2005 IMO Shortlist Problems/A1

Problem

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

This was also the last problem of the final round of the 2006 Polish Mathematics Olympiad.

Solution

Since the constant term of $p(x)q(x)$ is $\pm 1$ , and $p(x)$ and $q(x)$ both have integral constant terms, the constant term of $p(x)$ must be $\pm 1$ .

We note that for $|z| \ge 2$ , $n \ge 2$ ( $n \in \mathbb{N}$ ), we have

$|z|^n > \frac{|z|^n -1}{|z|-1} = \sum_{i=0}^{n-1}|\pm z|^i \ge \left| \sum_{i=0}^{n-1} \pm z^i \right|$

Since we must have $|z^n| = | p(x)q(x) - z^n |$ when $n$ is the degree of $p(x)q(x)$ and $z$ is a root thereof, this means that $p(x)q(x)$ cannot have any roots of magnitude greater than or equal to 2.

Now, if $p(x) = x^2 + kx + 1$ , then we cannot have $|k| \ge 3$ , for then one of the roots would have magnitude $\frac{|k| + \sqrt{k^2 - 4}}{2} \ge \frac{3 + \sqrt{3^2 - 4}}{2} > 2$ , and similarly, if $p(x) = x^2 +kx - 1$ , then we cannot have $|k| \ge 2$ , for then one of the roots would have magnitude $\frac{|n| + \sqrt{n^2+4}}{2} \ge \frac{2 + \sqrt{2^2 + 4}}{2} > 2$ .

This leaves us only the possibilities $p(x) = x^2 \pm 1,\; x^2 \pm x \pm 1,\; x^2 + 2x + 1,\; x^2 - 2x + 1$ . For these we have respective solutions $q(x) = x+1,\; 1,\; x - 1,\; x + 1$ . These are therefore the only solutions, Q.E.D.

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

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2005 IMO Shortlist Problems/A1

Problem

Solution

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