Selection Examination

In order the positive integer $n$ to have exactly 6 positive divisors, it must be of the form $n^2-9=q^5$ or $n^2-9=p^{2}q$, where $p$, $q$ are primes mutually different.

In the first case we have $n^2-9=q^5\Rightarrow (n-3)(n+3)=q^5$ (1) It gives that $$n-3=q^s\text{ and }n+3=q^t,\text{ with }t>s\text{ and }t+s=5.$$ By subtracting we get $q^t-q^s=6\iff q^s(q^{t-s}-1)=2\cdot 3$ and hence $q\in \{2,3\}$. For $q=2$ (1) gives $n^2=41$, (it has no solutions in $\mathbb{Z}$), while for $q=3$, (1) gives $n^2=252$, (impossible in $\mathbb{Z}$).

In the second case $n^2-9=p^{2}q$, it follows that $(n-3)(n+3)=p^{2}q$, and so $$\begin{array}{r}(n+3=pq\text{ and }n-3=p)\text{ or }(n+3=q\text{ and }n-3=p^2)\\ \text{or }(n+3=p^2\text{ and }n-3=q).\end{array}$$ From the first case we get $p(q-1)=6$, and hence ($p=3$ and $q=3$), which must be rejected because $p\ne q$. From the last two cases we conclude that $\text{gcd}(n-3,n+3)=1$.