Find all prime numbers $p, q, r$ such that $p$ divides $1+q^{r}$, $q$ divides $1+r^{p}$, and $r$ divides $1+p^{q}$.

## - Correction -
We start by examining the condition "$p$ divides $1+q^{r}$". It can be rewritten as $q^{r} \equiv -1 \pmod{p}$, which implies, in particular, $q^{2r} \equiv 1 \pmod{p}$. Thus, the order of $q$ modulo $p$ is a divisor of $2r$. Since $r$ is assumed to be prime, it is therefore an element of the set $\{1, 2, r, 2r\}$. If we further assume that $p \neq 2$, then $q^{r} \neq 1 \pmod{p}$, and thus the order of $q$ modulo $p$ must be 2 or $r$. In the first case, using the fact that $p$ is prime, we get $q \equiv -1 \pmod{p}$, while in the second case, we deduce that $2r$ divides $p-1$. By permuting the numbers $p, q$, and $r$, we naturally obtain analogous consequences from the other two conditions "$q$ divides $1+r^{p}$" and "$r$ divides $1+p^{q}$".

Now, suppose that $p, q$, and $r$ are all odd, and to begin with, that we are in the case where $q \equiv -1 \pmod{p}$. The prime number $p$ cannot therefore divide $q-1$ (since it already divides $q+1$ and is not 2). From the results of the first paragraph, the condition "$q$ divides $1+r^{p}$" implies that $r \equiv -1 \pmod{q}$. Applying the same argument again, we find that $p \equiv -1 \pmod{r}$. However, the three preceding congruences are not compatible. Indeed, for example, they imply $q \geq p-1$, $r \geq q-1$, and $p \geq r-1$, which cannot occur, given that $p, q$, and $r$ are odd primes, unless $p = q = r$; in which case, we clearly have $q \not\equiv -1 \pmod{p}$. We deduce that, still in the case where $p, q$, and $r$ are assumed to be odd, $2r$ divides $p-1$. By cyclically permuting the variables, we similarly prove that $2p$ divides $q-1$ and $2q$ divides $r-1$. Thus, $8pqr$ divides $(p-1)(q-1)(r-1)$, which is clearly not possible given that $8pqr > (p-1)(q-1)(r-1)$. Therefore, there is no solution when $p, q$, and $r$ are all odd.

We now consider the case where one of these three numbers is equal to 2. By cyclically permuting $p, q$, and $r$ again, we can assume that it is $p$. The conditions of the statement then say that $q$ is odd, that $r^{2} \equiv -1 \pmod{q}$, and that $2^{q} \equiv -1 \pmod{r}$. From what was done in the first paragraph, this last congruence implies that $r = 3$ or that $2q$ divides $r-1$. The first case leads to $9 \equiv -1 \pmod{q}$, which only occurs if $q = 5$ since we have already ruled out the case $q = 2$. We also verify that the triplet $(2, 5, 3)$ is indeed a solution. In the second case, now, the product $2q$ divides $r-1$, but also $2(r^{2} + 1)$ since we know that $r^{2} \equiv -1 \pmod{q}$. Thus, $2q$ divides $2(r^{2} + 1) - 2(r + 1)(r - 1) = 4$, which cannot happen.

In conclusion, there are exactly three solutions, which are the triplets $(2, 5, 3)$, $(5, 3, 2)$, and $(3, 2, 5)$.