164. Find a perfect number of the form $p^{2} q$.
164. From the equation $\left(p^{2}+p+1\right)(q+1)=2 p^{2} q$, we find $q=1+\frac{2(p+1)}{p^{2}-p-1}$, from which we obtain:
1) $p^{2}-p-1=1, p=2, q=7$
2) $p^{2}-p-1=2$, no solutions.

Moreover, the numbers $p+1$ and $p^{2}-p-1$ have no common divisors, since $\gcd \left(p+1, p^{2}-p-1\right)=\gcd[p(p+1), p^{2}-p-1]=\gcd(2 p+1, p+1)=\gcd(p, p+1)=1$.

Answer. 28.