Find all triplets of natural numbers $(a, b, k)$ such that:

$$
2^{a} 3^{b}=k(k+1)
$$

## 2 Solution
Since $k$ and $k+1$ are coprime, they cannot both be divisible by 3 nor can they both be divisible by 2. Therefore, we have 4 different cases: The first case is $2^{a} 3^{b}=k$ and $1=k+1$ but the last equation leads to $k=0$ which is not possible according to the first equation.

The second case is $2^{a} 3^{b}=k+1$ and $1=k$ but then $2^{a} 3^{b}=2$ so $a=1, b=0$ and $k=1$ but we verify that this triplet is indeed a solution.

The third case consists of

$$
2^{a}=k \quad 3^{b}=k+1
$$

which can be rewritten as

$$
2^{a}+1=3^{b}
$$

If $a=0$, we have $3^{b}=2$. Absurd.

If $a=1$, we have $3^{b}=3$ so $b=1$ so $k=2$. Conversely, the triplet $(1,1,1)$ satisfies the initial equation.

We now assume that $a \geq 2$. We look at the last equation modulo 4, we find that $3^{b} \equiv 2^{a}+1 \equiv 1 \bmod 4$ so $b$ is even and there exists a positive integer $b^{\prime}$ such that $b=2 b^{\prime}$. We can rewrite the equation as

$$
2^{a}=3^{2 b^{\prime}}-1=\left(3^{b^{\prime}}+1\right)\left(3^{b^{\prime}}-1\right)
$$

We have a product of 2 factors whose difference is 2 and which are both powers of 2. The only possibility is $3^{b^{\prime}}+1=4$ and $3^{b^{\prime}}-1=2$ which leads to $b^{\prime}=1$ so $b=2$ so $a=3$ so $k=8$. Conversely, the triplet $(3,2,8)$ satisfies the equation in the statement.

The fourth case consists of

$$
3^{b}=k \quad 2^{a}=k+1
$$

which leads to

$$
2^{a}-1=3^{b}
$$

If $b=0$ then $2^{a}=2$ so $a=1$ which leads to $k=1$. Conversely, the triplet $(1,0,1)$ satisfies the equation in the statement.

We now assume that $b \geq 1$ and we look at the equation modulo 3. We have that $2^{a}-1 \equiv$ $0 \bmod 3$ which implies that $a$ is even. So there exists a positive integer $a^{\prime}$ such that $a=2 a^{\prime}$. The equation becomes

$$
3^{b}=2^{2 a^{\prime}}-1=\left(2 a^{\prime}-1\right)\left(2^{a^{\prime}}+1\right)
$$

The product on the left is a product of 2 powers of 3 whose difference is 2. This forces $2 a^{\prime}-1=1$ and $2^{a^{\prime}}+1=3$ so $a^{\prime}=1$ so $a=2$ and $b=1$ and $k=3$. Conversely, the triplet $(2,1,3)$ satisfies the equation in the statement.

The solution triplets are therefore $\{(1,0,1),(1,1,1),(2,1,3),(3,2,8)\}$.

## 3 Group C

## 1 Statement