We want to plant our seedlings in such a way that as many seedlings go into each row as there are rows. In our first attempt, 39 seedlings turned out to be excessive. If we increase the number of rows (and thus the number of trees per row) by 1, we find that 50 seedlings are missing. How many seedlings do we have available?
I. solution: If we plant $x$ seedlings in each row, and there are $x$ rows, then $x^{2}$ seedlings are planted, leaving 39 remaining. If we plant $x+1$ seedlings in each row, and there are $x+1$ rows, then $(x+1)^{2}$ seedlings are planted, leaving 50 spaces empty. Thus,

$$
x^{2}+39=(x+1)^{2}-50, \quad \text { from which } \quad x=44 .
$$

Since we have $x^{2}+39$ seedlings, this means $44^{2}+39=1975$ seedlings are available.

György Komlóssy (Szolnok, Verseghy g. I. o. t.)

![](https://cdn.mathpix.com/cropped/2024_05_02_2965b58b76ff0a9c9995g-1.jpg?height=436&width=422&top_left_y=481&top_left_x=838)

II solution: In the first planting attempt, we would plant the seedlings in $x$ rows, and in the second attempt, in $x+1$ rows. Therefore, on the shaded area in the diagram, 89 seedlings must be placed. Since $2 x+1$ seedlings fit in this area, $2 x+1=89$, $x=44$.

Gábor Fenyó (Bp. V., Eötvös g. I. o. t.)

III. solution: If the number of seedlings is $x$, then in the first attempt, $\sqrt{x-39}$ seedlings are planted in each row, and in the second attempt, $\sqrt{x+50}$ seedlings are planted in each row. However, the number of rows in the second planting is 1 more than in the first, so

$$
\sqrt{x-39}+1=\sqrt{x+50}
$$

which means

$$
\begin{aligned}
& x-39+2 \sqrt{x-39}+1=x+50 \\
& \sqrt{x-39}=44 \\
& x=44^{2}+39=1975
\end{aligned}
$$

Árpád Bokor (Szombathely, Nagy L, g. II. o. t.)