Find the smallest positive integers $a$ and $b$ such that

$$
7 a^{3}=11 b^{5}
$$

(A. Bohiniková)

Hint. It is worth focusing on prime numbers.
From the given equation, it follows that among the prime factors of $a$ and $b$, there must be 7 and 11. Since we are looking for the smallest suitable numbers $a$ and $b$, we can assume that these numbers have no other prime factors. Thus, we are looking for numbers of the form

$$
a=7^{*} \cdot 11^{*}, \quad b=7^{*} \cdot 11^{*}
$$

where the asterisks represent unknown (positive integer) exponents. The smallest suitable numbers correspond to the smallest suitable exponents.

The given equation holds if and only if the prime factors appear on both sides in the same quantities (exponents). We will consider the possible factors, starting from the smallest counts, beginning with the right side:

- If $b=7 \cdot 11^{*}$, then on the right side there is $7^{5}$ (and some power of 11). However, the same power of 7 cannot be obtained on the left side: $7 \cdot 7^{3}=7^{4} \neq 7^{5}$.
- If $b=7^{2} \cdot 11^{*}$, then on the right side there is $\left(7^{2}\right)^{5}=7^{10}$ (and some power of 11). The same power of 7 can be obtained on the left side: $7 \cdot\left(7^{3}\right)^{3}=7^{10}$.

Thus, the sought numbers are of the form

$$
a=7^{3} \cdot 11^{*}, \quad b=7^{2} \cdot 11^{*} \text {. }
$$

- If $b=7^{*} \cdot 11$, then on the right side there is $11 \cdot 11^{5}=11^{6}$ (and some power of 7). The same power of 11 can be obtained on the left side: $\left(11^{2}\right)^{3}=11^{6}$.

Thus, the sought numbers are

$$
a=7^{3} \cdot 11^{2}=41503, \quad b=7^{2} \cdot 11=539 .
$$

Notes. If we denote the unknown exponents in $(*)$ as

$$
a=7^{k} \cdot 11^{m}, \quad b=7^{l} \cdot 11^{n},
$$

then after substitution and simplification of the equation from the problem, we get

$$
\begin{aligned}
7 \cdot\left(7^{k} \cdot 11^{m}\right)^{3} & =11 \cdot\left(7^{l} \cdot 11^{n}\right)^{5} \\
7^{3 k+1} \cdot 11^{3 m} & =7^{5 l} \cdot 11^{5 n+1} .
\end{aligned}
$$

By comparing the corresponding factors, we obtain linear equations

$$
3 k+1=5 l, \quad 3 m=5 n+1 .
$$

The smallest positive integer solutions to these equations are $k=3, m=2, l=2, n=1$, which corresponds to $(* *)$.

For interest, we add that all integer solutions to the given equation are of the form

$$
a=41503 \cdot c^{5}, \quad b=539 \cdot c^{3},
$$

where $c$ is any integer.