Product of 4 Consecutive Integers is One Less than Square

Theorem
Let $a$, $b$, $c$ and $d$ be consecutive integers.

Then:
 * $\exists n \in \Z: a b c d = n^2 - 1$

That is, the product of $a$, $b$, $c$ and $d$ is one less than a square.

Proof
If $0 \in \set {a, b, c, d}$ then $a b c d = 0$ which is $1$ less than $1^2$.

If $a$, $b$, $c$ and $d$ are all negative, then:
 * $a b c d = \paren {-a} \paren {-b} \paren {-c} \paren {-d}$

Hence it is sufficient to consider $a$, $b$, $c$ and $d$ all strictly positive.

As they are all consecutive, we can express them as:
 * $a$, $a + 1$, $a + 2$ and $a + 3$

where $a \ge 1$.

Hence:

Hence the result.