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

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==Solution==
 
==Solution==
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{{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>.   
 
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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Now, we have:
 
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+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>-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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<cmath>L^2+LN+N^2=K^2-KM+M^2</cmath>
 
So, we have:
 
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+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>=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>=KML^2+KMN^2+K^2LN+LM^2N</cmath>
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<cmath>&=&(KL+MN)(KN+LM)</cmath>
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<cmath>=(KL+MN)(KN+LM)</cmath>
 
Thus, it follows that <cmath>(KM+LN) \mid (KL+MN)(KN+LM).</cmath>   
 
Thus, it follows that <cmath>(KM+LN) \mid (KL+MN)(KN+LM).</cmath>   
 
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.
 
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==
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{{IMO box|num-b=5|num-a=6|year=2001}}
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{{IMO box|num-b=5|after=Last Problem|year=2001}}
  
 
[[Category: Olympiad Number Theory Problems]]
 
[[Category: Olympiad Number Theory Problems]]

Latest revision as of 23:23, 18 November 2023

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

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: \[(K+L-M+N)(-K+L+M+N)=KM+LN\] \[-K^2+KM+L^2+LN+KM-M^2+LN+N^2=KM+LN\] \[L^2+LN+N^2=K^2-KM+M^2\] So, we have: \[(KM+LN)(L^2+LN+N^2)=KM(L^2+LN+N^2)+LN(L^2+LN+N^2)\] \[=KM(L^2+LN+N^2)+LN(K^2-KM+M^2)\] \[=KML^2+KMN^2+K^2LN+LM^2N\] \[=(KL+MN)(KN+LM)\] 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
Last Problem
All IMO Problems and Solutions
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