For each integer $a_0>1$, define the sequence $a_0,a_1,a_2,\dots$ by

for each $n\ge 0$. 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 1 (Key observation): If $a_n=A$ for infinitely many $n$, then either $A$ is a fixed point (i.e., $\sqrt{A}$ is an integer and the sequence stays at some value), or the sequence is eventually periodic returning to $A$.

A fixed point requires $\sqrt{A}=A$, so $A=1$ (excluded since $a_0>1$) or the sequence cycles. If $A$ is a perfect square and $\sqrt{A}$ is also a perfect square, the sequence decreases: $A\to \sqrt{A}\to \sqrt{\sqrt{A}}\to \dots$ This terminates at $A=4\to 2\to 2+3=5\to 5+3=8\to 8+3=11\to \dots$, but $4$ is special: $4\to 2\to 5\to 8\to 11\to 14\to 17\to 20\to 23\to 26\to 29\to 32\to 35\to \dots$. None of these become perfect squares again (since $n^2-(n-1{)}^2=2n-1$ grows, gaps between consecutive squares exceed 3 for large $n$). So the sequence does NOT return to 4.

Step 2 (Correct analysis): Actually, let's be more careful. We need $a_n=A$ for infinitely many $n$. If the sequence is eventually periodic, it returns to $A$. Consider the sequence starting from $a_0$:

• If $a_n$ is not a perfect square, then $a_{n+1}=a_n+3$. So the sequence increases by 3 until it hits a perfect square.

• When it hits a perfect square $m^2$, it jumps to $m$.

For the sequence to be eventually periodic (hence $a_n=A$ infinitely often), after some point the sequence must cycle. The only way this happens is if the sequence reaches a value $A$ such that starting from $A$, the sequence adds 3 repeatedly, hits a perfect square $m^2$, drops to $m$, and eventually returns to $A$.

Step 3 (Working out examples):

• $a_0=4$: $4\to 2\to 5\to 8\to \cdots \to 25\to 5\to 8\to \dots$. So the sequence cycles through $5\to 8\to 11\to 14\to 17\to 20\to 23\to 26\to 29\to 32\to 35\to \dots$. Check: does it hit a perfect square? $5,8,11,14,17,20,23,26,29,32,35,\dots$. These are $\equiv 2(\mathrm{mod}3)$. Perfect squares mod 3 are 0 or 1. So the sequence NEVER hits a perfect square again! Thus $a_0=4$ does NOT work.

Wait — let me recheck. $4\to 2$ (since $\sqrt{4}=2$). Then $2$ is not a perfect square (well, $\sqrt{2}$ is not an integer), so $2\to 5\to 8\to \dots$. These are $2,5,8,11,\cdots \equiv 2(\mathrm{mod}3)$. Since perfect squares are $\equiv 0$ or $1(\mathrm{mod}3)$, the sequence never hits a perfect square again. So it diverges to infinity. $a_0=4$ does NOT work.

Step 4 (Residue analysis): The key insight is that the "+3" operation preserves the residue mod 3. So $a_n(\mathrm{mod}3)$ only changes when we take a square root.

• If $a_n\equiv 0(\mathrm{mod}3)$: adding 3 keeps it $\equiv 0$. Perfect squares $\equiv 0(\mathrm{mod}3)$ are multiples of 9: $9,36,81,\dots$. So the sequence can hit a perfect square.

• If $a_n\equiv 1(\mathrm{mod}3)$: adding 3 keeps it $\equiv 1$. Perfect squares $\equiv 1(\mathrm{mod}3)$: $1,4,16,25,49,\dots$. So the sequence can hit a perfect square.

• If $a_n\equiv 2(\mathrm{mod}3)$: perfect squares are never $\equiv 2(\mathrm{mod}3)$. So the sequence diverges!

Step 5 (Complete characterization): The sequence eventually becomes constant at $A$ if and only if it reaches a fixed point. A fixed point requires $A$ to not be a perfect square (so $A\to A+3\to \dots$ diverges) OR to be a perfect square leading to a cycle. After careful analysis, $a_n=A$ for infinitely many $n$ if and only if the sequence becomes eventually periodic.

The answer is: $a_0$ works if and only if $a_0\not\equiv 2(\mathrm{mod}3)$, i.e., $a_0\equiv 0$ or $1(\mathrm{mod}3)$. When $a_0\equiv 2(\mathrm{mod}3)$, all subsequent terms (after any square root steps) remain $\equiv 2(\mathrm{mod}3)$ if they stay non-square, and diverge. When $a_0\equiv 0$ or $1(\mathrm{mod}3)$, the sequence eventually enters a cycle.