Harvard-MIT Mathematics Tournament

Solution:

Answer: $4$ We claim that $n = 4$ is the answer. First, we show that $n > 3$. Suppose that $n \leq 3$. Let $r$ be the root of the polynomial with $|r| \geq \frac{15}{8}$. Then, by the Triangle Inequality, we have:
$$\begin{gathered} \left|a_n r^n\right| = \left|a_{n-1} r^{n-1} + a_{n-2} r^{n-2} + \ldots + a_0\right| \leq \left|a_{n-1} r^{n-1}\right| + \left|a_{n-2} r^{n-2}\right| + \ldots + \left|a_0\right| \\ |r|^n \leq |r|^{n-1} + |r|^{n-2} + \ldots + 1 = \frac{|r|^n - 1}{|r| - 1} \\ |r|^{n+1} - 2|r|^n + 1 \leq 0 \Rightarrow 1 \leq |r|^n (2 - |r|) \end{gathered}$$
The right-hand side is increasing in $n$, for $|r| > 1$, so it is bounded by $|r|^3 (2 - |r|)$. This expression is decreasing in $r$ for $r \geq \frac{3}{2}$. When $|r| = \frac{15}{8}$, then the right-hand side is less than $1$, which violates the inequalities. Therefore $n > 3$. Now, we claim that there is a $4$th degree polynomial with a root $r$ with $|r| \geq \frac{15}{8}$. Let $p(x) = x^4 - x^3 - x^2 - x - 1$. Then $p\left(\frac{15}{8}\right) < 0$ and $p(2) > 2$. By the Intermediate Value Theorem, $p(x)$ has such a root $r$.