### IMO 2001

Problem B3

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 K2 - KM + M2 = L2 + LN + N2. Hence (KM + LN)(L2 + LN + N2) = KM(L2 + LN + N2) + LN(K2 - KM + M2) = KML2 + KMN2 + K2LN + LM2N = (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!

Note that it is not hard to find K, L, M, N satisfying the condition in the question. For example 11, 9, 5, 1.