## Task 16/65

The difference between the difference of the cubes and the difference of the squares of two consecutive natural numbers is 114. Which numbers are these?
The two adjacent numbers can be expressed by a symbol. If $a$ is the larger of the two, then $a-1$ is the smaller. The difference of the cubes is $a^{3}-(a-1)^{3}=3 a^{2}-3 a-1$, and the difference of the squares is $a^{2}-(a-1)^{2}=2 a-1$. The difference between both is therefore

$$
\left|\left(3 a^{2}-3 a+1\right)-(2 a-1)\right|=\left|3 a^{2}-5 a+2\right|
$$

According to the conditions of the problem, the following two equations therefore apply

$$
3 a^{2}-5 a+2=114 \rightarrow a^{2}-\frac{5}{3} a-\frac{112}{3}=0 \quad \text { and } \quad-3 a^{2}+4 a-2=114 \rightarrow a^{2}-\frac{5}{3} a+\frac{116}{3}=0
$$

The right equation has no real solutions, while the left one yields the solutions $a_{1}=7$ and $a_{2}=-\frac{32}{6}$. The solution $a_{2}$ is not a solution in the sense of the problem, as it is not an integer. The numbers sought are therefore 7 and 6.

The same result is obtained if $a$ is used to denote the smaller of the two numbers. The result is therefore unique.