---
type: "problem"
title: "Four consecutive integers"
slug: "four-consecutive-integers"
author: "ancient-tree"
tags: ["algebra", "factorization", "identities", "integers"]
difficulty: 8
qualityStatus: "unreviewed"
listed: true
origin: "Original Math Woods problem"
originChapter: ""
originPage: ""
originNote: "Original starter problem written for Math Woods. The mathematical idea may be classical; the wording and framing are original."
license: "CC BY-SA 4.0"
---

Show that the product of four consecutive integers is always one less than a square.

In other words, if $n$ is an integer, show that

$$
n(n+1)(n+2)(n+3)
$$

can always be written in the form $m^2-1$, where $m$ is an integer.

## Proof 1

_By @ancient-tree_

The idea is to multiply $n$ and $n+3$ together, then $n+1$ and $n+2$ together:

$$
n(n+1)(n+2)(n+3)=\left(n^2+3n\right)\left(n^2+3n+2\right).
$$

If we set $x=n^2+3n$, the product becomes

$$
x(x+2).
$$

Now

$$
x(x+2)=x^2+2x=(x+1)^2-1.
$$

Therefore

$$
n(n+1)(n+2)(n+3)=\left(n^2+3n+1\right)^2-1,
$$

which is indeed one less than a square.