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