KöMaL (Metresis) 1894, Problem 2

Problem:
Find all positive integers $N$ divisible only by $2$ and by $3$ such that $N^2$ has three times the number of divisors of $N$.

Solution:
Let $N=2^a 3^b$, where $a,b \in \mathbb{N}^*$. Then $N^2 = 2^{2a}3^{2b}$. So, it must be $$3(a+1)(b+1)=(2a+1)(2b+1) \iff ab-a-b-2=0 \iff (a-1)(b-1)=3.$$ Since $a-1 \geq 0$ and $b-1 \geq 0$, we obtain $$\left\{ \begin{array}{rcl} a-1 & = & 1 \\ b-1 & = & 3 \end{array} \right. \qquad \textrm{or} \qquad \left\{ \begin{array}{rcl} a-1 & = & 3 \\ b-1 & = & 1 \end{array} \right.$$ i.e. $a=2, b=4$ or $a=4, b=2$. Then all the positive integers required are $N_1=2^2 3^4=324$ and $N_2=2^4 3^2=144$.