Four consecutive integers

The idea is to multiply $n$ and $n+3$ together, then $n+1$ and $n+2$ together:

$$n(n+1)(n+2)(n+3)=\left(n^2+3n\right)\left(n^2+3n+2\right).$$

If we set $x=n^2+3n$, the product becomes

$$x(x+2).$$

Now

$$x(x+2)=x^2+2x=(x+1)^2-1.$$

Therefore

$$n(n+1)(n+2)(n+3)=\left(n^2+3n+1\right)^2-1,$$

which is indeed one less than a square.