PI MU EISJLON JOURNAL FALL 1997 NUMBER7 521 523 528 532 540 (Cootitnaetf oa iaside back cover)
PI MU EPSILON JOURNAL THE OFFICIAL PUBLICATION OF THE NATIONAL HONORARY MATHEMATICS SOCIETY Editor: Problems Editor: Business Manager: Russell Euler Clayton W. Dodge Joan Weiss Officers of the Society President: Richard L. Poss. St. <strong>No</strong>rbert College President-Elect: J. Douglas Faires, Youngstown State University Secretary-Treasurer: Robert M. Woodside, East Carolina University Past-President: Robert C. Eslinger, Hendrix College Councilors S. Brent Morris, National Security Agency Brigitte Servatius, Worcester Polytechnic Institute Robert S. Smith, Miami University Joan Weiss, Fairfield University Editorial correspondence, including manuscripts and news items, should be sent to Russell Euler, Department of Mathematics and Statistics, <strong>No</strong>rthwest Missouri State University, Maryville, Missouri 64468 (e-mail address, 0<strong>10</strong>0120@acad.nwmissouri.edu). Guidelines for authors are found on page 441 ofthe <strong>Vol</strong>ume <strong>10</strong>, Number 5, Falll996 issue. Problems for solution and solutions to problems should be sent directly to Clayton W. Dodge, Mathematics Department, 5752 Neville Hall, University of Maine, Orono, Maine 04469. Correspondence regarding subscriptions, back issues or changes of address should be addressed to Joan Weiss, Department of Mathematics and U.oputer Science. Fairfield University, Fairfield, CT 06430. ...._.JlDu &PSILON JOURNAL is published twice a year- Fall and fi e years ( te1ussues). Some Solutions of the Generalized Fermat Equation Herbert E. Salzer The Diophantine equation pq(p + q) =an, n2: 3, (1) is here called the "generalized Fermat equation" because it is equivalent to Fermat's equation (2) only whenever a) (p, q) = 1, n ;e 3k, or b) for any (p, q), n = 3k. (1) For a), in (1) when (p, q) = I, also (p, p + q) = 1 and (q, p + q) = 1, so that p, q and p + q are nth powers, implying Fermat's equation (2) which, conversely, implies (1) for p = xn, q =~and a= xyz. For b), divide (I) by (p, q) 3 = d 3 to get p'q'(p' + q') = a 3 k/d 3 = (ak/d) 3 = a 13 , (p 1 , q 1 ) = I, to which we employ the argwnent in a) applied to n = 3. It was shown in [ 1] that ( 1 ), for (p, q) * 1, n * 3k, does have solutions. Following is an additional multiple infinitude of solutions to (l):t In the simpler case of(1) where p = q, 2p 3 = an, it is easily seen, for n * 3k, that all solutions are given by (2 to nt1 ntk)3 _ ( s 3t1 3\)0 2 Pt ... Pk - 2 Pt ... Pk ' (3) whenever 3to + 1 = sn, R is any odd prime and ~ is any integer, i = 1, ... ,k for any k. To satisfy (1) when p * q, (p, q) = d > 1, for n = 3k + m, m = 1 or 2, and pq(p + q) =ad· pd(ad + Pd) = ap(a + P)d 3 , (a, P) = 1: When m = 1, choose a= ap(a + p) and d =[ap(a + P>t, to get pq(p + q) = [ap(a + p)pk+l = a3k+1 =an. Whenm = 2, choose a= [ap(a + p)] 112 , which is an integer when a, P and a + p are squares satisfying the Pythagorean equation, and d =[ ap( a + P jk/ 2 , to get pq(p + q) = ap(a + p)d 3 = [ap(a + p)pklz+I = a 3 k+z =an. Another solution is had by choosing d = [ap(a + P)fk+I and a= [ap(a + P)f, so that pq(p + q) = ap(a + P)d3 = [ap(a + p)]6k+4 = a3k+z. The writer's original approach to the solutions for p * q did not go beyond r The present simplification of the original versions of the solutions was suggested by the referee. 521