A MARGINAL 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 (2.1)
xn + y n = z n
is called a Pythagorean Theorem 1.
triple
of degree n. The Fermat's
There is no Pythagorean triple of degree
Last Theorem
states that
n > 2.
We rst observes that since all positive integers can be reduced to a multiple of prime numbers, it is necessary to prove Fermat's Last Theorem for all odd prime numbers n. We also observe that if (x, y, z) is a Pythagorean triple and if d > 1 divides each of x, y, z , then xd , yd , dz is also a Pythagorean triple. In what follows, we x an odd prime number n and we assume that x, y , and z are relatively prime and form a Pythagorean triple of degree n. Then x and y cannot be both even. We are therefore left to discuss two cases. Case 1: Both x and y are odd positive integers. n In such a case, z must be even. Let hP = z − x. Since z n = xn + y n < (x + y) , we infer that h < y . Equation (2.1) n n is then equivalent to xn + y n = xn + p=1 p xn−p hp . From which, we obtain Proof.
(2.2)
n
y =
n X n p=1
p
xn−p hp .
Thus y , and hence y , is divisible by h. We have a contradiction since then the left-hand side of (2.2) is divisible by hn (n > 2) while its right-hand side is not divisible by h2 : indeed, after division by h the right-hand side of (2.2) n
2010 Mathematics Subject Classi cation. 11A-99. Key words and phrases. Fermat's last theorem.
1
A MARGINAL PROOF OF FERMAT'S LAST THEOREM
2
becomes n n X X n n−p p−1 n n−p p−1 x h = nxn−1 + x h ; p p p=1 p=2
(2.3)
xn−p hp−1 is not divisible by h because all of its terms is a mutiple of h except the rst one: nxn−1 . Case 2: Only one of the positive integers x and y is odd. Switching, if necessary the role of x and y , we assume that x is odd and y is even. Then z must be odd. We write z = x + 2h γ (respectively y = 2k β ) where h (respectively k ) is a positive integer and γ (respectivelyβ ) is an odd Pn
p=1
n p
positive integer. Then (2.1) can be written as
xn + 2kn β n = xn +
n X n
p
p=1
2hp xn−p γ p ,
or equivalently n X n hp n−p p 2 β = γ . 2 x p p=1
(2.4)
kn n
If kn < h, we divide both sides of (2.4) by 2kn . We obtain (2.5)
n
β =
n X n p=1
p
2hp−kn xn−p γ p .
This is not possible since the left-hand side of (2.5) is odd while its right-hand side is even. If kn > h, we divide both sides of (2.4) by 2h . We obtain (2.6)
2kn−h β n = nxn−1 γ +
n X n
p
p=2
2h(p−1) xn−p γ p .
This is not possible since the left-hand side of (2.6) is even while its right-hand side is odd. If h = kn, we divide both sides of (2.4) by 2h . We obtain (2.7)
n
n−1
β = nx
n X n h(p−1) n−p p γ+ 2 x γ . p p=2
Since n > 2, we can write (2.7) in the following equivalent form (2.8)
β n − γ n − nxn−1 γ =
n−1 X p=2
n h(p−1) n−p p 2 x γ . p
Again, we arrive at another contradiction since the left-hand side of (2.8) is odd while its right-hand side is even. The proof is complete. 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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