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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Solution by Ancient Tree

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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.

Solution by Lemins CateFR

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L’astuce est de considérer $m^{2}-1 = (m-1)(m+1)$ et de trouver $m$ en fonction de $n$ .

Après quelques essais, on se rend compte que $n(n+3) = n^{2}+3n$ et $(n+1)(n+2)=n^{2} + 3n+2$ que l’on peut écrire sous la forme $(n^{2}+3n+1)-1$ et $(n^{2}+3n+1)+1$ .

Ainsi on peut écrire $n(n+1)(n+2)(n+3) = ((n^{2}+3n+1)-1)((n^{2}+3n+1)+1) = (m-1)(m+1) = m^{2}-1$ avec $m = n^{2} + 3n + 1$ .

Cas concret : $10\times 11\times 12\times 13 = 17160 = 131^2- 1 = (10^{2}+3\times 10 + 1)^{2} - 1$ .

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