Let $f(x)=a_0+a_1x+a_2x^2+\cdots +a_nx^n$ be a $n$-degree polynomial and all its coefficients $a_i$ $(0\le i\le n)$ be either $1$ or $-1$. If $f(x)$ has only real roots, what is the maximum value of $n$?
The answer is $\boxed{3}$.
Let all the roots of $f(n)$ be $r_1$, $r_2$, $\cdots$, $r_n$ and $$\left\{ \begin{array}{rl} a = & \displaystyle\sum_{i=1}^n r_i \\ b=& \displaystyle\sum_{1\le i, j\le n}r_ir_j \\ c=&\displaystyle\prod_{i=1}^n r_i \end{array} \right.$$
Then we have $a$, $b$, and $c$ can be only $1$ or $-1$. Now by Vieta's theorem, $$\sum_{i=1}^nr_i^2 = a^2 - 2b = 1-2b \ge 0 \implies b=-1\quad\text{and}\quad \sum_{i=1}^nr_i^2=3 $$
Meanwhile, by the AM-GM inequality, $$\frac{1}{n}\sum_{i=1}^n r_i^2 \ge \left(\prod_{i=1}^nr_i^2\right)^{\frac{1}{n}}\implies \frac{3}{n}\ge c^{\frac{2}{n}} = 1 \implies n\le 3$$
When $n=3$, there indeed exists at least one polynomial satisfying the requirements as show below. Therefore, the answer is $3$. $$f(x)= x^3 - x^2 - x + 1 = (x-1)^2(x+1)$$