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 |
$\blacksquare$
Sources
- 1980: David M. Burton: Elementary Number Theory (revised ed.) ... (previous) ... (next): Chapter $2$: Divisibility Theory in the Integers: $2.2$ The Greatest Common Divisor: Problems $2.2$: $8 \ \text {(b)}$