I am having trouble finding all positive integers $n$ that have two prime factors, and such that $n$ has three times fewer divisors than $n^2$. I wasn't sure if I need to use Tau or if it can be just figured out using basic number theory. Any help?
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Let $n=p^rq^s$ with $p\ne q$ prime and $r,s\ge 1$. Then $n$ has $\tau(n)=(r+1)(s+1)$ divisors. Similarly $n^2$ has $(2r+1)(2s+1)$ divisors. You need to solve $$ (2r+1)(2s+1) = 3(r+1)(s+1).$$ This is equivalent to $rs-r-s-2=0$, i.e. $(r-1)(s-1)=3$. hence $r=2,s=4$ (or vice versa, which corresponds to switching $p$ and $q$). Thus the smallest example is $n=144$.