The difference between the cubes of two integers is 127. Which are these numbers?
Let the two sought numbers be $x$ and $y$. According to the given data,

$$
x^{3}-y^{3}=(x-y)\left(x^{2}+x y+y^{2}\right)=127 \text {. }
$$

Since 127 is a prime number, this equation can only hold if

$$
x-y=1 \quad \text { and } \quad x^{2}+x y+y^{2}=127
$$

Substituting $x=y+1$ yields:

$$
(y+1)^{2}+(y+1) y+y^{2}=127, \quad \text { i.e. } \quad 3 y^{2}+3 y-126=0
$$

Simplifying:

$$
\begin{gathered}
y^{2}+y-42=0 \\
y=\frac{1}{2}(-1 \pm \sqrt{1+168})=\frac{1}{2}(-1 \pm 13)
\end{gathered}
$$

Thus,

$$
y_{1}=6, \quad y_{2}=-7 \quad \text { and hence } \quad x_{1}=7, \quad x_{2}=-6
$$

The problem has two pairs of numbers that satisfy it:

$$
x_{1}=7, \quad y_{1}=6 \quad \text { and } \quad x_{2}=-6, y_{2}=-7
$$

Gábor Várnagy (Dobó István g. VI. o. Eger.).