Fermat's Last Theorem and its another Proof
aa r X i v : . [ m a t h . G M ] F e b FERMAT’S LAST THEOREM AND ITS ANOTHERPROOF
KIM YANGGON, KIM SOOGON, KIM SEUNGKON, KIM CHANGKON
Abstract.
We announce here that Fermat’s Last theorem was solved,but there is an easy proof of it on the basis of elemetary undergraduatemathematics. We shall disclose such an easy proof. introduction It is a well known fact that Fermat’s Last theorem wassolved by Andrew Wiles(1953.4.11 - ), but his paper is lengthyand hard. So probably many mathematicians still want aneasy one even for undergraduates or civilians to understandwithout difficulty.This paper could be a sort of an answer to their request.Hence it is hoped that this paper could be really helpful forthem.In this paper we shall proceed in the following order.
Mathematics Subject Classification.
Primary 12E99,13-02, secondary 11T99.
First, we are going to explain what the Fermat’s Last the-orem is.Secondly, we shall give an easy solution even for undergrad-uates or ordinary people to comprehend.Finally we finish this paper with a concluding remark deal-ing with some Diophantine problem which is equivalent to theFermat’s Last theorem .2.
What is the Fermat’s Last theorem?
Let us consider the quadratic equation X + Y = Z de-fined in the set of natural numbers. It has a solution X =3 , Y = 4 , Z = 5 as an example. As is well known,this equationis obtained from the Pythagorus theorem.Here we ask a question. What about the equation(2.1) X n + Y n = Z n for any integer value n > Z ?Does it have a solution at all with X = 0 , Y = 0 , Z = 0?The answer is ’No’ according to the French mathematicianPierre de Fermat(1601.8.17- 1665.1.12).But his proof was not given except that he said ” I have dis-covered a truly remarkable proof of this theorem which this ERMAT’S LAST THEOREM AND ITS ANOTHER PROOF 3 margin is too small to contain.”In 1995 the first proof was given by Andrew Wiles. Hisproof uses the Taniyama-Shimura conjecture relating to el-liptic curves and modular forms of them, and his paper isthought to be hard and long.We intend to give a short and easy proof for this problemby making use of reduction to absurdity.It suffices to consider only positive integer solutions for thisequation.We may also assume that n is any fixed prime number be-cause the exponent has a unique factorization of prime num-bers and X + Y = Z was solved by Ferma himself. Further-more we may assume that X, Y , and Z are relatively prime .If n is a prime number , then one of X and Y cannot be amultiple of n by our assumption.Suppose that we have a solution for the equation (2.1). Ifwe divide both sides by X n which may be assumed to be nomultiple of n without loss of generality,we get the equation ofthe form 1 + ( Y /X ) n = ( Z/X ) n .Now if we put A = Y /X, B = Z/X , where A and B are rational num-bers in the field Q , then there arises an equation of the form KIM YANGGON, KIM SOOGON, KIM SEUNGKON, KIM CHANGKON (2.2) 1 + A n = (1 + B ) n ,where A and B must be positive rational numbers. Ofcourse A must be exactly greater than B for the equality of(2.2) to hold,i.e., A>B > Q , then there must be a solution of integers for the equation(2.1).There are infinitely many solutions for n = 1 , n = 1, there are obviously many solutions.For n = 2, we consider the equation of the form 1 + A n =( A + α ) n . We then have 1 + A = A + 2 Aα + α , so that wemay get solutions A = (1 − α ) / α for any rational number α .For n ≥
3, however, there is no solution at all. We shallprove this fact in the next section 3 .3.
Proof of the Fermat’s last theorem
Now we assume henceforth once and for all that there is asolution in Q for the equation(3.1) 1 + A n = (1 + B ) n = B n + (cid:0) n (cid:1) B n − + · · · + (cid:0) nn − (cid:1) B + 1, ERMAT’S LAST THEOREM AND ITS ANOTHER PROOF 5 where A and B are positive rational numbers as manifestedjust above. Theorem 3.1.
We have no solution of the designated equa-tion (2.1).Proof.
We note first thing that the solution of (2.1), if any,may be obtained roughly from the congruence equation(3.2) 1 + A n ≡ (1 + A ) n ≡ A ≡ (1 + B ) n ≡ B n ≡ B modulo ( n )as follows.(3.3) A = ts , B = ( t + ln ) / ( s + kn ) for some integers l, k, s, and t because A n ≡ A and B n ≡ B in the multiplicative group Z × n and so in Z n .In particular we may assume without loss of generality that s, t are positive integers..Here the fractions of A and B are fixed to be irreducibleones and we would like to know the relationship of these in(3.3) with n in more detail.As a matter of fact we are dealing here with the localization S − Z of the ring of integers Z at a prime ideal ( n ), where S denotes a multiplicative set Z − ( n ). At the same time we areconsidering the composite map: Z → S − Z → Z n ,where Z n denotes the finite field consisting of n numbers { , , , · · · , n } . KIM YANGGON, KIM SOOGON, KIM SEUNGKON, KIM CHANGKON
From (2.2), we see that s + nk must be a multiple of s andthe numerator t of A must be a multiple of the numerator t + nl of B , Thus we may put(3.4) v ( t + ln ) = t, s + kn = ws for some integers v and w . So we must have that both t + ln and s + kn are negative integers. We put c = vw for brevity.On the other hand we have the following equations succes-sively;1 + A n = (1+ Ac ) n ⇒ c n (1 + A n ) = ( c + A ) n ⇒ c n { v ( t + ln ) /s ) n } = { c + ( v ( t + ln ) /s ) } n ⇒ c n { s n + ( v ( t + ln )) n } =( cs + v ( t + ln )) n ⇒ c | v ( t + ln ) ,which means that c must di-vide v ( t + ln ) = t by (3.4).It follows that B = ( t + ln ) / ( s + kn ) = v ( t + ln ) /v ( s + kn ) = v ( t + ln ) /vws = v ( t + ln ) /cs = t/cs = d/s ,where cd = v ( t + ln )for some positive integer d .Hence we get s + kn = − s and t + ln = − d , so that 2 s = − kn and t + d = − ln are obtained respectively.It follows that(3.5) n | s since n ≥ ERMAT’S LAST THEOREM AND ITS ANOTHER PROOF 7 which is a contradiction to our assumption that X = s should not be a multiple of n .We may still have another contradiction granting that (3.5)is right.If so, then we thus obtain that s n = 1 , ( s + kn ) = 1 alongwith the fact s n = ( s + kn ) n ( if n is odd).But then from (3.1) we get at the equation ( s + kn ) n ( s n + t n ) = s n { ( s + kn ) + ( t + ln ) } n by substitution of the fractions, so that ( − s ) n + ( − t ) n = ( − s + t + ln ) n . Hence in the field Z n we have − s − t ≡ − s + t + ln , which reults in 2 t ≡ n ), and hence(3.6) n | t because n ≥ t/s = A is irreducible.Hence we have completely proved the Fermat’s Last theo-rem.Note that we should have k = 0 , l = 0, w = − v ≤ − c ≡ n ) as a byproduct. (cid:3) Corollary 3.2.
We have no solution of the designated equa-tion (2.1) even in the field Q of rational numbers. KIM YANGGON, KIM SOOGON, KIM SEUNGKON, KIM CHANGKON
Proof.
Immediate consequence of the above theorem. (cid:3)