(N2 2010)

Find all pairs $(n, m)$ of natural numbers satisfying $m^{2}+2 \cdot 3^{n}=m\left(2^{n+1}-1\right)$.
We handle the small cases for $n$ by noting that we reduce to a quadratic equation in $m$. We find the solutions $(3,6),(3,9)$ and $(6,9),(6,54)$. Let's show that these are the only ones. We now assume $n \geq 6$.

We note that $m \mid 2 \cdot 3^{n}$. Thus, there exists an integer $p \leq n$ such that $m=3^{p}$ or $m=2 \cdot 3^{p}$. Let $q=n-p$. In the first case, we get $2^{n+1}-1=m+\frac{2 \cdot 3^{n}}{m}=3^{p}+2 \cdot 3^{q}$, and in the second $2^{n+1}-1=2 \cdot 3^{p}+3^{q}$. These two equations are therefore equivalent, up to swapping $p$ and $n-p$, so there is no need to treat both cases. We therefore only consider the first.

Let's try to bound $p$. We have $3^{p}<2^{n+1}=8^{\frac{n+1}{3}}<9^{\frac{n+1}{3}}=3^{\frac{2(n+1)}{3}}$, and similarly $2 \cdot 3^{n-p}<2 \cdot 3^{\frac{2(n+1)}{3}}$. Therefore, $\frac{n-2}{3}<p, q<\frac{2(n+1)}{3}$. Let $h=\min (p, q)$. Then $9\left|3^{h}\right| 3^{p}+2 \cdot 3^{q}$, so $9 \mid 2^{n+1}-1$. Since the order of 2 modulo 9 is 6, we have $6 \mid n+1$, so $n+1=6 r$. But then we have $2^{n+1}-1=4^{3 r}-1=$ $\left(4^{2 r}+4^{r}+1\right)\left(2^{r}+1\right)\left(2^{r}-1\right)$. We observe that $4^{2 r}+4^{r}+1$ is always divisible by 3 but never by 9, and $2^{r}-1$ and $2^{r}+1$ are coprime, so $3^{h-1}$ divides one of the two numbers $2^{r}-1$ or $2^{r}+1$.

We return to our inequalities: $3^{h-1} \leq 2^{r}+1 \leq 3^{r}$. Thus, $\frac{n-2}{3}-1<h-1 \leq \frac{n+1}{6}$, or $2(n-2)-6<$ $n+1$ or again $n<11$, but we assumed $n \geq 6$ and we showed $n \equiv-1[6]$, so $n \geq 11$, which is absurd. Therefore, there are no other solutions than those mentioned at the beginning.

## 4 End-of-course training

## Statements

