Determine the smallest positive integer $n$ for which there exists a polynomial

$$
P(X)=a_{2 n} X^{2 n}+a_{2 n-1} X^{2 n-1}+\ldots+a_{1} X+a_{0}
$$

with real coefficients that satisfies the following two properties:

- For $i=0,1, \ldots, 2 n$, we have $2014 \leq a_{i} \leq 2015$.
- There exists a real number $\xi$ such that $P(\xi)=0$.
It is $n=2014$. First, define for a specific choice of coefficients $a_{k}$, namely 2014 for even $k$ and 2015 for odd $k$, the polynomial $Q_{n}(X)=2014 X^{2 n}+2015 X^{2 n-1}+2014 X^{2 n-2}+\ldots+2015 X+2014$. It is $Q_{2014}(-1)=0$, so the desired number is at most 2014. It remains to show that $P(X)$ has no real roots for $n \leq 2013$. For $x \geq 0$, $P(x) \geq a_{0}>0$, since the coefficients are positive; for $x<0$ and $m \leq 2013$.

1. Proof: It can be shown that for real $x$,

$$
Q_{m}(x)=1007\left(\sum_{\nu=0}^{m-1}\left(x^{\nu}+x^{\nu+1}\right)^{2}\right)+\frac{1}{2}\left(\sum_{\nu=0}^{m-1}\left(x^{2 \nu+1}+1\right)\left(x^{2 m-2 \nu-1}+1\right)\right)+\left(1007-\frac{m}{2}\right)\left(x^{2 m}+1\right)
$$

The terms of the first sum are non-negative as squares. Since $x^{2 k+1}+1$ is positive, zero, or negative for all $k \geq 0$ for $x>-1$, $x=-1$, or $x< -1$ respectively, the terms of the second sum are non-negative. The third term is non-negative for $m \leq 2013$. Therefore, $Q_{m}(x) > 0$ for $m \leq 2013$.

2. Proof (using calculus): Define for $x < -1$ and $m \leq 2013$, and it is

$$
\begin{aligned}
f_{m}^{\prime}\left(x_{m}\right) & =2014(2 m+2) x_{m}^{2 m+1}+2015(2 m+1) x_{m}^{2 m}-2015= \\
& =\left(2014(2 m+2) x_{m}+2015(2 m+1)\right) x_{m}^{2 m}-2015=2015 \cdot x_{m}^{2 m}-2015-1 \text{ and negative for } x<0.
\end{aligned}
$$

## Remarks:

1. The polynomial function $P(x)$ can have no real roots outside the interval ] $-\frac{2015}{2014},-\frac{2014}{2015}$ [.
2. Often, terms $g_{k}(x)=2014 x^{2 k}+2015 x^{2 k-1}$ of $Q_{m}(x)$ for individual $k$ were isolated and incorrectly estimated against $g_{k}(-1)$. As shown by differentiation, $g_{k}$ is strictly monotonically increasing or decreasing for $0>x>-\frac{2015(2 k-1)}{2014 \cdot 2 k}$ or $x<-\frac{2015(2 k-1)}{2014 \cdot 2 k}$ respectively. Thus, $g_{k}(x) > g_{k}(-1)$ for $k \leq 1007$ and $x<-\frac{2015(2 k-1)}{2014 \cdot 2 k}$, and $g_{k}(x) < g_{k}(-1)$ for $k > 1007$ and $-1>x>-\frac{2015(2 k-1)}{2014 \cdot 2 k}$. Therefore, $Q_{m}(x) \geq Q_{m}(-1)$ can only be shown for $x<-1$ and $m \leq 1007$ in this way.

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