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QUESTION

Explain why, for any $n,\ n(n+1)$ is divisible by 2, and
$n(n+1)(n+2)$ is divisible by 3.

Use these ideas to show that if $n$ is an odd integer then $n^3-n$
is divisible by 24.



ANSWER

Given $n\in Z$, either $n$ is even or $n$ is odd (i.e. $n=2q$ or
$n=2q+1$). If $n$ is odd, $n+1$ is even, so in all cases either
$n$ or $n+1$ is even, so $n(n+1)$ is even.

In a similar way, any $n\in Z$ can be written in one of the forms
$3q,\ 3q+1,\ 3q+2$. If $n=3q$, then $3|n$. If $n=3q+1$, then
$3|n+2$, and if $n=3q+2$, then $3|n+1$. Thus in all cases, $n$
divides one of $n,\ n+1,\ n+2$ and hence $3|n(n+1)(n+2)$, that is
to say, 3 divides the product of any 3 consecutive integers.

Now $n^3-n=n(n^2-1)=n(n-1)(n+1)=(n-1)n(n+1)$,\ so $n^3-n$ is a
product of 3 consecutive integers, and so, by the above, it is
divisible by 3.

Now given $n$ odd, we may write $n=2q+1$, and then
$(n-1)n(n+1)=2q(2q+1)(2q+2)=4(2q+1)q(q+1),$ and by the first part
of the question $q(q+1)$ is even, so $n^3-n$ is divisible by 8.

Thus $n^3-n$ is divisible by 8 and 3, and hence by 24. (see
cor.1.7).




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