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### IMO $2017$ P$1$

Posted: Wed Jul 19, 2017 12:25 am
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 $a_0$ so that there exists a number $A$ such that $a_n = A$ for infinitely many values of $n$.

### Re: IMO $2017$ P$1$

Posted: Fri Aug 04, 2017 1:25 pm
Anyone??

### Re: IMO $2017$ P$1$

Posted: Fri Sep 15, 2017 3:11 am

### Re: IMO $2017$ P$1$

Posted: Sun Oct 01, 2017 10:25 pm
Case 1: $a_0\equiv 0\pmod{3}$.
We have $a_m\equiv 0\pmod{3}\,\,\forall m\geq 0$.
If $a_0=3$ then $a_{3m}=3\,\,\forall m\geq 0$, therefore $a_0=3$ satisfying the condition of the problem.
If $a_0=3k$ for some $k>1$. We will prove that there is an index $m_0$ such that $a_{m_0}<a_0$, and therefore (by replace $a_0$ by $a_{m_0}$) $a_0$ satisfying the condition of the problem. If $a_0$ is a square then $a_1<a_0$ else $m^2<3k<(m+1)^2$ for some positive integer $m$, exactly one of $m+1$, $m+2$, $m+3$ is divisible $3$, assume that it is $m+x\,\, (x\in\{0,1,2\})$, for some $m_0$ we have $a_{m_0}=m+x\leq m+3<3k=a_0$, we're done.

Case 2: $a_0\equiv 2\pmod{3}$.
In this case we have $a_n=a_0+3n\,\,\forall n\geq 0$ therefore $a_0$ does not meet the condition of the problem.

Case 3: $a_0\equiv 1\pmod{3}$.
We have $a_m\not\equiv 0\pmod{3}\,\,\forall m\geq 0$.
If $a_m\equiv 2\pmod{3}$ for some $m$ then by case 2, the sequence is unbound, therefore $a_0$ does not meet the condition of the problem.
If $a_m\equiv 1\pmod{3}\,\,\forall m\geq 0$ then assume that $a_k$ is smallest in $\{a_n\}$. Similary case 1, we can find $a_l$ such that $a_l<a_k$, contradiction! Therefore $a_0$ does not meet the condition of the problem.

Conclude, $a_0\equiv 0\pmod{3}$.[/

### Re: IMO $2017$ P$1$

Posted: Mon Oct 02, 2017 3:00 pm