2017 IMO Problems/Problem 1

Problem

For each integer $a_0 > 1$ , define the sequence $a_0, a_1, a_2, \ldots$ for $n \geq 0$ as $a_{n+1} = \begin{cases} \sqrt{a_n} & \text{if } \sqrt{a_n} \text{ is an integer,} \\ a_n + 3 & \text{otherwise.} \end{cases}$ Determine all values of $a_0$ such that there exists a number $A$ such that $a_n = A$ for infinitely many values of $n$ .

Solution

First we notice the following:

When we start with $a_0=3$ , we get $a_1=6$ , $a_2=9$ , $a_3=3$ and the pattern repeats.

When we start with $a_0=6$ , we get $a_1=9$ , $a_2=3$ , $a_3=6$ and the pattern repeats.

When we start with $a_0=9$ , we get $a_1=3$ , $a_2=6$ , $a_3=9$ and the pattern repeats.

When we start with $a_0=12$ , we get $a_1=15$ , $a_2=15$ ,..., $a_8=36$ , $a_9=6$ , $a_{10}=9$ , $a_{11}=3$ and the pattern repeats.

When this pattern $3,6,9$ repeats, this means that there exists a number $A$ such that $a_n = A$ for infinitely many values of $n$ and that number $A$ is either $3,6,$ or $9$ .

When we start with any number $a_0\not\equiv 0 mod 3$ , we don't see a repeating pattern.

Therefore the claim is that $a_0=3k$ where $k$ is a positive integer and we need to prove this claim.

When we start with $a_0=3k$ , the next term if it is not a square is $3k+3$ , then $3k+6$ and so on until we get $3k+3p$ where $p$ is an integer and $(k+p)=3q^2$ where $q$ is an integer. Then the next term will be $\sqrt{9q^2}=3q$ and the pattern repeats again when $q=k$ or when $q=3$ or $6$ . In order for these patterns to repeat, any square in the sequence need to be a multiple of 3. This will not work with any number or square that is not a multiple of 3.

So, the answer to this problem is $a_0=3k\;\forall k \in \mathbb{Z}^{+}$

~Tomas Diaz. orders@tomasdiaz.com

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

2017 IMO (Problems) • Resources
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2017 IMO Problems/Problem 1

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