3. Determine all pairs of positive integers $m$ and $n$ for which $37+27^{m}=n^{3}$.

The written part of the school round in category B takes place

## on Thursday, January 21, 2010

so that it starts in the morning and the participants have 4 hours of pure time to solve the problems. For each problem, participants can earn 6 points, and a successful solver is a student who scores 10 points or more. Allowed aids are writing and drawing supplies, school MF tables, and calculators without a graphical display. These details will be communicated to the students before the start of the competition.

## Solutions to the problems of the written part of the school round in category $B$
3. We rewrite the equation in the form $37=n^{3}-27^{m}$ and factor the difference of cubes:

$$
37=\left(n-3^{m}\right)\left(n^{2}+n \cdot 3^{m}+9^{m}\right) .
$$

The number 37 is a prime number, and on the right side of the equation, there is a product of two integers, with the second factor being greater than 1. Therefore, it must be true that

$$
n-3^{m}=1
$$

and

$$
n^{2}+n \cdot 3^{m}+9^{m}=37 \text {. }
$$

For $m \geqq 2$, $n^{2}+n \cdot 3^{m}+9^{m}>9^{2}>37$, so the only remaining possibility is $m=1$; from (3) it then follows that $n=1+3^{m}=4$. By checking, we can verify that $37+27^{1}=4^{3}$, or we can confirm that the pair $m=1, n=4$ satisfies condition (4).

For a complete solution, award 6 points, of which 2 points for rewriting the equation in the form $37=\left(n-3^{m}\right)\left(n^{2}+n \cdot 3^{m}+9^{m}\right)$, 2 points for the system of equations (3), (4), one point for justifying the equality $m=1$, and one point for determining the value of $n$ and verifying the correctness. Do not award any points for merely guessing the solution.