2001 IMO Problems/Problem 6

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: $(K+L-M+N)(-K+L+M+N)=KM+LN$ $-K^2+KM+L^2+LN+KM-M^2+LN+N^2=KM+LN$

\[L^2+LN+N^2&=&K^2-KM+M^2\] (Error compiling LaTeX. Unknown error_msg)

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. Unknown error_msg)

\[&=&KM(L^2+LN+N^2)+LN(K^2-KM+M^2)\] (Error compiling LaTeX. Unknown error_msg)

\[&=&KML^2+KMN^2+K^2LN+LM^2N\] (Error compiling LaTeX. Unknown error_msg)

\[&=&(KL+MN)(KN+LM)\] (Error compiling LaTeX. Unknown error_msg)

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.

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

Problem 6

Solution

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