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Chapter 4: Problem 48

Show that if \(n=p_{1} p_{2} \cdots p_{k},\) where \(p_{1}, p_{2}, \ldots, p_{k}\) are distinct primes that satisfy \(p_{j}-1 | n-1\) for \(j=1,2, \ldots, k\) then \(n\) is a Carmichael number.

Short Answer

Expert verified
If \(p_{j} - 1\) divides \(n-1\) for distinct primes \(p_{1}, p_{2}, ... , p_{k}\) forming \(n\), then \(n\) is a Carmichael number.

Step by step solution

01

Understand Carmichael Numbers

Recall that a Carmichael number is a composite number n which satisfies the modular arithmetic condition: for every integer a that is coprime with n, we have \(a^{n-1} \equiv 1 \pmod{n}\).
02

Decompose n

Given \(n = p_{1} p_{2} \cdots p_{k}\), where \(p_{1}, p_{2}, \ldots, p_{k}\) are distinct primes.
03

Use the Given Condition

Given that \(p_{j} - 1 | n - 1\) for \(j = 1, 2, \ldots, k\). This implies that: \(n-1\) is a multiple of \(p_{j} - 1\) for each j.
04

Apply Fermat鈥檚 Little Theorem

Fermat鈥檚 Little Theorem states that for any integer a such that \(\gcd(a, p_{j}) = 1\), we have \(a^{p_{j} - 1} \equiv 1 \pmod{p_{j}}\). Given that \(p_{j} - 1 | n - 1\), we can write \(n - 1 = k(p_{j} - 1)\) for some integer k.
05

Prove the Condition for a Carmichael Number

Since \(n - 1\) is a multiple of \(p_{j} - 1\), we get: \((a^{p_{j} - 1})^{k} \equiv 1^{k} \equiv 1 \pmod{p_{j}}\). This implies \(a^{n-1} \equiv 1 \pmod{p_{j}}\). Since this holds for each \(p_{j}\) and since the \(p_{j}\) are distinct primes, by the Chinese Remainder Theorem, \(a^{n-1} \equiv 1 \pmod{n}\).
06

Conclude the Proof

Therefore, for any integer \(a\) coprime with \(n\), the condition \(a^{n-1} \equiv 1 \pmod{n}\) is satisfied. Thus, \(n\) is a Carmichael number.

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Key Concepts

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

Carmichael numbers
First, let's understand what a Carmichael number is. It is a special type of composite number. A number n is called a Carmichael number if for every integer a that is coprime with n, the following congruence holds: \(a^{n-1} \equiv 1 \pmod{n}\). In simple terms, no matter which number a you pick, as long as it doesn't share any prime factors with n, raising a to the power of n-1 and then taking the remainder when divided by n will always yield 1. These numbers can sometimes fool Fermat's Little Theorem into thinking they are primes, even though they are not.
Fermat's Little Theorem
Fermat's Little Theorem is a fundamental principle in number theory. It states that for any integer a, if p is a prime number and a is not divisible by p, then \(a^{p-1} \equiv 1 \pmod{p}\). This theorem is often used to test if a number is prime. If you pick a number a (which is not divisible by p) and raise it to the power of p-1, the result, after taking the remainder when divided by p, should be 1. In our proof, by knowing that \(p_j - 1\) divides \(n - 1\), we can extend Fermat's Little Theorem to the Carmichael number setting.
Chinese Remainder Theorem
The Chinese Remainder Theorem (CRT) is a key concept in modular arithmetic. It states that if you have several congruences with pairwise coprime moduli, you can find a unique solution modulo the product of these moduli. In our proof, since the n in question is the product of distinct primes \(p_1, p_2,..., p_k\), and we have shown that \(a^{n-1} \equiv 1 \pmod{p_j}\) for each j, the CRT allows us to conclude that \(a^{n-1} \equiv 1 \pmod{n}\) under the combined modulus.
Modular arithmetic
Modular arithmetic is a system of arithmetic for integers, where numbers wrap around after reaching a certain value called the modulus. In modular arithmetic, we write equations like \(a \equiv b \pmod{n}\), which means that when a and b are divided by n, they give the same remainder. This concept is crucial in cryptography and helps us simplify large number calculations. In the context of our proof, all of our congruences are modulo n, and understanding modular arithmetic helps us follow the steps clearly, especially when we apply the Chinese Remainder Theorem.

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Most popular questions from this chapter

The Vigen猫re cipher is a block cipher, with a key that is a string of letters with numerical equivalents \(k_{1} k_{2} \ldots k_{m},\) where \(k_{i} \in \mathbf{Z}_{26}\) for \(i=1,2, \ldots, m .\) Suppose that the numerical equivalents of the letters of a plaintext block are \(p_{1} p_{2} \ldots p_{m} .\) The corresponding numerical ciphertext block is \(\left(p_{1}+k_{1}\right)\) mod 26 \(\left(p_{2}+k_{2}\right) \bmod 26 \ldots\left(p_{m}+k_{m}\right)\) mod \(26 .\) Finally, we translate back to letters. For example, suppose that the key string is RED, with numerical equivalents \(1743 .\) Then, the plaintext ORANGE, with numerical equivalents \(141700130604,\) is encrypted by first splitting it into two blocks 141700 and 13 \(0604 .\) Then, in each block we shift the first letter by 17 , the second by \(4,\) and the third by \(3 .\) We obtain 52103 and 0410 \(07 .\) The cipherext is FVDEKH. The ciphertext OIKYWVHBX was produced by encrypting a plaintext message using the Vigen猫re cipher with key HOT. What is the plaintext message?

Decrypt these messages encrypted using the shift cipher \(f(p)=(p+10) \bmod 26 .\) a) CEBBOXNOB XYG b) LO WIPBSOXN c) DSWO PYB PEX

In Exercises \(24-27\) first express your answers without com- puting modular exponentiations. Then use a computational aid to complete these computations. Encrypt the message ATTACK using the RSA system with \(n=43.59\) and \(e=13\) , translating each letter into integers and grouping together pairs of integers, as done in Example 8 .

Describe the steps that Alice and Bob follow when they use the Diffie-Hellman key exchange protocol to generate a shared key. Assume that they use the prime \(p=101\) and take \(a=2,\) which is a primitive root of \(101,\) and that Alice selects \(k_{1}=7\) and \(\mathrm{Bob}\) selects \(k_{2}=9 .\) (You may want to use some computational aid.)

Show that if \(p\) is an odd prime and \(a\) is an integer not divisible by \(p\) , then the congruence \(x^{2} \equiv a(\bmod p)\) has either no solutions or exactly two incongruent solutions modulo \(p .\)
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