Product of 4 Consecutive Integers is One Less than Square/Proof 1

From ProofWiki
Jump to navigation Jump to search

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

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.