K > L > M > N are positive integers such that KM + LN = (K + L - M + N)(-K + L + M + N). Prove that KL + MN is composite.

**Solution**

Note first that KL+MN > KM+LN > KN+LM, because (KL+MN) - (KM+LN) = (K - N)(L - M) > 0 and (KM+LN) - (KN+LM) = (K - L)(M - N) > 0.

Multiplying out and rearranging, the relation in the question gives K^{2} - KM + M^{2} = L^{2} + LN + N^{2}. Hence (KM + LN)(L^{2} + LN + N^{2}) = KM(L^{2} + LN + N^{2}) + LN(K^{2} - KM + M^{2}) = KML^{2} + KMN^{2} + K^{2}LN + LM^{2}N = (KL + MN)(KN + LM). In other words (KM + LN) divides (KL + MN)(KN + LM).

Now suppose KL + MN is prime. Since it greater than KM + LN, it can have no common factors with KM + LN. Hence KM + LN must divide the smaller integer KN + LM. Contradiction.

*Comment. This looks easy, but in fact I found it curiously difficult. It is easy to go around in circles getting nowhere. Either I am getting older, or this is harder than it looks!
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Note that it is not hard to find K, L, M, N satisfying the condition in the question. For example 11, 9, 5, 1.
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© John Scholes

jscholes@kalva.demon.co.uk

12 Aug 2001

Last corrected/updated 3 Feb 2004