Suppose $n=p_1p_2\cdots p_k$ be a product of distinct primes. Given $p_i-1\mid n-1$ for $i=1,2,\ldots,k$, show that $n$ is a Carmichael number.

Question

Suppose $n=p_1p_2\cdots p_k$ be a product of distinct primes. Given $p_i-1\mid n-1$ for $i=1,2,\ldots,k$, show that $n$ is a Carmichael number.

This is a question on a past exam that I find difficulty to answer. Can anyone help?

Answer 1

Hint: Let $a$ and $n$ be relatively prime. Then $p_i$ does not divide $a$. Thus by Fermat's Theorem we have $a^{p_i-1}\equiv 1\pmod{p_i}$, and therefore $a^{n-1}\equiv 1\pmod{p_i}$.

Answer 2

Let, $c_i = \frac{n-1}{p_i-1}$. $c_i$ is an integer $\Leftarrow$ $p_i-1\mid n-1, $ for $i=1,2,…,k$.

$a^{c_i*(p_i-1)}\equiv 1\pmod{p_i} \Leftarrow$ Fermat's little theorem

$\Rightarrow a^{n-1}\equiv 1\pmod{p_i}$ ------- (1)

(1) is true for $i = 1, 2, 3, ... k$. Therefore, it can be written as below,

$a^{(n-1)}\equiv 1\pmod{n}$ ------- (2)

Here, $n (= p_1p_2p_3 ... p_k)$ is a composite number.

Hence, according to (2), $n$ is a Carmichael number.