Difference between revisions of "2002 Indonesia MO Problems/Problem 1"

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[[Category:Intermediate Number Theory Problems]]
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[[Category:Introductory Number Theory Problems]]

Revision as of 12:10, 14 July 2018

Problem

Show that $n^4 - n^2$ is divisible by $12$ for any integers $n > 1$.

Solution

In order for $n^4 - n^2$ to be divisible by $12$, it must be divisible by $4$ and $3$. Note that $n^4 - n^2$ can be factored into $n^2 (n+1)(n-1)$.

If $n$ is even, then $n^2 \equiv 0 \pmod{4}$. If $n \equiv 1 \pmod{4}$, then $n-1 \equiv 0 \pmod{4}$, and if $n \equiv 3 \pmod{4}$, then $n+1 \equiv 0 \pmod{4}$. That means for all positive $n$, $n^4 - n^2$ is divisible by $4$.

If $n \equiv 0 \pmod{3}$, then $n^2 \equiv 0 \pmod{3}$. If $n \equiv 1 \pmod{3}$, then $n-1 \equiv 0 \pmod{3}$. If $n \equiv 2 \pmod{3}$, then $n+1 \equiv 0 \pmod{3}$. That means for all positive $n$, $n^4 - n^2$ is divisible by $3$.

Because $n^4 - n^2$ is divisible by $4$ and $3$, it must be divisible by $12$.

See Also

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