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Fermat’s Last Theorem (FLT) has fascinated and stimulated many professional and amateur mathematicians. This theorem inspired several authors who have seen in this problem many hidden mysteries waiting to be unveiled [
In the first case, we have the following results with different hypotheses.
- If z = y + 1 , FLT holds, then the first case of Abel conjecture is also proved. [
- For y = x + 1 , FLT has been proved by Dittmann [
- In 1823, Sophie Germain has demonstrated FLT with 2 p + 1 a prime (pp. 109-112) [
- In 1977, Terjanian proved FLT for n = 2 p (p. 209) [
However, the second case of those results has not been directly proved. Note that some intermediate problems related to the FLT have not been stated or treated. For example, in the first case, we have:
- If FLT is false, y − x , x + z , y + z are not divisible by p.
- If FLT is false for the exponent p, then 2 p ≡ 1 [ p 3 ] .
- A direct proof that FLT is true for the non-prime odd exponent.
In this paper, we give efficient tools to solve these problems mentioned above and a direct proof of Fermat Last Theorem. So, we prove the following results.
Theorem 1. Let p > 2 (p prime) and ( a , b , c ) ∈ F p be a primitive triple with ( d , e , f ) its principal divisors. We have:
a b c ≡ 0 [ p ] ⇒ { 2 a = d p − e p + f p 2 b = − d p + e p + f p 2 c = d p + e p + f p with d = gcd ( a , c − b ) .
Theorem 2. Let p > 2 (p prime) and ( a , b , c ) ∈ F p be a primitive triple with ( d , e , f ) its principal divisors. We have:
a ≡ 0 [ p ] ⇒ { 2 a = d p p − e p + f p 2 b = − d p p + e p + f p 2 c = d p p + e p + f p , b ≡ 0 [ p ] ⇒ { 2 a = d p − e p p + f p 2 b = − d p + e p p + f p 2 c = d p + e p p + f p ,
c ≡ 0 [ p ] ⇒ { 2 a = d p − e p + f p p 2 b = − d p + e p + f p p 2 c = d p + e p + f p p
We prove these main results based on Fermat divisors in the following plan: First, preliminary and second, the proofs of the theorems.
In this sub-section, we introduce for the first time the Fermat divisors associated with a hypothetical solution of the equation E H . Then, we effectively compute the Fermat divisors in the case of the Pythagorean equation where we call them Pythagorean divisors. Fermat’s divisors are efficient tools in the implementation of a direct algebraic proof of FLT. Here, we are laying the foundations of these tools.
Definition 2.1. Let p > 2 (p prime), ( a , b , c ) ∈ F p a primitive triple, d = gcd ( a , c − b ) , e = gcd ( b , c − a ) , f = gcd ( c , a + b ) , a = d α , b = e β and c = f γ .
The triples ( d , e , f ) and ( α , β , γ ) are respectively defined as primary divisors and secondary divisors of Fermat associated to the triple ( a , b , c ) .
Example 2.1. Let F 2 be the set of non-trivial primitive Pythagorean triples of positive integers. Some of these divisors are illustrated very well with Pythagorean triples. In this case, we speak about Pythagorean divisors. The following table gives the values of these divisors for some Pythagorean triples. The data for
Lemma 2.1. Let ( a , b , c ) ∈ F 2 a primitive triple.
f = gcd ( c , a + b ) ⇒ f 2 = gcd ( c 2 , ( a + b ) 2 )
Proof.
f = gcd ( c , a + b ) ⇒ c = f q a n d a + b = f r a n d gcd ( q , r ) = 1
⇒ c 2 = f 2 q 2 e t ( a + b ) 2 = f 2 r 2 e t gcd ( q , r ) = 1
⇒ gcd ( c 2 , ( a + b ) 2 ) = gcd ( f 2 q 2 , f 2 r 2 )
⇒ gcd ( c 2 , ( a + b ) 2 ) = f 2 gcd ( q 2 , r 2 )
⇒ gcd ( c 2 , ( a + b ) 2 ) = f 2 b e c a u s e gcd ( q 2 , r 2 ) = 1
Proposition 2.1. Let ( a , b , c ) a primitive triple.
( a , b , c ) ∈ F 2 ⇒ f = 1 a n d γ = c .
Proof:
Suppose that f > 1 , then we have:
f = gcd ( c , a + b ) ⇒ f 2 = gcd ( c 2 , ( a + b ) 2 )
⇒ f 2 = gcd ( a 2 + b 2 , ( a + b ) 2 )
⇒ f 2 = gcd ( a 2 + b 2 , a 2 + b 2 + 2 a b )
⇒ 2 a b ≡ 0 [ f ]
⇒ a b ≡ 0 [ f ] because f is odd
⇒ □ because gcd ( a b , c ) = 1
Thus, we indeed have f = 1 so c = γ f = γ as announced.
Remark 2.1. For any primitive triples ( a , b , c ) ∈ F 2 , the divisors ( d , e , 1 ) and ( α , β , c = γ ) are defined as Pythagorean divisors.
| ( a , b , c ) ∈ F 2 | d | e | f | α | β | γ | |||
|---|---|---|---|---|---|---|---|---|---|
| 3 | 4 | 5 | 1 | 2 | 1 | 3 | 2 | 5 | |
| 5 | 12 | 13 | 1 | 4 | 1 | 5 | 3 | 13 | |
| 15 | 8 | 17 | 3 | 2 | 1 | 5 | 4 | 17 | |
| 20 | 21 | 29 | 4 | 3 | 1 | 5 | 7 | 29 | |
| 20 | 99 | 101 | 2 | 9 | 1 | 10 | 11 | 101 | |
| 28 | 45 | 53 | 4 | 5 | 1 | 7 | 9 | 53 | |
| 35 | 12 | 37 | 5 | 2 | 1 | 7 | 6 | 37 | |
Definition 2.2. Let n > 2 an integer. We define the Terjanian polynomial of degree n, as follow
T n ( x , y ) = y n − x n y − x with y ≠ x .
Remark 2.2. Note that
T n ( x , y ) = ∑ k = 0 n − 1 C n k ( y − x ) n − k − 1 x k
Lemme 2.2. Let p > 2 a prime.
( a , b , c ) ∈ F p ⇒ T p ( a , b ) = ( b − a ) p − 1 + ∑ k = 1 p − 1 C p k ( b − a ) p − 1 − k a k .
Proof.
( a , b , c ) ∈ F p ⇒ b p − a p = ( b − a + a ) p − a p
⇒ ( b − a ) T p ( a , b ) = ( b − a ) p + ∑ k = 1 p − 1 C p k ( b − a ) p − k a k
⇒ T p ( a , b ) = ( b − a ) p − 1 + ∑ k = 1 p − 1 C p k ( b − a ) p − 1 − k a k
⇒ T p ( a , b ) = ( b − a ) p − 1 + ∑ k = 1 p − 2 C p k ( b − a ) p − 1 − k a k + p a p − 1
Lemme 2.3. Let p > 2 a prime and ( a , b , c ) ∈ F p .
a b c ≡ 0 [ 3 ] ⇒ b − a ≡ 0 [ 3 ]
Proof. Let p > 2 and ( a , b , c ) ∈ F p . We have:
a b c ≡ 0 [ 3 ] ⇒ a p + b p = c p
⇒ a + b = c [ 3 ] because p is odd
⇒ ( b − a ) ( a + b ) = ( b − a ) c [ 3 ]
⇒ b 2 − a 2 ≡ ( b − a ) c [ 3 ]
⇒ 0 ≡ ( b − a ) c [ 3 ]
⇒ b − a ≡ 0 [ 3 ] b e c a u s e c ≡ 0 [ 3 ]
Proposition 2.2. Let ( a , b , c ) ∈ F p ,
a b c ≡ 0 [ 3 ] ⇒ T p ( a , b ) ≡ p [ 3 ]
Proof. Let ( a , b , c ) ∈ F p and a b c ≡ 0 [ 3 ]
b − a ≡ 0 [ 3 ] ⇒ T p ( a , b ) ≡ p a p − 1 [ 3 ] [lemmas 2.2 & 2.3]
⇒ T p ( a , b ) ≡ p [ 3 ]
⇒ □
Proposition 2.3. Let p > 2 a prime, ( a , b , c ) ∈ F p and a b c ≡ 0 [ p ]
b − a ≡ 0 [ p ] ⇒ T p ( a , b ) ≡ p [ p 2 ]
Proof. Let p > 2 a prime, ( a , b , c ) ∈ F p and a b c ≡ 0 [ p ] .
b − a ≡ 0 [ p ] ⇒ T p ( a , b ) ≡ p a p − 1 [ p 2 ] [lemma 2.2]
⇒ T p ( a , b ) ≡ p [ p 2 ] because a p − 1 ≡ 1 [ p 2 ] ( [
⇒ □
In this section, we prove the following relations which are like Barlow’s relations [4, p.100] for a hypothetical primitive solution ( a , b , c ) of Fermat’s equation.
Lemma 2.4. Let ( a , b , c ) ∈ F p a primitive triple. We have:
a b c ≡ 0 [ p ] ⇒ d e f ≡ 0 [ p ]
Proof. Suppose that d e f ≡ 0 [ p ] and d ≡ 0 [ p ] . Then a ≡ 0 [ p ] so a b c ≡ 0 [ p ] . That contradicts the assumption a b c ≡ 0 [ p ] . We have the same results with the cases e ≡ 0 [ p ] and f ≡ 0 [ p ] .
Lemma 2.5. Let p > 2 a prime and ( a , b , c ) ∈ F p a primitive triple with ( d , e , f , α , β , γ ) its Fermat divisors. We have:
a b c ≡ 0 [ p ] ⇒ { c − b = d p , T p ( b , c ) = α p , a = d α c − a = e p , T p ( a , c ) = β p , b = e β a + b = f p , T p ( − a , b ) = γ p , c = f γ .
Proof. Firstly,
( a , b , c ) ∈ F p ⇒ a p = c p − b p ⇒ a p = ( c − b ) T p ( b , c ) and gcd ( c − b , T p ( b , c ) ) = 1
⇒ d p α p = ( c − b ) T p ( b , c ) and gcd ( c − b , T p ( b , c ) ) = 1
Where, because c − b ≡ 0 [ d ] and gcd ( c − b , α ) = 1
d p α p = ( c − b ) T p ( b , c ) a n d gcd ( c − b , T p ( b , c ) ) = 1 ⇒ c − b ≡ 0 [ d p ]
and,
d p α p = ( c − b ) T p ( b , c ) a n d gcd ( c − b , T p ( b , c ) ) = 1 ⇒ d p ≡ 0 [ c − b ]
so
d p α p = ( c − b ) T p ( b , c ) a n d gcd ( c − b , T p ( b , c ) ) = 1 ⇒ c − b = d p .
Secondly,
( a , b , c ) ∈ F p a n d c − b > 1 ⇒ c − b = d p a n d a p = c p − b p
⇒ c − b = d p a n d d p α p = ( c − b ) T p ( b , c )
⇒ d p α p = d p T p ( b , c ) because c − b = d p
⇒ α p = T p ( b , c )
In Fermat’s equation, a and b can play symmetrical roles. The case of c − a is proved in the same way as that of c − b . The case of a + b is not completely different. We factorize Fermat’s equation as follows
( a + b ) T p ( − a , b ) = c p ,
and we proceed in the same way as before.
Remark 2.3. We have
{ c − a > c − b c − b ≥ 1 ⇒ c − a > 1 ⇒ e > 1.
In the second case of this problem, we have the following results.
Lemma 2.6. Let p > 2 be a prime and ( a , b , c ) ∈ F p a primitive triple with ( d , e , f , α , β , γ ) its Fermat divisors. We have:
a ≡ 0 [ p ] ⇒ { c − b = d p p , T p ( b , c ) = p α p , a = d α c − a = e p , T p ( a , c ) = β p , b = e β a + b = f p , T p ( a , − b ) = γ p , c = f γ
Proof. Let ( a , b , c ) ∈ F p
a ≡ 0 [ p ] ⇒ a p = c p − b p
⇒ d p α p = ( c − b ) T p ( b , c ) a n d g c d ( c − b , T p ( b , c ) ) = p
⇒ p v p d 1 p α p = p 2 c − b p × T p ( b , c ) p
⇒ p v p − 2 d 1 p α p = c − b p × T p ( b , c ) p
⇒ d 1 p α p = c − b p v p − 1 × T p ( b , c ) p .
As gcd ( c − b p p − 1 , α ) = 1 and gcd ( T p ( b , c ) p , d 1 ) = 1 , we have firstly
d 1 p α p = c − b p p − 1 × T p ( b , c ) p ⇒ T p ( b , c ) p ≡ 0 [ α p ] and c − b p p − 1 ≡ 0 [ d 1 p ]
and secondly
d 1 p α p = c − b p p − 1 × T p ( b , c ) p ⇒ α p ≡ 0 [ T p ( b , c ) p ] and d 1 p ≡ 0 [ c − b p p − 1 ]
we deduce that:
a ≡ 0 [ p ] ⇒ α p = T p ( b , c ) p and d 1 n = c − b p p − 1
⇒ T p ( b , c ) = p α p and c − b = p v p − 1 d 1 p = p v p d 1 p p
⇒ T p ( b , c ) = p α p and c − b = p v p − 1 d 1 p = d p p
In Fermat’s equation, a and b can play symmetrical roles, we get the same result with c − a and a + b .
Remark 2.4.
An approach like the above allows us to obtain the following results.
b ≡ 0 [ p ] ⇒ { c − b = d p , T p ( b , c ) = α p , a = d α c − a = e p p , T p ( a , c ) = p β p , b = e β a + b = f p , T p ( − a , b ) = γ p , c = f γ
and,
c ≡ 0 [ p ] ⇒ { c − b = d p , T p ( b , c ) = α p , a = d α c − a = e p , T p ( a , c ) = β p , b = e β a + b = f p p , T p ( − a , b ) = p γ p , c = f γ
In this subsection, we demonstrate theorem 1 mentioned in the introduction.
Proof.
Let p > 2 (p prime) and ( a , b , c ) ∈ F p a primitive triple with ( d , e , f , α , β , γ ) its Fermat divisors.
a b c ≡ 0 [ p ] ⇒ { 2 a = c − b − ( c − a ) + a + b 2 b = − ( c − b ) + ( c − a ) + a + b 2 c = ( c − b ) + ( c − a ) + ( a + b )
⇒ { 2 a = d p − e p + f p 2 b = − d p + e p + f p 2 c = d p + e p + f p [lemmes 2.5]
Here, we prove theorem 2 mentioned in the introduction.
Proof
Let p > 2 be a prime and ( a , b , c ) ∈ F p a non-trivial primitive triple.
b ≡ 0 [ p ] ⇒ { 2 a = ( c − b ) − ( c − a ) + a + b 2 b = − ( c − b ) + ( c − a ) + a + b 2 c = ( c − b ) + ( c − a ) + ( a + b )
⇒ { 2 a = d p − e p p + f p 2 b = − d p + e p p + f p 2 c = d p + e p p + f p [Remark 2.4]
Remark 3.1.
Similarly, we show that
a ≡ 0 [ p ] ⇒ { 2 a = d p p − e p + f p 2 b = − d p p + e p + e p p + f p 2 c = d p p + e p + f p and c ≡ 0 [ p ] ⇒ { 2 a = d p − e p + f p p 2 b = − d p + e p + f p p 2 c = d p + e p + f p p
We introduced, for the first time, the Fermat divisors, in the study of the Fermat Last Theorem (FLT). Then, we expressed the eventual solutions in terms of their principal divisors. We have improved and implemented, respectively, Barlow relations and efficient tools for direct evidence of FLT. In the future, we plan to:
- Express Pythagorean triples in terms of its Pythagorean divisors.
- Express hypothetical solution of x 2 p + y 2 p = z 2 p in term of its Fermat divisors.
- We will use these expressions to solve Diophantine equations in general and in particular to prove FLT and its sub-problems as mentioned in the introduction.
I would like to thank Dr TANOE François, Teacher-Researcher at Félix Houphouet-Boigny University and Dr KOUAKOU Vincent, Teacher-Researcher at Nangui Abrogoua University, Abidjan, RCI, for their wise eyes and constructive criticism during this research.
The author declares no conflicts of interest regarding the publication of this paper.
Kimou, P.K. (2023) On Fermat Last Theorem: The New Efficient Expression of a Hypothetical Solution as a Function of Its Fermat Divisors. American Journal of Computational Mathematics, 13, 82-90. https://doi.org/10.4236/ajcm.2023.131002