In 1637, Fermat wrote in Latin the following statement translated into modern language: there does not exist a non-zero natural number $x,y,z$ and $n>2$ such that,

$x^n+y^n=z^n$ (1)

He claimed to hold a marvellous proof of it that the margin could not contain. In 1670, Samuel de Fermat reissued Bachet’s arithmetica augmented with his father’s annotations. Then, Fermat’s words were made public and popular. Subsequently, they proved to be a powerful stimulus in number theory research and contributed to the development of modern mathematics. We note the following significant facts attached to the problem posed by Fermat.

Since that time, the hope of finding a direct or elementary proof at FLT disappeared to give way to indirect and sophisticated methods. After a period of euphoria maintained by the proof of Wiles, the idea of a direct and simple proof is reborn but in isolation [7] [8] .

It’s well known that the proof of FLT would be shortened if the abc conjecture were proven. By application of the abc conjecture, there exists a number such that if $n>n_0$ then FLT is true ( [1] , pp. 364-365; [9] , p. 1436). But this evidence is indirect, and today inadmissible because the abc conjecture has still not been fully demonstrated.

In 2023 Kimou Prosper and Tanoe François introduced Fermat divisors, quotients and Diophantine remainders and demonstrated results attached to them. These efficient tools have given new possibilities to indeed compute and analyse the Fermat equation and other Diophantine type equations [10] [11] [12] . Our tools have enabled us to find new properties of Fermat’s equation or its hypothetical solution that were previously unsuspected. In this paper, we use an original asymptotic approach and Fermat divisors, Diophantine quotients and Diophantine remainders to prove the following main results.

Theorem 1. Let $n\ge 1$ be a positive integer and let be a prime. Consider that there exist a sequence of triple ${\left(a_n,b_n,c_n\right)}_{n\ge 1}$ such that $\left(a_n,b_n,c_n\right)$ is a non-trivial primitive triple solution of Equation (1) with exponent for any $n\ge 1$ and $a_n<b_n<c_n$. Then

$\{\begin{array}{c}\underset{n\to \infty }{\lim }\frac{b_n}{a_n}=\underset{n\to \infty }{\lim }\frac{c_n}{a_n}=q_{\infty }+lif{q}_{2,n}isbound\\ \underset{n\to \infty }{\lim }\frac{b_n}{a_n}=\underset{n\to \infty }{\lim }\frac{b_n}{a_n}=+\infty if{q}_{2,n}i{s}^{\prime }ntbound\\ \underset{n\to \infty }{\lim }\frac{c_n}{b_n}=1\end{array}$

Where $b_n=a_n{q}_{1,n}+r_{1,n},c_n=a_n{q}_{1,n}+r_{2,n},\underset{n\to +\infty }{\lim }{q}_{2,n}=q_{\infty }$ and

$l=,\underset{n\to +\infty }{\lim }\frac{r_{1,n}}{a_n}$

Add this text between them.

The 4-uple $\left(q_{1,n},q_{2,n},r_{1,n},r_{2,n}\right)$ is define as the Diophantine Quotients and Remainders of $\left(a_n,b_n,c_n\right)$ [11] .

Theorem 2. Let $n\ge 1$ be a positive integer and let $p_n$ be a prime. Consider that there exists a sequence of triple ${\left(a_n,b_n,c_n\right)}_{n\ge 1}$ such that $\left(a_n,b_n,c_n\right)$ is a non-trivial primitive triple solution of Equation (1) with exponent $p_n$ for any $n\ge 1$ and $q_{2,n}$ the quotient of the Euclidean division of $c_n$ by $a_n$. Then

$q_{2,n}>1\Rightarrow \{\begin{array}{c}\underset{n\to +\infty }{\lim }\frac{e_n}{f_n}\\ \underset{n\to +\infty }{\lim }\frac{{\beta }_n}{{\gamma }_n}\end{array}$

where $e_n=\gcd \left(b_n,c_n-a_n\right),f_n=\gcd \left(c_n,b_n+a_n\right),b_n=e_n{\beta }_n$ and $c_n=f_n{\gamma }_n$

where $e_p=\gcd \left(b_p,c_p-a_p\right),f_p=\gcd \left(c_p,a_p+b_p\right),b_p=e_p{\beta }_p,c_p=f_p{\gamma }_p$

Theorem 3. Let $n\ge 1$ be a positive integer and let be a prime. Consider that there exist a sequence of triple ${\left(a_n,b_n,c_n\right)}_{n\ge 1}$ such that $\left(a_n,b_n,c_n\right)$ is a non-trivial primitive triple solution of Equation (1) with exponent $p_n$ for any $n\ge 1$, $a_n<b_n<c_n$ and $c_n=q_{2,n}{a}_n+r_{2,n},b_n=q_{1,n}{a}_n+r_{1,n}$ with $r_{1,n},r_{2,n}<a$. Then,

$q_{2,n}=1,\frac{r_{1,n}}{a_n}<\frac{1}{p_n^2}\Rightarrow \frac{r_{2,n}}{a_n}\underset{n\to \infty }{~}\frac{\ln 2}{p_n}$.

Theorem 4. Let $p>2$ be a prime and let $\left(a,b,c\right)$ be a non-trivial primitive triple solution of Equation (1) with exponent p such that $a<b<c$. Consider $\left(e,f\right)$ the Main Kimou’s Divisors of $\left(b,c\right)$. Then

$b\equiv 0\left[p\right],c=b+1,f>e\Rightarrow \exists N>2,\forall p>N,a^p+b^p\ne c^p$

We give definitions, lemmas, and theorems with their proofs in the following secund section. In the third section, we give complementary results. In the section 4, we prove the main results. We conclude and give some perspectives at the end.

In this section, we give definitions, lemmas and theorems known in the literature and useful to establish our main results.

Theorem 2.1. (Bolzano-Weierstrass). Every bounded sequence has a convergent sub-sequence.

Definition 2.1. Let ${\left(u_n\right)}_{n\ge 1}$ and ${\left(v_n\right)}_{n\ge 1}$ two positive real sequences. We say that ${\left(u_n\right)}_{n\ge 1}$ and ${\left(v_n\right)}_{n\ge 1}$ are equivalent near infinity and we note $u_n\underset{n\to \infty }{~}{v}_n$ if only if:

$\underset{n\to \infty }{\lim }\frac{u_n}{v_n}=1$.

Lemma 2.1. Let ${\left(u_n\right)}_{n\ge 1}$ and ${\left(v_n\right)}_{n\ge 1}$ two positive real sequences such that $\underset{n\to \infty }{\lim }\left(u_n+v_n\right)=0$ Then

${\left(u_n\right)}_{n\ge 1}isbound\Rightarrow \exists s\left(n\right),\underset{n\to \infty }{\lim }{u}_{s\left(n\right)}=\underset{n\to \infty }{\lim }{v}_{s\left(n\right)}=0$

Proof.

If ${\left(u_n\right)}_{n\ge 1}$ is bound then,

$\begin{array}{l}\underset{n\to \infty }{\lim }{u}_{s(n)}>0\Rightarrow \exists s\left(n\right)andl>0,\underset{n\to \infty }{\lim }{u}_{s\left(n\right)}=l\left[Theorem\mathrm{2.1.}\right]\\ \Rightarrow \underset{n\to \infty }{\lim }\left(u_{s(n)}+v_{s(n)}\right)=0\\ \Rightarrow \underset{n\to \infty }{\lim }{u}_{s\left(n\right)}+\underset{n\to \infty }{\lim }{v}_{s(n)}=0\\ \Rightarrow \underset{n\to \infty }{\lim }{v}_{s\left(n\right)}=-l\\ \Rightarrow Absurdbecause{v}_{s\left(n\right)}>0\forall n>1.\end{array}$

Notation 2.1.

We denote by $F_p$ the set of non-trivial hypothetical primitive solutions of Equation (1) with degree $p$:

$F_p=\left\{\left(a,b,c\right)\in {\mathbb{N}}^{*3}:a^p+b^p=c^p,\gcd \left(a,b,c\right)=1\right\}$.

Definition 2.2. Let $p>2$ a prime and $\left(a,b,c\right)\in F_p$. We recall that the Kimou’s Divisors (or Fermat Divisors) is the sextuple $\left(d,e,f,\alpha ,\beta ,\gamma \right)$ defined that:

$\begin{array}{l}d=\gcd \left(a,c-b\right),e=\gcd \left(b,c-a\right),\\ f=\gcd \left(c,a+b\right)anda=d\alpha ,b=e\beta ,c=f\gamma ,\end{array}$

where $\left(d,e,f\right)$ is the Main Kimou’s Divisors (MKD) and $\left(\alpha ,\beta ,\gamma \right)$ the Auxiliary Kimou’s divisors (AKD).

Remark 2.1. The elements of the 6-uplet $\left(d,e,f,\alpha ,\beta ,\gamma \right)$ were originally called Fermat’s divisors. We now call them Kimou’s Divisors after the author who first introduced them in 2023 [10] . This new name will distinguish them from Fermat’s divisors, which are all the integers that divide the polynomial $x^n-x$.

Theorem 2.2. Let $p>2$ be a prime and let $\left(a,b,c\right)\in F_p$. Consider $\left(d,e,f\right)$ the Main Kimou’s Divisors of $\left(a,b,c\right)$. Then

$abc\cancel{\equiv }0\left[p\right]\Rightarrow \{\begin{array}{c}2a=d^p-e^p+f^p\\ 2b=-d^p+e^p+f^p\\ 2c=d^p+e^p+f^p\end{array},c\equiv 0\left[p\right]\Rightarrow \{\begin{array}{c}2a=d^p-e^p+\frac{f^p}{p}\\ 2b=-d^p+e^p+\frac{f^p}{p}\\ 2c=d^p+e^p+\frac{f^p}{p}\end{array}.$

Proof. See [10] , Theorem 1.

Theorem 2.3. Let $p>2$ be a prime and let $\left(a,b,c\right)\in F_p$ . Consider $\left(d,e,f\right)$ the Main Kimou’s Divisors of $\left(a,b,c\right)$. Then

$a\equiv 0\left[p\right]\Rightarrow \{\begin{array}{c}2a=\frac{d^p}{p}-e^p+f^p\\ 2b=-\frac{d^p}{p}+e^p+f^p\\ 2c=\frac{d^p}{p}+e^p+f^p\end{array},b\equiv 0\left[p\right]\Rightarrow \{\begin{array}{c}2a=d^p-\frac{e^p}{p}+f^p\\ 2b=-d^p+\frac{e^p}{p}+f^p\\ 2c=d^p+\frac{e^p}{p}+f^p\end{array}.$

Proof. See [10] , Theorem 1.

Lemma 2.2. Let $p>2$ be a prime and let $\left(a,b,c\right)\in F_p$. Consider $\left(d,e,f\right)$ the Main Kimou’s Divisors of $\left(a,b,c\right)$. We have:

$abc\cancel{\equiv }0\left[p\right]\Rightarrow \{\begin{array}{c}c-b=d^p\\ c-a=e^p\\ a+b=f^p\end{array},c\equiv 0\left[p\right]\Rightarrow \{\begin{array}{c}c-b=d^p\\ c-a=e^p\\ a+b=\frac{f^p}{p}\end{array}.$

Proof. See [10] , Theorem 1.

Lemma 2.3. Let $p>2$ be a prime and let $\left(a,b,c\right)\in F_p$. Consider $\left(d,e,f\right)$ the Main Kimou’s Divisors of $\left(a,b,c\right)$. We have:

$a\equiv 0\left[p\right]\Rightarrow \{\begin{array}{c}c-b=\frac{d^p}{p}\\ c-a=e^p\\ a+b=f^p\end{array},b\equiv 0\left[p\right]\Rightarrow \{\begin{array}{c}c-b=d^p\\ c-a=\frac{e^p}{p}\\ a+b=f^p\end{array}.$

Proof. See [10] , Theorem 1.

Remark 2.2. We also have,

$\begin{array}{c}\left(a,b,c\right)\in F_p\Rightarrow e=\gcd \left(b,c-a\right)\\ \Rightarrow \{\begin{array}{c}e=\gcd \left(\beta e,\frac{e^p}{p}\right)\text{if}b\equiv 0\left[p\right]\\ e=\gcd \left(\beta e,e^p\right)\text{otherwise}\end{array}\\ \Rightarrow \{\begin{array}{c}1=\gcd \left(\beta ,\frac{e^{p-1}}{p}\right)\text{if}b\equiv 0\left[p\right]\\ 1=\gcd \left(\beta ,e^{p-1}\right)\text{otherwise}\end{array}\\ \Rightarrow \gcd \left(\beta ,e\right)=1\end{array}$

Lemma 2.4. Let $p>2$ a prime. If $\left(a,b,c\right)\in F_p$ such that $a<b<c$ then, there exist $\left(q_1,r_1\right)$ and $\left(q_2,r_2\right)$ such that $b=a{q}_1+r_1$ and $c=a{q}_2+r_2$ with $r_1,r_2<a.$

Proof. Immediate consequence of the Euclidean Division Theorem.

Theorem 2.5. Let $p>2$ a prime and let $a,b$ and c a pairwise relativity prime such that $a<b<c$. Consider $\left(q_1,r_1\right)$ and $\left(q_2,r_2\right)$ the respective quotient and the remainder of Euclidean divisibility of $b$ and $c$ by $a$. Then

$\left(a,b,c\right)\in F_p\Rightarrow q_2=q_{1}or{q}_2=q_1+1.$

Proof. See [11] (p. 201).

Theorem 2.6. Let $p>2$ a prime and let $\left(a,b,c\right)\in F_p$ such that $a<b<c$. Consider $\left(q_1,r_1\right)$ and $\left(q_2,r_2\right)$ the respective quotient and the remainder of Euclidean divisibility of $b$ and $c$ by $a$. Then

$c-b=1\Rightarrow q_2=q_1.$

Proof. See [11] (p. 201).

In this section, we establish news results necessary to prove our main results. A little further, we use the asymptotic approach whose principle is as follow. Let $p>2$ be a prime parameter. We assume that for all $p>2$ it is possible to construct primitive solutions $\left(a_p,b_p,c_p\right)$ or $\left(a_{s\left(n\right)},b_{s\left(n\right)},c_{s\left(n\right)}\right)$ of Equation (1) where ${\left(s\left(n\right)\right)}_{n\ge 1}$ is a sub-sequence of prime numbers. We consider the set of prime numbers and define the sequence $s$ of primes number as follow:

$\left(a,b,c\right)\in F_p\Rightarrow \{\begin{array}{c}d<e<fifabc\cancel{\equiv }0\left[p\right]\\ \frac{d}{p}<e<fifa\equiv 0\left[p\right]\\ d<\frac{e}{p}<fifb\equiv 0\left[p\right]\\ d<e<\frac{f}{p}ifc\equiv 0\left[p\right]\end{array}.$

where $\left(d,e,f\right)$ is the Main Kimou’s Divisors of $\left(a,b,c\right)$.

Proof.

$\left(a,b,c\right)\in F_p\Rightarrow c-b<c-a<a+b$

According to lemmas 2.1., 2.2. and taking account each case, we have:

$\left(a,b,c\right)\in F_p\Rightarrow \{\begin{array}{c}{d}^p<e^p<f^{p}ifabc\cancel{\equiv }0\left[p\right]\\ \frac{d^p}{p}<e^p<f^{p}ifa\equiv 0\left[p\right]\\ d^p<\frac{e^p}{p}<f^{p}ifb\equiv 0\left[p\right]\\ d^p<e^p<\frac{f^p}{p}ifc\equiv 0\left[p\right]\end{array}$

$\Rightarrow \{\begin{array}{c}d<e<fifabc\cancel{\equiv }0\left[p\right]\\ \frac{d}{p}<e<fifa\equiv 0\left[p\right]\\ d<\frac{e}{p}<fifb\equiv 0\left[p\right]\\ d<e<\frac{f}{p}ifc\equiv 0\left[p\right]\end{array}$

Lemma 3.4. Let $p>2$ be a prime and let $\left(a,b,c\right)$ a triple of positive integers such that $a<b<c$. Then,

$\left(a,b,c\right)\in F_p\Rightarrow \{\begin{array}{c}f<\sqrt[p]{3p}eifc\equiv 0\left[p\right]\\ f<\sqrt[p]{\frac{3}{p}}eifb\equiv 0\left[p\right]\\ f<\sqrt[p]{3}eotherwise\end{array}$

where $\left(d,e,f\right)$ is the Main Kimou’s Divisors of $\left(a,b,c\right)$.

Proof.

In the one hand,

$\begin{array}{l}c\equiv 0\left[p\right],q_2>1\Rightarrow \frac{e^p}{a}>1,\frac{e^p}{a}=\frac{2e^p}{2a}\left[Lemma3.2\right]\\ \Rightarrow \frac{e^p}{a}>1,\frac{e^p}{a}=\frac{2e^p}{d^p-e^p+\frac{f^p}{p}}\left[Theorem2.2\right]\\ \Rightarrow \frac{e^p}{a}>1,\frac{e^p}{a}=\frac{2}{\frac{d^p}{e^p}-1+\frac{f^p}{p{e}^p}}\\ \Rightarrow 0<\frac{d^p}{e^p}-1+\frac{f^p}{p{e}^p}<2\\ \Rightarrow \frac{d^p}{e^p}+\frac{f^p}{p{e}^p}<3\\ \Rightarrow \frac{f^p}{p{e}^p}<3\\ \Rightarrow f<\sqrt[p]{3p}e.\end{array}$

In the other hand,

$b\equiv 0\left[p\right],q_2>1\Rightarrow \frac{e^p}{pa}>1and\frac{e^p}{pa}=\frac{2e^p}{2pa}\left[Lemma3.2,Theorem2.2\right]$

$\begin{array}{l}\Rightarrow \frac{e^p}{pa}>1,\frac{e^p}{pa}=\frac{2}{p\frac{d^p}{e^p}-1+p\frac{f^p}{e^p}}\left[Theorem2.2\right]\\ \Rightarrow 0<p\frac{d^p}{e^p}-1+p\frac{f^p}{e^p}<2\\ \Rightarrow p\frac{d^p}{e^p}+p\frac{f^p}{e^p}<3\\ \Rightarrow p\frac{f^p}{e^p}<3\\ \Rightarrow f<\sqrt[p]{\frac{3}{p}}e.\end{array}$

For the other cases ( $abc\cancel{\equiv }0\left[p\right]anda\equiv 0\left[p\right]$) we proceed in the same way and obtain

$abc\cancel{\equiv }0\left[p\right],a\equiv 0[p]\Rightarrow f<\sqrt[p]{3}e.$

Remark 3.1. We also have,

$b\equiv 0\left[p\right],q_2>1\Rightarrow f<\sqrt[p]{\frac{3}{p}}e\Rightarrow f<e.$

Lemma 3.5. Let $p>2$ a prime. We have,

$\left(a,b,c\right)\in F_p\Rightarrow p\left(c-b\right){\left(\frac{b}{a}\right)}^{p-1}<a<p\left(c-b\right){\left(\frac{c}{a}\right)}^{p-1}\text{.}$

Proof. Let $p>2$ a prime.

$\begin{array}{l}\left(a,b,c\right)\in F_p\Rightarrow a^p+b^p=c^p\\ \Rightarrow a^p=c^p-b^p\\ \Rightarrow a^p=\left(c-b\right)\underset{k=0}{\overset{p-1}{{\displaystyle \sum }}}{c}^k{b}^{p-1-k}\\ \Rightarrow (c-b)\underset{k=0}{\overset{p-1}{{\displaystyle \sum }}}{b}^k{b}^{p-1-k}<a^p<\left(c-b\right)\underset{k=0}{\overset{p-1}{{\displaystyle \sum }}}{c}^k{c}^{p-1-k}\\ \Rightarrow \left(c-b\right)\underset{k=0}{\overset{p-1}{{\displaystyle \sum }}}{b}^{p-1}<a^p<\left(c-b\right)\underset{k=0}{\overset{p-1}{{\displaystyle \sum }}}{c}^k{c}^{p-1-k}\\ \Rightarrow p\left(c-b\right)b^{p-1}<a^p<p\left(c-b\right)c^{p-1}\\ \Rightarrow p\left(c-b\right){\left(\frac{b}{a}\right)}^{p-1}<a<\left(c-b\right){\left(\frac{c}{a}\right)}^{p-1}\text{.}\end{array}$

Lemma 3.6. Let $p>2$ a prime, $\left(a,b,c\right)\in F_p$ and $\left(q_1,q_2\right)$ the quotient of $\left(b,c\right)$ by $a$. Then

$p{q}_1^{p-1}\left(c-b\right)<a<p{\left(1+q_2\right)}^{p-1}\left(c-b\right).$

Proof. Let $\left(a,b,c\right)\in F_p$ such that $a\cancel{\equiv }0\left[p\right]$, we have:

$\begin{array}{l}\left(a,b,c\right)\in F_p\Rightarrow p\left(c-b\right){\left(\frac{b}{a}\right)}^{p-1}<a<\left(c-b\right){\left(\frac{c}{a}\right)}^{p-1}\left[Lemma3.5\right]\\ \Rightarrow p\left(c-b\right){\left(q_1+\frac{r_1}{a}\right)}^{p-1}<a<(c-b){\left(q_2+\frac{r_2}{a}\right)}^{p-1}\left[Lemma\mathrm{2.3.}\right]\end{array}$

$\Rightarrow p\left(c-b\right)q_1^{p-1}<a<\left(c-b\right){\left(q_2+1\right)}^{p-1}\text{because}\frac{r_1}{a},\frac{r_2}{a}<1.$

Remark 3.2. According to Lemmas 2.1, 2.2 and 3.6. we deduce that,

$\left(a,b,c\right)\in F_p\Rightarrow \{\begin{array}{c}{d}^p{q}_1^{p-1}<a<d^p{\left(q_2+1\right)}^{p-1}ifa\equiv 0\left[p\right]\\ p{d}^p{q}_1^{p-1}<a<p{d}^p{\left(q_2+1\right)}^{p-1}otherwise\end{array}.$

Lemma 3.7. Let $p>2$ be a prime and let $\left(a,b,c\right)$ be a triple of integers such that $a<b<c$. Then

$\left(a,b,c\right)\in F_p\Rightarrow \frac{c-b}{a}<\frac{1}{p}.$

Proof. Let $\left(a,b,c\right)\in F_p$ and $q_1$ the quotient of Euclidean division of $b$ by $a$. We have

$\begin{array}{l}a\equiv 0\left[p\right]\Rightarrow a>q_1^{p-1}\left(c-b\right)\left[lemma3.6\right]\\ \Rightarrow a>p\left(c-b\right)\\ \Rightarrow a>p\left(c-b\right)\\ \Rightarrow \frac{c-b}{a}<\frac{1}{p}.\end{array}$

Lemma 3.8. Let $p>2$be a prime and let $\left(a,b,c\right)$ a triple of positive integers such that $a<b<c$. Then,

$\left(a,b,c\right)\in F_p\Rightarrow {\left(\frac{c-b}{a}\right)}_{\infty }=0.$

Proof.

$\begin{array}{l}\left(a,b,c\right)\in F_p\Rightarrow 0<\frac{c-b}{a}<\frac{1}{p}\left[Lemma3.4\right]\\ \Rightarrow 0\le {\left(\frac{c-b}{a}\right)}_{\infty }\le 0\\ \Rightarrow {\left(\frac{c-b}{a}\right)}_{\infty }=0.\end{array}$

Lemma 3.9. Let $p>2$ be a prime and let $\left(a,b,c\right)$ be a triple of positive integers such that $a<b<c$. Consider the Main Kimou’s Divisors $\left(d,e,f\right)$ of $\left(a,b,c\right)$. Then,

$\left(a,b,c\right)\in F_p\Rightarrow \{\begin{array}{c}{\left(q_2-q_1\right)}_{\infty }=0\\ {\left(\frac{r_2-r_1}{a}\right)}_{\infty }=0\end{array}.$

Proof.

$\begin{array}{l}\left(a,b,c\right)\in F_p\Rightarrow \frac{c-b}{a}=q_2-q_1+\frac{r_2-r_1}{a}\\ \Rightarrow 0<q_2-q_1+\frac{r_2-r_1}{a}<\frac{c-b}{a}\\ \Rightarrow {\left(q_2-q_1+\frac{r_2-r_1}{a}\right)}_{\infty }=0\end{array}$

$\begin{array}{l}\Rightarrow {\left(q_2-q_1\right)}_{\infty }=0and{\left(\frac{r_2-r_1}{a}\right)}_{\infty }\\ =0\text{because}\frac{r_2-r_1}{a}<1.\end{array}$

Theorem 3.10. Let $p>2$ be a prime and let $\left(a,b,c\right)$ be a triple of positive integers such that $a<b<c$. Consider the quotients $q_1$ and $q_2$ of the Euclidean division of b and c by a. Then,

$\left(a,b,c\right)\in F_p\Rightarrow \exists N,p>N,q_2=q_1.$

Proof.

$\begin{array}{l}\left(a,b,c\right)\in F_p,q_2=q_1+1\Rightarrow q_2-q_1=1\\ \Rightarrow {\left(q_2-q_1\right)}_{\infty }=1\\ \Rightarrow 0=1\left[Lemma3.9\right]\\ \Rightarrow \text{Absurd}\end{array}$

Hence there exist a rank N such that $q_2=q_1$ for any $p>N$.

Lemma 3.11. Let $p>2$ be a prime and let ${\left(a_p,b_p,c_p\right)}_{n\ge 1}$ be an infinite sequence such that $\left(a_p,b_p,c_p\right)\in F_p$. Consider $\left(q_{1,p},q_{2,p}\right)$ and (r_1,p, r_2,p) the Diophantine quotients and remainders of $\left(a_p,b_p,c_p\right)$. We construct a new sequence

rational numbers ${\left(\frac{r_1}{a}\right)}_p$, ${\left(\frac{r_2}{a}\right)}_p$. Then, there exist two subsequence’s ${\left(\frac{r_2}{a}\right)}_{p_n}$ and ${\left(\frac{r_1}{a}\right)}_{p_n}$ converging to the same limit $l$:

$\left(a_p,b_p,c_p\right)\in F_p\Rightarrow \{\begin{array}{c}\exists l>0,{\left(\frac{r_1}{a}\right)}_{\infty }={\left(\frac{r_2}{a}\right)}_{\infty }=l\\ {\left(q_1\right)}_{\infty }={\left(q_2\right)}_{\infty }\end{array}.$

Proof. Let ${\left(\frac{r_1}{a}\right)}_{S\left(n\right),n\ge 1}$ be a converging subsequence extracted from the sequence ${\left(\frac{r_1}{a}\right)}_{p\ge 3}$. Let’s simplify notations by adopting the following conventions:

$\underset{n\to \infty }{\lim }{\left(\frac{r_1}{a}\right)}_{S(n)}=\underset{n\to \infty }{\lim }{\left(\frac{r_1}{a}\right)}_{p_n}=\underset{n\to \infty }{\lim }\frac{r_{1,n}}{a_n}=l$

We’ll do the same with ${\left(\frac{r_2}{a}\right)}_{S\left(n\right),n\ge 1}$.

In the one hand

$\begin{array}{l}\left(a,b,c\right)\in F_p\Rightarrow \frac{c-b}{a}=\frac{r_2-r_1}{a}\left[Theorem3.10\right]\\ \Rightarrow {\left(\frac{c-b}{a}\right)}_{\infty }={\left(\frac{r_2-r_1}{a}\right)}_{\infty }\\ \Rightarrow {\left(\frac{r_2-r_1}{a}\right)}_{\infty }=0\left[\text{Lemma}\mathrm{3.9.}\right]\end{array}$

In the other hand,

$\begin{array}{l}\left(a,b,c\right)\in F_p\Rightarrow 0<r_1,r_2<a\\ \Rightarrow 0<\frac{r_1}{a},\frac{r_2}{a}<1\\ \Rightarrow {\left(\frac{r_1}{a}\right)}_{p}and{\left(\frac{r_2}{a}\right)}_p\text{are bound}.\end{array}$

So according to Theorem 2.1. There exist convergent subsequences ${\left(\frac{r_2}{a}\right)}_{S\left(n\right)}$, ${\left(\frac{r_1}{a}\right)}_{S\left(n\right)}$. Hence

$\begin{array}{l}\left(a,b,c\right)\in F_p\Rightarrow \underset{n\to \infty }{\lim }{\left(\frac{r_2}{a}\right)}_{p_n}-\underset{n\to \infty }{\lim }{\left(\frac{r_1}{a}\right)}_{p_n}=0\\ \Rightarrow \underset{n\to \infty }{\lim }{\left(\frac{r_2}{a}\right)}_{p_n}=\underset{n\to \infty }{\lim }{\left(\frac{r_1}{a}\right)}_{p_n}\\ \Rightarrow {\left(\frac{r_1}{a}\right)}_{\infty }={\left(\frac{r_2}{a}\right)}_{\infty }\end{array}$

Furthermore

$\begin{array}{l}\left(a,b,c\right)\in F_p\Rightarrow {\left(q_2-q_1\right)}_{\infty }=0\left[\text{Lemma}\mathrm{3.9.}\right]\\ \Rightarrow \{\begin{array}{c}{\left(q_1\right)}_{\infty }={\left(q_2\right)}_{\infty }if{q}_2<+\infty \\ {\left(q_1\right)}_{\infty }={\left(q_2\right)}_{\infty }=+\infty otherwise\end{array}\\ \Rightarrow {\left(q_1\right)}_{\infty }={\left(q_2\right)}_{\infty }.\end{array}$

Lemma 3.12. Let $p>2$ be a prime and let ${\left(a_p,b_p,c_p\right)}_{p\ge 1}$ be an infinite sequence such that $\left(a_p,b_p,c_p\right)\in F_p$. Consider $\left(q_{1,p},q_{2,p}\right)$ and $\left(r_{1,p},r_{2,p}\right)$ the Diophantine quotients and Remainders of $\left(a_p,b_p,c_p\right)$ and the

convergent subsequence ${\left(\frac{r_2}{a}\right)}_{p_n}$ and ${\left(\frac{r_1}{a}\right)}_{p_n}$. Then

$\underset{n\to \infty }{\lim }{\left(\frac{r_2}{a}\right)}_{p_n}=\underset{n\to \infty }{\lim }{\left(\frac{r_1}{a}\right)}_{p_n}=l\Rightarrow l\in \left[0,1\right].$

Proof.

$\begin{array}{l}\left(a_p,b_p,c_p\right)\Rightarrow 0<r_{1,p},r_{2,p}<a\\ \Rightarrow 0<\frac{r_{1,p}}{a_p},\frac{r_{2,p}}{a_p}<1\end{array}$

$\begin{array}{l}\Rightarrow 0\le \underset{n\to \infty }{\lim }{\left(\frac{r_2}{a}\right)}_{p_n}\le 1\\ \Rightarrow 0\le l\le 1.\end{array}$

Remark 3.3. Figure 1 shows the Flowchart of the proof of Lemma 3.11. it’s shows that the basic assumption $b=a{q}_1+r_1$, $c=a{q}_2+r_2$ has three possible

consequences: $\frac{1}{p{\left(q_2+1\right)}^{p-1}}<\frac{c-b}{a}<\frac{1}{p}{q}_1^{p-1}$ , $\frac{c-b}{a}<\frac{1}{p}$ or $\frac{r_1}{a},\frac{r_2}{a}<1$.

Proof of Theorem 1

For the proof of this theorem, we will need the following Theorem.

Theorem 4.1. Let $p>2$ a prime and let $\left(a_p,b_p,c_p\right)\in F_p$ such that $a_p<b_p<c_p$. Consider the Diophantine Quotient and Remainders $\left(q_{1,p},q_{2,p},r_{1,p},r_{2,p}\right)$ of $\left(a_p,b_p,c_p\right)$. Then there exist $l\in \left[0,1\right]$ such that

$\{\begin{array}{c}\underset{p\to \infty }{\lim }\frac{b_p}{a_p}=\underset{p\to \infty }{\lim }\frac{c_p}{a_p}=q_{\infty }+lif{q}_{\infty }<\infty \\ \underset{p\to \infty }{\lim }\frac{b_p}{a_p}=\underset{p\to \infty }{\lim }\frac{c_p}{a_p}=+\infty otherwise\end{array}.$

with $l=\underset{p\to \infty }{\lim }\frac{r_{2,p}}{a_p}=\underset{p\to \infty }{\lim }\frac{r_{1,p}}{a_p}$.

Proof.

$\begin{array}{l}\left(a_p,b_p,c_p\right)\in F_p\Rightarrow b_p=q_{1,p}{a}_p+r_{1,p},c_p=q_{2,p}{a}_p+r_{2,p}where0<r_{1,p},r_{2,p}<a\\ \Rightarrow \frac{b_{p_n}}{a_{p_n}}=q_{1,p}+\frac{r_{1,p}}{a_p},\frac{c_p}{a_p}=q_{2,p}+\frac{r_{2,p}}{a_p}where0<r_{1,p},r_{2,p}<a\\ \Rightarrow \exists N,p>N\frac{b_p}{a_p}=q_{1,p}+\frac{r_{1,p}}{a_p},\frac{c_p}{a_p}=q_{1,p_n}+\frac{r_{2,p}}{a_p}\\ \Rightarrow \{\begin{array}{c}\underset{p\to \infty }{\lim }{\left(\frac{b}{a}\right)}_p=\underset{p\to \infty }{\lim }{\left(\frac{c}{a}\right)}_p=q_{\infty }+lif{q}_{\infty }<\infty \\ \underset{p\to \infty }{\lim }{\left(\frac{b}{a}\right)}_p=\underset{p\to \infty }{\lim }{\left(\frac{c}{a}\right)}_p=l+(+\infty )=+\infty otherwise\end{array}\\ \Rightarrow \{\begin{array}{c}\underset{p\to \infty }{\lim }{\left(\frac{b}{a}\right)}_p=\underset{p\to \infty }{\lim }{\left(\frac{c}{a}\right)}_p=q_{\infty }+lif{q}_{\infty }<\infty \\ \underset{p\to \infty }{\lim }{\left(\frac{b}{a}\right)}_p=\underset{p\to \infty }{\lim }{\left(\frac{c}{a}\right)}_p=+\infty otherwise\end{array}.\end{array}$

Proof of Theorem 1. Let $p>2$ a prime and let $\left(a_p,b_p,c_p\right)\in F_p$ a non-trivial primitive triple of positive integers, $\left(q_{1,p},q_{2,p},r_{1,p},r_{2,p}\right)$ the Diophantine Quotients and Remainders of $\left(a_p,b_p,c_p\right)$ and a positive fix real M.

On the one hand, according to theorem 4.1.

$\{\begin{array}{c}\underset{p\to \infty }{\lim }\frac{a_p}{b_p}=\underset{p\to \infty }{\lim }\frac{1}{\frac{b_p}{a_p}}=\frac{1}{q_{\infty }+l}if{q}_{\infty }<\infty \\ \underset{p\to \infty }{\lim }\frac{a_p}{b_p}=\underset{p\to \infty }{\lim }\frac{1}{\frac{b_p}{a_p}}=0otherwise\end{array},$

we proceed as the same for the sequence ${\left(\frac{a}{c}\right)}_p$.

In the one hand, we have

$\begin{array}{l}{q}_2\gg 1whenn\gg 1\Rightarrow \frac{c}{b}=\frac{a{q}_2+r_2}{a{q}_1+r_1}=\frac{q_2+\frac{r_2}{a}}{q_1+\frac{r_1}{a}}\left[Lemma2.4\right]\\ \Rightarrow \frac{c_p}{b_p}\underset{p\to \infty }{~}\frac{q_{1,p}+l}{q_{1,p}+l}\left[Lemma3.12\right]\\ \Rightarrow \{\begin{array}{c}\frac{c_p}{b_p}\underset{p\to \infty }{~}1if{q}_{1,p}isbound\\ \frac{c_p}{b_p}\underset{p\to \infty }{~}\frac{q_{1,p}}{q_{1,p}}\underset{p\to \infty }{~}1otherwise\end{array}\\ \Rightarrow {\left(\frac{c}{b}\right)}_{\infty }=1\end{array}$

Remark 4.1. When q1 is bound, we can give another proof of theorem 1.

$\begin{array}{l}{q}_2<M\Rightarrow \frac{c}{b}=\left(\frac{c}{a}\right)\left(\frac{a}{b}\right)\\ \Rightarrow {\left(\frac{c}{b}\right)}_{\infty }={\left(\frac{c}{a}\right)}_{\infty }{\left(\frac{a}{b}\right)}_{\infty }\\ \Rightarrow {\left(\frac{c}{b}\right)}_{\infty }=\left(q_{\infty }+l\right)\frac{1}{q_{\infty }+l}\\ \Rightarrow {\left(\frac{c}{b}\right)}_{\infty }=1.\end{array}$