Consider $$ \left(\dfrac{n(n+1)}{2}\right)^2-1. $$ Is is possible to say something about the lower bound on the greatest prime divisor of the above expression depending only on $n$? I surfed through some articles and sites on the internet but have not managed to find anything about this concrete situation.
Obviously, we can write it as $$ \left(\dfrac{n(n+1)}{2}+1\right)\left(\dfrac{n(n+1)}{2}-1\right). $$ It is also evident that the greatest common divisor of these two terms can be at most $2$. I do not know if this help, but at least we can say that the greatest prime divisor divides exactly one of these terms...
Also it is one less then the square of a triangular number. I added both elementary and standard number theory as tags, because I am not able to decide the hardness of the problem.