# 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 1

### Lemma

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

Let us wish to prove that the product of $a$, $b$, $c$ and $d$ is one less than a square.

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

$\Box$

As $a$, $b$, $c$ and $d$ are all consecutive, we can express them as:

$a$, $a + 1$, $a + 2$ and $a + 3$

where $a \ge 1$.

Hence:

 $\ds a \paren {a + 1} \paren {a + 2} \paren {a + 3} + 1$ $=$ $\ds a^4 + 6 a^3 + 11 a^2 + 6 a + 1$ $\ds$ $=$ $\ds \paren {a^2 + 3 a + 1}^2$ by inspection

Hence the result.

## Proof 2

### Lemma

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

Let us wish to prove that the product of $a$, $b$, $c$ and $d$ is one less than a square.

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

$\Box$

As $a$, $b$, $c$ and $d$ are all consecutive, we can express them as:

$a$, $a + 1$, $a + 2$ and $a + 3$

where $a \ge 1$.

Then:

 $\ds a \paren {a + 3}$ $=$ $\ds a^2 + 3 a$ $\ds \paren {a + 1} \paren {a + 2}$ $=$ $\ds a^2 + 3 a + 2$ $\ds \leadsto \ \$ $\ds a \paren {a + 1} \paren {a + 2} \paren {a + 3}$ $=$ $\ds \paren {a^2 + 3 a} \paren {a^2 + 3 a + 2}$ $\ds$ $=$ $\ds \paren {n - 1} \paren {n + 1}$ where $n = a^2 + 3 a + 1$ $\ds$ $=$ $\ds n^2 - 1$ Difference of Two Squares

$\blacksquare$