Number theory problem! Prove the following using the method that relies on “Universal Generalization”. If $n$ is the product of four consecutive integers then $n+1$ is a perfect square. Domain is all natural numbers
What I got so far:
Let $a$ be an element of natural numbers selected arbitrarily
$a(a+1)(a+2)(a+3) = n$
$a(a+3)(a+1)(a+2)= n$
$(a^2+3a)(a^2+3a+2)=n$
I am stuck after this and don't know where to go. Help appreciated!

$a(a+1)(a+2)(a+3)+1 = (a^2+3a+2)(a^2+3a)+1 = (a^2+3a)^2+ 2(a^2+3a) + 1 = (a^2+3a+1)^2$
