jmc

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 $$\begin{array}{rl}{r}^3& =r^2+r\\ & =(r+1)+r\\ & =2r+1.\end{array}$$Repeating this process twice more, we have $$\begin{array}{rl}{r}^4& =r(2r+1)\\ & =2r^2+r\\ & =2(r+1)+r\\ & =3r+2\end{array}$$and $$\begin{array}{rl}{r}^5& =r(3r+2)\\ & =3r^2+2r\\ & =3(r+1)+2r\\ & =5r+3.\end{array}$$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.)