IMC 2014 · Day 1 · P3

We proceed by induction on $n$. The statement is trivial for $n = 1$. Thus assume that we have some $a_0, \dots, a_n$ which satisfy the conditions for some $n$. Consider now the polynomials $$\tilde{P}(x) = \pm a_n x^{n+1} \pm a_{n-1} x^n \pm \dots \pm a_1 x^2 \pm a_0 x$$ By induction hypothesis and $a_0 \ne 0$, each of these polynomials has $n + 1$ distinct zeros, including the nonzero roots of $\pm a_n x^n \pm a_{n-1} x^{n-1} \pm \dots \pm a_1 x \pm a_0$ and $0$. In particular none of the polynomials has a root which is a local extremum. Hence we can choose some $\varepsilon > 0$ such that for each such polynomial $\tilde{P}(x)$ and each $s$ of its local extrema we have $|\tilde{P}(s)| > \varepsilon$. We claim that then each of the polynomials $$P(x) = \pm a_n x^{n+1} \pm a_{n-1} x^n \pm \dots \pm a_1 x^2 \pm a_0 x \pm \varepsilon$$ has exactly $n + 1$ distinct zeros as well. As $\tilde{P}(x)$ has $n + 1$ distinct zeros, it admits a local extremum at $n$ points. Call these local extrema $-\infty = s_0 < s_1 < s_2 < \dots < s_n < s_{n+1} = \infty$. Then for each $i \in \{0, 1, \dots, n\}$ the values $\tilde{P}(s_i)$ and $\tilde{P}(s_{i+1})$ have opposite signs (with the obvious convention at infinity). By choice of $\varepsilon$ the same holds true for $P(s_i)$ and $P(s_{i+1})$. Hence there is at least one real zero of $P(x)$ in each interval $(s_i, s_{i+1})$, i.e. $P(x)$ has at least (and therefore exactly) $n + 1$ zeros. This shows that we have found a set of positive reals $a'_{n+1} = a_n, a'_n = a_{n-1}, \dots, a'_1 = a_0, a'_0 = \varepsilon$ with the desired properties.