Post Number:**#4** by **ahmedittihad** » 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, $a_0\equiv 0\pmod{3}$.[/

Frankly, my dear, I don't give a damn.