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Journal of Mathematical Cryptology

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An elementary proof of Fermat’s Last Theorem for all even exponents

Sudhangshu B. Karmakar
Published Online: 2017-05-17 | DOI: https://doi.org/10.1515/jmc-2017-2000

This Erratum corrects the original online version which can be found here: https://doi.org/10.1515/jmc-2016-0018

Abstract

An elementary proof that the equation x2n+y2n=z2n can not have any nonzero 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.

Keywords: Rational triangle; parametric solutions; Fermat’s Last Theorem; even exponents

MSC 2010: 11D61; 11D41; 11A99; 11D99

There was a necessary restructuring of the references and corresponding changes in the main body of the first version of the paper. This has been corrected in the following version.

Ever since the French mathematician Pierre de Fermat [2] stated the conjecture in 1637 that the equation an+bn=cn cannot have any solutions if a, b, and c are integers >0 and n is an integer >2, the equation has been a subject of intense and often heated discussion amongst mathematicians and non-mathematicians alike. This conjecture is known as Fermat’s Last Theorem. The fact that Fermat claimed to have a proof but never wrote it down has put researchers in a quandary since nobody has yet been able to duplicate the proof the way Fermat originally claimed. Perhaps the strong appeal of the problem is the simplicity and elegance of its statement contrasted with the apparent hopelessness [6] of finding an elementary way to establish it. Finally, around 1994 based on a property called modularity of an elliptic curve, Andrew Wiles [11, 10] offered a proof of the theorem. His paper incorporates by reference [7, 3] a vastly larger body of mathematical work developed over the last several decades. But it requires an extraordinary arsenal [9] of mathematical tools to understand Wiles’s complex and very lengthy proof. The proof vastly differs in scope and complexity from the proof Fermat originally envisioned. Consequently, the quest for a simple and short proof continues. Of course any general proof of the theorem will also imply the proof of any special case. In this paper, based on elementary principles, a simple proof of the theorem is given for even exponents >2.

Theorem.

Equation (1) has no nonzero integer solutions when the exponent n is an integer >1:

x2n+y2n=z2n.(1)

Simplification of the theorem.

Any integer >2 is either divisible by 4 or by an odd prime. Fermat’s Last Theorem is already known [9] to be true when n is a multiple of 3 or 4. Again, x, y, and 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 is equivalent to 2 (mod 4) and the right-hand side 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>3 and none of the variables [2] is a prime.

Previous works.

Equation (1) has been of great interest to number theorists for a long time. In 1837, E. E. Kummer [2, 8] proved that if (1) has integer solutions then n 1 (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=4t+3 or one of the variables x, y, z is a prime. In 1977 Terjanian [8] offered a surprisingly simple proof that if (1) is satisfied for nonzero 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.

Search for integer solutions.

Throughout the paper all the variables are positive reals, (x,y,z)=1 means that x, y, z are coprime integers, 2|y and (a,b)=1 means that a, b are coprime integers and 2|b.

Hypothesis.

Fermat’s equation with an even exponent has integer solutions.

Equation (1) can be written as (2) where X=xk, Y=yk, Z=z2:

X2+Y2=Zk.(2)

Conditions.

(X,Y,Z)=1, Z is an odd square, X, Y are k-th powers, k is a prime >3. It is noted that (2) can have integer solutions as seen from the example 412+382=55. The objective here is to show that the solutions of (2) cannot be of the form X=Uk, Y=Vk, Z=z2 where (U,V,z)=1. Since integer solutions of (1) are assumed by using Terjanian’s result [8] one notes 2k|Y.

To investigate the integer solutions of (2) under the stated conditions, two equations (3) and (4) are introduced [1, p. 536] such that (g,h)=1:

X+iY=(g+ih)k,(3)X-iY=(g-ih)k.(4)

Equations (5) and (6) are respectively obtained from (3) and (4) where tanH=h/g, 0<H<π/2:

X+iY=(g2+h2)k/2[cos(kH)+isin(kH)],(5)X-iY=(g2+h2)k/2[cos(kH)-isin(kH)].(6)

Equation (7) is obtained by multiplying the corresponding sides of (5) and (6):

(X2+Y2)=(g2+h2)k[cos2(kH)+sin2(kH)].(7)

Since cos2(kH)+sin2(kH)=1, equation (8) is obtained from (7):

(X2+Y2)=(g2+h2)k.(8)

Equation (9) is obtained by comparing (2) and (8):

z2=g2+h2.(9)

Under the assumptions z, g, and h are all integers >0, equation (9) represents a right triangle ZGH whose sides and area are integers, and z is the hypotenuse. Therefore, ZGH is a rational right triangle [5]. Equivalently, (g,h,z) is a Pythagorean triple. Consequently, equations (10), (11), (12), and (13) are obtained where tanH=h/g, and 0<H<π/2:

X=zkcos(kH),(10)Y=zksin(kH),(11)x=z(coskH)1/k,(12)y=z(sinkH)1/k.(13)

By substituting the values of x and y as obtained from (12) and (13) into (1), equation (14) is obtained by replacing n with k:

x2k+y2k=z2k[cos2(kH)+sin2(kH)].(14)

Since cos2(kH)+sin2(kH)=1, it is concluded that x and y as obtained in (12) and (13) are indeed the parametric solutions of (1).

From (3) it is noted that X=Real[(g+ih)k] and Y=Imag[(g+ih)k]. Thus one gets (15) and (16), and then (17) and (18), where j=(k-1)/2:

X=i=0j(-1)i(k2i)gk-2ih2i,(15)Y=i=0j(-1)i(k2i+1)gk-2i-1h2i+1,(16)X=g(gk-1-C1gk-3h2+C2gk-5h4++(-1)(k-1)/2khk-1),(17)Y=sh(hk-1-C1hk-3g2+C2hk-5g4++(-1)(k-1)/2kgk-1),(18)

where C1,C2, are nonzero integers, each divisible by k. Moreover, s=+1 if k1 (mod 4), and s=-1 otherwise. The sign of s will influence only the orientation of X and Y but will have no impact on the integer solutions of (1). Also (X,Y)=1 since (g,h)=1.

Equations (17) and (18) are rewritten as (19) and (20), respectively, where Q and R are integers:

X=gQ,(19)Y=hR.(20)

Since (g,h)=1 and k|h, one concludes that (g,Q)=1 and (h,R)=k. Therefore, if X and Y are k-th powers, then g, Q, h and R must assume values of the forms g=uk, Q=wk, h=krk-1vk, and R=kdk, where u, r, v, and d are integers >0, and w is an integer >1.

One thus obtains (21) and (22) from (19) and (20):

tankH=Y/X=(h/g)(kdk/wk),(21)tankH/tanH=k(d/w)k.(22)

The impossibility of (22) will imply the impossibility of (1). By expanding tankH in terms of tanH (see [4, p. 111]), one gets (23), where U and V are given by U=kdk and V=wk. It is noted that tanH=h/g and (g,h)=1. One thus gets (24) and (25):

k(d/w)k=U/V,(23)kdk=hk-1-Ck-1hk-2+Ck-2hk-3-+C1h2-kgk-1,(24)V=gk-1-Dk-1gk-2+Dk-2gk-3++D1g+D0.(25)

In (24) and (25) the coefficients Ck-1,Ck-2, and Dk-1,Dk-2, are all divisible by k.

Assertion.

Under the assumption (g,h)=1 equation (24) can not be satisfied.

Justification.

Here the coefficients Ck-1,Ck-2,,C1 are all divisible by k; g and h are the two sides of a right triangle ZGH such that z2=h2+g2. Equation (24) can be written as (26):

kdk=M/N,(26)

where M=zk-1cos(kq), N=cos(q), and q=arctan(g/h). Here z is the hypotenuse of the right triangle ZGH, hence has linear dimension and cos(kq)/cos(q) is dimensionless. It is seen that the left-hand side of (26) has dimension k and the right-hand side has dimension k-1. This leads to an inconsistency and justifies the assertion.

One numerical example.

Let k=7. Then (27) is obtained from (24), and (28) is obtained from (26):

7d7=h6-21h4g2+35h2g4-7g6,(27)7d7=z6[cos(7q)/cos(q)],(28)

where q=arctan(g/h). It is recognized that the two sides of (28) are dimensionally inconsistent.

Therefore the hypothesis is contradicted and this proves Fermat’s Last Theorem for all even exponents.

Acknowledgements

The author wishes to thank the referee for his/her valuable comments on the initial version of the manuscript.

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About the article


Received: 2016-04-04

Revised: 2016-08-05

Published Online: 2017-05-17


Citation Information: Journal of Mathematical Cryptology, ISSN (Online) 1862-2984, ISSN (Print) 1862-2976, DOI: https://doi.org/10.1515/jmc-2017-2000.

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