HMMT February

Let $t_{n}$ be the larger real such that $a_{n}=t_{n}+\frac{1}{t_{n}}$. Then $t_{1}=\frac{3+\sqrt{5}}{2}$. We claim that $t_{n}=2 t_{n-1}$. Writing the recurrence as a quadratic polynomial in $a_{n}$, we have: $$4 a_{n}^{2}-10 a_{n-1} a_{n}+4 a_{n-1}^{2}+9=0$$ Using the quadratic formula, we see that $a_{n}=\frac{5}{4} a_{n-1}+\frac{3}{4} \sqrt{a_{n-1}^{2}-4}$. (We ignore the negative square root, since $a_{n}$ is the largest real number satisfying the polynomial.) Substituting $t_{n-1}+\frac{1}{t_{n-1}}$ for $a_{n-1}$, we see that $\sqrt{a_{n-1}^{2}-4}=\sqrt{t_{n-1}^{2}-2+\frac{1}{t_{n-1}^{2}}}$, so we have: $$a_{n}=\frac{5}{4}\left(t_{n-1}+\frac{1}{t_{n-1}}\right)+\frac{3}{4} \sqrt{\left(t_{n-1}-\frac{1}{t_{n-1}}\right)^{2}}=2 t_{n-1}+\frac{1}{2 t_{n-1}}$$ so $t_{n}=2 t_{n-1}$, as claimed. Then $a_{8}=\frac{128(3+\sqrt{5})}{2}+\frac{2}{128(3+\sqrt{5})}$. The second term is vanishingly small, so $\left\lfloor a_{8}\right\rfloor=\lfloor 64(3+\sqrt{5})\rfloor$. We approximate $\sqrt{5}$ to two decimal places as 2.24, making this expression $\lfloor 335.36\rfloor=335$. Since our value of $\sqrt{5}$ is correct to within 0.005, the decimal is correct to within 0.32, which means the final answer is exact.