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

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==Problem==
 
==Problem==
  
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Show that <math>n^4 - n^2</math> is divisible by <math>12</math> for any integers <math>n > 1</math>.
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Show that <math>n^4 - n^2</math> is divisible o by <math>12</math> for any integers <math>n > 1</math>.
  
 
==Solution==
 
==Solution==
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<br>
 
<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>.
 
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>.
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 +
==Solution 2==
 +
 +
<math>n^4 - n^2</math> = <math>m^2 - m</math> if <math>n^2 = m</math>. The case 0 is trivial.
 +
The squares of 1, 2, 3, 4, 5, 6 mod 12 are 1, 4, 9, 4, 1, 0. The rest are symmetric. For m = 0 (0), 1 (1), 4 (12), and 9 (72) <math>m^2 - m</math> is divisible by 12. So <math>n^4 - n^2</math> is divisible by 12.
  
 
==See Also==
 
==See Also==
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{{Indonesia MO 7p box
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{{Indonesia MO box
 
|year=2002
 
|year=2002
 
|before=First Problem
 
|before=First Problem
 
|num-a=2
 
|num-a=2
 +
|eight=
 
}}
 
}}
  
 
[[Category:Intermediate Number Theory Problems]]
 
[[Category:Intermediate Number Theory Problems]]

Latest revision as of 20:12, 13 March 2025

Problem

Show that $n^4 - n^2$ is divisible o 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 $12$.

Solution 2

$n^4 - n^2$ = $m^2 - m$ if $n^2 = m$. The case 0 is trivial. The squares of 1, 2, 3, 4, 5, 6 mod 12 are 1, 4, 9, 4, 1, 0. The rest are symmetric. For m = 0 (0), 1 (1), 4 (12), and 9 (72) $m^2 - m$ is divisible by 12. So $n^4 - n^2$ is divisible by 12.

See Also

2002 Indonesia MO (Problems)
Preceded by
First Problem 		1 • 2 • 3 • 4 • 5 • 6 • 7 Followed by
Problem 2
All Indonesia MO Problems and Solutions
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