On a Diophantine Proofs of FLT: The First Case and the Secund Case z≡0 (mod p) and Significant Related Problems

Applied Mathematics and Sciences: An International Journal (MathSJ) Vol.12, No.1, March 2025
DOI : 10.5121/mathsj.2025.12101 1
ON A DIOPHANTINE PROOFS OF FLT: THE FIRST
CASE AND THE SECOND CASE 𝒛 ≡ 𝟎 (𝒎𝒐𝒅𝒑) AND
SIGNIFICANT RELATED PROBLEMS
Kimou Kouadio Prosper 1
, Kouakou Kouassi Vincent 2
1
UMRI MSN, Felix Houphouet-Boigny National Polytechnic Institute, Yamoussoukro,
Ivory Coast
2
Nangui Abrogoua University, Applied Fundamental Sciences Department, Abidjan,
Ivory Coast
ABSTRACT
In this paper, we study Fermat's equation,
𝑥𝑛
+ 𝑦𝑛
= 𝑧𝑛
(1)
with 𝑛 > 2, 𝑥, 𝑦, 𝑧 non-zero positive integers. Let (𝑎, 𝑏, 𝑐) be a triple of non-zero positive integers relativity
prime. Consider the equation (1) with prime exponent 𝑝 > 2. We establish the following results:
- 𝑎𝑝
+ 𝑏𝑝
≠ (𝑏 + 1)𝑝
. This completes the general direct proof of Abel's conjecture only prove in
the first case 𝑎𝑏(𝑏 + 1) ≢ 0 (𝑚𝑜𝑑 𝑝).
- 𝑎2𝑝
+ 𝑏2𝑝
≠ 𝑐2𝑝
. This completes the direct proof of Terjanian Theorem only prove in the first
case 𝑎𝑏𝑐 ≢ 0 (𝑚𝑜𝑑 𝑝)).
- 𝑎𝑛
+ 𝑏𝑛
≠ 𝑐𝑛
with𝑛 is a non-prime integer.A new result almost absent in the literature of this
problem.
- If 𝑎𝑏 ≢ 0 (𝑚𝑜𝑑 𝑝) then𝑎𝑝
+ 𝑏𝑝
≠ 𝑐𝑝
. This provides simultaneous Diophantine evidence for the
first case oand the second case𝑐 ≡ 0 (𝑚𝑜𝑑 𝑝) of FLT.
We analyse each of the evidence from the previous results and propose a ranking in order of increasing
difficulty to establish them.
KEYWORDS
Fermat Last Theorem (FLT), Fermat equation, Abel conjecture, the first case, the secund case, prime
exponent, non-prime exponent, even exponent, the principal Kimou divisors.
1. INTRODUCTION.
In 1670 Fermat wrote that "It is impossible for a cube to be written as the sum of two cubes or for
a fourth power to be written as the sum of two fourth powers or, in general, for any number equal
to a power greater than two to be written as the sum of two powers" [1] p.1-2.Fermat claimed to
have "woven" a wonderful proof of his problem. He gave the principle of is proof, the infinite
descent, and illustrated it by proving the exponent 4 of his problem. For a little more than three
centuries, Fermat's proposition, hitherto called Fermat's conjecture, had not yet been
demonstrated in generality, even for the first case. However, non-obvious elementary proofs
based on the principle of Fermat's infinite descent or not have been obtained for the small
exponents of 3, 5, … ,100 (first case) and 3, …,14 (general case) [1] p. 64. Using computer tools,

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these limits had been pushed to 57*109
(Morishima and Gunderson, 1948) for the first case and to
125 000 (Wagstaff,1976) for the general case [1] p.19. Apart from these results concerning
precise values of the exponents or its programming, there are other partial results involving
families of prime exponents and based on relatively elementary theories [2] p. 109-122,203-
211,360-361:
- In 1823, Sophie Germain and Legendre established the first case of FLT for exponents
n less than 100. It also states that if n = p is prime number such that 2p + 1 is still
prime, then the first case of FLT for exponent p is true.
- In 1846, Kummer used the theory of cyclotomic fields to obtain some very remarkable
results: The impossibility of Fermat's equation for regular prime number n and deduce
that the first case of FLT failsfor all prime exponents less than 100 except 37, 59 and 67.
- In 1977, Terjanian proved the first case of even exponent of FLT. He considered𝑥2𝑝
+
𝑦2𝑝
= 𝑧2𝑝
with pa prime and he used the law of reciprocity to prove an important lemma
involving quotients
𝑧𝑝−𝑦𝑝
𝑧−𝑦
,
𝑧𝑞−𝑦𝑞
𝑧−𝑦
and Jacobi's symbols.
Despite these results, general proof was still slow to be found.It was in 1985 that Andrews Wiles
provided the first recognized proof by the scientific community of Fermat's conjecture, which
would become the Fermat-Wiles theorem [2] [3]. In 2023, Kimou K. P. took the decisive step by
introducing Kimou 's divisors for a hypothetical solution of xn
+ yn
= zn
with n = 4, p, 2pand
proposing new proofs of FLT for exponent 4, the first case of the Abel conjecture, and proved
some properties related to Fermat problem [4]-[15]. Then, he proved new fundamental and
decisive results for this problem: A crucial relationship and a fundamental theorem that will
allow him to reach the "Heart" of the problem [10]-[11]. Then, in oral communication, he used
them to prove the first and second cases 𝑧 ≡ 0 (𝑚𝑜𝑑𝑝) of FLT [15].A solution (x, y, z) to the
equation (1) will be called primitive if gcd(x, y, z) = 1. This solution will be called trivial if
xyz = 0. Let n > 2 a natural number. Consider the set Fnof triples of non-trivial positive integers
solution to equation (1) define as follow:
𝐹𝑛 = {(𝑥, 𝑦, 𝑧) ∈ ℕ∗3
, 𝑥𝑛
+ 𝑦𝑛
= 𝑧𝑛}.
The objective of this paper is to give a Diophantine proof for following main results.
Theorem 1.1. Let p > 2 be a primenumber and (a, b, c) be a triple of non-zero positive integers
relatively prime. Then.
𝑐 − 𝑏 = 1 ⟹ 𝑎𝑝
+ 𝑏𝑝
≠ 𝑐𝑝
.
Theorem 1.2. Let pbe a prime number. Then,
𝑝 > 2 ⟹ 𝐹2𝑝 = ∅.
Theorem 1.3. Let nbe a nonprimepositive integer. Then,
𝑛 > 2 ⟹ 𝐹𝑛 = ∅.
Theorem 1.4. Let p > 2be a primenumber and let (a, b, c) be a triple of non-null positive
integers relatively prime. Then
𝑎𝑏 ≢ 0 (𝑚𝑜𝑑 𝑝) ⟹ 𝑎𝑝
+ 𝑏𝑝
≠ 𝑐𝑝
.

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We present our work by organizing it as follows. Section 2 preliminaries, we recall the theorems
of principal Kimou’s divisors and Diophantine quotients and remainders, the Fundamental
Relation of the Fermat equation and its corollary. In Section 3, we prove our main results. In
section 4, we give a classification in increasing order of the difficulty of the Fermat problems
studied here. In section 5, we conclude this work with a conclusion with perspectives.
2. PRELIMINARIES
In this section we define commonly used terms, state and prove theorems and lemmas necessary
for the proofs of our main results.
Definitions 2.1.
1. Diophantine proof is direct proof based on the natural integers, using only the properties
of addition, multiplication, Euclidean division, the order relation in ℕ and the
fundamental theorem of arithmetic to analyze a Diophantine equation.
2. A hypothetical solution (𝑎, 𝑏, 𝑐)of Fermat's equation is primitive if gcd(𝑎, 𝑏, 𝑐) = 1.
Remark 2.1.
1. In our research on FLT, we use classical tools such as Newton's binomial formula,
factorization, the fundamental theorem of arithmetic (implicitly), Fermat’s little theorem
and intensively modular arithmetic. We have developed some very effective new tools
for analyzing the Fermat equation. These tools are all Diophantine [Definition 2.1].
2. If (𝑎, 𝑏, 𝑐) is a non-trivial primitive solution of Fermat equation, then:
gcd(𝑎, 𝑏) = gcd(𝑎, 𝑐) = gcd(𝑏, 𝑐) = 1.
Notation 2.1.
1. We use the symbol ◻to represent the empty clause. It is the proposition that is always
false or absurd.
2. Let 𝑝 > 2 a prime number, (𝑎, 𝑏, 𝑐) ∈ 𝐹𝑝 such that 𝑎 < 𝑏 < 𝑐. Let 𝑇𝑝(𝑥, 𝑦) be the
quantity defined by
𝑇𝑝(𝑥, 𝑦) =
𝑦𝑝
− 𝑥𝑝
𝑦 − 𝑥
with 𝑥, 𝑦 ∈ {𝑏, ±𝑎, 𝑐}, 𝑦 > 𝑥.
𝑇𝑝(𝑥, 𝑦)is a positive integer.
Theorem 2.1. (Fermat’s little theorem). If 𝑝 is a prime number, then for any integer 𝑎, where 𝑝
does not divide 𝑎 (𝑎 ≢ 0 (𝑚𝑜𝑑 𝑝)) the following holds
𝑎𝑝−1
≡ 1 (𝑚𝑜𝑑 𝑝)
Proof. See [12] p.33
Theorem2.2.Let 𝑛 > 2 be an odd integer and let 𝑝 > 2be a prime number. Then
𝐹𝑝 = ∅ ⟹ 𝐹𝑛 = ∅.
Proof. Proving Theorem 2.2. is equivalent to proving that if 𝐹𝑛 ≠ ∅ then 𝐹𝑝 ≠ ∅. We proceed by
contraposed reasoning. Let us consider that 𝐹𝑛 ≠ ∅. We distinguish two cases.

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On the one hand, if 𝑛 = 2𝑙
with𝑙 ≥ 2. Consider the case 𝑙 = 2. In that case 𝑛 = 4 and equation
(1) becomes
𝑥4
+ 𝑦4
= 𝑧4
.
This is Fermat's biquadraticequation and we all know that it does not admit non-trivial solutions
[2] p.13 (2C).
Consider the case where 𝑙 > 2 then 𝑛 = 2𝑙
≡ 0 (𝑚𝑜𝑑 4). Therefore, there exists a natural
number k such that 𝑛 = 4𝑘. Equation (1) becomes 𝑥4𝑘
+ 𝑦4𝑘
= 𝑧4𝑘
. As a result
𝑥4𝑘
+ 𝑦4𝑘
= 𝑧4𝑘
⟹ (𝑥𝑘)
4
+ (𝑦𝑘)
4
= (𝑧𝑘)
4
⟹◻.
Hence 𝑛 ≠ 2𝑙
, with 𝑙 > 2. In short 𝑛 ≠ 2𝑙
with 𝑙 ≥ 2.
On the other hand, if 𝑛 ≠ 2𝑙
then 𝑛 admits a prime factor 𝑞 > 2. There exist𝑘 ≥ 2such as 𝑛 =
𝑘𝑞. Then
𝐹𝑛 ≠ ∅ ⟹ ∃(𝑎, 𝑏, 𝑐) ∈ 𝐹𝑛, 𝑎𝑏𝑐 ≠ 1
⟹ (𝑎, 𝑏, 𝑐) ∈ 𝐹𝑘𝑞 ⟹ 𝑎𝑘𝑞
+ 𝑏𝑘𝑞
= 𝑐𝑘𝑞
⟹ (𝑎𝑘)
𝑞
+ (𝑏𝑘)
𝑞
= (𝑐𝑘)
𝑞
𝑤𝑖𝑡ℎ 𝑞 > 2 𝑎 𝑝𝑟𝑖𝑚𝑒
⟹ (𝑎𝑘
,𝑏𝑘
, 𝑐𝑘) ∈ 𝐹𝑞, 𝑎𝑘
𝑏𝑘
𝑐𝑘
≠ 0
⟹ 𝐹𝑞 ≠ ∅.
Hence if 𝐹𝑝 = ∅ then 𝐹𝑛 = ∅.
Lemma 2.1. Let 𝑝 > 2be a prime number and let (𝑎, 𝑏, 𝑐) ∈ 𝐹𝑝 such that 𝑏 > 𝑎. Consider
(𝑞1, 𝑞2) and (𝑟1, 𝑟2) the quotients and the remainders of the Euclidean division of 𝑏 and 𝑐 by 𝑎:
𝑏 = 𝑎𝑞1 + 𝑟1 and 𝑐 = 𝑎𝑞2 + 𝑟2. Then,
𝑏 = 𝑎 + 1 ⟹ 𝑞1 = 𝑞2 = 1.
Proof. See [8].
Lemma 2.2. Let 𝑝 > 2be a prime number and let (𝑎, 𝑏, 𝑐) ∈ 𝐹𝑝 be a triple of primitive solution
such that 𝑏 > 𝑎. Consider 𝑐 = 𝑎𝑞2+𝑟2 𝑤𝑖𝑡ℎ 𝑟2 < 𝑎and 𝑒 = gcd(𝑏, 𝑐 − 𝑎). Then,
𝑞2 = 1 ⟹ {
𝑟2 =
𝑒𝑝
𝑝
if 𝑏 ≡ 0 (𝑚𝑜𝑑 𝑝)
𝑟2 = 𝑒𝑝
otherwise.
.
Proof.See [8].
Theorem 2.3. (Kimou-Fermat). Let 𝑝 > 2be a prime number and let (𝑎, 𝑏, 𝑐) be a triple of
positive integers relativity prime such that 𝑏 > 𝑎. Then,
(𝑎, 𝑏, 𝑐) ∈ 𝐹𝑝 ⟹ {
𝑏 − 𝑎 = 1 if 𝑐 ≡ 1 (𝑚𝑜𝑑 2)
𝑏 − 𝑎 = 2 otherwise.
.
Proof. See [10].
Lemma 2.3. Let 𝑝 > 2be a prime number and let (𝑎, 𝑏, 𝑐) ∈ 𝐹𝑝 be a triple of primitive solution
such that 𝑏 > 𝑎. Then

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𝑏 − 𝑎 = 1 ⟺ 𝑐 ≡ 1 (𝑚𝑜𝑑 2).
Proof. According to the assumptions of the previous lemma, we have:
On the one hand, let us show that if𝑏 − 𝑎 = 1 then𝑐 ≡ 1 (𝑚𝑜𝑑 2). We have:
𝑏 − 𝑎 = 1 ⟹ 𝑏 = 𝑎 + 1
⟹ 𝑎, 𝑏 are opposite parity
⟹ 𝑐 is odd because gcd(𝑎, 𝑏, 𝑐) = 1.
Reciprocally, let us show that if𝑐 ≡ 1 (𝑚𝑜𝑑 2)then 𝑏 − 𝑎 = 1. We proceeded by reasoning by
absurd:
𝑐 ≡ 1 (𝑚𝑜𝑑 2) , 𝑏 − 𝑎 = 2 ⟹ 𝑏, 𝑎 have the same parity
⟹ 𝑏, 𝑎 are odd because gcd(𝑎, 𝑏, 𝑐) = 1.
⟹ 𝑐is even
⟹◻.
Hence 𝑏 − 𝑎 = 1.
Lemma 2.4. Let 𝑝 > 2be a prime and let (𝑎, 𝑏, 𝑐) ∈ 𝐹𝑝be a triple of primitive solution such that
𝑏 > 𝑎. Then
𝑏 − 𝑎 = 2 ⟺ 𝑐 ≡ 0 (𝑚𝑜𝑑 2).
Proof.Can be deduced by contraposition of the previous lemma.
Lemma 2.5. Let 𝑝 > 2be a prime number and let(𝑎, 𝑏, 𝑐) be a triple of relativity primeintegers.
Then
(𝑎, 𝑏, 𝑐) ∈ 𝐹𝑝 ⟹
{
𝑐 − 𝑏 =
𝑑𝑝
gcd(𝑑, 𝑝)
, 𝑇𝑝(𝑏, 𝑐) = gcd(𝑑, 𝑝)𝛼𝑝
, 𝑎 = 𝑑𝛼
𝑐 − 𝑎 =
𝑒𝑝
gcd(𝑒, 𝑝)
, 𝑇𝑝(𝑎, 𝑐) = gcd(𝑒, 𝑝)𝛽𝑝
, 𝑏 = 𝑒𝛽
𝑎 + 𝑏 =
𝑓𝑝
gcd(𝑓, 𝑝)
, 𝑇𝑝(−𝑎, 𝑏) = gcd(𝑓, 𝑝)𝛾𝑝
, 𝑐 = 𝑓𝛾.
where the sextuple (𝑑, 𝑒, 𝑓, 𝛼, 𝛽, 𝛾) of positive integers is the Kimou divisors of (𝑎, 𝑏, 𝑐).
Proof. See [7], [9].
Remark 2.2.
1. The triple (𝑑, 𝑒, 𝑓) of non-zero positive integers is called Kimou primaries divisors of
(𝑎, 𝑏, 𝑐) and defined by follow:
𝑑 = gcd(𝑎, 𝑐 − 𝑏) , 𝑒 = gcd(𝑏, 𝑐 − 𝑎) 𝑎𝑛𝑑 𝑓 = gcd(𝑐, 𝑎 + 𝑏).
2. Lemma 2.5 is the concise and unified version of Lemmas 2.4, 2.5 and Remarks 2.3,2.4.
in [8] pp. 87-89.
3. If (𝑎, 𝑏, 𝑐) ∈ 𝐹𝑝 then

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{
gcd(𝑑, 𝑝) = gcd(𝑒, 𝑝) = gcd(𝑓, 𝑝) = 1 𝑖𝑓 𝑎𝑏𝑐 ≢ 0 (𝑚𝑜𝑑 𝑝)
gcd(𝑑, 𝑝) = 𝑝, gcd(𝑒, 𝑝) = gcd(𝑓, 𝑝) = 1 𝑖𝑓 𝑎 ≡ 0 (𝑚𝑜𝑑 𝑝)
gcd(𝑒, 𝑝) = 𝑝, gcd(𝑑, 𝑝) = gcd(𝑓, 𝑝) = 1 𝑖𝑓 𝑏 ≡ 0 (𝑚𝑜𝑑 𝑝)
gcd(𝑓, 𝑝) = 𝑝, gcd(𝑑, 𝑝) = gcd(𝑒, 𝑝) = 1 𝑖𝑓 𝑐 ≡ 0 (𝑚𝑜𝑑 𝑝).
Lemma 2.6. Let 𝑝 > 2be a prime number, let (𝑎, 𝑏, 𝑐) be a triple of relativity prime integers.
Then
(𝑎, 𝑏, 𝑐) ∈ 𝐹𝑝 ⟹ gcd(𝑑, 𝛼) = gcd(𝑒, 𝛽) = gcd(𝑓, 𝛾) = 1
where the sextuple (𝑑, 𝑒, 𝑓, 𝛼, 𝛽, 𝛾) of positive integers is the Kimou’s divisors of (𝑎, 𝑏, 𝑐)
[Lemma 2.5].
Proof.Under the assumptions of the previous lemma, we have:
(𝑎, 𝑏, 𝑐) ∈ 𝐹𝑝 ⟹ {
𝑑 = gcd(𝑎, 𝑐 − 𝑏)
𝑒 = gcd(𝑏, 𝑐 − 𝑎)
𝑓 = gcd(𝑐, 𝑎 + 𝑏)
[𝑅𝑒𝑚𝑎𝑟𝑘 2.2. ][7]𝑝. 84
⟹
{
𝑑 = gcd (𝑎,
𝑑𝑝
gcd(𝑑, 𝑝)
)
𝑒 = gcd (𝑏,
𝑒𝑝
gcd(𝑒, 𝑝)
)
𝑓 = gcd (𝑐,
𝑓𝑝
gcd(𝑓, 𝑝)
)
[𝐿𝑒𝑚𝑚𝑎 2.5]
⟹
{
𝑑 = gcd (𝑑𝛼,
𝑑𝑝
gcd(𝑑, 𝑝)
)
𝑒 = gcd (𝑒𝛽,
𝑒𝑝
gcd(𝑒, 𝑝)
)
𝑓 = gcd (𝑓𝛾,
𝑓𝑝
gcd(𝑓, 𝑝)
)
⟹
{
1 = gcd (𝛼,
𝑑𝑝−1
gcd(𝑑, 𝑝)
)
1 = gcd (𝛽,
𝑒𝑝−1
gcd(𝑒, 𝑝)
)
1 = gcd (𝛾,
𝑓𝑝−1
gcd(𝑓, 𝑝)
)
⟹ {
1 = gcd(𝛼, 𝑑)
1 = gcd(𝛽, 𝑒)
1 = gcd(𝛾, 𝑓).
Lemma 2.7. Let 𝑝 > 2be a prime number and let (𝑎, 𝑏, 𝑐) ∈ 𝐹𝑝 be a triple of primitive solution.
Then
𝑎𝑏𝑐 ≢ 0 (𝑚𝑜𝑑 𝑝) ⟹ (𝑐 − 𝑏)(𝑐 − 𝑎)(𝑎 + 𝑏) ≢ 0 (𝑚𝑜𝑑 𝑝)
Proof. Let 𝑝 > 2be a prime number and let (𝑎, 𝑏, 𝑐) ∈ 𝐹𝑝 be a triple of non-zero primitive
positive integers. Consider the sextuple (𝑑, 𝑒, 𝑓, 𝛼, 𝛽, 𝛾)of positive integers, its Kimou divisors.
We have
𝑎𝑏𝑐 ≢ 0 (𝑚𝑜𝑑 𝑝) ⟹ 𝑑𝑒𝑓 ≢ 0 (𝑚𝑜𝑑 𝑝)otherwise 𝑎𝑏𝑐 ≡ 0 (𝑚𝑜𝑑 𝑝)

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⟹ 𝑑 ≢ 0 (𝑚𝑜𝑑 𝑝),𝑒 ≢ 0 (𝑚𝑜𝑑 𝑝),𝑓 ≢ 0 (𝑚𝑜𝑑 𝑝)
⟹ 𝑑𝑝
≢ 0 (𝑚𝑜𝑑 𝑝),𝑒𝑝
≢ 0 (𝑚𝑜𝑑 𝑝), 𝑓𝑝
≢ 0 (𝑚𝑜𝑑 𝑝)
⟹ 𝑐 − 𝑏 ≢ 0 (𝑚𝑜𝑑 𝑝),𝑐 − 𝑎 ≢ 0 (𝑚𝑜𝑑 𝑝),𝑎 + 𝑏 ≢ 0 (𝑚𝑜𝑑 𝑝) [𝐿𝑒𝑚𝑚𝑎 2.5. ]
Lemma 2.8. Let 𝑝 > 2be a prime number and (𝑎, 𝑏, 𝑐) ∈ 𝐹𝑝 be a triple of primitive solution.
Consider the triple (𝛼, 𝛽, 𝛾) theauxiliary Kimou divisors of (𝑎, 𝑏, 𝑐).Then
𝛼 ≡ 𝛽 ≡ 𝛾 ≡ 1 (𝑚𝑜𝑑 𝑝)
Proof. Let's deal with the first case of this problem. We have
𝑎𝑏𝑐 ≢ 0 (𝑚𝑜𝑑 𝑝) ⟹ 𝑇𝑝(𝑎, 𝑐) =
𝑐𝑝
− 𝑎𝑝
𝑐 − 𝑎
⟹ 𝑇𝑝(𝑎, 𝑐) ≡
𝑐𝑝
− 𝑎𝑝
𝑐 − 𝑎
(𝑚𝑜𝑑 𝑝)[𝐿𝑒𝑚𝑚𝑎 2.7. ]
⟹ 𝑇𝑝(𝑎, 𝑐) ≡
𝑐 − 𝑎
𝑐 − 𝑎
(𝑚𝑜𝑑 𝑝) [𝑇ℎ𝑒𝑜𝑟𝑒𝑚 2.1. ]
⟹ 𝑇𝑝(𝑎, 𝑐) ≡ 1 (𝑚𝑜𝑑 𝑝)
⟹ 𝛼𝑝
≡ 1 (𝑚𝑜𝑑 𝑝) [𝐿𝑒𝑚𝑚𝑎 2.5]
⟹ 𝛼 ≡ 1 (𝑚𝑜𝑑 𝑝) [𝑇ℎ𝑒𝑜𝑟𝑒𝑚 2.1. ]
The same approach is followed to show that 𝛽 ≡ 𝛾 ≡ 1 (𝑚𝑜𝑑 𝑝).
In the second case, let us illustrate the evidence on the case 𝑎 ≡ 0 (𝑚𝑜𝑑 𝑝). On the one hand,
𝑎 ≡ 0 (𝑚𝑜𝑑 𝑝) ⟹ 𝑎𝑝
+ 𝑏𝑝
= 𝑐𝑝
⟹ 𝑏𝑝
≡ 𝑐𝑝
(𝑚𝑜𝑑 𝑝𝑝
) ⟹ 𝑏 ≡ 𝑐 (𝑚𝑜𝑑 𝑝𝑝−1
)
⟹ 𝑏 ≡ 𝑐 (𝑚𝑜𝑑 𝑝2) 𝑏𝑒𝑐𝑎𝑢𝑠𝑒 𝑝 ≥ 3)
On the other hand,
𝑎 ≡ 0 (𝑚𝑜𝑑 𝑝) ⟹ 𝑇𝑝(𝑏, 𝑐) = 𝑝𝛼𝑝
[𝐿𝑒𝑚𝑚𝑎 2.5, 𝑅𝑒𝑚𝑎𝑟𝑘 2.2. ]
⟹ 𝑇𝑝(𝑏, 𝑐) ≡ 𝑝𝛼𝑝
(𝑚𝑜𝑑 𝑝2
)
⟹ 𝑝𝑏𝑝−1
≡ 𝑝𝛼𝑝(𝑚𝑜𝑑 𝑝2), 𝑢𝑠𝑖𝑛𝑔 𝑡ℎ𝑒 𝑝𝑟𝑒𝑣𝑖𝑜𝑢𝑠 𝑟𝑒𝑠𝑢𝑙𝑡
⟹ 𝑏𝑝−1
≡ 𝛼𝑝
(𝑚𝑜𝑑 𝑝)
⟹ 1 ≡ 𝛼𝑝(𝑚𝑜𝑑 𝑝) [𝑇ℎ𝑒𝑜𝑟𝑒𝑚 2.1]
⟹ 1 ≡ 𝛼 (𝑚𝑜𝑑 𝑝) [𝑇ℎ𝑒𝑜𝑟𝑒𝑚 2.1].
A similar approach is followed to deal with cases 𝑏𝑐 ≡ 0 (𝑚𝑜𝑑 𝑝).
Remark 2.3. Let 𝑝 > 2be a prime number and let (𝑎, 𝑏, 𝑐)be a triple of relativity prime integers.
Then
(𝑎, 𝑏, 𝑐) ∈ 𝐹𝑝 ⟹ 𝛼 ≥ 𝑝, 𝛽 ≥ 𝑝, 𝛾 ≥ 𝑝 ⟹ 𝛼 > 2, 𝛽 > 2, 𝛾 > 2.
3. PROOF OF OUR MAIN RESULTS
3.1. Proof of Theorem 1.1.
Conjecture (Abel). Let 𝑝 > 2be a prime number and let (𝑎, 𝑏, 𝑐) ∈ 𝐹𝑝 be a triple of primitive
solution. Then none of the 𝑎, 𝑏and𝑐 is the power of a prime number.

8
The cases where 𝑏 and 𝑐 are powers of a prime number have been proved by Moller [14]. He also
proved that if 𝑎 is a prime power, then 𝑐 − 𝑏 = 1 [13], [14].The first case of this conjecture was
proved by Abel himself. New evidence was given by Kimou P. in 2023 [6]. The second case has
yet to receive direct proof. That's precisely the aim of this subsection.In what follows, we prove
this conjecture in full.
Lemma 3.1. Let 𝑝 > 2be a prime and let (𝑎, 𝑏, 𝑐) ∈ 𝐹𝑝be a triple of primitive solution such that
𝑎𝑏𝑐 ≡ 0 (𝑚𝑜𝑑 𝑝). If ∀𝑥 ∈ {𝑎, 𝑏, 𝑐}, ∄(𝜋, 𝑚) ∈ ℕ∗2
with 𝜋is a prime number and 𝑥 = 𝜋𝑚
then
𝑥 ≢ 0 (𝑚𝑜𝑑 𝑝).
Proof.Under the assumptions of the previous lemma, we will proceed by absurdity, assuming that
𝑥 ≡ 0 (𝑚𝑜𝑑 𝑝).
On the one hand,
𝑏 = 𝜋𝑚
,𝑏 ≡ 0 (𝑚𝑜𝑑 𝑝) ⟹ 𝜋𝑚
≡ 0 (𝑚𝑜𝑑 𝑝)
⟹ 𝜋 ≡ 0 (𝑚𝑜𝑑 𝑝)
⟹ 𝜋 = 𝑝 𝑜𝑟 𝑝 = 1 ⟹ 𝜋 = 𝑝.
On the other hand, according to the above, we have:
𝑏 = 𝜋𝑚
,𝑏 ≡ 0 (𝑚𝑜𝑑 𝑝) ⟹ 𝑒𝛽 = 𝑝𝑚
⟹ 𝑒𝛽 ≡ 0 (𝑚𝑜𝑑 𝑝)
⟹ 𝑒 ≡ 0 (𝑚𝑜𝑑 𝑝) [𝐿𝑒𝑚𝑚𝑎 2.8])
⟹ 𝑒 ≡ 0 (𝑚𝑜𝑑 𝑝𝑚
)
⟹ 𝑒 = 𝑘𝑏 = 𝑏 𝑤𝑖𝑡ℎ 𝑘 ≥ 1 𝑖𝑠 𝑎 𝑖𝑛𝑡𝑒𝑔𝑒𝑟
⟹ 𝑒 = 𝑏, 𝛽 = 1
⟹◻ [𝑅𝑒𝑚𝑎𝑟𝑘 2.3].
Hence𝑏 ≢ 0 (𝑚𝑜𝑑 𝑝).We proceed in the same way with 𝑎 and 𝑐.
Remark 3.1. When 𝑏 is even, it is treated as follows.If 𝑏 is even, then 𝑏 = 2𝑚
and consequently
𝑏 ≢ 0 (𝑚𝑜𝑑𝑝). According to lemma 3.1. we treat in the following the triplets (𝑎, 𝑏, 𝑐) ∈ 𝐹𝑝 such
that
𝑏𝑐 ≢ 0 (𝑚𝑜𝑑𝑝) with 𝑐 = 𝑏 + 1.
Lemma 3.2. Let 𝑝 > 2 be a prime and let (𝑎, 𝑏, 𝑐) ∈ 𝐹𝑝 be a triple of primitive solution such that
𝑒 = gcd(𝑏, 𝑐 − 𝑎) and 𝑏 = 𝑒𝛽. Then
𝛽 > 𝑒𝑝−1
.
Proof.
(𝑎, 𝑏, 𝑐) ∈ 𝐹 ⟹ 𝑎 + 𝑏 − 𝑐 = 𝑏 − (𝑐 − 𝑎) = 𝑒𝛽 − 𝑒𝑝
⟹ 𝑎 + 𝑏 − 𝑐 = 𝑒(𝛽 − 𝑒𝑝−1
)
⟹ 𝛽 − 𝑒𝑝−1
> 0 ⟹ 𝛽 > 𝑒𝑝−1
.
Lemma 3.3. Let 𝑝 > 2 be a prime and let (𝑎, 𝑏, 𝑐) ∈ 𝐹𝑝 be a triple of primitive solution such that
𝑒 = gcd(𝑏, 𝑐 − 𝑎) and 𝑏 = 𝑒𝛽. Then

9
{
𝛾 >
𝑓𝑝−1
2𝑝
𝑖𝑓 𝑐 ≡ 0 (𝑚𝑜𝑑 𝑝)
𝛾 >
𝑓𝑝−1
2𝑝
𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
.
Proof. Consider the assumptions of the previous lemma and (𝑎, 𝑏, 𝑐) ∈ 𝐹𝑝 a triple of primitive
solution. On the one hand
𝑐 ≡ 0 (𝑚𝑜𝑑 𝑝) ⟹ 2𝑐 = 𝑑𝑝
+ 𝑒𝑝
+
𝑓𝑝
𝑝
[7]
⟹ 2𝑐 >
𝑓𝑝
𝑝
⟹ 2𝑓𝛾 >
𝑓𝑝
𝑝
⟹ 𝛾 >
𝑓𝑝−1
2𝑝
.
On the other hand
𝑐 ≢ 0 (𝑚𝑜𝑑 𝑝) ⟹ 2𝑐 =
𝑑𝑝
𝑝
+ 𝑒𝑝
+ 𝑓𝑝
𝑜𝑟 2𝑐 = 𝑑𝑝
+
𝑒𝑝
𝑝
+ 𝑓𝑝
[7]
⟹ 2𝑐 > 𝑓𝑝
⟹ 2𝑓𝛾 > 𝑓𝑝
⟹ 𝛾 >
𝑓𝑝−1
2
.
Remark. If 𝑐 ≡ 0 (𝑚𝑜𝑑 𝑝) then 𝑐 ≡ 0 (𝑚𝑜𝑑 𝑝2
)[2] p. Hence 𝑓 ≡ 0 (𝑚𝑜𝑑𝑝2
)as a result:
𝑓 ≥ 𝑝2
> 4. When𝑐 ≢ 0 (𝑚𝑜𝑑𝑝)let consider the following case:
𝑏 ≡ 0 (𝑚𝑜𝑑 𝑝) ⟹ 𝑐 ≡ 𝑎 (𝑚𝑜𝑑 𝑝2
) ⟹ 𝑓 ≡ 𝑑 (𝑚𝑜𝑑 𝑝2
)
⟹ 𝑓 = 𝑑 + 𝑘𝑝2
⟹ 𝑓 > 𝑝2
The same procedure is followed for the other cases.
Lemma 3.4. Let 𝑝 > 2be a prime and let (𝑎, 𝑏, 𝑐) ∈ 𝐹𝑝. Consider 𝑒 = gcd(𝑏, 𝑐 − 𝑎). Then
𝑏 ≢ 0 (𝑚𝑜𝑑 𝑝) ⟹ ∄(𝜋, 𝑚) ∈ ℕ∗2
, 𝑏 = 𝜋𝑚
with 𝜋 is prime number.
Proof. Let 𝑏 ≢ 0 (𝑚𝑜𝑑 𝑝). We proceed by reasoning by the absurd. Let’s assume that (𝑎, 𝑏, 𝑐) ∈
𝐹𝑝 𝑎𝑛𝑑𝑏 = 𝜋𝑚
.We have:
𝑏 = 𝜋𝑚
⟹ 𝑒𝛽 = 𝜋𝑚
⟹ 𝑒 = 1 𝑐𝑎𝑟 𝛽 > 𝑒 [𝐿𝑒𝑚𝑚𝑎 3.2]
⟹ 𝑒𝑝
= 1 ⟹ 𝑐 − 𝑎 = 1 [𝐿𝑒𝑚𝑚𝑎 2.5]
⟹ 𝑐 = 𝑎 + 1
⟹◻ because 𝑎 < 𝑏 < 𝑐.
Hence ∄(𝜋, 𝑚) ∈ ℕ∗2
, 𝑏 = 𝜋𝑚
with 𝜋 a prime number.
Lemma3.5. Let 𝑝 > 2be a prime and let (𝑎, 𝑏, 𝑐) ∈ 𝐹𝑝 be a triple of primitive solution. Then
𝑐 ≢ 0 (𝑚𝑜𝑑 𝑝) ⟹ ∄(𝜋, 𝑚) ∈ ℕ∗2
, 𝑐 = 𝜋𝑚
with π is a prime

10
Proof. Let (𝑎, 𝑏, 𝑐) ∈ 𝐹𝑝 and 𝑐 ≢ 0 (𝑚𝑜𝑑 𝑝). We reason from the absurd by supposing that 𝑐 =
𝜋𝑚
.We have
(𝑎, 𝑏, 𝑐) ∈ 𝐹𝑝, 𝑐 ≢ 0 (𝑚𝑜𝑑 𝑝) ⟹ 𝑓𝛾 = 𝜋𝑚
⟹ 𝑓 = 1 𝑏𝑒𝑐𝑎𝑢𝑠𝑒 𝛾 > 𝑓 [𝐿𝑒𝑚𝑚𝑎 3.3. ]
⟹ 𝑓𝑝
= 1
⟹ 𝑎 + 𝑏 = 1 [𝐿𝑒𝑚𝑚𝑎 2.5]
⟹ 𝑎𝑏 = 0 ⟹◻.
Hence ∄(𝜋, 𝑚) ∈ ℕ∗2
, 𝑐 = 𝜋𝑚
with 𝜋 a prime number.
Lemma 3.6. Let 𝜋 > 2be a prime and let (𝑎, 𝑏, 𝑐) ∈ 𝐹𝑝be a triple of primitive positive integers
solution of equation (1).Then,
𝑎 = 𝜋𝑚
⟹ 𝑐 − 𝑏 = 1.
Proof. Under the assumptions of lemma 3.4. we proceedby absurd supposing that 𝑐 − 𝑏 > 1. We
have:
𝑐 − 𝑏 > 1 ⟹ 𝑑𝛼 = 𝜋𝑚
,𝑑𝑝
> 1 [𝐿𝑒𝑚𝑚𝑎𝑠 2.5, 2.6]
⟹ 𝑑𝛼 = 𝜋𝑚
,𝑑 > 1
⟹ 𝑑𝛼 = 𝜋𝑚
, 𝑑 > 1, 𝛼 > 𝑑 > 1.
⟹◻ because gcd(𝑑, 𝛼) = 1 [𝐿𝑒𝑚𝑚𝑎 2.6].
Hence 𝑐 − 𝑏 = 1.
Lemme 3.7. Let 𝑝 > 2be a prime number and let (𝑎, 𝑏, 𝑐) ∈ 𝐹𝑝be a triple of primitive solution.
Then
𝑐 − 𝑏 = 1 ⟹ 𝑎 ≡ 1 (𝑚𝑜𝑑 2).
Proof.
𝑐 − 𝑏 = 1 ⟹ 𝑐 = 𝑏 + 1
⟹ 𝑐 or 𝑏 is even
⟹ 𝑎 is odd.
Lemme 3.8. Let 𝑝 > 2 be a prime number and let (𝑎, 𝑏, 𝑐) ∈ 𝐹𝑝 be a triple of primitive solution.
Then
𝑐 − 𝑏 = 1 ⟹ {
𝑐 − 𝑎 = 2 if 𝑐 ≡ 1 (𝑚𝑜𝑑 2)
𝑐 − 𝑎 = 3 otherwise
Proof. Under the assumptions of the previous lemma, we have:
𝑐 − 𝑏 = 1 ⟹ 𝑏 = 𝑐 − 1
⟹ {
𝑏 − 𝑎 = 1 if 𝑐 ≡ 1 (𝑚𝑜𝑑 2)
𝑏 − 𝑎 = 2 otherwise
[𝑇ℎ𝑒𝑜𝑟𝑒𝑚 2.3. ]
⟹ {
𝑐 − 1 − 𝑎 = 1 if 𝑐 ≡ 1 (𝑚𝑜𝑑 2)
𝑐 − 1 − 𝑎 = 2 otherwise
⟹ {
𝑐 − 𝑎 = 2 if 𝑐 ≡ 1 (𝑚𝑜𝑑 2)

11
Lemma 3.9. Let 𝑝 > 2 be a prime number and let (𝑎, 𝑏, 𝑐) a triple of non-null positive integers
relativity primesuch that 𝑐 − 𝑏 = 1. Then
𝑏 ≢ 0 (𝑚𝑜𝑑 𝑝) ⟹ 𝑎𝑝
+ 𝑏𝑝
≠ 𝑐𝑝
Proof. Under the assumptions of the previous lemma, we have if 𝑏 ≢ 0 (𝑚𝑜𝑑 𝑝). Consider 𝑣1
and 𝑣2 respectively the 2-adic and 3-adic valuations of 𝑒. Then
𝑐 − 𝑏 = 1 ⟹ {𝑐 − 𝑎 = 2 if 𝑐 ≡ 1 (𝑚𝑜𝑑 2)
⟹ {
𝑒𝑝
= 2 𝑖𝑓 𝑐 ≡ 1 (𝑚𝑜𝑑 2)
𝑒𝑝
= 3 otherwise
[𝐿𝑒𝑚𝑚𝑎 2.5, 𝑅𝑒𝑚𝑎𝑟𝑘 2.2]
⟹ {
𝑘1
𝑝
2𝑣1𝑝−1
= 1 if 𝑐 ≡ 1 (𝑚𝑜𝑑 2)
𝑘2
𝑝
3𝑣2𝑝−1
= 1 otherwise
with 𝑒 = 𝑘12𝑣1 = 𝑘23𝑣2
⟹ {
2𝑣1𝑝−1
= 1 if 𝑐 ≡ 1 (𝑚𝑜𝑑 2)
3𝑣2𝑝−1
= 1 otherwise
⟹
{
𝑝 =
1
𝑣1
if 𝑐 ≡ 1 (𝑚𝑜𝑑 2)
𝑝 =
1
𝑣2
otherwise
⟹ {
𝑝 = 1 if 𝑐 ≡ 1 (𝑚𝑜𝑑 2)
𝑝 = 1 otherwise
⟹ 𝑝 = 1 ⟹◻ because 𝑝 > 2.
Hence 𝑐 − 𝑏 > 1.
Remark 3.2. Let 𝑝 > 2 be a prime number and let (𝑎, 𝑏, 𝑐) a triple of non-null positive integers
relativity prime such that 𝑐 − 𝑏 = 1. When 𝑐 ≡ 0 (𝑚𝑜𝑑𝑝), we have 𝑏 ≢ 0 (𝑚𝑜𝑑𝑝) and
consequently, according to the preceding Lemma, 𝑎𝑝
+ 𝑏𝑝
≠ 𝑐𝑝
.
Lemma 3.10. Let 𝑝 > 2 be a prime number and let (𝑎, 𝑏, 𝑐) a triple of non-null positive integers
relativity prime such that 𝑐 − 𝑏 = 1. Then
𝑏 ≡ 0 (𝑚𝑜𝑑 𝑝) ⟹ 𝑎𝑝
+ 𝑏𝑝
≠ 𝑐𝑝
Proof. Under the assumptions of the previous lemma, we have≡ 0 (𝑚𝑜𝑑 𝑝). Consider 𝑣1, 𝑣2and
𝑣3 respectively the 2-adic, 3-adic and 𝑝-adic valuations of 𝑒. Then
𝑏 ≡ 0 (𝑚𝑜𝑑 𝑝) ⟹ {
𝑐 − 𝑎 = 2 if 𝑐 ≡ 1 (𝑚𝑜𝑑 2)
⟹
{
𝑒𝑝
𝑝
= 2 if 𝑐 ≡ 1 (𝑚𝑜𝑑 2)
𝑒𝑝
𝑝
= 3 otherwise
[𝐿𝑒𝑚𝑚𝑎 2.5, 𝑅𝑒𝑚𝑎𝑟𝑘 2.2]
⟹ {
𝑘1
𝑝
2𝑣1𝑝−1
= 𝑝 if 𝑐 ≡ 1 (𝑚𝑜𝑑 2)
𝑘2
𝑝
3𝑣2𝑝−1
= 𝑝 otherwise
⟹ {
𝑘1
𝑝
2𝑣1𝑝−1
𝑝𝑣3𝑝−1
= 1 if 𝑐 ≡ 1 (𝑚𝑜𝑑 2)
𝑘2
𝑝
3𝑣2𝑝−1
𝑝𝑣3𝑝−1
= 1 otherwise
⟹ {
𝑘1 = 𝑘2 = 1
2𝑣1𝑝−1
𝑝𝑣3𝑝−1
= 1 if 𝑐 ≡ 1 (𝑚𝑜𝑑 2)
3𝑣2𝑝−1
𝑝𝑣3𝑝−1
= 1 otherwise

12
⟹ {
𝑣1𝑝 − 1 = 0, 𝑣3𝑝 − 1 = 0 if 𝑐 ≡ 1 (𝑚𝑜𝑑 2)
𝑣2𝑝 − 1 = 0, 𝑣3𝑝 − 1 = 0 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
⟹
{
𝑝 =
1
𝑣1
, 𝑝 =
1
𝑣3
if 𝑐 ≡ 1 (𝑚𝑜𝑑 2)
𝑝 =
1
𝑣2
, 𝑝 =
1
𝑣3
, 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
⟹ {
𝑣1 = 𝑣2 = 𝑣3 = 1
𝑝 = 1
⟹ 𝑝 = 1 ⟹◻.
Hence 𝑐 − 𝑏 > 1.
Remark 3.3. Let 𝑝 > 2 be a prime number and let (𝑎, 𝑏, 𝑐) a triple of non-null positive integers
relativity prime such that 𝑐 − 𝑏 = 1. When 𝑐 ≢ 0(𝑚𝑜𝑑𝑝), we have 𝑏 ≡ 0 (𝑚𝑜𝑑𝑝)or
𝑏 ≢ 0 (𝑚𝑜𝑑𝑝) . In both cases, lemmas 3.9 and 3.10 confirm that 𝑎𝑝
+ 𝑏𝑝
≠ 𝑐𝑝
.
Proof of Theorem 1.1. Immediate consequences of Lemmas 3.9 and 3.10, and Remark 3.2 and
3.3.
3.2. Proof of Theorem 1.2
Proof. Let 𝑝 > 2 be a prime number and let (𝑎, 𝑏, 𝑐) be a triple of relativity prime integers. Then
(𝑎, 𝑏, 𝑐) ∈ 𝐹2𝑝 ⟹ (𝑎2
, 𝑏2
, 𝑐2) ∈ 𝐹𝑝, 𝑐 ≡ 1 (𝑚𝑜𝑑 2)
⟹ 𝑏2
− 𝑎2
= 1 [Theorem 2.3.]
⟹ (𝑏 − 𝑎)(𝑎 + 𝑏) = 1
⟹ 𝑎 + 𝑏 = 1 𝑒𝑡 𝑏 − 𝑎 = 1
⟹ 𝑏 = 1 ⟹◻ because 𝑏 > 1.
Hence the result.
Remark 3.4. Because ofTheorem1.2 and [1] p.13 (2C), Fermat's theorem is true for all even
exponents.
3.3.1. Proof of FLT for Odd No-Prime Exponent
Theorem 3.3. Let 𝑚 > 2be a positive integer. Then.
𝑚 𝑖𝑠 𝑎𝑛 odd nonprime integer ⟹ 𝐹𝑚 = ∅.
Proof. Let 𝑚 > 2be an odd no-prime number. (𝑎, 𝑏, 𝑐) ∈ 𝐹𝑚. Then:
𝑚 is odd nonprime integer ⟹ ∃(𝑠, 𝑘), 𝑠 > 2, 𝑘 > 2 are odd prime, 𝑎𝑘𝑠
+ 𝑏𝑘𝑠
= 𝑐𝑘𝑠
⟹ (𝑎𝑘)
𝑠
+ (𝑏𝑘)
𝑠
= (𝑐𝑘)
𝑠
⟹ 𝑏𝑘
− 𝑎𝑘
= 1 or𝑏𝑘
− 𝑎𝑘
= 2 [𝑇ℎ𝑒𝑜𝑟𝑒𝑚 2.3]
⟹ (𝑏 − 𝑎)𝑇𝑘(𝑎, 𝑏) = 1 or 𝑏 − 𝑎 = 2 because 𝑘 is odd
⟹ 1 > 𝑘(𝑏 − 𝑎)𝑎𝑘−1
or 2 > 𝑘(𝑏 − 𝑎)𝑎𝑘−1
⟹ 𝑏 = 𝑎 ⟹◻;

13
Hence 𝑚 cannot be an odd nonprime integer and consequently 𝐹𝑚 = ∅.
Remark 3.5. FLT is true for odd nonprime exponent.
3.3.2. Proof FLT for Nonprime Exponent
Under the assumptions of Theorem 1.3. let consider(𝑎, 𝑏, 𝑐) ∈ 𝐹𝑛 be a triple of primitive integers
with 𝑛a nonprime positive integer. We proceed by the absurd:
𝑛 ≡ 0 (𝑚𝑜𝑑2) ⟹ ∃(𝑙, 𝑞) ∈ ℕ2
, 𝑙 > 1, 𝑞 > 3 𝑎𝑛𝑜𝑑𝑑𝑖𝑛𝑡𝑒𝑔𝑒𝑟, 𝑛 = 2𝑞𝑜𝑟2𝑙
𝑞
⟹ 𝑎2𝑞
+ 𝑏2𝑝
= 𝑐2𝑞
𝑜𝑟 𝑎2𝑙𝑞
+ 𝑏2𝑙𝑝
= 𝑐2𝑙𝑞
⟹ ∃ 𝑝 > 3 𝑎 𝑝𝑟𝑖𝑚𝑒, 𝑙1 ≥ 1, 𝑎2𝑝𝑞1 + 𝑏2𝑝𝑞1 = 𝑐2𝑝𝑞1 𝑜𝑟 𝑎4𝑙1𝑞
+ 𝑏4𝑙1 = 𝑐4𝑙1𝑞
⟹ (𝑎𝑞1)2𝑝
+ (𝑎𝑞1)2𝑝
= (𝑎𝑞1)2𝑝
𝑜𝑟 (𝑎𝑙1𝑞)
4
+ (𝑎𝑙1𝑞)
4
= (𝑐𝑙1𝑞)
4
⟹◻. [Theorem 1.2] [2] p.13 (2C).
Hence 𝑛 ≢ 0 (𝑚𝑜𝑑2). Par suite 𝑛 ≡ 1 (𝑚𝑜𝑑2). Traitons ce cas :
𝑛 ≡ 1 (𝑚𝑜𝑑2) ⟹ 𝑛 𝑖𝑠 𝑎𝑛 𝑜𝑑𝑑 𝑛𝑜𝑛𝑝𝑟𝑖𝑚𝑒 𝑖𝑛𝑡𝑒𝑔𝑒𝑟
⟹ 𝐹𝑛 = ∅.
Hence if n is nonprime integer FLT is true. This proves Theorem 1.3.
In this section we prove the first case of FLT and the second case 𝑧 ≡ 0 (𝑚𝑜𝑑 𝑝). We distinguish
two new cases: The case 𝑐 ≡ 1 (𝑚𝑜𝑑 2) or 𝑐 ≡ 0 (𝑚𝑜𝑑 2)
Lemma 3.8. Let 𝑝 > 2 be a prime number and let (𝑎, 𝑏, 𝑐) be a triple of relativity prime
integers. Then
(𝑎, 𝑏, 𝑐) ∈ 𝐹𝑝 ⟹
{
𝑏 − 𝑎 = 𝑒𝑝
− 𝑑𝑝
𝑖𝑓 𝑎𝑏 ≢ 0 (𝑚𝑜𝑑 𝑝)
−
𝑑𝑝
𝑝
𝑖𝑓 𝑎 ≡ 0 (𝑚𝑜𝑑 𝑝)
𝑏 − 𝑎 =
𝑒𝑝
𝑝
− 𝑑𝑝
𝑖𝑓 𝑏 ≡ 0 (𝑚𝑜𝑑 𝑝)
where (𝑑, 𝑒) is the couple of Kimou’s primaries divisors of (𝑎, 𝑏).
Proof. According to [7], [9], we have
If(𝑎, 𝑏, 𝑐) ∈ 𝐹𝑝 then
𝑏 − 𝑎 = (−
𝑑𝑝
gcd(𝑑, 𝑝)
+
𝑒𝑝
gcd(𝑒, 𝑝)
+
𝑓𝑝
gcd(𝑓, 𝑝)
) − (
𝑑𝑝
gcd(𝑑, 𝑝)
−
𝑒𝑝
gcd(𝑒, 𝑝)
+
𝑓𝑝
gcd(𝑓, 𝑝)
)
=
𝑒𝑝
gcd(𝑒, 𝑝)
−
𝑑𝑝
gcd(𝑑, 𝑝)
.
Hence,

14
(𝑎, 𝑏, 𝑐) ∈ 𝐹𝑝 ⟹
{
− 𝑑𝑝
𝑖𝑓 𝑎𝑏 ≢ 0 (𝑚𝑜𝑑 𝑝)
−
𝑑𝑝
𝑝
𝑖𝑓 𝑎 ≡ 0 (𝑚𝑜𝑑 𝑝)
𝑏 − 𝑎 =
𝑒𝑝
𝑝
− 𝑑𝑝
𝑖𝑓 𝑏 ≡ 0 (𝑚𝑜𝑑 𝑝)
[𝑅𝑒𝑚𝑎𝑟𝑘 2.2. ]
Proof of Theorem 1.4. Under the assumptions of the Theorem 1.4. we proceed to a proof by the
absurd by assuming that (𝑎, 𝑏, 𝑐) ∈ 𝐹𝑝.
We have:
𝑎𝑏 ≢ 0 (𝑚𝑜𝑑 𝑝) ⟹ {
𝑏 − 𝑎 = 1 𝑖𝑓 𝑐 ≡ 1 (𝑚𝑜𝑑 2)
𝑏 − 𝑎 = 2 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
[𝑇ℎ𝑒𝑜𝑟𝑒𝑚 2.2. ]
⟹ {
𝑒𝑝
− 𝑑𝑝
= 1 𝑖𝑓 𝑐 ≡ 1 (𝑚𝑜𝑑 2)
𝑒𝑝
− 𝑑𝑝
= 2 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
= 1 [𝐿𝑒𝑚𝑚𝑒 3.8]
⟹ {
1 > 𝑝(𝑒 − 𝑑) 𝑑𝑝−1
𝑖𝑓 𝑐 ≡ 1 (𝑚𝑜𝑑 2)
2 > 𝑝(𝑒 − 𝑑) 𝑑𝑝−1
𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
⟹ 2 > 𝑝(𝑒 − 𝑑) 𝑑𝑝−1
⟹ (𝑒 − 𝑑) 𝑑𝑝−1
<
2
𝑝
< 1
⟹ 𝑒 = 𝑑 ⟹◻.
Hence the result.
Remark 3.6. We have just proved Fermat's last theorem with the even exponent, in its first case
and in the second case where 𝑐 ≡ 0 (𝑚𝑜𝑑 𝑝) . However, when 𝑎𝑏 ≡ 0 (𝑚𝑜𝑑 𝑝) we have:
𝑎 ≡ 0 (𝑚𝑜𝑑 𝑝) ⟹
{
𝑒𝑝
−
𝑑𝑝
𝑝
= 1 𝑖𝑓 𝑐 ≡ 1 (𝑚𝑜𝑑 2)
𝑒𝑝
−
𝑑𝑝
𝑝
= 2 otherwise
,
and
𝑏 ≡ 0 (𝑚𝑜𝑑 𝑝) ⟹
{
𝑒𝑝
𝑝
− 𝑑𝑝
= 1 𝑖𝑓 𝑐 ≡ 1 (𝑚𝑜𝑑 2)
𝑒𝑝
𝑝
− 𝑑𝑝
= 2 otherwise.
These new Diophantine equations promise to be difficult despite their simple appearances.
4. ANALYSIS OF THE DIFFICULTY OF ESTABLISHING RESULTS
At this stage we propose a classification by increasing difficulty of solving the problems dealt
with in this paper. First place is occupied by the second FLT case with the odd exponent: it is
obvious that if(𝑎, 𝑏, 𝑐) ∈ 𝐹2𝑝, with 𝑝a prime number, then 𝑏2
− 𝑎2
= 1is impossible by simple
making factorisation. The second position is occupied simultaneously by the first and second
FLT cases 𝑐 ≡ 0 (𝑚𝑜𝑑 𝑝): if (𝑎, 𝑏, 𝑐) ∈ 𝐹𝑝Then 𝑒𝑝
− 𝑑𝑝
= 1. Then you'll have to factor and
major. You'll conclude that this relationship is impossible. This case is more difficult than the
previous one. The third place is occupied by FLT with nonprime exponent. To prove this, we had
to distinguish two sub-problems: Prove that FLT is true for the odd non-prime exponent and then
for the even exponent.In the last position is the second case of Abel's conjecture. Indeed, it turned
out to be a little more difficult than previous problem because it was necessary to use the

15
relations 𝑏 − 𝑎 = 1, 𝑐 − 𝑏 = 1, the Kimou's principal divisors and 𝑝 adic valuations to establish
a contradiction. As surprising as it may seem, it explains the difficulty of prove this problem.
5. CONCLUSION
In this paper we establish Diophantine proofs of Abel's conjecture, Fermat Last Theorem for the
exponents even, non-prime exponent, the first case and the second case 𝑧 ≡ 0 (𝑚𝑜𝑑 𝑝). We
analyse these proofs and establish a ranking in order of increasing difficulty in solving the Fermat
problems treated.In perspective, we intend to:
- establish a Diophantine proof of the second remaining cases, i.e. to prove that if
𝑥𝑦 ≡ 0 (𝑚𝑜𝑑 𝑝) then 𝑥𝑝
+ 𝑦𝑝
≠ 𝑧𝑝
.
- extend methods to broader classes of equations: Catalan’s equation, Beal problem and
others General Fermat problem.
- introduce new concepts such as the universe and Diophantine galaxies, as well as the
similarity principle, and then find applications for them in astronomy, astrophysics,
cosmology and artificial intelligence.
REFERENCES
[1] P. Rimbenboim (1979), 13 Lectures on Fermat Last Theorem, ISBN-0-387-90432-8, Springer-
Verlag New York Inc, 1999.
http://www.numdam.org/item?id=SB_1984-1985__27__309_0
[2] P. Rimbenboim (1999), Fermat’s last Theorem for amateurs, ISBN-0-387-98508-7, Springer-
Verlag New York Inc, 1999.
http://www.numdam.org/item?id=SB_1984-1985__27__309_0
[3] A. J. Wiles*, Modular elliptic curves and Fermat’s Last Theorem, 1995, Annals of Mathematics,
141, pp. 443-551.
[4] Kimou, P. K.,Tanoé, F.E. and Kouakou, K. V. (2023). Fermat and Pythagoras Divisors for a New
Explicit Proof of Fermat's Theorem: a ^4 + b^ 4 = c ^4 . Part I, Advances in Pure Mathematics,
14,303-319. https://doi.org/10.4236/apm.2024.144017.
[5] Kimou, P. K.,Tanoé, F.E. and Kouakou, K. V. (2023). A new proof of Fermat Last Theorem for
exponent 4
using Fermat Divisors (2023)
https://www.researchgate.net/publication/371159864
[6] Kimou, P. K. (2023) A efficient proof of the first case of Abel’s Conjecture using new tools (2023)
https://www.researchgate.net/publication/37262223
[7] 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
[8] Kimou, P.K. and Tanoé, F.E. (2023). Diophantine Quotients and Remainders with Applications to
Fermat and Pythagorean Equations. American Journal of Computational Mathematics, 13, 199-210.
https://doi.org/10.4236/ajcm.2023.131010.
[9] Kimou, P. K. (2024) New Kimou Unified theorem for principal divisors of x^p+y^p =z^p, p a prime
Research Gate
[10] Kimou P. K. (2024), On Direct Proof of FLT: A fundamental Surprising Theorem
Research Gate.
[11] Kimou P., K. (2024), On Direct Proof of FLT: A crucial Relation,
Recherche Gate.
[12] Nicolas B. (2018) Théorie des nombres, Université de Saint Boniface.
[13] Zhong Chuixiang (1989), Fermat's Last Theorem: A Note about Abel's Conjecture, C.R. Hath. Rep.
Acad. Sci. Canada - Vol. XI, No. 1, February 1989 février
[14] Moller, K., (1955) UntereSchianke fur die Anzahl der Piimazahlen, ausdenenx,y,z der Fer
matshenOdchung𝑥𝑛
+ 𝑦𝑛
= 𝑧𝑛
° besteden muss. Math. Nachr., 14, 1955,25-28.

16
[15] Kimou Kouadio Prosper, Kouakou Kouassi Vincent (2024), On Direct Proof of Fermat Last
Theorem: The Abel conjecture, the even and nonprime exponent, and the first case, 2nd
International
Conference on Mathematics, Computer Sciences & Engineering (MATHCS2024), 28-29/12/2024,
DUBAI, ERU.
AUTHORS
Kimou Kouadio Prosper,,
I am an Ivorian. I am a teacherresearcher at the Institute
Polytechnique Felix Houphouet-Boigny of Yamoussoukro, RCI (INPHB) . Since
May 2011. I carry out my teaching and research activities there. My research
work is mainly focused on artificial intelligence, number theory and computer
security, especially cryptography.

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