1. a) Decide whether there exists a natural number $n$ such that $2n$ is the square of a natural number and $3n$ is the cube of a natural number.

b) Decide whether there exists a natural number $n$ that satisfies both conditions from part a) and additionally $4n$ is the fourth power of a natural number.

(Josef Tkadlec)
SOLUTION. a) Such a number exists. For example, the number $n=72$ satisfies both conditions, since for it $2 n=144=12^{2}$ and $3 n=216=6^{3}$.

However, let's also see how we arrive at (the smallest suitable) number 72. Consider the prime factorization of the sought number $n$ in the form

$$
n=2^{a_{2}} \cdot 3^{a_{3}} \cdot \ldots \cdot p^{a_{p}},
$$

where $p$ is the largest prime dividing $n$, and the numbers $a_{2}, \ldots, a_{p}$ are non-negative integers. Then the number $2 n$ has the prime factorization

$$
2 n=2^{a_{2}+1} \cdot 3^{a_{3}} \cdot \ldots \cdot p^{a_{p}}.
$$

We note that this number is a square of a natural number if and only if all exponents $a_{2}+1, a_{3}, \ldots, a_{p}$ are even. The prime factorization of the number $3 n$ is

$$
3 n=2^{a_{2}} \cdot 3^{a_{3}+1} \cdot \ldots \cdot p^{a_{p}}.
$$

This number is a cube if and only if all exponents $a_{2}, a_{3}+1, \ldots, a_{p}$ are divisible by three. Therefore, the number $n$ satisfies both conditions from the problem if and only if: the number $a_{2}$ is odd and divisible by three, the number $a_{3}$ is even and gives a remainder of 2 when divided by three, and finally, for each prime $q>3$, the number $a_{q}$ is divisible by both two and three (i.e., it is divisible by six). We see that the exponents $a_{2}=3$ and $a_{3}=2$ are the smallest that satisfy the corresponding conditions, and at the same time in (1) we can choose $p=3$ (i.e., set $a_{q}=0$ for each $q>3$). This gives us the smallest possible $n=2^{3} \cdot 3^{2}=72$, which satisfies the conditions of part a). At the same time, from our considerations, it follows that all suitable numbers $n$ are of the form

$$
n=2^{6 a-3} \cdot 3^{6 b-4} \cdot M^{6}, \quad \text { or more simply } \quad n=2^{3} \cdot 3^{2} \cdot N^{6},
$$

where $a, b, M, N$ are positive integers (we can also assume that the number $M$ is coprime with the number 6).

b) We will prove by contradiction that such a number $n$ does not exist. Its prime factorization (1) would have to satisfy the conditions determined in part a) of our solution and, in addition, the number

$$
4 n=2^{a_{2}+2} \cdot 3^{a_{3}} \cdot \ldots \cdot p^{a_{p}}
$$

would have to be a fourth power, i.e., all exponents $a_{2}+2, a_{3}, \ldots, a_{p}$ would have to be divisible by four. In particular, the number $a_{2}+2$ would have to be divisible by four, even though the number $a_{2}$ is, as we know from the solution of part a), necessarily odd. This contradiction completes the proof of the non-existence of the number $n$.

ANOTHER SOLUTION to part b) without considering prime factorizations: The sought number $n$ must satisfy the equation $2 n=k^{2}$ for some natural $k$. Since $\sqrt{4 n}=\sqrt{2 k^{2}}=k \cdot \sqrt{2}$ and the number $\sqrt{2}$ is known to be irrational, the radicand $4 n$ is not a square, and thus not a fourth power of a natural number.

For a complete solution, award 6 points, 3 points for part a) and 3 points for part b). For a minor flaw in the argument for part b), deduct 1 point.