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!

You can take $$n = x(x+1)(x+2)(x+3)$$ $$n+1 = x(x+1)(x+2)(x+3) + 1 = x^4 +6x^3 +11x^2 + 6x +1 =$$ 
$$(x^4 + 9x^2 + 6x^3) + (2x^2 + 6x) + 1 =  (x^2 + 3x)^2 + 1 + 2(x^2+3x) = (x^2+3x+1)^2 $$
Since x is a natural number, n+1 is a perfect square of a natural number.
