SOLVED

Let $P(z)=\sum_{1\leq k\leq n}a_kz^k$ for some $a_k\in \mathbb{C}$ with $\lvert a_k\rvert=1$ for $1\leq k\leq n$. Does there exist a constant $c>0$ such that, for $n\geq 2$, we have
\[\max_{\lvert z\rvert=1}\lvert P(z)\rvert \geq (1+c)\sqrt{n}?\]

The lower bound of $\sqrt{n}$ is trivial from Parseval's theorem. The answer is no (contrary to Erdős' initial guess). Kahane [Ka80] constructed 'ultraflat' polynomials $P(z)=\sum a_kz^k$ with $\lvert a_k\rvert=1$ such that
\[P(z)=(1+o(1))\sqrt{n}\]
uniformly for all $z\in\mathbb{C}$ with $\lvert z\rvert=1$, where the $o(1)$ term $\to 0$ as $n\to \infty$.

For more details see the paper [BoBo09] of Bombieri and Bourgain and where Kahane's construction is improved to yield such a polynomial with \[P(z)=\sqrt{n}+O(n^{\frac{7}{18}}(\log n)^{O(1)})\] for all $z\in\mathbb{C}$ with $\lvert z\rvert=1$.

See also [228].