7th Irish 1994 Problem B2 p, q, r are distinct reals such that q = p(4-p), r = q(4-q), p = r(4-r). Find all possible values of p+q+r.
6,7 Solution By inspection p = 3 gives q = 3, r = 3 and p = 0 gives q = 0, r = 0. These are disallowed by the requirement that p, q, r be distinct, but they are likely to be solutions of the equations. It is convenient to put P = p-2, Q = q-2, R = r-2, so Q = 2 - P 2 , R = 2 - Q 2 , P = 2 - R 2 . Substituting gives P = 2 - R 2 = -2 + 4Q 2 - Q 4 = -2 + 4(2 - P 2 ) 2 - (2 - P 2 ) 4 = -(P 8 - 8P 6 + 20P 4 - 16P 2 + 2). Taking out the factors P = 1, -2 already noticed we get: (P - 1)(P + 2)(P 6 - P 5 - 5P 4 + 3P 3 + 7P 2 - P - 1) = 0. It is not so obvious how to factor this, but with some trial we get (P 3 - 3P - 1)(P 3 - P 2 - 2P + 1). Note that P, Q, R satisfy the same equation. So we hope that if P satisfies P 3 - 3P - 1 = 0, then Q and R satisfy the same cubic (rather than the other one). Substituting P = 2 - R 2 gives R 6 - 6R 4 + 9R 2 - 1 = 0, which factorises as (R 3 - 3R - 1)(R 3 - 3R + 1) = 0, so if R does not satisfy the same cubic as P, then it satisfies R 3 - 3R + 1 = 0. But then it also satisfies the cubic R 3 - R 2 - 2R + 1 = 0, so subtracting it satisfies R 2 - R = 0, so R = 0 or 1. Those do not satisfy R 3 - 3R + 1 = 0. Hence R does satisfy the same cubic as P. Similarly so does Q. Note that we cannot have P = Q because then P 2 + P - 2 = 0, so P = 1 or -2 and it is easy to check that they do not satisfy P 3 - 3P - 1 = 0. Similarly for the Q = R or R = P. Thus the roots of P 3 - 3P - 1 = 0 are all distinct. We can check this more directly by noting that at -2 it is negative, at -1 positive, at 0 negative, at 2 positive. Hence there is a root in each of the intervals (-2,-1), (-1,0) and (0,2). We have P + Q + R = 0. Hence p + q + r = 6. The three roots of the other equation must form another possible P, Q, R family. Again it is easy to check directly that the cubic is negative at -2, positive at -1, positive at 0, negative at 1 and positive at 2, so it has a root in each of (-2,-1), (0,1) and (1,2). P + Q + R = 1, so p + q + r = 7. 7th Irish 1994 © John Scholes jscholes@kalva.demon.co.uk 17 Dec 2003 Last updated/corrected 17 Dec 03