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

Revision as of 12:23, 27 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 $12$ , $n^4 - n^2$ must be divisible by $4$ and $3$ .

Lemma 1: $n^4 - n^2$ is divisible by 4
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 $n$ , $n^2 (n+1)(n-1)$ is divisible by $4$ .

Lemma 2: $n^4 - n^2$ is divisible by 3
Again, note that $n^4 - n^2$ can be factored into $n^2 (n+1)(n-1)$ . 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 $n$ , $n^2 (n+1)(n-1)$ is divisible by $3$ .

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

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

Revision as of 12:23, 27 July 2018

Problem

Solution

See Also