Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 6.9 National Olympiad Prove it United States

Problem:

An integer $c$ is square-friendly if it has the following property: For every integer $m$ , the number $m^{2}+18 m+c$ is a perfect square. (A perfect square is a number of the form $n^{2}$ , where $n$ is an integer. For example, $49=7^{2}$ is a perfect square while $46$ is not a perfect square. Further, as an example, $6$ is not square-friendly because for $m=2$ , we have $(2)^{2}+(18)(2)+6=46$ , and $46$ is not a perfect square.)

In fact, exactly one square-friendly integer exists. Show that this is the case by doing the following:

a. Find a square-friendly integer, and prove that it is square-friendly.

b. Prove that there cannot be two different square-friendly integers.

Solution

Solution:

a. $c=81$ is square-friendly, since $m^{2}+18 m+81=(m+9)^{2}$ which is an integer whenever $m$ is an integer.

b. Suppose there are two different square-friendly integers $c$ and $c^{\prime}$ . In that case, for every integer $m$ , we have that $m^{2}+18 m+c$ and $m^{2}+18 m+c^{\prime}$ are perfect squares. Thus we have infinitely many pairs of perfect squares that differ by $c-c^{\prime}$ . However, this is not possible, because the differences between consecutive squares are $1,3,5,7,9,11, \ldots$ . Eventually these differences are greater than $c-c^{\prime}$ , so no perfect square beyond that point differs from another perfect square by $c-c^{\prime}$ . Thus we have a contradiction, and so there cannot be two different square-friendly integers.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.