Fermat’s Last Theorem: A Componential Algebraic Proof Benson Schaeffer∗ Portland, OR, USA
Abstract
In this paper I offer a componential algebraic proof by contradiction of Fermat’s Last Theorem. Using an alternative to the standard binomial exn ∑ pansion, (b + c)n = bn + c bn−i (b + c)i−1 , b and c distinct nonzero integers, i=1
n a positive integer, I show that a componential rewrite of the equation stating the theorem, (F + E)p + (F + G)p = (F + E + G)p , F, E, G and p positive integers, G > E, and p ≥ 3, entails the contradiction of the sum of an irreducible fraction and several integers equal to zero, E 2Gp+1 + g p−1 Gp1 E 2Gp+1 + F1 (Sum 1) + G1 (Sum 2) = 0, g g, G1 and F1 positive integers, g relatively prime to E, F1 (Sum 1) and G1 (Sum 2) integers.
AMS 2010 subject classification: Number theory Keywords Diophantine equations, Fermat’s Last Theorem.
∗
The corresponding author. E-mail:
[email protected]
1
1
Introduction To prove Fermat’s Last Theorem, it suffices to show that the equation Ap + B p = C p
(1)
has no solution for A, B, C and p positive integers, A, B and C pairwise relatively prime and p ≥ 3 [1, pp. 1]. Wiles [4], and Taylor and Wiles [3], used Galois representations, Frey’s elliptic curves and the modular forms associated with them, and Wiles’ demonstration of the validity of the Shimura-Taniyama Conjecture, to prove the theorem [1, pp. 366-374]. Much of the mathematics involved was developed after World War II. Excellent earlier work, beginning in the late 18th century [1, pp. 24], resulted in proofs of the theorem for many exponent powers, but not a general proof, and “led to the creation of the theory of algebraic numbers” [1, pp. 366]. The componential algebraic proof by contradiction I offer below employs algebra and number theory concepts that would have been available to Fermat in the 17th century. Usn ∑ ing an alternative to the standard binomial expansion, (b + c)n = bn + c bn−i (b + c)i−1 , i=1
b and c distinct nonzero integers, n a positive integer, I show that a componential rewrite of equation (1), (F + E)p + (F + G)p = (F + E + G)p , F, E, G and p positive integers, G > E, and p ≥ 3, entails the contradiction of the sum of an irreducible fraction and several integers equal to zero, E 2Gp+1 + g p−1 Gp1 E 2Gp+1 + F1 (Sum 1) + G1 (Sum 2) = 0, g g, G1 and F1 positive integers, g relatively prime to E, F1 (Sum 1) and G1 (Sum 2) integers. This contradiction implies that both the componential rewrite and its source, equation (1), have no positive integer solutions for p ≥ 3 and proves Fermat’s Last Theorem.
2
2
Proof Overview I begin by demonstrating the validity of an alternative to the standard binomial
expansion formula, n
n
(b + c) = b + c
n ∑
bn−i (b + c)i−1 ,
i=1
b and c distinct nonzero integers, n a positive integer, Lemma 1, then use the alternative binomial formula to uncover the contradiction. In order, I develop a componential rewrite of equation (1) with A, B and C broken ∑ into parts, (F + E)p + (F + G)p = (F + E + G)p , with (F + E)p = E pi=1 (F + G)p−1 (F + ∑ E + G)i−1 and (F + G)p = G pi=1 (F + E)p−i (F + E + G)i−1 , Lemma 2; demonstrate that
[ i−1 ] p ∑ ∑ Fp = F p−i (Ai−1−j + B i−1−j )C j−1 , EG i=2 j=1
and, hence, since F contains all of the prime factors of E and G, that E and G are relatively prime, and that F = gF1 and G = gG1 , Lemma 3; and finally demonstrate that the compential rewrite entails the contradiction of an irreducible fraction integers equal to zero, (Gp + 1)E 2G p +1
E 2G g
p +1
+ F (Sum 1) + G(Sum 2) = 0, p +1
+ g p−1 Gp1 E 2G
+ F1 (Sum 1) + G1 (Sum 2) = 0,
g relatively prime to E.
3 3.1
Proof Lemma 1: Alternative Binomial Formula Ribenboim (1999) notes [2, pp.80] for a and b distinct nonzero integers, n a positive
integer, k a positive integer or zero, an − bn ∑ k n−1−k = a b a−b k=0 n−1
3
Letting a = (b + c), c a nonzero integer, and k = i − 1, then gives, n
n
(b + c) = b + c
n ∑
bn−i (b + c)i−1 ,
(2)
i=1
an alternative to the standard binomial formula [1, pp.9].
3.2
Lemma 2: (F + E)p + (F + G)p = (F + E + G)p
Below I use the alternative binomial formula, equation (2), Lemma 1, to rewrite equation (1) componentially as, (F + E)p + (F + G)p = (F + E + G)p . Since A, B and C are pairwise relatively prime, equation (1) can without loss of generality be rewritten as, Ap + (A + D)p = (A + D + E)p ,
(3)
A, D and E positive integers, A, (A + D) = B and (A + D + E) = C pairwise relatively prime. By the alternative binomial formula, then, Ap = (A + D + E)p − (A + D)p , p ∑ p A =E (A + D)p−i (A + D + E)i−1 . i=1
If E ≥ A, the last term in the sum E(A + D + E)p−1 > Ap . Therefore, A = (F + E), F a positive integer, and equation (3) can be written as, Ap + (A + D)p = (A + D + E)p , (F + E)p + (F + E + D)p = (F + E + D + E)p , and then, letting G = E + D, as (F + E)p + (F + G)p = (F + E + G)p ,
(4)
(F + E) = A, (F + G) = B, (F + E + G) = (A + G) = (B + E) = C pairwise relatively prime. By the alternative binomial formula, thus, (F + E) = (F + E + G) − (F + G) = E
p ∑
(F + G)p−i (F + E + G)i−1 ,
(5)
p ∑ (F + E)p−i (F + E + G)i−1 . (F + G) = (F + E + G) − (F + E) = G
(6)
p
p
p
i=1 p
p
p
i=1
4
3.3
Lemma 3: E and G Relatively Prime, F = gF1 , G = gG1
I now use the alternative binomial formula, Lemma 1, to solve for F p and show that F contains all of the prime factors of E and G and, hence, that E and G are relatively prime. By the alternative binomial formula, then, p
p
p
A = (F + E) = F + E
p ∑
F p−i Ai−1 ,
i=1
B p = (F + G)p = F p + G p
p
p
C = (F + E + G) = F + E
p ∑
p ∑
F p−i B i−1 ,
i=1
F
p−i
C
i−1
+G
i=1
p ∑
F p−i C i−1 .
i=1
Substitution of the above results into equation (3) then gives, (F + E)p + (F + G)p = (F + E + G)p , [ p
F +E
p ∑
] [ F
p−i
A
i−1
p
+ F +G
i=1
p ∑
] F
p−i
B
i−1
[ p
= F +E
i=1
p ∑
F
p−i
C
i−1
+G
i=1
p ∑
] F
p−i
C
i=1
and with regrouping,
p
F =E
p ∑
F
p−i
(
C
i−1
−A
i−1
)
+G
i=2
p ∑
( ) F p−i C i−1 − B i−1 ,
(7)
i=2
i ≥ 2 because when i = 1, C 1−1 − A1−1 = C 1−1 − B 1−1 = 0. Application of the alternative binomial formula to the terms in parenthesis then gives, since C = (A + G) = (B + E), i−1 i−1 ∑ ∑ i−1 i−1 i−1 i−1 i−1−j j−1 C − A = (A + G) − A = G A (A + G) =G Ai−1−j C j−1 , C i−1 − B i−1 = (B + E)i−1 − B i−1 = E
j=1 i−1 ∑
B i−1−j (B + E)j−1 = E
j=1
Substitution of these results into equation (7) results in,
F
p
= E
p ∑ ( i=2
C
i−1
−A
i−1
)
p ∑ ) ( , F p−i C i−1 − B i−1 , +G i=2
5
j=1 i−1 ∑ j=1
B i−1−j C j−1 .
i−1
,
= E
p ∑
[ F p−i G
i=2
= GE
i−1 ∑
] Ai−1−j C j−1 + G
j=1
p ∑
F p−i
[ i−1 ∑
i=2
]
F p = EG
F p−i
i=2
∑ Fp = F p−i EG i=2 p
[ i−1 ∑(
[ F p−i E
i=2
Ai−1−j C j−1 + EG
j=1 p ∑
p ∑
F p−i
i=2
[ i−1 ∑ j=1
) i−1−j
Ai−1−j + B
B i−1−j C j−1
) i−1−j
]
B i−1−j C j−1 .
]
C j−1 .
j=1
[ i−1 ∑(
]
j=1
p ∑
Ai−1−j + B
i−1 ∑
] C j−1 .
(8)
j=1
Equation (8) shows that F contains all of the prime factors of E and G, because EG divides F p . For this reason, it also shows that E and G are relatively prime, otherwise (F + E) = A, (F + G) = B and (F + E + G) = C would not be pairwise relatively prime. Further, since F contains all of the prime factors of G, I can without loss of generality rewrite, F as F = gF1 and G as G = gG1 ,
(9)
g a prime factor of G raised to the first power and relatively prime to E, F1 the multiplicative remainder of F , and G1 the multiplicative remainder of G.
3.4
Contradiction I now demonstrate that the componential rewrite of Fermat’s equation, equation
(4), Lemma 2, entails the contradiction of the sum of an irreducible fraction and several integers equal to zero, E 2p+2x−1 + g p−1 Gp1 E 2p+2x−1 + F1 (Sum 1) + G1 (Sum 2) = 0, g g, a prime factor of G, relatively prime to E, x = Gp − p + 1. Under the assumption that Fermat’s equation (1) has a positive integer solution, and by application of the alternative binomial formula, equation (2), Lemma 1, (Ap+x + B p+x + C p+x )(Ap+x + B p+x − C p+x ) = 2Ap+x B p+x + A2p+2x + B 2p+2x − C 2p+x = 2Ap+x B p+x +Ap ·Ap+2x +B p ·Ap+2x −C p ·Ap+2x +B p (B p+2x −Ap+2x )−C p (C p+2x −Ap+2x ) = 6
2Ap+x B p+x + B p (B p+2x − Ap+2x ) − (Ap + B p )(C p+2x − Ap+2x ) = 2Ap+x B p+x + B p (B p+2x − C p+2x ) − Ap (C p+2x − Ap+2x ), and since A = (F + E), B = (F + G), and C = (F + E + G), (Ap+x + B p+x + C p+x )(Ap+x + B p+x − C p+x ) = 2Ap+x B p+x p+2x p+2x ∑ ∑ p p+2x−i i−1 p B C −A G Ap+2x−i C i−1 . −B E i=1
(10)
i=1
Without the assumption that equation (1) has a positive integer solution, however, (Ap+x + B p+x + C p+x )(Ap+x + B p+x − C p+x ) = p+x ∑ p+x p+x (2A +B +G Ap+x−i C i−1 )(Ap+x + B p+x − C p+x ) = i=1
2A2p+2x + 2Ap+x B p+x − 2Ap+x C p+x + Ap+x B p+x + B 2p+2x − B p+x C p+x p+x p+x p+x ∑ ∑ ∑ p+x p+x−i i−1 p+x p+x−i i−1 p+x +A G A C +B G A C −C G Ap+x−i C i−1 = i=1
3A
p+x
B
p+x
+B
p+x
G
p+x ∑
i=1
A
p+x−i
C
i−1
− 2A
p+x
i=1
p+x ∑
−B p+x E
p+x ∑
B p+x−i C i−1 − G
i=1 p+x
3A
p+x
B
p+x
+B
p+x
G
G
A
p+x−i
C
i−1
i=1
−G2
−A [ p+x ∑
i=1 p+x−i
A
C
i=1
p+x
G
p+x ∑
i−1
p+x
+A
G
A
C
i−1
−B
p+x
E
]
i−1
Ap+x−i C i−1
Ap+x−i C i−1 =
i=1 p+x−i
p+x ∑ i=1
p+x ∑
Ap+x−i C i−1 · G
i=1
∑
p+x ∑
p+x ∑
B p+x−i C i−1
i=1
Ap+x−i C i−1 .
(11)
i=1
Therefore, since result (11) minus result (10) equals zero, p+x
3A
B
p+x
+B
p+x
G
p+x ∑
A
C
i−1
p
+B E
i=1 p+2x
−2Ap+x B p+x + Ap G
∑
Ap+2x−i C i−1 − B p+x E
p+x
−Ap+x G B
p+x
G
p+x ∑
Ap+x−i C i−1 − G2
[ p+x ∑
∑
B p+x−i C i−1 ]2
i=1
Ap+x−i C i−1
i=1
i=1
∑
= 0,
p+2x p+x−i
A
C
i−1
+A
p+x
B
p+x
i=1
∑
p+2x
+Ap G
B p+2x−i C i−1
i=1 p+x
i=1
∑
∑
p+2x p+x−i
Ap+2x−i C i−1 − G2
p
+B E [ p+x ∑ i=1
i=p+x+1
7
B p+2x−i C i−1
i=p+x+1
Ap+x−i C i−1
]2 = 0.
(12)
Insertion of results (5) and (6) then gives, B
p+x
G
p+x ∑
p+x−i
A
C
i−1
p+x
+A
B G
i=1
∑
p−i
A
C
i−1
+G
p ∑
i=1 p+2x
∑
p
+E
p ∑
x
B
p−i
C
i−1
·G
i=1
i=1 p+2x−i
A
C
i−1
−G
2
∑
p+2x p−i
A
C
[ p+x ∑
i=p+x+1
i−1
·E
B p+2x−i C i−1
]
i=p+x+1 p+x−i
A
C
i−1
= 0,
i=1
and after division by G, B
p+x
p+x ∑
p+x−i
A
C
i−1
p+x
+A
B
p ∑
x
i=1
A
p−i
C
i−1
+
p ∑
i=1
+E
p ∑
∑
A
i=1
p+2x
B p−i C i−1 ·
i=1
Ap+2x−i C i−1 − G
i=p+x+1
∑
p+2x p−i
C
[ p+x ∑
i−1
·E
B p+2x−i C i−1
i=p+x+1
]2
Ap+x−i C i−1
= 0.
(13)
i=1
An exhaustive count of E 2p+2x−1 terms, however, permits a radical simplification of equation (13). The third sum, p ∑
A
C
i−1
·E
i=1
∑
Ap+2x−i C i−1
i=p+x+1 p+2x
p
=E
∑
p+2x p−i
∑
(F + E)p−i (F + E + G)i−1 ·
i=1
(F + G)p+2x−i (F + E + G)i−1 ,
i=p+x+1
generates pE 2p+2x−1 , and the fourth sum, E
p ∑
B
i=1
∑
C
i−1
·
Ap+2x−i C i−1
i=p+x+i p+2x
p
=E
∑
p+2x p−i
∑
(F + G)p−i (F + E + G)i−1 ·
i=1
(F + E)p+2x−i (F + E + G)i−1 ,
i=p+x+1
xE 2p+2x−1 , for a total of (p + x)E 2p+2x−1 . Given x = Gp − p + 1, this translates to (p + Gp − p + 1)E 2p+2(G
p −p+1)
= (Gp + 1)E 2G
p +1
,
Therefore, equation (13) can be rewritten in radically simplified form as, (Gp + 1)E 2G
p +1
+ F (Sum 1) + G(Sum 2) = 0,
(14)
F (Sum 1) the sum of all terms in equation (13) containing F but not G, G(Sum 2) the sum of all terms containing G, Insertion of result (9), which presumes equation (1) has a positive integer solution, into equation (14) then yields,
8
E 2G
p +1
+ g p Gp1 E 2Gp+1 + gF1 (Sum 1) + gG1 (Sum 2) = 0,
and after division of all terms by g, p +1
E 2G g
p =1
+ g p−1 Gp1 E 2G
+ F1 (Sum 1) + G1 (Sum 2) = 0.
(15)
Because g, a prime factor G, is relatively prime to E, however, equation (15) posits the p E 2G +1 sum of an irreducible fraction, generated by , plus several integers, the remainder g of the equation, equal to zero. This contradiction invalidates equation (15) and its sources, equations (4) and (1), and proves Fermat’s Last Theorem: equation (1) no positive integer solutions. It is possible that the above proof corresponds to the “truly marvelous proof” Fermat was alluding to in his tantalizing marginal comment [1 pp. 1].
References [1] D.M.Burton, Elementary Number Theory, Fifth Edition, McGraw-Hill, New York, 2002. [2] P. Ribenboim, Fermat’s Last Theorem for Amateurs, Springer-Verlag, New York, 1999. [3] R. Taylor and A. Wiles, Ring-theoretic properties of certain Hecke algebras, Annals Math., 142 (1995), 553-572. [4] A. Wiles, Modular elliptic curves and Fermat’s Last Theorem, Annals Math., 142 (1995), 443-451.
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