LVII OM - III - Problem 6

Determine all pairs of integers $ a $, $ b $, for which there exists a polynomial $ P(x) $ with integer coefficients, such that the product $ (x^2 + ax + b)\cdot P(x) $ is a polynomial of the form



where each of the numbers $ c_0,c_1,\dots ,c_{n-1} $ is equal to 1 or -1.
If the quadratic polynomial $x^2 + ax + b$ is a divisor of some polynomial $Q(x)$ with integer coefficients, then of course the constant term $b$ is a divisor of the constant term of the polynomial $Q(x)$. Therefore, in the considered situation, $b=\pm 1$.
No polynomial $Q(x)$ of the form $x^n \pm x^{n-1} \pm x^{n-2} \pm \ldots \pm x \pm 1$ has real roots with an absolute value greater than or equal to 2. Suppose, for instance, that a number $r$ with $|r| \geq 2$ satisfies the equation $Q(r) = 0$. Then



skipping some steps, we conclude that $|r| < 2$. This is a contradiction.
On the other hand, every root of the quadratic polynomial $T(x) = x^2 + ax + b$ is also a root of the polynomial $T(x) \cdot P(x)$.
Therefore, the quadratic polynomial $T(x)$ has no roots in the set $(-\infty, -2] \cup [2, +\infty)$.
Thus, $T(-2) > 0$ and $T(2) > 0$, which means $4 - 2a + b > 0$ and $4 + 2a + b > 0$.
If $b = 1$, then we have $5 > 2a$ and $5 > -2a$, which implies that $a$ is one of the numbers $-2, -1, 0, 1, 2$. If, however, $b = -1$, then $3 > 2a$ and $3 > -2a$, and the cases to consider are when $a$ is one of the numbers $-1, 0, 1$.
Direct verification shows that all the obtained pairs $(a, b)$ satisfy the conditions of the problem:







Answer: The conditions of the problem are satisfied by the following eight pairs $(a, b)$: (-2,1), (-1,-1), (-1,1), (0,-1), (0,1), (1,-1), (1,1), (2,1).