Note: $m$ is a root of $x^2+x-4=0$. Since $m$ is not a rational number, and $P(x)-2018=0$ we get, for some polynomial $Q(x)$ with integer coefficients: \begin{align*} Q(x)(x^2+x-4) = P(x) - 2018 \Longrightarrow a_0 + a_1 + \dots + a_n = P(1) = 2018-2Q(1) \equiv 0(mod2) \end{align*}

parmenides51 wrote: Let $m=\frac{-1+\sqrt{17}}{2}$. Let the polynomial $P(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0$ is given, where $n$ is a positive integer, the coefficients $a_0,a_1,a_2,...,a_n$ are positive integers and $P(m) =2018$ . Prove that the sum $a_0+a_1+a_2+...+a_n$ is divisible by $2$ . Please Check if this is the correct way to solve (I know it's ugly) but still For every $k\in N$ we have $2^{k-1}(x+y\sqrt{17})=(\sqrt{17}-1)^k$ for some odd integer $x,y$ So we have $a_n*(x_n +y_n\sqrt{17})+.... +a_1*(x_1+y_1\sqrt{17})+2*a_0=2*2018$ here $x_i, y_i$ is odd integer Hence from above equation we can get that $4|a_n+a_{n-1}+... a_1, 2|a_0$ so $2|a_0+a_1+... +a_{n-1}+a_n $