37 views
Find all monic polynomials $p(x)$ with integer coefficients of degree two for which there exists a polynomial $q(x)$ with integer coefficients such that $p(x) q(x)$ is a polynomial having all coefficients $\pm 1$.
| 37 views

0 like 0 dislike
Clearly, $p(x)$ has to be of the form $p(x)=x^2+a x \pm 1$, where $a$ is an integer. For $a=\pm 1$ and $a=0$, polynomial $p$ has the required property: it suffices to take $q=1$ and $q=x+1$, respectively.
Suppose now that $|a| \geq 2$. Then $p(x)$ has two real roots, say $x_1, x_2$, which are also roots of $p(x) q(x)=x^n+a_{n-1} x^{n-1}+\cdots+a_0, a_i=\pm 1$. Thus
$1=\left|\frac{a_{n-1}}{x_i}+\cdots+\frac{a_0}{x_i^n}\right| \leq \frac{1}{\left|x_i\right|}+\cdots+\frac{1}{\left|x_i\right|^n}<\frac{1}{\left|x_i\right|-1},$
which implies $\left|x_1\right|,\left|x_2\right|<2$. This immediately rules out the case $|a| \geq 3$ and the polynomials $p(x)=x^2 \pm 2 x-1$. The remaining two polynomials $x^2 \pm 2 x+1$ satisfy the condition for $q(x)=x \mp 1$.
Therefore, the polynomials $p(x)$ with the desired property are $x^2 \pm x \pm 1, x^2 \pm 1$, and $x^2 \pm 2 x+1$
by Platinum (138,598 points)