Find all integers $a, b, k$ such that $2^{b} 3^{a}=k(k+1)$
The only prime divisors of $k$ and $k+1$ are 2 and 3. Since they are coprime, one is a power (possibly multiplied by -1) of 2 and the other is a power of 3 (possibly multiplied by -1). We are thus reduced exactly to Exercise 19 which provides the solutions $(a, b, k):(0,1,1),(0,1,-2),(1,1,2),(1,1,-3),(1,2,3),(1,2,-4),(2,3,8),(2,3,-9)$.