## Task 34/65

There are four consecutive natural numbers for which the following is true:

The sum of the cubes of the two smaller numbers is equal to the difference of the cubes of the two larger numbers.

These four numbers are to be found.
When $w, x, y$ and $z$ are four consecutive natural numbers, the claim states that

$$
w^{3}+x^{3}=z^{3}-y^{3} \rightarrow w^{3}+x^{3}+y^{3}=z^{3}
$$

Furthermore, if $w$ denotes the smallest of the four numbers, then $x=w+1, y=w+2, z=w+3$, so

$$
w^{3}+(w+1)^{3}+(w+2)^{3}=(w+3)^{3} \rightarrow 2 w^{3}-12 w=18 \rightarrow w\left(w^{2}-6\right)=9
$$

Since $w$ must be a natural number, $w^{2}$ is also an integer, and thus the factor in the parentheses is an integer as well. The number 9 can therefore be factored into two integers in the following ways:

1. $9=1 \cdot 9$ : this possibility is ruled out because neither $w=1$ nor $w=9$ satisfies the equation.
2. $9=3 \cdot 3$ : $w=3$ satisfies the equation

If $w=3$, then $x=4, y=5, z=6$. Substituting these values into the original equation yields $27+64=216-125$. Thus, the problem is solved.