First up, if say $p=q$, then $p=q=r=2$, so assume all the primes are distinct (and odd), WLOG let $r$ be the largest. Then $r$ divides $(p-2)(p+2)$ and is also larger than $p-2$, so $r \mid p+2 \leqslant r+2$. Since $r$ is greater than two, $p$ must be equal to $r-2 > q-2$ (since the primes are distinct). We also have: $$r-2 = p \mid (q-2)(q+2) \Rightarrow r-2 \mid q+2.$$Therefore: $$q \mid (r-2)(r+2) \mid (q+2)(r+2),$$so $q \mid r+2$, since $q$ is odd. We then have $r-4 \leqslant q \leqslant r$. Since $p=r-2$ and $p,q,r$ are distinct, it must be the case that $q=r-4$. One of the three primes must then be divisble by three and hence equal to three. It is the smallest, or $q = 3, r=7, p=5$. Summing up, all the solutions are $\boxed{(2,2,2), (5,3,7), (7,5,3), (3,7,5)}$.

Fun fact: The olympiad was hosted (virtually) by Colombia. In Colombia parcero means 'mate' or 'bro'. In general is slang for friend.

What a cool problem, I'm glad I was able to solve it during the competition. An ordered triple $(p,q,r)$ of prime numbers is called parcera if $p$ divides $q^2-4$, $q$ divides $r^2-4$ and $r$ divides $p^2-4$. Find all parcera triples. First, note that it's cyclic. If $p=q \implies p|p^2-4, p|r^2-4 \implies p|4 \implies p=q=2 \implies 2|r^2-4 \implies 2|r^2 \implies r=2 \implies p=q=r=2 \implies$ we have $(p,q,r)=(2,2,2)$ and if $p,q,r>2$ they are all distinct primes. $p|q^2-4, q|r^2-4, r|p^2-4 \implies p|(q+2)(q-2), q|(r+2)(r-2), r|(p+2)(p-2) \implies$ since $p,q,r$ are distinct primes $p$ is a factor of $q+2$ or $q-2$, $q$ is a factor of $r+2$ or $r-2$, and $r$ is a factor of $p+2$ or $p-2$ $\implies p\leq q+2, q\leq r+2, r\leq p+2 \implies$ 1. $p-2\leq q \implies p-2\leq r+2 \implies p-4\leq r\leq p+2$ 2. $q-2\leq r \implies q-2\leq p+2 \implies q-4\leq p\leq q+2$ 3. $r-2\leq p \implies r-2\leq q+2 \implies r-4\leq q\leq r+2$ If $p,q,r>2 \implies$ $p,q,r$ are distinct primes and $p \equiv q\equiv r\equiv 1 \mod (2)$ $\implies$ 1. $p=q-4,q-2,q+2$ 2. $q=r-4,r-2,r+2$ 3. $r=p-4,p-2,p+2$ Without loss of generality let's assume that $p>q,r$ $\implies p=q+2$ and $r=p-2,p-4 \implies r=q,q-2$. Since $p,q,r$ are distinct primes $\implies r=q-2 \implies (p,q,r)=(q+2,q,q-2)$. Since for primes $x>3 \implies x\equiv 1,5 \mod (6) \implies$ if $p,q,r>3$ one of them would not be prime $\implies$ the smallest one (in this case $r$) equals 3 $\implies$ we have $(p,q,r)=(7,5,3)$ and its VALID combinations (moving the numbers to the right in a cycle). So we get $(p,q,r)=(2,2,2),(3,7,5),(5,3,7),(7,5,3)$ Q.E.D. I hope you like the solution

jbaca wrote: An ordered triple $(p, q, r)$ of prime numbers is called parcera if $p$ divides $q^2-4$, $q$ divides $r^2-4$ and $r$ divides $p^2-4$. Find all parcera triples. Obviously $(2,2,2)$ works;now WLOG $p \le q \le r$ since it is cyclic. By quadratic reciprocity,we get $$q \equiv \pm 2 \pmod p,r \equiv \pm 2 \pmod q,p \equiv \pm 2 \pmod r$$Since $p \ge q $ so $p=q-2$ and similarly $q=r-2$ therefore since $(3,5,7)$ is the only set of three twin primes:so $\boxed{(p,q,r)=(2,2,2),(3,7,5),(5,3,7),(7,5,3)}$

1.p=q=r then p=2=q=r 2.WLOG $p \geq q \geq r$ then $r=p-4,q=p-2,p$ assume that p-4 isn't 3 then p-4=1,2( mod 3) p-2=1 mod 3 p=0 mod 3 so contradicition so p-4=3 and we are done