Find all natural numbers $a, b, c$ that are solutions to the Diophantine equation: $2^{a} 3^{b}+9=c^{2}$
Let $(a, b, c)$ be a solution triplet. We factorize directly:

$$
2^{a} 3^{b}=(c-3)(c+3)
$$

so $c-3 \mid 2^{a} 3^{b}$ and $c+3 \mid 2^{a} 3^{b}$, and therefore there exist natural numbers $w, x, y, z$ such that:

$$
c-3=2^{w} 3^{x} \text { and } c+3=2^{y} 3^{z}
$$

with $a=w+y$ and $b=x+z$. We then have:

$$
6=2^{y} 3^{z}-2^{w} 3^{x}
$$

If $a=0$: only $b=3$ works, and thus $(0,3,6)$ works.

If $b=0$: $(4,0,5)$.

Otherwise, $w, x, y, z>0$ and $y=1$ or $w=1$ because $4 \nmid 6$; $x=1$ or $z=1$ because $9 \nmid 6$:

If $y=1, x=1$:

$$
1=3^{z-1}-2^{w-1}
$$

so $z=2, w=2$ or $z=3, w=4$. Then $(3,3,15)$ and $(5,4,51)$ work.

If $y=1, z=1$:

$$
1=1-2^{w-1} 3^{x-1}
$$

$c'$ is absurd.

If $w=1, z=1$:

$$
1=2^{y-1}-3^{x-1}
$$

so $y=3, x=2$. Then $(4,3,12)$ works.

If $w=1, x=1$:

$$
1=2^{y-1} 3^{z-1}-1
$$

$c'$ is absurd.

Conversely, these solutions work.