Find natural numbers $a$ and $b$ such that $7^{3}$ is a divisor of $a^{2}+a b+b^{2}$, but 7 is not a divisor of either $a$ or $b$.
First, we show that

$$
7^{3}\left|a^{2}+a b+b^{2} \Longleftrightarrow 7^{3}\right|(18 a-b)(18 b-a)
$$

Indeed, $(18 a-b)(18 b-a)=-18 a^{2}-18 b^{2}+325 a b=7^{3} a b-18\left(a^{2}+a b+b^{2}\right)$, from which the above equivalence is obvious.

If we now consider numbers $a, b$ such that $7 \nmid a$ and $7 \nmid b$ hold, then

$$
7^{3}\left|(18 a-b)(18 b-a) \Longleftrightarrow 7^{3}\right| 18 a-b \quad \text { or } \quad 7^{3} \mid 18 b-a
$$

also holds. Indeed, suppose that $7^{3} \mid(18 b-a)(18 a-b)$, but $7^{3} \nmid 18 a-b$ and $7^{3} \nmid 18 b-a$. Then necessarily $7 \mid 18 a-b$ and $7 \mid 18 b-a$, from which $7 \mid 18 a-b+18 b-a=17 a+17 b$ follows. From this, $7 \mid a+b$, and then $7 \mid(18 a-b)+(a+b)=19 a$ follows, so $7 \mid a$ holds.

This proves one direction of the implication, the other is obvious. Therefore,

$$
7^{3} \mid a^{2}+a b+b^{2}, \quad 7 \nmid a, \quad 7 \nmid b \Longleftrightarrow 7 \nmid a, \quad 7 \nmid b \quad \text { and } \quad 7^{3} \mid 18 a-b \quad \text { or } \quad 7^{3} \mid 18 b-a .
$$

Accordingly, the pairs of numbers that satisfy the conditions are those for which

$$
7 \nmid a, \quad 7 \nmid b \quad \text { and } \quad 7^{3} \mid 18 a-b \quad \text { or } \quad 7^{3} \mid 18 b-a
$$

and all such pairs are good. Thus, we have given all solutions; for example, $a=1, b=18$:

$$
a^{2}+a b+b^{2}=1+18+18^{2}=343=7^{3}
$$

Remark. Of course, it was not necessary to provide all solutions to get 5 points on the exam - the task was only to provide one pair of numbers.