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

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.

@@ Line 25: / Line 25: @@
 ==See also==
-{{IMO box|num-b=5|num-a=6|year=2001}}
+{{IMO box|num-b=5|after=Last Problem|year=2001}}
 [[Category: Olympiad Number Theory Problems]]

2001 IMO (Problems) • Resources
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Difference between revisions of "2001 IMO Problems/Problem 6"

Latest revision as of 23:23, 18 November 2023

Problem 6

Solution

See also