Which are those two-digit numbers, when the sum of them is squared, we get the same four-digit number as when we place the two two-digit numbers next to each other?
I. Solution: Let the two-digit numbers we are looking for be denoted by $x$ and $y$. According to the problem,

$$
(x+y)^{2}=100 x+y=99 x+x+y
$$

thus, if we replace $x+y$ with $z$,

$$
99 x=z^{2}-z=z(z-1)
$$

where, since $z^{2}$ is a four-digit number,

$$
32 \leq z \leq 99
$$

and according to the condition,

$$
10 \leq x \leq 99
$$

From (1),

$$
x=\frac{z(z-1)}{99}
$$

Since $z$ and $z-1$ are consecutive integers, they cannot both be divisible by 3, and the assumption $z=99$ leads to $x=98$ and $y=1$, where the latter is not a two-digit number. Therefore, $x$ can only be a two-digit integer that meets our problem if one of $z$ and $z-1$ is divisible by 9 and the other by 11. (We note that if we allow 0 as the first digit, then $x=98, y=01$ is also a solution to the problem.) Let's first consider the case $z=9 u$ and $z-1=11 v$, i.e.,

$$
9 u=11 v+1
$$

where, as we have seen,

$$
31<9 u<99 \quad \text { and } \quad 30<11 v<98
$$

It is easy to find a solution to the indeterminate equation (2): $u=5, v=4$ is, for example, a solution, then as we know (see Surányi's article "Integer solutions of first-degree equations" in the 3-4 issue of November 1953, page 72, point 14).

$$
u=5+11 t \quad \text { and } \quad v=4+9 t
$$

is the parametric form of all solutions.

Thus,

$$
\begin{aligned}
& z=9 u=99 t+45 \text {, where } \\
& \qquad 31<99 t+45<99
\end{aligned}
$$

From this,

$$
-\frac{14}{99}<t<\frac{54}{99}
$$

or

$$
t=0, \quad z=9 u=45, \quad z-1=11 v=44
$$

thus

$$
x=\frac{45 \cdot 44}{9 \cdot 11}=5 \cdot 4=20, \quad \text { and } \quad y=z-x=45-20=25
$$

Indeed, $(20+25)^{2}=45^{2}=2025$.

Considering the case $z=11 u$ and $z-1=9 v$,

$$
11 u-9 v=1
$$

from which

$$
u=9 w-4, \quad v=11 w-5
$$

and thus

$$
\begin{gathered}
11 u=99 w-44, \quad 9 v=99 w-45 . \\
31<99 w-44<99 \text {, so } \\
w=1 \quad z=99-44=55, \quad z-1=54 . \\
x=\frac{z(z-1)}{99}=\frac{55 \cdot 54}{11 \cdot 9}=5 \cdot 6=30, \quad v=z-x=55-30=25 .
\end{gathered}
$$

Indeed, $(30+25)^{2}=55^{2}=3025$.

Apart from the solutions given here, no other solution is possible.

II. Solution: Keeping the notation,

$$
(x+y)^{2}=100 x+y=100 x+100 y-99 y=100(x+y)-99 y
$$

thus

$$
z^{2}=100 z-99 y
$$

from which

$$
y=\frac{z(100-z)}{99}
$$

Since $z=99$ again leads to $y=1$, which contradicts the fact that $y$ is two-digit, and $z=33$ results in $y$ not being an integer, either $z$ or $100-z$ must be divisible by 11 and the other by 9. This only follows when, after 8 attempts: $88,77, \ldots, 11$, we can be convinced that if

$$
z_{1}=55 \quad \text { and } \quad 100-z_{1}=45
$$

or

$$
100-z_{2}=55 \quad \text { and } \quad z_{2}=45
$$

In both cases,

$$
y_{1,2}=\frac{45 \cdot 55}{9 \cdot 11}=5 \cdot 5=25
$$

and thus

$$
x_{1}=z_{1}-y=55-25=30, \quad x_{2}=z_{2}-y=45-25=20
$$

László Eördögh (Bp., VIII., Apáczai Csere János g. IV. o. t.)