An elementary proof of Fermat’s last theorem for all even exponents

Abstract An elementary proof that the equation x2n + y2n = z2n can not have any non-zero positive integer solutions when n is an integer ≥ 2 is presented. To prove that the equation has no integer solutions it is first hypothesized that the equation has integer solutions. The absence of any integer solutions of the equation is justified by contradicting the hypothesis.


Theorem. The equation
x n + y n = z n (1) has no non-zero integer solutions when the exponent n is an integer > .
Previous works. Equation (1) has been of great interest to number theorists for a long time. In 1837, E. E. Kummer [2,9] proved that if (1) has integer solutions then n ≡ (mod 8). Rothholtz [2] extended Kummer's result to prove that (1) has no integer solution if the exponent n is a prime of the form n = t + or one of the variables x, y, z is a prime. In 1977, Terjanian [9] offered a surprisingly simple proof that if (1) is satisfied for non-zero integers then n divides x or y. Equivalently, Terjanian proved Fermat's last theorem for the first case with even exponents. In this paper, a simple proof of the theorem is offered for all even exponents.
Simplification of the theorem. Any integer > is either divisible by 4 or an odd prime. Fermat's last theorem is already known to be true when n is a multiple of 3 or 4 (see [10]). Again, x, y, z must not have any common factor. Otherwise, both sides of the equation can be divided by the common factor to obtain a smaller solution. Also for consistency only one of the variables can be even. When z is even, the left-hand side of (1) is equivalent to 2 (mod 4) and the right-hand side of (1) is equivalent to 0 (mod 4). This leads to an inconsistency. Again since Fermat's equation deals with the situation where all three variables have like powers, it is enough to prove the theorem when the three variables x, y, z are relatively prime integers, y is even, the exponent n is a prime k > and none of the variables is a prime (see [2]).

Search for integer solutions.
Throughout the paper all the variables are positive integers. By (x, y, z) = we mean that x, y, z are coprime integers and |y. By (a, b) = we mean that a, b are coprime integers and |b.
Hypothesis. Fermat's equation with an even exponent has integer solutions.
Equation (1) can be written as where (2) can have integer solutions as seen from the example + = . The objective here is to show that the solutions of (2) cannot be of the form Since integer solutions of (1) are assumed by using Terjanian's result [9], one notes k|Y.
Under the assumption that z, g, h are all integers > , equation (7) represents a right triangle ZGH whose sides and area are integers, and z is the hypotenuse. Therefore ZGH is a rational right triangle [6]. Equivalently, (g, h, z) is a Pythagorean triple. Consequently, we get where tan H = h/g and < H < π/ . Substituting (8) and (9) in (1) and n = k, we get x k + y k = z k [cos (kH) + sin (kH)].
Since cos (kH) + sin (kH) = , we conclude that x and y as obtained in (8) and (9) are indeed the parametric solutions of (1).
From (3) we get X = Real[(g + ih) k ] and Y = Imag[(g + ih) k ]. Thus we get where j = (k − )/ , and thus where C , C , . . . are integers, each divisible by k, f = + if k ≡ (mod 4). Otherwise, f = − . The sign of f will influence only the orientation of X and Y but will have no impact on the integer solutions of (1). Equations (10) and (11) are rewritten as respectively, where Q, R are real integers. Since (g, h) = and k|h, we conclude that (g, Q) = and (h, R) = k. Therefore, if X and Y are k-th powers, then g, Q, h, and R must take values of the forms where (u, v) = , and u, r, v, d are integers > , and w is an integer > . From (12) and (13) we thus obtain The impossibility of (14) will imply the impossibility of (1). By expanding tan kH in terms of tan H (see [5, p. 111]), we get k(d/w) k = U/V, (U, V) = , U = kd k , V = w k , e = h , f = g . Thus we get where the coefficients C , C , . . . , C p− and D , D , . . . , D p− are non-zero integers. It will be enough to prove that e is not an integer given f is an integer. This will imply that h is not an integer given g is an integer. With this assumption, equations (15) and (16) are transformed into To prove that both g and h cannot be integers, it is enough to prove that at least one of (17) and (18)  Consequently, (14) cannot be satisfied under the given conditions. Therefore, the hypothesis is contradicted. This proves Fermat's last theorem for all even exponents.