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

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==Solution==
 
==Solution==
  
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In order for <math>n^4 - n^2</math> to be divisible by <math>12</math>, it must be divisible by <math>4</math> and <math>3</math>.  Note that <math>n^4 - n^2</math> can be factored into <math>n^2 (n+1)(n-1)</math>.
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In order for <math>n^4 - n^2</math> to be divisible by <math>12</math>, <math>n^2 (n+1)(n-1)</math> must be divisible by <math>4</math> and <math>3</math>.
  
−
If <math>n</math> is even, then <math>n^2 \equiv 0 \pmod{4}</math>.  If <math>n \equiv 1 \pmod{4}</math>, then <math>n-1 \equiv 0 \pmod{4}</math>, and if <math>n \equiv 3 \pmod{4}</math>, then <math>n+1 \equiv 0 \pmod{4}</math>.  That means for all positive <math>n</math>, <math>n^4 - n^2</math> is divisible by <math>4</math>.
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<br>
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'''Lemma 1:''' <math>n^2 (n+1)(n-1)</math> is divisible by 4<br>
 +
Note that <math>n^4 - n^2</math> can be factored into <math>n^2 (n+1)(n-1)</math>.  If <math>n</math> is even, then <math>n^2 \equiv 0 \pmod{4}</math>.  If <math>n \equiv 1 \pmod{4}</math>, then <math>n-1 \equiv 0 \pmod{4}</math>, and if <math>n \equiv 3 \pmod{4}</math>, then <math>n+1 \equiv 0 \pmod{4}</math>.  That means for all positive <math>n</math>, <math>n^2 (n+1)(n-1)</math> is divisible by <math>4</math>.
  
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If <math>n \equiv 0 \pmod{3}</math>, then <math>n^2 \equiv 0 \pmod{3}</math>.  If <math>n \equiv 1 \pmod{3}</math>, then <math>n-1 \equiv 0 \pmod{3}</math>.  If <math>n \equiv 2 \pmod{3}</math>, then <math>n+1 \equiv 0 \pmod{3}</math>.  That means for all positive <math>n</math>, <math>n^4 - n^2</math> is divisible by <math>3</math>.
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<br>
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'''Lemma 2:''' <math>n^2 (n+1)(n-1)</math> is divisible by 3<br>
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Again, note that <math>n^4 - n^2</math> can be factored into <math>n^2 (n+1)(n-1)</math>.  If <math>n \equiv 0 \pmod{3}</math>, then <math>n^2 \equiv 0 \pmod{3}</math>.  If <math>n \equiv 1 \pmod{3}</math>, then <math>n-1 \equiv 0 \pmod{3}</math>.  If <math>n \equiv 2 \pmod{3}</math>, then <math>n+1 \equiv 0 \pmod{3}</math>.  That means for all positive <math>n</math>, <math>n^2 (n+1)(n-1)</math> is divisible by <math>3</math>.
  
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Because <math>n^4 - n^2</math> is divisible by <math>4</math> and <math>3</math>, it must be divisible by <math>12</math>.
+
<br>
 +
Because <math>n^4 - n^2</math> is divisible by <math>4</math> and <math>3</math>, <math>n^4 - n^2</math> must be divisible by <math>12</math>.
  
 
==See Also==
 
==See Also==

Revision as of 12:20, 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 $12$, $n^2 (n+1)(n-1)$ must be divisible by $4$ and $3$.


Lemma 1: $n^2 (n+1)(n-1)$ 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: $n^2 (n+1)(n-1)$ 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 $12$.

See Also

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