Difference between revisions of "2001 IMO Problems/Problem 6"

m (Problem 6)
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==Solution==
 
==Solution==
 
{{solution}}
 
{{solution}}
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First, <math>(KL+MN)-(KM+LN)=(K-N)(L-M)>0</math> as <math>K>N</math> and <math>L>M</math>.  Thus, <math>KL+MN>KM+LN</math>. 
  
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Similarly, <math>(KM+LN)-(KN+LM)=(K-L)(M-N)>0</math> since <math>K>L</math> and <math>M>N</math>.  Thus, <math>KM+LN>KN+LM</math>. 
 +
 +
Putting the two together, we have
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<cmath>KL+MN>KM+LN>KN+LM</cmath>
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 +
Now, we have:
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<cmath>(K+L-M+N)(-K+L+M+N)&=&KM+LN</cmath>
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<cmath>-K^2+KM+L^2+LN+KM-M^2+LN+N^2&=&KM+LN</cmath>
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<cmath>L^2+LN+N^2&=&K^2-KM+M^2</cmath>
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So, we have:
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<cmath>(KM+LN)(L^2+LN+N^2)&=&KM(L^2+LN+N^2)+LN(L^2+LN+N^2)</cmath>
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<cmath>&=&KM(L^2+LN+N^2)+LN(K^2-KM+M^2)</cmath>
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<cmath>&=&KML^2+KMN^2+K^2LN+LM^2N</cmath>
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<cmath>&=&(KL+MN)(KN+LM)</cmath>
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Thus, it follows that <cmath>(KM+LN) \mid (KL+MN)(KN+LM).</cmath> 
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Now, since <math>KL+MN>KM+LN</math> if <math>KL+MN</math> is prime, then there are no common factors between the two.  So, in order to have <cmath>(KM+LN)\mid (KL+MN)(KN+LM),</cmath> we would have to have <cmath>(KM+LN) \mid (KN+LM).</cmath> This is impossible as <math>KM+LN>KN+LM</math>.  Thus, <math>KL+MN</math> must be composite.
 
==See also==
 
==See also==
 
{{IMO box|num-b=5|num-a=6|year=2001}}
 
{{IMO box|num-b=5|num-a=6|year=2001}}
  
 
[[Category: Olympiad Number Theory Problems]]
 
[[Category: Olympiad Number Theory Problems]]

Revision as of 10:43, 10 February 2015

Problem 6

$K > L > M > N$ are positive integers such that $KM + LN = (K + L - M + N)(-K + L + M + N)$. Prove that $KL + MN$ is not prime.

Solution

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First, $(KL+MN)-(KM+LN)=(K-N)(L-M)>0$ as $K>N$ and $L>M$.  Thus, $KL+MN>KM+LN$.  

Similarly, $(KM+LN)-(KN+LM)=(K-L)(M-N)>0$ since $K>L$ and $M>N$. Thus, $KM+LN>KN+LM$.

Putting the two together, we have \[KL+MN>KM+LN>KN+LM\]

Now, we have:

\[(K+L-M+N)(-K+L+M+N)&=&KM+LN\] (Error compiling LaTeX. ! Misplaced alignment tab character &.)
\[-K^2+KM+L^2+LN+KM-M^2+LN+N^2&=&KM+LN\] (Error compiling LaTeX. ! Misplaced alignment tab character &.)
\[L^2+LN+N^2&=&K^2-KM+M^2\] (Error compiling LaTeX. ! Misplaced alignment tab character &.)

So, we have:

\[(KM+LN)(L^2+LN+N^2)&=&KM(L^2+LN+N^2)+LN(L^2+LN+N^2)\] (Error compiling LaTeX. ! Misplaced alignment tab character &.)
\[&=&KM(L^2+LN+N^2)+LN(K^2-KM+M^2)\] (Error compiling LaTeX. ! Misplaced alignment tab character &.)
\[&=&KML^2+KMN^2+K^2LN+LM^2N\] (Error compiling LaTeX. ! Misplaced alignment tab character &.)
\[&=&(KL+MN)(KN+LM)\] (Error compiling LaTeX. ! Misplaced alignment tab character &.)

Thus, it follows that \[(KM+LN) \mid (KL+MN)(KN+LM).\] Now, since $KL+MN>KM+LN$ if $KL+MN$ is prime, then there are no common factors between the two. So, in order to have \[(KM+LN)\mid (KL+MN)(KN+LM),\] we would have to have \[(KM+LN) \mid (KN+LM).\] This is impossible as $KM+LN>KN+LM$. Thus, $KL+MN$ must be composite.

See also

2001 IMO (Problems) • Resources
Preceded by
Problem 5
1 2 3 4 5 6 Followed by
Problem 6
All IMO Problems and Solutions
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