## Task 25/75

We are looking for all two-digit natural numbers whose square is equal to the cube of the sum of their digits.
The two-digit number is $10 a+b=z^{\prime}$ with $a \neq 0 ; a, b \leq 9$, a, b natural. Then the condition of the problem is:

$$
z=(10 a+b)^{2}=(a+b)^{3}=z
$$

The number $z$ is thus both a square and a cube of a natural number, and the number $z'$ is therefore a cube of a natural number. The only two-digit cubes are $z_{1}^{\prime}=27$ and $z_{2}^{\prime}=64$. Upon checking, it turns out that only $z_{1}^{\prime}=27$ satisfies the required conditions: $27^{2}=729=(2+7)^{2}$.

## Solution by Dr. Sander:

The digits are $x$ and $y$ with $1 \leq x \leq 9,0 \leq y \leq 9$; then it holds that

$$
(10+x+y)^{2}=(x+y)^{3} \quad \rightarrow \quad 10 x+y=(x+y) \sqrt{x+y}
$$

Since $x$ and $y$ are natural numbers, the radicand must be a square number; it further follows that $4 \leq x+y \leq 18$. From $x+y=4$ it follows that $x=0$, so this possibility is excluded. The remaining possibilities are:

$$
x+y=9 \quad ; \quad x+y=16
$$

with

$$
(10 x+y)^{2}=729, \quad 10 x+y=27 \quad \text { and } \quad(10 x+y)^{2}=4096, \quad 10 x+y=64
$$

where only the first one satisfies the conditions of the problem.

The only number with the required properties is thus the number 27.