SAUDI ARABIAN MATHEMATICAL COMPETITIONS

Since $r_1,\dots ,r_5$ 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, $P(i)P(-i)=(2+i)(2-i)=4-i^2=5$.