Proof of statement with modular arithmetics

Question

I have the following question:

Let $p$ and $q$ are distinct prime numbers. Let $a \in \mathbb{N}$ such that $p \nmid (a^{pq}-1)$ and $q \nmid (a^{pq}-1)$. Prove that $\exists$ prime number $r$ such that $r \mid (a^{pq} - 1)$ and $pq \mid (r-1)$.

My steps to solve it:

1. Firstly, I understood that the first statement is definitely true: if $p,q \nmid (a^{pq}-1)$ then some prime $r \neq p, r \neq q,r \neq p \cdot q$ definitely exists as a prime factor of $(a^{pq}-1)$.

2. Secondly, by little Fermat theorem we get $a^{r-1} \equiv 1~(\textrm{mod}~r)$ (if $gcd(a,r) = 1$ and we don't know this thing also). From the 1) we get $a^{pq} \equiv 1~(\textrm{mod}~r)$. It means that $ord(a) \mid (r-1)$. So $p$ and $q$ could be factors of $r-1$. There are three possible variants: $p \mid (r-1)$ or $q \mid (r-1)$ or $pq \mid (r-1)$.

Here I am stuck. I don't know how to prove that exists $r$ for which $pq \mid (r-1)$.

Answer 1

Note that Zsigmondy's theorem could be used to prove this. Instead, we can continue with your idea of using the multiplicative order. For each prime $r$ where $r \mid a^{pq}-1$, and their multiplicative orders $m = \operatorname{ord}_r(a)$, we have

$$m \mid pq \;\;\to\;\; m \in \{1, p, q, pq\}$$

With $m = 1$, consider

$$a^{pq}-1=(a-1)(a^{pq-1}+a^{pq-2}+\ldots+a+1)$$

Since $r \mid a - 1 \;\to\; a \equiv 1 \pmod{r}$, then

$$a^{pq - 1} + a^{pq - 2} + \ldots + a + 1 \equiv 1 + 1 + \ldots + 1 + 1 \equiv pq \pmod{r}$$

but $r \neq p$ and $r \neq q$, so

$$r \nmid a^{pq - 1} + a^{pq - 2} + \ldots + a + 1$$

Thus, the product of all primes $r$ (including multiplicities), where $r \mid a^{pq} - 1$ with $m = 1$, divides $a - 1$ (note the product is actually $a - 1$). Next, with $m = p$, use

$$(a^{p})^q - 1 = (a^{p} - 1)\left((a^{p})^{q - 1} + (a^{p})^{q - 2} + \ldots + a^{p} + 1 \right)$$

Since $r \mid a^{p}-1 \;\to\; a^{p} \equiv 1 \pmod{r}$, then

$$(a^{p})^{q - 1} + (a^{p})^{q - 2} + \ldots + a^{p} + 1 \equiv 1 + 1 + \ldots + 1 + 1 \equiv q \pmod{r}$$

but $r \neq q$, so

$$r \nmid (a^{p})^{q - 1} + (a^{p})^{q - 2} + \ldots + a^{p} + 1$$

This means that the product of all primes $r$ (including multiplicities), with $m = p$, divides $a^{p} - 1$. Similarly, for all primes $r$ where $m = q$, we get

$$r \nmid (a^{q})^{p - 1} + (a^{q})^{p - 2} + \ldots + a^{p} + 1$$

Since $a - 1 \mid a^p - 1$, and the primes $r$ where $m = 1$ and those where $m = p$ are distinct (because $p \neq 1$), then the product of all primes $r$, including multiplicities, with $m = 1$ or $m = p$, divide $a^p - 1$. Also, the product of all primes $r$, including multiplicities, with $m = q$, divide $a^q - 1$. Thus, the product of all the primes $r$ (including multiple instances of them) where $r \mid a^{pq} - 1$ with $m \in \{1, p, q\}$ can at the most be $(a^p - 1)(a^q - 1)$. However, with $p$ and $q$ being distinct primes, we have $pq \gt p + q$ so, since $a \gt 1$, then

$$\begin{equation}\begin{aligned} (a^p - 1)(a^q - 1) &= a^{p + q} - a^p - a^q + 1 \\ & \lt a^{p+q} - 1 \\ & \lt a^{pq} - 1 \end{aligned}\end{equation}$$

Thus, there must be at least one prime $r$ where $m = pq$. With (as you've already indicated) $m \mid r - 1$, we therefore have

$$pq \mid r - 1$$