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 exactlythe 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 needto 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, thenx=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, thenz = 1+ b₁-2¹ + b₂ · 2²+.... + b 2 = 2y + 1,= 2t-1.for some integer y, i.e., z is an odd number. Let m =

Given: x is a number in the set {0, 1, . . . , 2t} x written as: x = b0 · 20 + b1 · 21 + b2 · 22 + ·…

Answered: Show that if b0 = 0, then (g^xm)≡ 1…

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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).

Elements Of Modern Algebra

8th Edition

ISBN:9781285463230

Author:Gilbert, Linda, Jimmie

Publisher:Gilbert, Linda, Jimmie

Chapter3: Groups

Section3.3: Subgroups

Problem 21E: 21. Let Be the special linear group of order over .Find the inverse of each of the following...

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