Let's find two different natural numbers whose sum of cubes is equal to the square of their sum.
I. solution. Let the two numbers be $x$ and $y$, thus

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

and since $x+y>0$,

$$
x+y=x^{2}-x y+y^{2}
$$

Consider $y$ as a parameter and solve the equation for $x$. Rearrange the equation in terms of $x$ and complete the square on the left side, first multiplying by 4:

$$
\begin{gathered}
x^{2}-(y+1) x+\left(y^{2}-y\right)=0 \\
(2 x-y-1)^{2}=-3 y^{2}+6 y+1=4-3(y-1)^{2}
\end{gathered}
$$

Accordingly, the right side is not greater than 4, and only when $y-1=0$ and $y-1= \pm 1$ does it not yield a negative value. In the first case, $y=1$, and the positive value of $x$ is $x=2$. In the second case, only the value $y=2$ is positive, from which $x=1$, again yielding the same pair of numbers.

Accordingly, only the pair of numbers 1, 2 meets the requirements.

Béla Sásdy (Szentendre, Ferences g. III. o. t.)

II. solution. Divide (1) by the (positive) product $x y$. After rearranging,

$$
\frac{1}{y}+\frac{1}{x}=\frac{x}{y}-1+\frac{y}{x}, \quad \frac{x-1}{y}+\frac{y-1}{x}=1
$$

and neither fraction on the left side is negative. Choose the notation so that $x>y$, then the first term

$$
\frac{x-1}{y} \geq \frac{y}{y}=1, \quad \text { hence } \quad \frac{y-1}{x} \leq 0
$$

This is only satisfied when $y=1$, and then $x=2$.

Mihály Medveczky (Szombathely, Latinka S. gépip. t. III. o. t.)

Note. We get a solution not essentially different from the above by subtracting $x y$ from (1): $x+y-x y=(x-y)^{2} \geq 1$, or $(x-1)(y-1) \leq 0$, but the product cannot be negative because $x, y$ are positive integers, so one of $x, y$ is 1, and the other is 2.

Péter Farkas (Budapest, I. István g. I. o. t.)