Problem 1380

1. Given that $p, q,$ and $r$ are distinct positive prime numbers, we need to show that if
$$pqr \mid (pq)^r + (qr)^p + (rp)^q - 1,$$
then
$$(pqr)^3 \mid 3((pq)^r + (qr)^p + (rp)^q - 1).$$

2. First, let's analyze the given condition $pqr \mid (pq)^r + (qr)^p + (rp)^q - 1$. This means that
$$(pq)^r + (qr)^p + (rp)^q \equiv 1 \pmod{pqr}.$$

3. We need to show that
$$(pqr)^3 \mid 3((pq)^r + (qr)^p + (rp)^q - 1).$$

4. Let's consider the expression modulo $p$, $q$, and $r$ separately.

- Modulo $p$:
$$(pq)^r \equiv 0 \pmod{p}, \quad (qr)^p \equiv (qr)^p \pmod{p}, \quad (rp)^q \equiv 0 \pmod{p}.$$
Therefore,
$$(pq)^r + (qr)^p + (rp)^q \equiv (qr)^p \pmod{p}.$$
Since $p \mid (pq)^r + (qr)^p + (rp)^q - 1$, we have
$$(qr)^p \equiv 1 \pmod{p}.$$

- Modulo $q$:
$$(pq)^r \equiv (pq)^r \pmod{q}, \quad (qr)^p \equiv 0 \pmod{q}, \quad (rp)^q \equiv 0 \pmod{q}.$$
Therefore,
$$(pq)^r + (qr)^p + (rp)^q \equiv (pq)^r \pmod{q}.$$
Since $q \mid (pq)^r + (qr)^p + (rp)^q - 1$, we have
$$(pq)^r \equiv 1 \pmod{q}.$$

- Modulo $r$:
$$(pq)^r \equiv 0 \pmod{r}, \quad (qr)^p \equiv 0 \pmod{r}, \quad (rp)^q \equiv (rp)^q \pmod{r}.$$
Therefore,
$$(pq)^r + (qr)^p + (rp)^q \equiv (rp)^q \pmod{r}.$$
Since $r \mid (pq)^r + (qr)^p + (rp)^q - 1$, we have
$$(rp)^q \equiv 1 \pmod{r}.$$

5. Now, we need to show that
$$(pqr)^3 \mid 3((pq)^r + (qr)^p + (rp)^q - 1).$$

6. Since $pqr \mid (pq)^r + (qr)^p + (rp)^q - 1$, we can write
$$(pq)^r + (qr)^p + (rp)^q - 1 = kpqr$$
for some integer $k$.

7. Therefore,
$$3((pq)^r + (qr)^p + (rp)^q - 1) = 3kpqr.$$

8. We need to show that $(pqr)^3 \mid 3kpqr$. Since $p, q,$ and $r$ are distinct primes, $pqr$ is a product of distinct primes, and thus $(pqr)^3 \mid 3kpqr$ if and only if $pqr \mid 3k$.

9. Since $p, q,$ and $r$ are distinct primes, $pqr$ does not divide 3. Therefore, $k$ must be a multiple of $pqr$.

10. Hence, we have shown that
$$(pqr)^3 \mid 3((pq)^r + (qr)^p + (rp)^q - 1).$$

$\blacksquare$