Three consecutive odd numbers, the sum of the squares of which is a four-digit number, all of whose digits are equal. Which are these numbers?
I. Solution: Let the middle odd number be denoted by $x$, then according to the problem,

\[
(x-2)^{2}+x^{2}+(x+2)^{2}=3 x^{2}+8=1111 y
\]

where $y$ is a single-digit odd number.

Following the method for solving Diophantine equations,

\[
x^{2}=\frac{1111 y-8}{3}=370 y-2+\frac{y-2}{3}=370 y-2+u
\]

where

\[
0<y=3 u+2 \leq 9
\]

From this,

\[
0<u \leq \frac{7}{3}, \quad \text { or } \quad u=1, \quad \text { or } \quad 2 \text {. }
\]

$u=2$ does not fit, because in this case $y$ would be even, so only $u=1, y=5$ can provide a solution, if one exists at all. For $y=5$,

\[
x^{2}=370 \cdot 5-2+1=1849=43^{2}
\]

from which

\[
x= \pm 43
\]

and thus

\[
41, 43, 45 \quad \text { or } \quad-45,-43,-41
\]

so our problem has two solutions.

Indeed, $41^{2}+43^{2}+45^{2}=1681+1849+2025=5555$.

Marianna Katona (Bp. VIII., Apáczai Csere lg. III. o. t.)

II. Solution: The Diophantine equation

\[
3 x^{2}+8=1111 y
\]

in the first solution can be solved more simply under the conditions of the problem. The left side, when divided by 3, gives the same remainder as 8, which is 2. The right side must also give the same remainder. The right side can be written in the form $3 \cdot 370 y+y$, and it is clear that $y$ must give a remainder of 2 when divided by 3. However, among the single-digit odd numbers, only 5 does this, so $y=5$, and thus $3 x^{2}+8=5555$, from which $x^{2}=\frac{5547}{3}=1849 \mathrm{~s} . \mathrm{i} . \mathrm{t} .$, as in the first solution.

Sándor Frivaldszky (Bp. II., Rákóczi g. III. o. t.)