Problem (Number Theory):
Find all pairs $(m, n)$ of nonnegative integers for which

$$
m^{2}+2 \cdot 3^{n}=m\left(2^{n+1}-1\right)
\tag{1}
$$

Solution:
For fixed values of $n$, the equation (1) is a simple quadratic equation in $m$. For $n \leq 5$ the solutions are listed in the following table.

| case | equation | discriminant | integer roots |
| :--- | :--- | :--- | :--- |
| $n=0$ | $m^{2}-m+2=0$ | -7 | none |
| $n=1$ | $m^{2}-3 m+6=0$ | -15 | none |
| $n=2$ | $m^{2}-7 m+18=0$ | -23 | none |
| $n=3$ | $m^{2}-15 m+54=0$ | 9 | $m=6$ and $m=9$ |
| $n=4$ | $m^{2}-31 m+162=0$ | 313 | none |
| $n=5$ | $m^{2}-63 m+486=0$ | $2025=45^{2}$ | $m=9$ and $m=54$ |

We prove that there is no solution for $n \geq 6$.

Suppose that $(m, n)$ satisfies (1) and $n \geq 6$. Since $m \mid 2 \cdot 3^{n}=m\left(2^{n+1}-1\right)-m^{2}$, we have $m=3^{p}$ with some $0 \leq p \leq n$ or $m=2 \cdot 3^{q}$ with some $0 \leq q \leq n$.

In the first case, let $q=n-p$; then

$$
2^{n+1}-1=m+\frac{2 \cdot 3^{n}}{m}=3^{p}+2 \cdot 3^{q}
$$

In the second case let $p=n-q$. Then

$$
2^{n+1}-1=m+\frac{2 \cdot 3^{n}}{m}=2 \cdot 3^{q}+3^{p}
$$

Hence, in both cases we need to find the nonnegative integer solutions of

$$
3^{p}+2 \cdot 3^{q}=2^{n+1}-1, \quad p+q=n .
\tag{2}
$$

Next, we prove bounds for $p, q$. From (2) we get

$$
3^{p}<2^{n+1}=8^{\frac{n+1}{3}}<9^{\frac{n+1}{3}}=3^{\frac{2(n+1)}{3}}
$$

and

$$
2 \cdot 3^{q}<2^{n+1}=2 \cdot 8^{\frac{n}{3}}<2 \cdot 9^{\frac{n}{3}}=2 \cdot 3^{\frac{2 n}{3}}<2 \cdot 3^{\frac{2(n+1)}{3}}
$$

so $p, q<\frac{2(n+1)}{3}$. Combining these inequalities with $p+q=n$, we obtain

$$
\frac{n-2}{3}<p, q<\frac{2(n+1)}{3}
\tag{3}
$$

Now let $h=\min (p, q)$. By (3) we have $h>\frac{n-2}{3}$; in particular, we have $h>1$. On the left-hand side of (2), both terms are divisible by $3^{h}$, therefore $9\left|3^{h}\right| 2^{n+1}-1$. It is easy check that $\operatorname{ord}_{9}(2)=6$, so $9 \mid 2^{n+1}-1$ if and only if $6 \mid n+1$. Therefore, $n+1=6 r$ for some positive integer $r$, and we can write

$$
2^{n+1}-1=4^{3 r}-1=\left(4^{2 r}+4^{r}+1\right)\left(2^{r}-1\right)\left(2^{r}+1\right)
\tag{4}
$$



Notice that the factor $4^{2 r}+4^{r}+1=\left(4^{r}-1\right)^{2}+3 \cdot 4^{r}$ is divisible by 3 , but it is never divisible by 9 . The other two factors in (4), $2^{r}-1$ and $2^{r}+1$ are coprime: both are odd and their difference is 2 . Since the whole product is divisible by $3^{h}$, we have either $3^{h-1} \mid 2^{r}-1$ or $3^{h-1} \mid 2^{r}+1$. In any case, we have $3^{h-1} \leq 2^{r}+1$. Then

$$
\begin{gathered}
3^{h-1} \leq 2^{r}+1 \leq 3^{r}=3^{\frac{n+1}{6}} \\
\frac{n-2}{3}-1<h-1 \leq \frac{n+1}{6} \\
n<11 .
\end{gathered}
$$

But this is impossible since we assumed $n \geq 6$, and we proved $6 \mid n+1$.

Final answer: $(6,3),(9,3),(9,5),(54,5)$