From outside the Bradley University community, correct solutions were submitted by Ron Welch, Aaron David Kahn, Jens Voss, Emanuele Macri, Lorenzo Pozzoli, Massimo Brigogne, Larry Baum, Maxsim Ovsjanikov, Khanh Ngo, Ariel Flat, Yves Thiry, Ivan Lisac, Dave L. Renfro, Olier Randschau, Brian Laughlin, and the proposer, Philippe Fondanaiche.

Let the roots be a, b, and 2a. Since the largest is twice the smallest, all the roots are positive.  Write the polynomial as p(x) = (x-a)(x-b)(x-2a).  Since p(B) = (B-a)(B-b)(B-2a) is prime, B-b and B-2a must equal 1.  Therefore b = 2a = B-1.  Similarly, since p(A) = (A-a)(A-b)(A-2a) is prime, A = b-1 = B-2.  Given that one of the numbers is thirteen, the possibilities for B, p(B), A, p(A) are 13,7,11,5; 25,13,23,11; 15,8,13,6; 29,15,27,13.  Only the first one fulfils all the requirements, yielding the roots 6,12,12 and the polynomial

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