IMC 2007 · Day 1 · P6

We show that the number of nonzero coefficients can be 0, 1 and 2. These values are possible, for example the polynomials $P_0(z) = 0$, $P_1(z) = 1$ and $P_2(z) = 1 + z$ satisfy the conditions and they have 0, 1 and 2 nonzero terms, respectively.

Now consider an arbitrary polynomial $P(z) = a_0 + a_1 z + \dots + a_n z^n$ satisfying the conditions and assume that it has at least two nonzero coefficients. Dividing the polynomial by a power of $z$ and optionally replacing $p(z)$ by $-p(z)$, we can achieve $a_0 > 0$ such that conditions are not changed and the number of nonzero terms is preserved. So, without loss of generality, we can assume that $a_0 > 0$.

Let $Q(z) = a_1 z + \dots + a_{n-1} z^{n-1}$. Our goal is to show that $Q(z) = 0$.

Consider those complex numbers $w_0, w_1, \dots, w_{n-1}$ on the unit circle for which $a_n w_k^n = |a_n|$; namely, let $$w_k = \begin{cases} e^{2k\pi i/n} & \text{if } a_n > 0 \\ e^{(2k+1)\pi i/n} & \text{if } a_n < 0 \end{cases} \quad (k = 0, 1, \dots, n).$$ Notice that $$\sum_{k=0}^{n-1} Q(w_k) = \sum_{k=0}^{n-1} Q(w_0 e^{2k\pi i/n}) = \sum_{j=1}^{n-1} a_j w_0^j \sum_{k=0}^{n-1} (e^{2j\pi i/n})^k = 0.$$ Taking the average of polynomial $P(z)$ at the points $w_k$, we obtain $$\frac{1}{n} \sum_{k=0}^{n-1} P(w_k) = \frac{1}{n} \sum_{k=0}^{n-1} \bigl( a_0 + Q(w_k) + a_n w_k^n \bigr) = a_0 + |a_n|$$ and $$2 \ge \frac{1}{n} \sum_{k=0}^{n-1} \bigl| P(w_k) \bigr| \ge \frac{1}{n} \left| \sum_{k=0}^{n-1} P(w_k) \right| = a_0 + |a_n| \ge 2.$$ This obviously implies $a_0 = |a_n| = 1$ and $P(w_k) = 2 + Q(w_k) = 2$ for all $k$. Therefore, all values of $Q(w_k)$ must lie on the circle $|2 + z| = 2$, while their sum is 0. This is possible only if $Q(w_k) = 0$ for all $k$. Then polynomial $Q(z)$ has at least $n$ distinct roots while its degree is at most $n - 1$. So $Q(z) = 0$ and $P(z) = a_0 + a_n z^n$ has only two nonzero coefficients.

Remark. From Parseval's formula (i.e. integrating $|P(z)|^2 = P(z) \overline{P(z)}$ on the unit circle) it can be obtained that $$\tag{4} |a_0|^2 + \dots + |a_n|^2 = \frac{1}{2\pi} \int_0^{2\pi} \left| P(e^{it}) \right|^2 dt \le \frac{1}{2\pi} \int_0^{2\pi} 4\,dt = 4.$$ Hence, there cannot be more than four nonzero coefficients, and if there are more than one nonzero term, then their coefficients are $\pm 1$.

It is also easy to see that equality in (4) cannot hold two or more nonzero coefficients, so it is sufficient to consider only polynomials of the form $1 \pm x^m \pm x^n$. However, we do not know (yet :-)) any simpler argument for these cases than the proof above.