31.8 Primality testing
31.8-1
Prove that if an odd integer \(n > 1\) is not a prime or a prime power, then there exists a nontrivial square root of \(1\) modulo \(n\).
(Omit!)
31.8-2 \(\star\)
It is possible to strengthen Euler's theorem slightly to the form
\(a^{\lambda(n)} \equiv 1 \pmod n\) for all \(a \in \mathbb Z_n^\*\),
where \(n = p_1^{e_1} \cdots p_r^{e_r}\) and \(\lambda(n)\) is defined by
\[\lambda(n) = \text{lcm}(\phi(p_1^{e_1}), \ldots, \phi(p_r^{e_r})). \tag{31.42}\]Prove that \(\lambda(n) \mid \phi(n)\). A composite number \(n\) is a Carmichael number if \(\lambda(n) \mid n - 1\). The smallest Carmichael number is \(561 = 3 \cdot 11 \cdot 17\); here, \(\lambda(n) = \text{lcm}(2, 10, 16) = 80\), which divides \(560\). Prove that Carmichael numbers must be both "square-free" (not divisible by the square of any prime) and the product of at least three primes. (For this reason, they are not very common.)
-
Prove that \(\lambda(n) \mid \phi(n)\).
We have
\[ \begin{aligned} n & = p_1^{e_1} \cdots p_r^{e_r} \\\\ \phi(n) & = \phi(p_1^{e_1}) \times \dots \times \phi(p_r^{e_r}). \end{aligned} \]Thus,
\[ \begin{aligned} \lambda(n) & \mid \phi(n) \\\\ \Rightarrow \text{lcm}(\phi(p_1^{e_1}, \ldots, \phi(p_r^{e_r})) & \mid (\phi(p_1^{e_1}) \times \dots \times \phi(p_r^{e_r})) \end{aligned} \] -
Prove that Carmichael numbers must be "square-free" (not divisible by the square of any prime).
Assume \(n = p^\alpha m\) is a Carmichael number, where \(\alpha \ge 2\) and \(p \nmid m\).
By the Chinese Remainder Theorem, the system of congruences
\[ \begin{aligned} x & \equiv 1 + p \pmod p^\alpha, \\\\ x & \equiv 1 \pmod m \end{aligned} \]has a solution \(a\). Note that \(\gcd(a, n) = 1\).
Since \(n\) is a Carmichael number, we have \(a^{n - 1} \equiv 1 \pmod n\). In particular, \(a^{n - 1} \equiv 1 \pmod {p^\alpha}\); therefore, \(a^n \equiv a \pmod {p^2}\).
So, \((1 + p)^n \equiv 1 + p \pmod {p^2}\). Expand \((1 + p)^n\) modulo \(p^2\) using the binomial theorem. We have
\[(1 + p)^n \equiv 1 \pmod {p^2},\]since the first two terms of the expansion are \(1\) and \(np\), and the rest of the terms are divisible by \(p^2\).
Thus, \(1 \equiv 1 + p \pmod {p^2}\). This is impossible.
-
Prove that Carmichael numbers must be the product of at least three primes.
Assume that \(n = pq\) is a Carmichael number, where \(p\) and \(q\) are distant primes and \(p < q\).
Then we have
\[ \begin{aligned} & q \equiv 1 \pmod{q - 1} \\\\ \Rightarrow & n \equiv pq \equiv p \pmod{q - 1} \\\\ \Rightarrow & n - 1 \equiv p - 1 \pmod{q - 1}, \end{aligned} \]Here, \(0 < p − 1 < q − 1\) , so \(n − 1\) is not divisible by \(q − 1\).
31.8-3
Prove that if \(x\) is a nontrivial square root of \(1\), modulo \(n\), then \(\gcd(x - 1, n)\) and \(\gcd(x + 1, n)\) are both nontrivial divisors of \(n\).
\(n \mid (x + 1)(x - 1)\), suppose \(\gcd(x - 1, n) = 1\), then \(n \mid (x + 1)\), then \(x \equiv -1 \pmod n\) which is trivial, it contradicts the fact that \(x\) is nontrivial, therefore \(\gcd(x - 1, n) \ne 1\), \(\gcd(x + 1, n) \ne 1\).
本页面的全部内容在 小熊老师 - 莆田青少年编程俱乐部 0594codes.cn 协议之条款下提供,附加条款亦可能应用