## Task B-4.1.

The first row of the theater has 15 seats, and each subsequent row has two more seats. The total number of seats in the theater is the square of some natural number. How many rows are there in this theater?
## First Solution.

Let $k$ be the number of rows in the theater. The total number of seats $S$ in the theater is equal to the sum of an arithmetic sequence:

$S=15+17+19+\cdots+(15+2(k-1))=k^{2}+14 k$.

For $k^{2}+14 k$ to be a perfect square, we need to add 49, so we have:

$k^{2}+14 k+49=(k+7)^{2}=n^{2}+49$,

$(k+n+7)(k-n+7)=49$.

Since $49=( \pm 1) \cdot( \pm 49)$ or $49=( \pm 7) \cdot( \pm 7)$, and $k+n+7>0$ and $k-n+7<k+n+7$, we conclude that the only possibility is $k-n+7=1$ and $k+n+7=49$.

From the above system of equations, we get $k=18$, meaning there are 18 rows in the theater.

Note: From $n^{2}+7^{2}=(k+7)^{2}$, a student can also find the solution by looking for the "smallest" Pythagorean triple where one of the members is 7, which is $(7,24,25)$. It follows that $k+7=25$, or $k=18$.