Rubaanov I.S.

A polynomial $P(x)$ of degree $n$ has $n$ distinct real roots. What is the maximum number of its coefficients that can be zero?

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Evaluation. Between any two roots of a differentiable function, there is a root of its derivative. This means that the polynomial $P'(x)$ (of degree $n-1$) has $n-1$ distinct real roots, i.e., it has no multiple roots. Continuing this, we find that all derivatives of the polynomial $P$ have the same property. From this, it follows that among any two consecutive coefficients of the polynomial $P$, at least one is not zero. Indeed, if the coefficients of $x^k$ and $x^{k+1}$ are both zero, then the constant term and the coefficient of $x$ in the derivative $P^{(k)}$ are zero. This means that 0 is a multiple root of $P^{(k)}$, which is not the case.

Let's divide the coefficients of the polynomial into pairs (leaving the leading coefficient without a pair if $n$ is even). By the proven result, the number of zero coefficients does not exceed the number of pairs, which is $n / 2$ for even $n$ and $(n+1) / 2$ for odd $n$.

Examples. The polynomials $\left(x^{2}-1\right)\left(x^{2}-2^{2}\right) \ldots\left(x^{2}-k^{2}\right)$ of degree $n=2k$ and $x\left(x^{2}-1\right)\left(x^{2}-2^{2}\right) \ldots\left(x^{2}-k^{2}\right)$ of degree $n=2k+1$ show that this result cannot be improved: the first has coefficients of all odd powers equal to zero, and the second has coefficients of all even powers equal to zero.

## Answer

$n / 2$ for even $n$, $(n+1) / 2$ for odd $n$.

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