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Chapter 8: Problem 18

Suppose that RSA is used to send a message \(m\) to three recipients, who have relatively prime encryption moduli \(n_{1}, n_{2}\), and \(n_{3} .\) All three recipients use the same encryption exponent \(e=3\), a once-popular choice as it makes encryption very fast. Show that someone who intercepts all three encrypted messages \(c_{1}=m^{3}\) \(\bmod n_{1}, c_{2}=m^{3} \bmod n_{2}\), and \(c_{3}=m^{3} \bmod n_{1}\) can efficiently decipher \(m .\) Hint: The Chinese remainder theorem implies that you can efficiently find a \(c\) such that \(c=c_{1} \bmod n_{1}, c=c_{2} \bmod n_{2}\), and \(c=c_{3} \bmod n_{3} .\) Assume this, and show that it implies \(c=m^{3} \bmod n_{1} n_{2} n_{3} .\) Then note \(m^{3}

Short Answer

Expert verified
Use CRT to solve \(c = m^3 \mod (n_1 \cdot n_2 \cdot n_3)\) and find \(m\) by taking the cube root of \(c\).

Step by step solution

01

Information Setup

Identify the given information. Encryption exponent is given as \(e=3\). We have three encryption moduli \(n_1, n_2, n_3\), all of which are relatively prime. The intercepted encrypted messages are \(c_1=m^3 \mod n_1\), \(c_2=m^3 \mod n_2\), and \(c_3=m^3 \mod n_3\).
02

Use Chinese Remainder Theorem (CRT)

According to the hint, use the Chinese Remainder Theorem to find \(c\) such that it satisfies: \(c \equiv c_1 \mod n_1\), \(c \equiv c_2 \mod n_2\), and \(c \equiv c_3 \mod n_3\). This theorem allows us to efficiently combine the congruences because the moduli \(n_1, n_2, n_3\) are relatively prime.
03

Combine Congruences

Using CRT, determine \(c\) such that: the value of \(c\) will satisfy: \[ c \equiv c_1 \mod n_1,\]\[ c \equiv c_2 \mod n_2,\]\[ c \equiv c_3 \mod n_3\]This combination can be found efficiently and uniquely modulo \(n_1 \cdot n_2 \cdot n_3\). Hence, we get \(c = m^3 \mod (n_1 \cdot n_2 \cdot n_3)\).
04

Inequality Condition

Note that since \(m^3 < n_1 \cdot n_2 \cdot n_3\), it implies that \(m^3\) is a positive integer that is less than the product of the three moduli.
05

Cube Root to Find Original Message

Since \(m^3 < n_1 \cdot n_2 \cdot n_3\), solving for \(m\) can be done by taking the integer cube root of \(c\) to retrieve the original message, i.e., \(m = \sqrt[3]{c}\). This step involves efficient computation since \(c\) is already known from the previous steps.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chinese Remainder Theorem
The Chinese Remainder Theorem (CRT) is a powerful tool in number theory. It helps to solve system of simultaneous congruences with different moduli. To understand CRT better, consider three numbers: n1, n2, and n3, which are pairwise relatively prime - meaning each pair of numbers share no common divisor other than 1. In RSA cryptography, instead of working modulo one large number, you can split the work across these smaller moduli using CRT for efficiency.
For our problem, we have three moduli n1, n2, and n3 and corresponding ciphertexts:
  • c1 = m^3 mod n1
  • c2 = m^3 mod n2
  • c3 = m^3 mod n3
By applying CRT, you can find a unique integer 'c' such that:
c 鈮� c1 (mod n1)
c 鈮� c2 (mod n2)
c 鈮� c3 (mod n3)
This 'c' helps to combine the modular equations into a single expression efficiently. We end up with c = m^3 mod (n1 * n2 * n3). CRT is critical in simplifying problems involving multiple moduli.
Encryption Moduli
In RSA cryptography, the encryption modulus is a crucial part. An encryption modulus 'n' is typically a product of two large prime numbers, p and q. Each RSA user has their own 'n', which must be kept secret. In this exercise, we deal with multiple recipients having their individual encryption moduli: n1, n2, and n3.
Key aspects of working with encryption moduli include:
  • They must be large enough to secure the encryption.
  • Each pair of moduli in our problem is relatively prime, ensuring CRT's effectiveness.
  • The exponent used (e=3 in this case) also factors into the encryption process.
These moduli determine the group in which operations are conducted. In our case, instead of one large number, the product of three individual moduli is taken, making decryption feasible using CRT.
Encryption moduli ensure that the encryption process is computationally hard to break unless the decryption key is known or special properties, like in this exercise, are leveraged.
Cubing and Cube Roots
In RSA encryption, the message 'm' is usually raised to the power of an encryption exponent 'e'. Here, e=3, meaning m^3. This cubing process makes encryption quick, thanks to its low exponent. However, decryption generally requires knowledge of the private key.
In this exercise, since the intercepted messages are already in cubed form modulo different primes, we use CRT to combine them efficiently. Then we have:
c = m^3 mod (n1 * n2 * n3)
Given the condition that m^3 is less than the product of these moduli (n1 * n2 * n3), finding 'm' is straightforward. We calculate the cube root of 'c'.
The steps can be summarized as:
  • Intercept messages c1, c2, and c3.
  • Use CRT to combine into c = m^3 mod (n1 * n2 * n3).
  • Confirm m^3 < n1 * n2 * n3.
  • Compute the cube root to get 'm'.
Cube roots bring back the original message 'm' from its encrypted form efficiently, rounding off the decryption process.

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Most popular questions from this chapter

Suppose a firewall is configured to allow outbound TCP connections but inbound connections only to specified ports. The FTP protocol now presents a problem: When an inside client contacts an outside server, the outbound TCP control connection can be opened normally but the TCP data connection traditionally is inbound. (a) Look up the FTP protocol in, for example, Request for Comments 959 . Find out how the PORT command works. Discuss how the client might be written so as to limit the number of ports to which the firewall must grant inbound access. Can the number of such ports be limited to one? (b) Find out how the FTP PASV command can be used to solve this firewall problem.

Estimate the probabilities of finding two messages with the same MD5 checksum, given total numbers of messages of \(2^{63}, 2^{64}\), and \(2^{65}\). Hint: This is the birthday problem again, as in Exercise 49 of Chapter 2, and again the probability that the \(k+1\) th message has a different checksum from each of the preceding \(k\) is \(1-k / 2^{128}\). However, the approximation in the hint there for simplifying the product fails rather badly now. So, instead, take the log of each side and use the approximation \(\log \left(1-k / 2^{128}\right) \approx-k / 2^{128}\).

Suppose two people want to play poker over the network. To "deal" the cards they need a mechanism for fairly choosing a random number \(x\) between them; each party stands to lose if the other party can unfairly influence the choice of \(x\). Describe such a mechanism. Hint: You may assume that if either of two bit strings \(x_{1}\) and \(x_{2}\) are random, then the exclusive-OR \(x=x_{1} \oplus x_{2}\) is random.

Suppose you are doing RSA encryption with \(p=101, q=113\), and \(e=3 .\) (a) Find the decryption exponent \(d\). (Hint: Although there are methodical ways to do this, trial and error is efficient for \(e=3 .\) ) (b) Encrypt the message \(m=9876\). Note that evaluating \(m^{3}\) with 32 -bit arithmetic results in overflow.

Prove that the RSA decryption algorithm recovers the original message; that is, \(m^{e d} \equiv m \bmod p q .\) Hint: You may assume that, because \(p\) and \(q\) are relatively prime, it suffices to prove the congruence \(\bmod p\) and \(\bmod q\).
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