(dur) Find all positive integers $a, b, c$ such that $2^{a} 3^{b}+9=c^{2}$
We then have $2^{a} 3^{b}=(c+3)(c-3)$. Therefore, there exist $x, y, u, v$ such that $x+u=a, y+v=b, c-3=2^{x} 3^{y}$
and $c+3=2^{u} 3^{v}$. Thus, $2^{x} 3^{y}+6=2^{u} 3^{v}$. If $y \geqslant 2$, then $2^{u} 3^{v} \equiv 6[9]$, so $v=1$. If $x \geqslant 2$, then $2^{u} 3^{v} \equiv 2[4]$, so $u=1$. Therefore, if $x, y \geqslant 2$, then $u=v=1$, hence $2^{x} 3^{y}=0:$ absurd. Thus, we necessarily have $y \leqslant 1$ or $x \leqslant 1$. By testing the remaining cases (and using Exercise 19), we find the solutions: $(a, b, c) \in\{(4,0,5),(4,5,51),(3,3,15),(4,3,21),(3,2,9)\}$