<s>[INST] Proving any product of four consecutive integers is one less than a perfect square Prove or disprove that : Any product of four consecutive integers is one less than a
perfect square.
OK so I start with $n(n+1)(n+2)(n+3)$ which can be rewritten $n(n+3)(n+1)(n+2)$
After multiplying we get $(n^2 + 3n)(n^2 + 3n + 2)$
How do I proceed from here to end up with something squared $- 1$?
 [/INST] Here's another approach you might use, a sort of brute-force approach.  You said you got as far as $$(n^2 + 3n)(n^2 + 3n + 2).$$  Continuing to multiply, you get: $$ n^4 + 6n^3 + 11n^2 + 6n.$$
The claim is that that this is one less than a perfect square, or equivalently, that $$n^4 + 6n^3 + 11n^2 + 6n + 1$$ is a perfect square.  If this last expression is a perfect square, it must be the square of something of the form $n^2 + an+b$ for some $a$ and $b$.  But what are $a$ and $b$?
Squaring $n^2 + an+b$, we get $$n^4 + 2an^3 +(a^2+2b)n^2 + (2ab)n + b^2$$
and equating the coefficients of the two polynomials we get $$\begin{align}
2a & = 6 \\
a^2+2b & = 11\\
2ab & = 6 \\
b^2 & = 1
\end{align}$$
which we can easily solve to obtain $a=3, b=1$.  So putting together the various steps we have $$n(n+1)(n+2)(n+3) = (n^2 + 3n + 1)^2 - 1$$ which is what we were looking for.
</s>