For which natural numbers $n$ is it true that the fraction

$$
\frac{n^{2}+2}{n(n+1)}
$$

is not simplifiable?
The fraction can be transformed as follows:

$$
\frac{n^{2}+2}{n^{2}+n}=\frac{n^{2}+n-n+2}{n^{2}+n}=1-\frac{n-2}{n(n+1)},
$$

thus it is irreducible if and only if the second fraction in the last form, denoted by $t$, is irreducible.

$t$ can only be simplified by a proper (greater than 1) divisor of $(n-2)$ that is either a divisor of $n$ or a divisor of $(n+1)$, but not both, since no proper divisor can be a divisor of both $n$ and $(n+1)$, as they are consecutive natural numbers and have no proper common divisor.

Now, the proper common divisors of $n$ and $(n-2)$ are also divisors of their difference, 2, so $\frac{n-2}{n}$ can only not be simplified if $n$ is not divisible by 2, i.e., if the remainder of $n: 2$ is 1.

Similarly, the only proper common divisor of $(n+1)$ and $(n-2)$ is 3, the only proper divisor of their difference, 3, so $\frac{n-2}{n+1}$ can only not be simplified if $n+1$ is not divisible by 3. In other words, if the remainder of $(n+1): 3$ is 1 or 2, or if the remainder of $n: 3$ is 0 or 1.

These two conditions can be combined by considering the remainder of $n$ when divided by $2 \cdot 3=6$. From this perspective, every integer falls into one of the following 6 forms: $6k, (6k+1), (6k+2), (6k+3), (6k+4), (6k+5)$, where $k$ is an integer. The first condition is met by the 2nd, 4th, and 6th forms, while the second condition is met by the 1st, 2nd, 4th, and 5th forms, so both conditions are met only by the 2nd and 4th forms. Therefore, the fraction in question is irreducible for natural numbers of the form $n=6k+1$ and $n=6k+3$, and reducible in all other cases.

Gáll Márton (Budapest, Arany J. Gymnasium and Secondary School, 8th grade)