Saudi Arabian Mathematical Competitions

Since $r_{1}, \ldots, r_{5}$ are the roots of $P(x) = x^{5} - x^{2} + 1$, by factorization theorem we have
$$P(x) = \prod_{i=1}^{5} (x - r_{i}) .$$
It follows that
$$\prod_{j=1}^{5} Q(r_{j}) = \prod_{j=1}^{5} (r_{j}^{2} + 1) = \prod_{j=1}^{5} (r_{j} + i) \prod_{j=1}^{5} (r_{j} - i) = P(i) P(-i),$$
where $i^{2} = -1$. This gives $P(i) = i^{5} - i^{2} + 1 = -i + 1 + 1 = 2 - i$ and $P(-i) = (-i)^{5} - (-i)^{2} + 1 = 2 + i$. Hence, $P(i) P(-i) = (2 + i)(2 - i) = 4 - i^{2} = 5$.