Problem 1

For each integer

a_0>1

, define the sequence

a_0,a_1,a_2,\ldots

for

n\ge 0

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

for which there exists a number

A

such that

a_n=A

for infinitely many values of

n

Step 3 of 5: If

3\nmid a_0

, the sequence is unbounded

Previous step Next step

a_n\equiv 2\ (\mathrm{mod}\ 3)\implies a_{n+1}=a_n+3

Detailed analysis

Perfect squares are $\equiv 0$ or $1\pmod 3$ , never $2$ . So a term $\equiv 2\pmod 3$ is never a square, forcing $a_{n+1}=a_n+3$ forever after, and the sequence grows without bound. If instead $a_n\equiv 1\pmod 3$ for every $n$ , follow the arithmetic progression $a_n,a_n+3,a_n+6,\ldots$ (which visits every integer $\equiv 1\pmod 3$ above $a_n$ ) until it meets its first perfect square $K^2$ ; taking $K$ to be the least integer $\ge\lceil\sqrt{a_n}\rceil$ with $3\nmid K$ gives $K\le \sqrt{a_n}+2$ , so the next term $K$ is strictly smaller than $a_n$ whenever $a_n\ge 7$ . Repeating this descent, the value must eventually reach $4=2^2$ , whose successor is $2\equiv 2\pmod 3$ . So sooner or later a term $\equiv 2\pmod 3$ appears, and the sequence is unbounded.