Turkish Mathematical Olympiad

The polynomial $x^4 + x^3 - 4x^2 + x + 1 = (x-1)^2(x^2 + 3x + 1)$ satisfies the conditions. Let us show that for $d \ge 5$ there is no polynomial satisfying given conditions.

Solution 1. Let $x_1, x_2, \dots, x_d$ be the roots, $S_k$ be the sum of all $k$-tuple products of roots. Using Vieta theorem, we get
$$S_1^2 - 2S_2 = \sum_{i=1}^{d} x_i^2 \geq 0.$$
When the first three coefficients are $1$, we have $S_1 = -1, S_2 = 1$ and $S_1^2 - 2S_2 = -1 < 0$, which is a contradiction. Hence at least one of the first three coefficients should not be equal to $1$. This means that all roots are non-zero and hence we obtain:
$$\left(\frac{S_{d-1}}{S_d}\right)^2 - 2\left(\frac{S_{d-2}}{S_d}\right) = \sum_{i=1}^{d} \frac{1}{x_i^2} > 0.$$
If the last three coefficients are $1$, we get $S_{d-2} = -S_{d-1} = S_d$ and hence we conclude that $\left(\frac{S_{d-1}}{S_d}\right)^2 < 2\left(\frac{S_{d-2}}{S_d}\right)$, which is again a contradiction.
For $d \ge 5$ case, as $d$ coefficients out of $d+1$ coefficients are $1$, we conclude that either the first three or the last three coefficients should be $1$. Hence, we are done.

The polynomial $x^4 + x^3 - 4x^2 + x + 1 = (x-1)^2(x^2 + 3x + 1)$ satisfies the conditions. Let us show that for $d \ge 5$ there is no polynomial satisfying given conditions.

Solution 2. We will find a contradiction for $d \ge 5$ case. We can express the polynomial as $P(x) = x^d + x^{d-1} + \dots + 1 + a x^b$ where $b$ is an integer satisfying $0 \le b \le d$ and $a$ is a real number. Since all roots of $P(x)$ are real numbers, all roots of the polynomial $Q(x) = (x-1)P(x) = x^{d+1} - 1 + a x^{b+1} - a x^b$ should be real numbers, too. If zero is a root of $Q(x)$, then we get $a = -1, b = 0$ and in this case all roots of $Q(x) = x^{d+1} - x$ cannot be real. So, all roots of $Q(x)$ should be non-zero.
The polynomial $Q(x)$ has at most $4$ non-zero coefficients. By Descartes' rule of signs, $Q$ has at most $3$ positive real roots. $Q(-x) = (-1)^{d+1} x^{d+1} - 1 + a(-1)^{b+1}(x^{b+1} + x^b)$ has also at most $4$ non-zero coefficients. If there are exactly $4$ non-zero coefficients, the coefficients of $x^{b+1}$ and $x^b$ should have the same sign. (Otherwise we would have at most $3$ non-zero coefficients.) Therefore, in both cases, using Descartes' rule of signs again, we conclude that $Q(-x)$ has at most $2$ positive real roots. As zero is not a root, $Q(x)$ can have at most $5$ real roots. The degree of $Q(x)$ is $d+1 \ge 6$ and hence all roots cannot be real. This finishes the proof.