For which natural numbers $n$ is $n^{2}+1983$ divisible by $n+1983$?
$\mathrm{Az}$

$$
n^{2}+1983=(n+1983)(n-1983)+1983 \cdot 1984
$$

According to this, $n^{2}+1983$ is divisible by $(n+1983)$ if and only if $1983 \cdot 1984$ is also divisible by it. The numbers $n$ we are looking for can thus be obtained by subtracting 1983 from the divisors of $A=1983 \cdot 1984$ that are not less than 1983.

The prime factorization of $A$ is: $A=2^{6} \cdot 3 \cdot 31 \cdot 661$. If a divisor of this number is at least 1983, then the prime factorization of the divisor must include at least one of 31 and 661, since the largest divisor not divisible by either of these primes, $2^{6} \cdot 3$, is also smaller than 1983. Let's group the sought divisors according to whether their largest prime factor is 661 or 31.

In the first case, the divisors of the form $m \cdot 661$ will be suitable, where $m \geq 1983 / 661=3$ and $m \mid 2^{6} \cdot 3 \cdot 31$. The divisors of this number are of the form $2^{a} \cdot 3^{b} \cdot 31^{c}$, where $0 \leq a \leq 6, 0 \leq b \leq 1, 0 \leq c \leq 1$, independently of each other. This results in a total of 28 numbers, of which only 1 and 2 are smaller than 3.

In the second case, the divisors of the form $m \cdot 31$ will be suitable, where $m \geq 1983 / 31=63.9$ and $m \mid 2^{6} \cdot 3$. It can be seen that only three suitable values of $m$ exist: $m=2^{6}, m=2^{5} \cdot 3$, and $m=2^{6} \cdot 3$.

In total, there are 29 natural numbers $n$ for which $n^{2}+1983$ is divisible by $(n+1983)$. These numbers are contained in the table below.

![](https://cdn.mathpix.com/cropped/2024_05_02_000563d866805c251558g-1.jpg?height=1667&width=1095&top_left_y=760&top_left_x=211)