Determine all possible triplet(s) (P, Q, R) of nonnegative integers, with P ≥ Q, such that each of P^{2} + 1 and Q^{2} + 1 is a prime, and: (P^{2} + 1)(Q^{2} + 1) = (R^{2} + 1).

Without the prime restriction, there are an infinite number of solutions with P=Q+1, R=Q^2+Q+1. But the only consecutive primes of the form X^2+1 are 2 and 5, from Q=1.

If there are other solutions in general then PQ+P/Q > R > PQ+Q/P. This implies if P and Q are not consecutive integers then P > 2Q. Also PQ+1 <= R <= PQ+P-Q.