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Chapter 1: Q24E (page 49)

If p is prime, how many elements of{0,1,...pn-1} have an inverse modulopn ?

Short Answer

Expert verified

The total number of inverses isPn–1(P-1). .

Step by step solution

01

Introduction

The modular inverse of AmodCis the B value that makesA*BmodC=1 . The modular multiplicative inverse can be defined as theGCD(a,b) must be equal to 1 and‘a’and‘b’ are relatively prime.

02

Inverse modulo of pn

Given elements are, 0,1,...pn-1.

If is a prime, then for all given elements i.e. 0,1,...pn-1, it is not the multiple of p and it contains the inverse of pn.

We have given that the range of elements is0,1,...pn-1

Now,pn-1 is the only element that get divisible bypn .

So, total number of inverses are,

Pn–Pn–1=Pn–1(P-1)

Therefore, total number of inverse isPn–1(P-1) .

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Most popular questions from this chapter

Alice and her three friends are all users of the RSA cryptosystem. Her friends have public keys (Ni,ei=3),i=1,2,3 where as always,Ni=piqi for randomly chosen n-bit primes piqi. Showthat if Alice sends the same n-bit message M (encrypted using RSA) to each of her friends, then anyone who intercepts all three encrypted messages will be able to efficiently recover M.
(Hint: It helps to have solved problem 1.37 first.)

In the RSA cryptosystem, Alice’s public key (N,e)is available to everyone. Suppose that her private key d is compromised and becomes known to Eve. Show that e=3if (a common choice) then Eve can efficiently factor N.

The grade-school algorithm for multiplying two n-bit binary numbers x and y consist of addingtogethern copies of r, each appropriately left-shifted. Each copy, when shifted, is at most 2n bits long.
In this problem, we will examine a scheme for adding n binary numbers, each m bits long, using a circuit or a parallel architecture. The main parameter of interest in this question is therefore the depth of the circuit or the longest path from the input to the output of the circuit. This determines the total time taken for computing the function.
To add two m-bit binary numbers naively, we must wait for the carry bit from position i-1before we can figure out the ith bit of the answer. This leads to a circuit of depthΟ(m). However, carry-lookahead circuits (seewikipedia.comif you want to know more about this) can add inΟ(logn)depth.

  1. Assuming you have carry-lookahead circuits for addition, show how to add n numbers eachm bits long using a circuit of depth Ο(lognlogm).
  2. When adding three m-bit binary numbers x+y+z, there is a trick we can use to parallelize the process. Instead of carrying out the addition completely, we can re-express the result as the sum of just two binary numbersr+s, such that the ith bits of r and s can be computedindependently of the other bits. Show how this can be done. (Hint: One of the numbers represents carry bits.)
  3. Show how to use the trick from the previous part to design a circuit of depthΟ(logn)for multiplying two n-bit numbers.

Digital signatures, continued.Consider the signature scheme of Exercise 1.45.

(a) Signing involves decryption, and is therefore risky. Show that if Bob agrees to sign anything he is asked to, Eve can take advantage of this and decrypt any message sent by Alice to Bob.

(b) Suppose that Bob is more careful, and refuses to sign messages if their signatures look suspiciously like text. (We assume that a randomly chosen message−that is, a random number in the range{1,...,N-1} is very unlikely to look like text.) Describe a way in which Eve can nevertheless still decrypt messages from Alice to Bob, by getting Bob to sign messages whose signatures look random.

Determine necessary and sufficient conditions on xandc so that the following holds: for anya,b, if ax≡bxmodc, thena≡bmodc .
See all solutions

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