$3025=(30+25)^{2}$? Is there another four-digit number with a similar property?
I. Solution. If the number formed by the first two digits of the four-digit number is $x$, and the number formed by the second two digits is $y$, then the requirement of the problem is that

$$
100 x + y = (x + y)^2 \quad \text{or} \quad (x + y)^2 - (x + y) = 99 x
$$

Let $(x + y) = z$, then

$$
z^2 - z = 99 x \quad \text{or} \quad x = \frac{z(z-1)}{99}
$$

where $z \leq 99$. If $z \geq 100$, then $x \geq 100$; this is impossible because $x$ is a two-digit number.
I. $x$ is an integer if $z = 99$; in this case, $x = 98$. Since $z = x + y = 99$, then $y = 01$. Indeed, $9801 = (98 + 01)^2 = 99^2$.

II. $x$ is an integer if 99 is a divisor of the product $z(z-1)$. $z$ and $z-1$ cannot both be multiples of 3. Therefore, either $z$ or $z-1$ must be a multiple of 9. If

$$
z = 9u, \quad \text{then} \quad z - 1 = 11v
$$

We need to find the positive integer solutions of the indeterminate equation $9u - 1 = 11v$ such that $z < 99$. Thus,

$$
\begin{gathered}
9u = 11v + 1, \quad u = v + \frac{2v + 1}{9} = v + w, \quad \text{if} \quad \frac{2v + 1}{9} = w \\
\text{From this,} \quad v = \frac{9w - 1}{2} = 4w + \frac{w - 1}{2} = 4w + t, \quad \text{if} \quad \frac{w - 1}{2} = t, \text{that is} \\
w = 2t + 1
\end{gathered}
$$

Now $u = v + w = 4w + t + w = 4(2t + 1) + t + 2t + 1 = 11t + 5$ and thus $z = 99t + 45$.

According to our condition, $z < 99$, so only $t = 0$ is possible, that is,

$$
z = x + y = 45, \quad z - 1 = 44, \quad x = \frac{45 \cdot 44}{99} = 20, \quad y = 25
$$

III. If $z$ is a multiple of 11 and $z-1$ is a multiple of 9, then

$$
z = 11u, \quad z - 1 = 9v, \quad \text{so} \quad 9v = 11u - 1
$$

2

From this, $\quad v = u + \frac{2u - 1}{9} = u + w, \quad$ if $\quad \frac{2u - 1}{9} = w, \quad u = \frac{9w + 1}{2}$.

$$
u = 4w + \frac{w + 1}{9} = 4w + t, \quad \text{if} \quad \frac{w + 1}{2} = t \quad \text{or} \quad w = 2t - 1
$$

Thus,

$$
\begin{gathered}
v = u + w = 4w + t + w = 4(2t - 1) + t + 2t - 1 = 11t - 5 \\
z = 9v + 1 = 99t - 45 + 1 = 99t - 44
\end{gathered}
$$

Since $z \leq 99$, only $t = 1$ is possible, so

$$
z = x + y = 55 \quad \text{and} \quad z - 1 = 54; \quad x = \frac{55 \cdot 54}{99} = 30, \quad y = 25
$$

The given property is true for three four-digit numbers:

$$
3025 = (30 + 25)^2 = 55^2, \quad 2025 = (20 + 25)^2 = 45^2 \quad \text{and} \quad 9801 = 99^2
$$

II. Solution. Part of the solution involves transforming the equation

$$
100x + y = (x + y)^2
$$

into the form

$$
x^2 + 2(y - 50)x + (y^2 - y) = 0
$$

and solving it for $x$, leading to the expression

$$
x = 50 - y \pm \sqrt{2500 - 99y}
$$

where the discriminant must be such that

$$
2500 - 99y = k^2 \quad \text{or} \quad y = \frac{2500 - k^2}{99} = \frac{(50 + k)(50 - k)}{99}
$$

The first requirement is: $k < 50$.
I. $y$ is an integer if $50 + k < 100$ is a multiple of 99, that is, $50 + k = 99, k = 49$ and thus $y = 1$, so

$$
x = 50 - 1 \pm 49, \quad \text{that is} \quad x_1 = 98, \quad x_2 = 0
$$

Clearly, only $x = 98$ is valid.

II. $50 + k$ and $50 - k$ cannot both be multiples of 3; one must be a multiple of 9 and the other a multiple of 11, so

$$
(50 + k) + (50 - k) = 9u + 11v = 100
$$

We need to find the positive integer solutions of this equation. There is only one: $u = 5$, $v = 54$. Thus,

$$
y = \frac{55 \cdot 45}{99} = 25 \quad \text{and} \quad x = 50 - 25 \pm 5
$$

that is, $x_1 = 30, x_2 = 20$.

[^0]:    ${ }^{1}$ If we list the multiples of 9 from 18 to 90, we find the one that is 1 greater than a multiple of 11.

    ${ }^{2}$ If we list the multiples of 9 from 18 to 90, we find the ones that are 1 less than a multiple of 11.

[^1]:    ${ }^{3}$ If $50 + k = 3m_1$ and $50 - k = 3m_2$, then $100 = 3(m_1 + m_2)$. This cannot be true!

    ${ }^{4}$ From the solutions of the form $u = 16 - 11t$ and $v = 9t - 4$, it follows that $\frac{4}{9} < t < \frac{16}{11}$, so only $t = 1$ is possible.