gnoka wrote: 1. Baron Munchhausen was told that some polynomial $P(x)=a_{n} x^{n}+\ldots+a_{1} x+a_{0}$ is such that $P(x)+P(-x)$ has exactly 45 distinct real roots. Baron doesn't know the value of $n$. Nevertheless he claims that he can determine one of the coefficients $a_{n}, \ldots, a_{1}, a_{0}$ (indicating its position and value). Isn't Baron mistaken? Since $P(x)+P(-x)$ is even and has an odd number of distinct real roots then one is zero and $P(0)=0$ and so $\boxed{a_0=0}$

Is not wrong. It is evident that $P(X)+P(-X)$ is an even polynomial. Therefore, there is a polynomial $Q(X)$ such that $Q(X^2)=P(X)+P(-X)$. Let's suppose $r_1$, $r_2$, ... , $r_k$ are all the roots (counting multiplicities) of the polynomial $Q(X^2)$, then $$P(X)+P(-X)=a(X^2-r_1)(X^2-r_2)\cdots (X^2-r_n).$$If for some index $i$, we have to $r_i\neq 0$, then the polynomial $X^2-r_i$ has $0$ or $2$ real roots, that is, it has an even number of real roots. This means that the multiplicity of each non-zero root is even. Then, since the polynomial $P(X)+P(-X)$ has exactly 45 different real roots, then one of such roots is $0$, so we have to $$a_0=\frac{P(0)+P(-0)}{2}=0.$$