HMMT November

Let $L_{n}$ denote the Lucas numbers given by $L_{0}=2, L_{1}=1$, and $L_{n+2}=L_{n+1}+L_{n}$. Note that $L_{n}^{2}-2=L_{2 n}$ when $n$ is even (one can show this by induction, or explicitly using $L_{n}=$ $\left.\left(\frac{1+\sqrt{5}}{2}\right)^{n}+\left(\frac{1-\sqrt{5}}{2}\right)^{n}\right)$.So, $f^{24}\left(L_{6}\right)=L_{3 \cdot 2^{25}}$. Now note that since $89 \equiv 4(\bmod 5)$, we have $5^{\frac{p-1}{2}} \equiv 1(\bmod 89)$ so $L_{89}=\left(\frac{1+\sqrt{5}}{2}\right)^{p}+\left(\frac{1-\sqrt{5}}{2}\right)^{p} \equiv L_{1}$ $(\bmod 89)$ and similarly $L_{90} \equiv L_{2}$, so the sequence $L_{n}(\bmod 89)$ is periodic with period 88. (Alternatively, reason by analog of Fermat's little theorem, since we can substitute an integer residue for $\sqrt{5}$.) We have $3 \cdot 2^{25} \equiv 3 \cdot 2^{5} \equiv 8(\bmod 11)$ and $\equiv 0(\bmod 8)$, so $L_{3 \cdot 2^{25}} \equiv L_{8}(\bmod 89)$. Computing $L_{8}=47$ gives the answer.