There are integers $b,c$ for which both roots of the polynomial $x^2-x-1$ are also roots of the polynomial $x^5-bx-c$. Determine the product $bc$.
Let $r$ be a root of $x^2-x-1$. Then, rearranging, we have
$$r^2 = r+1.$$Multiplying both sides by $r$ and substituting gives
\begin{align*}
r^3 &= r^2+r \\
&= (r+1)+r \\
&= 2r+1.
\end{align*}Repeating this process twice more, we have
\begin{align*}
r^4 &= r(2r+1) \\
&= 2r^2+r \\
&= 2(r+1)+r \\
&= 3r+2
\end{align*}and
\begin{align*}
r^5 &= r(3r+2) \\
&= 3r^2+2r \\
&= 3(r+1)+2r \\
&= 5r+3.
\end{align*}Thus, each root of $x^2-x-1$ is also a root of $x^5-5x-3$, which gives $bc = 5\cdot 3 = \boxed{15}$.

(It is left to the reader to investigate why this answer is unique.)