Shortlist JBMO, 2009

Solution:

We have an integer $x$ such that
$$x^2 = k + 9$$
where $k = 2^a 3^b$, $a, b \geq 0$, $a, b \in \mathbb{N}$.
Therefore,
$$(x-3)(x+3) = k.$$
If $b = 0$ then we have $k = 16$.

If $b > 0$ then we have $3 \mid k + 9$. Hence, $3 \mid x^2$ and $9 \mid k$.
Therefore, we have $b \geq 2$. Let $x = 3y$.
$$(y-1)(y+1) = 2^a 3^{b-2}$$
If $a = 0$ then $b = 3$ and we have $k = 27$.

If $a \geq 1$, then the numbers $y-1$ and $y+1$ are even. Therefore, we have $a \geq 2$, and
$$\frac{y-1}{2} \cdot \frac{y+1}{2} = 2^{a-2} 3^{b-2}$$
Since the numbers $\frac{y-1}{2}$ and $\frac{y+1}{2}$ are consecutive numbers, these numbers have to be powers of $2$ and $3$. Let $m = a-2$, $n = b-2$.

- If $2^m - 3^n = 1$ then we have $m \geq n$. For $n = 0$ we have $m = 1$, $a = 3$, $b = 2$ and $k = 72$. For $n > 0$ using $\bmod 3$ we have that $m$ is even number. Let $m = 2t$. Therefore,
$$\left(2^t - 1\right)\left(2^t + 1\right) = 3^n$$
Hence, $t = 1$, $m = 2$, $n = 1$ and $a = 4$, $b = 3$, $k = 432$.

- If $3^n - 2^m = 1$, then $m > 0$. For $m = 1$ we have $n = 1$, $a = 3$, $b = 3$, $k = 216$. For $m > 1$ using $\bmod 4$ we have that $n$ is even number. Let $n = 2t$.
$$\left(3^t - 1\right)\left(3^t + 1\right) = 2^m$$
Therefore, $t = 1$, $n = 2$, $m = 3$, $a = 5$, $b = 4$, $k = 2592$.

Set of solutions: $\{16, 27, 72, 216, 432, 2592\}$.