250 Problems in Elementary Number Theory - Sierpinski (1970)

SOLUTIONS 79154*. LEMMA. If a, b, c are positive, real, and not all equal, thenPROOF.( a+b+c)3 b3 >a c. (1)Suppose that the numbers a, b, c are positive and not all equal.Then there exist positive numbers u, v, and w, not all equal and such thata = u\ b = v 3 , and c = w 3 • We have the identityU3 +V 3 +w 3 -3uvw = l(u+v+w) [(u-v)2+(v-wl+(w-U)2].Since not all numbers u, v, ware equal, the last factor is strictly positive,and we havewhich, in view of u 3 = a, V = b, w 3 = c gives (1), and completes the proofof the lemma.Let now x, y, z be positive integers. It the numbers x/y, y/z, and z/x wereall equal, then, being positive and their product being equal to 1, they wouldhave to be all equal 1, and we would have~+L+~= 3 >2.y z xThus, not all numbers x, y, z are equal, and by the lemma we have[~(~+L+~)]3 > ~. L. ~= 1,3y z x y z xhencex y z-+-+->3.y z xThus, the equation x/y+y/z+z/x = 2 is impossible in positive integers x, y, z.155. Suppose that the positive integers x, y, z satisfy our equation. Ifnot all three numbers x/y, y/z, and z/x are equal, then from the solutionof Problem 154 it follows that~+L+~>3.y z xWe must have, therefore, x/y = y/z = z/x, and our equation implies thateach of these numbers is 1. Thus, x = y = z. In this case we havex y z-+-+- = 1+1+1 = 3,y z x