bo = 1, then z = 1+ b₁-2¹ + b₂ · 2²+.... + b 2 = 2y + 1, = 2t-1. for some integer y, i.e., z is an odd number. Let m =
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