Assume that there are primes $p,q,r$ and an integer $n$, such that $$(p^2+p)(q^2+q)(r^2+r)=n^2.$$WLOG $p\geq q,r$. Since $p|n^2\Rightarrow p^2|n^2$ we have $$p^2|(q^2+q)(r^2+r).$$If $p|q$, then $p=q$ and $r^2+r$ is a perfect square. Contradiction. Hence, $p|(q+1)(r+1)$. WLOG $p|q+1$. As $p\geq q$, this implies $p=q+1$, so $p=3, q=2$. Hence, $r=2$ or $r=3$. In both cases we get a contradiction, since $2^2+2$ and $3^2+3$ aren't perfect squares. Therefore, there are no solutions.

parmenides51 wrote: Decide if there are primes $p, q, r$ such that $(p^2 + p) (q^2 + q) (r^2 + r)$ is a square of an integer. $(p^2 + p) (q^2 + q) (r^2 + r)=pqr(p+1)(q+1)(r+1)$ This implies that $p,q,r$ are all distinct. WLOG let $p>q>r$ Now if $pqr(p+1)(q+1)(r+1)$ is a perfect square then , there exist $k,l,m$ such that $q+1=pk$ and $r+1=ql$ and $p+1=rm$ (where $k\cdot l\cdot m$ is a perfect square) $\implies \frac{p+1}{r} ; \frac{r+1}{q} ; \frac{q+1}{p}\in \mathbb{Z^{+}}$ But $p>q>r$ implies $q=r+1$ and $p=q+1$ Thus only possibility would be $(p,q,r)=(r+2,r+1,r)$ mod 3 gives no solution.