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 $- 1$?
You might note that for any $N$ at all, $$(N-1)(N+1) = N^2-1$$ and so is one less than a perfect square. I presume you have seen this before.

Then you could take $N = M+1$ in the formula above, and get $$M\cdot(M+2) = (M+1)^2 - 1$$ is one less than a perfect square.

And here you have $M = n^2+3n$, don't you?
It seems like the way to attack this that doesn't require guessing is this:  Start with $N (N-1) (N-2) (N-3) = N^4 + 6 N^3 + 11 N^2 + 6 N = M^2 - 1$  Since the product is "near" $N^4$, $M$ has to be "near" $N^2$.  And it's very likely that $M$ is a polynomial in $N$.  So set $M = N^2 + aN + b$.  Then $M^2-1 = N^4 + 2aN^3 + (2b+a^2)N^2 + 2abN + (b^2-1)$.  Setting that into the first equation and equating powers of $N$ gives $2a = 6, 2b+a^2 = 11, 2ab = 6, b^2-1 = 0$, which has the unique solution $a = 3, b = 1$.  So $M = N^2 + 3N + 1$ and $N (N-1) (N-2) (N-3) = (N^2 + 3N + 1)^2 -1$.