Can there exist three integers p, q and r with 2 ≤ p < q < r, that satisfy each of the following conditions?

(i) p^{2} -1 is divisible by each of q and r, and:

(ii) r^{2} -1 is divisible by each of p and q.

No, this is not possible.

Lemma) x divides y^{2} -1 only if x and y are relatively prime. This is because if they had any common factors, that common factor would divide y^{2} evenly, so it could not divide y^{2} -1 evenly.

Proof) Assume that p, q and r exists.

Since r^{2} -1 is divisible by q, q and r do not share a common factor.

Since q & r both divide p^{2} -1, and since they have no common factors, then q*r <= p^{2} -1, since they at most have all factors of p^{2} -1 accounted for between them, without any overlap. But q and r are both > p, so q*r >p*p > p^{2} -1, so a contradiction exists.

Therefore, p, q and r do not exist.