Let's determine the pairs of squares of two-digit (integer) numbers that differ only in the hundreds place (for example, 2401 and 2601).
The squares of two-digit numbers are three- or four-digit numbers. Let the two square numbers be $X^{2}$ and $Y^{2}$; we can assume that $X$ and $Y$ are positive and $X>Y$. According to the requirement, in the difference of the square numbers, 0 stands in the units, tens, and hundreds places, thus

$$
X^{2}-Y^{2}=(X-Y)(X+Y)=100 t=2^{2} \cdot 5^{2} t
$$

where $t$ is an integer and $1 \leqq t \leqq 9^{1}$

Since $X-Y$ and (2) implies $l \leqq 2$. Let's write $k$ in the form $l+m$, then $X=5(k+l)=10 l+5 m$, $Y=5(k-l)=5 m, X^{2}=100 l(l+m)+25 m^{2}, Y^{2}=25 m^{2}$. Thus $m \geqq 2($ because $Y$ is two-digit), and $l(l+m)<10$, and even $100 l(l+m)$ and $25 m^{2}$ together cannot reach 1000. Therefore, for $m=5,6,7,8$, $l(l+m)$ must be less than $4,1,-2,-6$ respectively, which is impossible, so $m \leqq 4$.

The following solutions are obtained with the possible $l$ values for the remaining $m$ values:

| $m$ | 1 | 2 | 3 | 4 |
| :---: | ---: | ---: | ---: | ---: |
| $l$ | 1 | 2 | 1 | 1 |
| $X$ | 20 | 30 | 25 | 30 |
| $Y$ | 10 | 10 | 15 | 20 |
| $X^{2}$ | 400 | 900 | 625 | 900 |
| $Y^{2}$ | 100 | 100 | 225 | 400 |

2,3. Then

$$
X=25 k+l, \quad Y=25 k-l
$$

$k l=t \leqq 9$, so $X$ and $Y$ are numbers symmetric to the numbers 25, 50, 75 on the number line, it is sufficient to determine the deviation $l$.

For $k=1$, $l$ can be at most 9, but due to the requirement of the equality of the thousands digits, only the numbers $l \leqq 6$ give solutions:

$$
\begin{array}{llllll}
X^{2}=676, & 729, & 784, & 841, & 900, & 961 \\
Y^{2}=576, & 529, & 484, & 441, & 400, & 361
\end{array}
$$

(The last solution in the table was obtained again because $t=5$ makes $X^{2}-Y^{2}$ a multiple of $5^{3}$.)

For $k=2$, $l$ can be at most 4, all four pairs of numbers are valid, for $k=3$, $l \leqq 3$, but $l=3$ is not valid:

$$
\begin{array}{llllll}
X^{2}=2601, & 2704, & 2809, & 2916, & 5776, & 5929 \\
Y^{2}=2401, & 2304, & 2209, & 2116, & 5476, & 5329
\end{array}
$$

According to all this, there are 15 valid pairs of square numbers, the square numbers 100, 400, and 900 each appear in two pairs.

Katona Mária (Bp. I., Szilágyi E. gyak. lg. I. o. t.)[^0]


[^0]:    ${ }^{1}$ With this, we only used the equality of the tens and units digits of $X^{2}$ and $Y^{2}$, as (1) is only a necessary, but not a sufficient condition for the fulfillment of our requirement. We will use the requirement of the equality of the thousands digits only to eliminate cases that satisfy (1) but do not meet the problem's requirement.