## Task 1 - 280921

Determine the smallest four numbers that are the square of a natural number and at the same time the cube of another natural number!
I. If a number is the square of a natural number $n$ and at the same time also the third power of a natural number $m$, then it is a natural number, in whose prime factorization each prime factor occurs in a number divisible by 2 and at the same time by 3.

Therefore (and because 2 and 3 are coprime), each prime factor must occur in a number divisible by 6, so the number must be the sixth power of a natural number. If the condition $n \neq m$ is also to be met, then 0 and 1 are excluded, as these numbers are squares and third powers of the same number only.

II. The numbers $2^{6}, 3^{6}, 4^{6}$, and $5^{6}$ all satisfy these conditions, as is evident from

$$
216=8^{2}=4^{3}, \quad 3^{6}=27^{2}=9^{3}, \quad 4^{6}=64^{2}=16^{3}, \quad 5^{6}=125^{2}=25^{3}
$$

With I. and II. (and because all $k^{6}$ with $k>5$ are greater than $2^{6}, \ldots, 5^{6}$), it is shown: The four sought numbers are $2^{6}=64, 3^{6}=729, 4^{6}=4096, 5^{6}=15625$.