Four consecutive integers

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.

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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.

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