A MARGINAL SIMPLE PROOF OF FERMAT'S LAST THEOREM
MANGATIANA A. ROBDERA
Abstract. We give a simple direct proof of the Fermat's last theorem.
1.
Introduction
Around 1637, Pierre de Fermat conjectured, in the most famous mathematical marginal note, that the equation xn + y n = z n has no solutions if n > 2 and if x, y , and z are required to be positive integers. Fermat wrote another marginal note that can be used to give a quick proof for the case n = 4 [1]. In 1753, Lenohard Euler gave a proof for the exponent n = 3 [1]. After 358 years of eorts and attempts by several mathematicians, the rst proof agreed upon as successful was released in 1994 by Andrew Wiles and formally published in 1995 [1]. However, Wiles' proof was extremely long and using highly complex mathematical developments, leaving the question as to whether a direct proof using only elementary algebras can be given. The purpose of this note is to present one such a proof. 2.
Proof of Fermat's Last Theorem
Given a positive integer n, a triple (x, y, z) of positive integers satisfying the equation xn + y n = z n is called a Pythagorean triple of degree n. The Fermat's Last Theorem (FLT) states that Theorem 1.
There is no Pythagorean triple of degree
n > 2.
We shall denote such a statement by (F LT )n . It is clear that (F LT )n implies (F LT )αn for any positive integer α. Since all positive integers can be reduced to a multiple of prime numbers, it su ces to prove (F LT )n for prime numbers n.(F LT )4 had been established (by Fermat himself). Since for every integer k ≥ 2, 2k can be written 4α for some integer α, (F LT )4 implies (F LT )2k for all positive integers k > 1. We also observe that if for some d > 1 ,(dx, dy, dz) is a Pythagorean triple of order n , then (x, y, z) is also a Pythagorean triple of order n. So in what follows, we x an odd prime number n and a relatively prime Pythagorean triple (x, y, z) of degree n. We let h = z − x. We note that the rst nth -power integer after xn is (x + 1)n . Thus if xn + y n were to be an nth -power integer then one must have (x + 1)n ≤ y n < xn + y n = n n n n (x + h)n < (x +P y) , and . Clearly, x + y = z nis equivalent to thus 1 < h n< yP n n n n−p p n n−p p n n n x + y = x + p=1 p x h or y = p=1 p x h . Thus y is divisible by h, and y and h cannot be relatively prime. Let k = gcd(y, h). We write h = αk r for
Proof.
Mathematics Subject Classi cation. 11A99. Key words and phrases. Fermat's last theorem.
2010
1
A MARGINAL SIMPLE PROOF OF FERMAT'S LAST THEOREM
2
some positive integer r and where α is a positive integer not divisible by k. Thus we can write y n = βkrn+q for some non-negative integer q and a positive integer β that does not divide k. Thus βk
rn+q
=
n−1 X p=1
n n−p p x h + hn = p
n−1 X p=1
! n n−p p rp x α k + αn k rn p
Dividing both sides of the above equation by kr , we obtain (2.1)
βk
r(n−1)+q
= nx
n−1
α+
n−1 X p=2
n n−p p r(p−1) x α k + αn k r(n−1) p
or equivalently (2.2)
−nxn−1 α =
n−1 X p=2
n n−p p r(p−1) x α k + αn k r(n−1) − βk r(n−1)+q p
We notice that x cannot have a common divisor with k because such a common divisor would be also a common divisor for x and y , and therefore x,y , and z , would not be relatively prime. Likewise, α and k are relatively prime because otherwise, k and x, and consequently x and y , would have a common divisor. We have two cases: • k 6= n: since n is odd, we see that the left-hand side of (2.2) is relatively prime with k. However on the other-hand, the right-hand side is clearly divisible by k. Contradiction. • k = n: the right-hand side is divisible by k 2 = n2 while the left-hand side can only be divided by k = n. Another contradiction. The proof is complete. I would like to acknowledge the persistent and pertinent remarks from several colleagues that really helped in the completion of the writing of this proof. References [1] Wiles,
A.,
Modular elliptic curves and Fermat's last theorem.
Annals
of
Mathematics.
1995;141(3):443-551.
Department of Mathematics, University of Botswana, 4775 Notwane Road, Gaborone, Botswana
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