On the inverse Fermat equation

Lenstra Jr, H.W., On the inverse Fermat equation, Discrete Mathematics 106/107 (1992) 329-33 1. In this paper the equation x”” + y”” = z “n is solved in positive integers x, y, z, n. If the nth roots are taken to be positive real numbers, then all solutions are known to be trivial in a certain sense. A very short proof of this is provided. The argument extends to give a complete description of all solutions when other nth roots are allowed. It turns out that up to a suitable equivalence relation there are exactly four nontrivial solutions. The inverse Fermat equation is the diophantine equation X I/n + yI/n = Zlln, to be solved in positive integers X, y, z, n. When the nth roots are interpreted as positive real numbers, then it is known that the only solutions are given by x = can y = cb”, z = c(a + b)“, where a, b, c are positive integers with gcd(a, I;) = 1; see [l, 21 and the references listed there. Equivalently, if (Y, /I are positive real numbers for which cr+p=1, a? and p” are rational, then (Y and /3 are rational. The following proof is so short that it might be called a one line proof, had it not employed two circles as well. It relies on a fact from Euclidean geometry: if two nonconcentric circles in the plane intersect in a point that is collinear with their centres, then they have no other intersection point. The rationality of LX” implies that the algebraic number (Y and all of its conjugates have the same absolute value, so that in the complex plane they are all located on a circle centred at 0; and since the same is true for /3 = 1 a, they also lie on a circle centred at 1. Thus, by the geometric fact just stated, (Y has no conjugates different from itself, which means that it is rational. Correspondence to: H.W. Lenstra Jr, Department of Mathematics, University of California, Berkeley, CA 94720, USA 0012-365X/92/$05.00 @ 1992 Elsevier Science Publishers B.V. All rights reserved 330 H. W. Lenstra Jr When other nth roots than positive real ones are allowed in the inverse Fermat equation, then there are a few special solutions. Namely, consider the identities 1 + 1: = 163, 1+ If = Id, 1+ 9: = 646, I+ Ia = 7296, where the roots are suitably chosen. The first identity leads to a solution x = y = 1, z = 16, n = 8 of the inverse Fermat equation. The others lead in a similar way to solutions, with 12 = 6, 12, 12, respectively. There are essentially no other solutions. To formulate this precisely, denote by G the multiplicative group of nonzero complex numbers 6 with the property that 6” is rational for some positive integer n. Consider the equation cu+p+y=O, cu,p, yeG. Each of the above four identities represents a solution; let the solutions obtained in this way be called special. In addition, there are trivial solutions, in which a; fi, and y are rational. Let two solutions be called equivalent if one is proportional to a permutation of the other, up to complex conjugation. With this terminology, each solution is equivalent either to a trivial one or to one of the four special solutions. Permuting a; /I, y one can achieve that IyI = max{lal, I/-II, Irl}, and dividing by -y one may assume that y = -1, so that (Y + /3 = 1. If a is real, then the same proof as above shows that the solution is trivial. Suppose that (Y is not real. Then the same reasoning leads to two circles that intersect in two nonreal points, so LY is imaginary quadratic. From IaJ s 1, II(~1 = IpI 6 1 one sees that the real part of (Y is strictly between 0 and 1. Also, from (Y E G it follows that the number C = (u/5 is a root of unity, and it is different from 51. Further, 5‘ belongs to the quadratic field generated by cy. The same statements are true for the number 17 = p/p = (1 w)/(l 5). H owever, the only quadratic fields that contain roots of unity different from fl are the Gaussian field, generated by a primitive fourth root of unity, and the Eisenstein field, generated by a primitive cube root of unity. If (Y generates the Gaussian field, then 5‘ has order 4, and the same is true for q, so that the triangle with vertices 0, 1, a has angles equal to n/4, JC/~, n/2; in this case the solution is equivalent to the first special one. If (Y generates the Eisenstein field, then c has order 3 or 6, and the same is true for n. If both 5 and r] have order 3, then the triangle with vertices 0, 1, (Y is equilateral, and the solution is equivalent to the second special one. If one of c and n has order 6, and the other has order 3 or 6, then one finds in a similar way one of the remaining two special solutions. On the inverse Fermat equation 331