smc

You may recognize that $r^2+s^2$ can be written as $(r+s{)}^2-2rs$. Then, by Vieta's formulas, $$r+s=-(-3)=3$$and $$rs=1.$$Therefore, plugging in the values for $r+s$ and $rs$, $$(r+s{)}^2-2rs=(3{)}^2-2(1)=9-2=7.$$Hence, we can say that the expression $r^2+s^2$ is $\boxed{\text{a positive integer}}$