You should be able to (very easily) show that n(n+1) lies between two consecutive squares for all n
OK, so the problem is Prove n(n+1) is never a square for all n>0 where n is an integer.
What i have so far is (n,(n+1))=1 ie. they are relatively prime. and that [n(n+1)]^(1/2) must divide evenly into n(n+1) so [n(n+1)]^(1/2) x C = n(n+1). I also know that the two integer solutions are n=-1 and n=0 but i don't think that is useful.
I think the problem i am having is i don't know where to start. I have a bunch of info on the problem, but no way of constructing a coherent proof. any help will be greatly appreciated.