9.3. Consider natural numbers $a, b$, and $c$ such that the fraction

$$
k=\frac{a b+c^{2}}{a+b}
$$

is a natural number less than $a$ and $b$. What is the smallest number of natural divisors that the number $a+b$ can have?

(P. Kozlov)
Answer. Three divisors.

First solution. Since the number $a+b$ is greater than one, it has at least two distinct divisors. We will prove that there cannot be exactly two, i.e., that the number $a+b$ cannot be prime. Multiplying the equality from the condition by the denominator, we get $a b+c^{2}=k a+k b$ or, equivalently, $a b-k a-k b+k^{2}=k^{2}-c^{2}$. Factoring both sides, we arrive at the relation

$$
(a-k)(b-k)=(k-c)(k+c)
$$

Since $k<a$ and $k<b$, both parentheses on the left side are positive, and thus, $c<k$. Then there exist natural numbers $x, y, z$ and $t$ such that

$$
a-k=x y, \quad b-k=z t, \quad k-c=x z \quad \text { and } \quad k+c=y t .
$$

For example, we can set $x=\text{GCD}(a-k, k-c), t=\text{GCD}(b-k, k+c), y=(a-k) / x$ and $z=(b-k) / t$. Then the first two equalities are satisfied by definition; on the other hand, $k-c$ divides $x z$, and $k+c$ divides $y t$, so the written equalities follow from the equality of the products.

Therefore,

$$
\begin{aligned}
& a+b=(a-k)+(b-k)+(k-c)+(k+c)= \\
& =x y+z t+x z+y t=(x+t)(y+z)
\end{aligned}
$$

Thus, the number $a+b$ can be represented as the product of two natural numbers greater than 1, and therefore, it is not prime.

Finally, it is not difficult to see that $a+b$ can have exactly three distinct divisors. For example, if $a=10, b=15, c=5$, then $k=\frac{10 \cdot 15+5^{2}}{10+15}=7$, and $a+b=25=5^{2}$ has three divisors.

Remark. For no $a$ and $b$ can the sum $a+b$ equal $2^{2}$ or $3^{2}$. But for any prime number $p \geqslant 5$, there exist numbers $a, b$ and $c$ satisfying the condition of the problem such that $a+b=p^{2}$. For the case $a \leqslant b$, all suitable $a, b, c$ and $k$ have the form $a=n p$, $b=(p-n) p, c=m p$ and $k=n p-n^{2}+m^{2}$, where $2 \leqslant n \leqslant \frac{p-1}{2}$, $1 \leqslant m \leqslant n-1$. In particular, for $p=5$ the example is unique up to the permutation of the numbers $a$ and $b$.

Second solution. We will provide another proof that the number $p=a+b$ cannot be prime. Assume the contrary.

Assume $a \leqslant b$. Then the number $k p=a b+c^{2}=a(p-a)+c^{2}=a p+c^{2}-a^{2}$ is divisible by $p$ and less than $a p$. Therefore, the number $a^{2}-c^{2}=(a-c)(a+c)$ is positive and divisible by $p$. Then the first parenthesis is positive and $a-c<a+b=p$, so it does not divide $p$. The second parenthesis is also positive and $a+c<2 a \leqslant a+b=p$, so it also does not divide $p$. We have reached a contradiction, so the assumption is false. Thus, $a+b$ is a composite number and, therefore, it has at least three divisors.

Comment. Only the answer - 0 points.

Only the answer and an example where $a+b$ is the square of a prime number (i.e., it has three divisors) - 2 points.

Only the proof that the number $a+b$ cannot be prime (i.e., it has more than two divisors) - 4 points.

It is well known that the number of divisors of a natural number $n$, having a prime factorization of the form $n=p_{1}^{\alpha_{1}} \ldots p_{k}^{\alpha_{k}}$, is $\left(\alpha_{1}+1\right)\left(\alpha_{2}+1\right) \ldots\left(\alpha_{k}+1\right)$. If this formula is used without proof - points are not deducted.

In the first solution, the representation (*) is known. If it is used without proof - points are not deducted. If it is used with an incorrect proof - 1 point is deducted.