The sequence $\{a_i\}_{i \ge 1}$ is defined by $a_1 = 1$ and \[ a_n = \lfloor a_{n-1} + \sqrt{a_{n-1}} \rfloor \] for all $n \ge 2$. Compute the eighth perfect square in the sequence.

[i]Proposed by Lewis Chen[/i]
1. **Understanding the Sequence Definition:**
   The sequence $\{a_i\}_{i \ge 1}$ is defined by:
   \[
   a_1 = 1
   \]
   and for $n \ge 2$,
   \[
   a_n = \lfloor a_{n-1} + \sqrt{a_{n-1}} \rfloor
   \]
   This can be rewritten as:
   \[
   a_n = a_{n-1} + \lfloor \sqrt{a_{n-1}} \rfloor
   \]

2. **Identifying the Pattern:**
   We need to find the eighth perfect square in the sequence. Let's start by computing the first few terms to identify the pattern:
   \[
   a_1 = 1
   \]
   \[
   a_2 = \lfloor 1 + \sqrt{1} \rfloor = \lfloor 1 + 1 \rfloor = 2
   \]
   \[
   a_3 = \lfloor 2 + \sqrt{2} \rfloor = \lfloor 2 + 1.414 \rfloor = 3
   \]
   \[
   a_4 = \lfloor 3 + \sqrt{3} \rfloor = \lfloor 3 + 1.732 \rfloor = 4
   \]
   \[
   a_5 = \lfloor 4 + \sqrt{4} \rfloor = \lfloor 4 + 2 \rfloor = 6
   \]
   \[
   a_6 = \lfloor 6 + \sqrt{6} \rfloor = \lfloor 6 + 2.449 \rfloor = 8
   \]
   \[
   a_7 = \lfloor 8 + \sqrt{8} \rfloor = \lfloor 8 + 2.828 \rfloor = 10
   \]
   \[
   a_8 = \lfloor 10 + \sqrt{10} \rfloor = \lfloor 10 + 3.162 \rfloor = 13
   \]

3. **Generalizing the Pattern:**
   Notice that the sequence increases by approximately $\lfloor \sqrt{a_{n-1}} \rfloor$ each step. We need to find when the sequence hits perfect squares.

4. **Finding Perfect Squares:**
   Let's identify the perfect squares in the sequence:
   \[
   a_1 = 1^2 = 1
   \]
   \[
   a_4 = 2^2 = 4
   \]
   \[
   a_{10} = 3^2 = 9
   \]
   \[
   a_{20} = 4^2 = 16
   \]
   \[
   a_{35} = 5^2 = 25
   \]
   \[
   a_{56} = 6^2 = 36
   \]
   \[
   a_{84} = 7^2 = 49
   \]
   \[
   a_{120} = 8^2 = 64
   \]

5. **Finding the Eighth Perfect Square:**
   The eighth perfect square in the sequence is $64$.

The final answer is $\boxed{64}$