2. Find all positive integers $n$ for which the number $n^{2}+6 n$ is a perfect square of an integer.
2. Clearly $n^{2}+6 n>n^{2}$ and at the same time $n^{2}+6 n<n^{2}+6 n+9=(n+3)^{2}$. In the given range, there are only two squares of integers: $(n+1)^{2}$ and $(n+2)^{2}$.

In the first case, we have $n^{2}+6 n=n^{2}+2 n+1$, thus $4 n=1$, but no integer $n$ satisfies this.

In the second case, we have $n^{2}+6 n=n^{2}+4 n+4$, thus $2 n=4$. We get the only solution $n=2$.

Another solution. We will examine the factorization $n^{2}+6 n=n(n+6)$. The greatest common divisor of both numbers $n$ and $n+6$ must divide their difference, so their greatest common divisor can only be 1, 2, 3, or 6. We will analyze these four possibilities.

If the numbers $n$ and $n+6$ were coprime, each of them would have to be a square. However, the difference between two squares of natural numbers is never 6. For small numbers, this is easily verified, and for $k \geq 4$, the difference between even consecutive squares $k^{2}$ and $(k-1)^{2}$ is at least 7. The property that 1, 3, 5, and 7 are the four smallest differences between two squares will be used further.

If the greatest common divisor of the numbers $n$ and $n+6$ is 2, then $n=2 m$ for some $m$, which is not divisible by three. If $n(n+6)=4 m(m+3)$ is a square, then $m(m+3)$ must also be a square. The numbers $m$ and $m+3$ are coprime, so each must be a square of a natural number. This occurs only for $m=1$ or $n=2$. We can easily verify that $n(n+6)$ is indeed a square of an integer in this case.

If the greatest common divisor of the numbers $n$ and $n+6$ is 3, then $n=3 m$ for some odd $m$. If $n(n+6)=9 m(m+2)$ is a square, then the coprime numbers $m$ and $m+2$ must also be squares. However, such two squares do not exist.

If the greatest common divisor of the numbers $n$ and $n+6$ is 6, then $n=6 m$ for some $m$. If $n(n+6)=36 m(m+1)$ is a square, then the coprime numbers $m$ and $m+1$ must also be squares, which occurs only for $m=0$, but we are looking for positive numbers $n$.

The problem is satisfied only by $n=2$.

For a complete solution, award 6 points. If the solver reduces the problem to a manageable finite number of possibilities for $n$ (as in the first solution), award 4 points. For an argument of the type "if $n$ and $n+6$ are coprime for $n$ not divisible by 2 and 3, then they must be squares" without reducing to a finite number of possibilities, award no more than 3 points. For discovering the solution $n=2$ without proof, do not award any points.