. Find the natural integers $k$ of the form $k=\frac{a^{2}+a b+b^{2}}{a b-1}$ where $a$ and $b$ are two natural integers such that $a b \neq 1$.
(arithmetic). If $a b=0$, the value of $k$ is strictly negative, which is excluded by the problem. We therefore only consider natural numbers $a, b$ such that $a b>1$. The relation $k=\frac{a^{2}+a b+b^{2}}{a b-1}$ can be written as $a^{2}+(1-k) a b+b^{2}+k=0$. First, let's show that $k \geqslant 4$. If we had $k \leqslant 3$, we would have

$$
a^{2}+(1-k) a b+b^{2}+k \geqslant a^{2}-2 a b+b^{2}+k \geqslant(a-b)^{2}+k>0
$$

which is excluded. Thus, $k \geqslant 4$.

Among the pairs $(a, b)$ of natural numbers such that $a b>1$ and $a^{2}+(1-k) a b+b^{2}+k=0$, consider the one where $a$ is minimal. The integer $b$ is a root of the quadratic polynomial with integer coefficients $f(x)=$ $x^{2}+(1-k) a x+a^{2}+k$. The sum of the roots of $f(x)$ is $(k-1) a$, and the product is $a^{2}+k$. The other root of $f(x)$ is therefore $c=(k-1) a-b$, which is an integer, and we have the factorization

$$
f(x)=x^{2}+(1-k) a x+a^{2}+k=(x-b)(x-c)
$$

With $b c=a^{2}+k$, we see that $c>0$. The integers $b$ and $c$ play the same role; without loss of generality, we can assume $b \leqslant c$. We also have $a \leqslant b$ because otherwise, by swapping $a$ and $b$, we would contradict the minimality of $a$. Thus $a \leqslant b \leqslant c$.

In the relation (IV.6), let $x=a$. It follows that $f(a)=(3-k) a^{2}+k=(b-a)(c-a) \geqslant 0$. It follows that $k-(k-3) a^{2} \geqslant 0$, so (remembering that $k \geqslant 4$)

$$
a^{2} \leqslant \frac{k}{k-3}=1+\frac{3}{k-3} \leqslant 1+3=4
$$

and thus $a \leqslant 2$, so $a=1$ or $a=2$.

If $a=1$, then $f(a)=3=(b-a)(c-a)$. Since 3 is prime and $b \leqslant c$, this implies $b-a=1$ and $c-a=3$, then $b=2$. Finally, the pair $(a, b)=(1,2)$ gives $k=7$. If $a=2$, the inequality $k-(k-3) a^{2} \geqslant 0$ implies $k \leqslant 4$, then $k=4$ (we have seen that $k \geqslant 4$) and $f(a)=0$, so $b=2$. Finally, the pair $(a, b)=(2,2)$ gives $k=4$. In conclusion, the possible values of $k$ are 4 and 7, and the minimal pairs $(a, b)$ that correspond are respectively $(2,2)$ and $(1,2)$.

Remark. It is important to note that the entire solution relies on the minimal pair $(a, b)$ associated with $k$; furthermore, it can be verified that each of the values 4 and 7 of $k$ can be obtained for an infinite number of other pairs $(a, b)$.