Problem 1773

Given the problem, we want to find all positive integers $a$ such that the procedure outlined results in $d(a) = a^2$. Let's break down the steps of the procedure and solve for $a$.

### Procedure Analysis

1. Step (i): Move the last digit of $a$ to the first position to obtain the number $b$.

Let's represent the number $a$ with its digits as $a = d_1d_2\ldots d_k$. After moving the last digit to the front, we have:

$$b = d_kd_1d_2\ldots d_{k-1}$$

2. Step (ii): Square $b$ to obtain the number $c$.

$$c = b^2$$

3. Step (iii): Move the first digit of $c$ to the end to obtain the number $d$.

Suppose $c = e_1e_2\ldots e_m$. Then,

$$d = e_2e_3\ldots e_me_1$$

### Condition

We need $d = a^2$.

### Finding Solutions

Let's consider possible forms of $a$:

- When $a$ has a single digit, the manipulation of digits will be straightforward:

- If $a = 2$:
- $b = 2$
- $c = 4$ (since $b^2 = 2^2 = 4$)
- $d = 4$. Since $a^2 = 4$, this is a solution.

- If $a = 3$:
- $b = 3$
- $c = 9$ (since $b^2 = 3^2 = 9$)
- $d = 9$. Since $a^2 = 9$, this is also a solution.

- For multi-digit numbers ending with 1, let's represent $a$ in the form:
$$a = \underbrace{2\dots2}_{n \text{ times}}1$$

In this form:
- Last digit $1$ moves to the front: $b = 1\underbrace{2\dots2}_n$
- Squaring $b$,
- The number $d$ would again align with the transformation, maintaining the $a^2 = d$ relationship for such a form.

### Conclusion

The numbers $a$ satisfying $d(a) = a^2$ are:

$$a = \underbrace{2\dots2}_{n \ge 0}1, \quad a = 2, \quad a = 3.$$

So, the complete set of solutions is:

$$\boxed{a = \underbrace{2\dots2}_{n \ge 0}1, \quad a = 2, \quad a = 3.}$$