Pythagorician Divisors and Applications to Some Diophantine Equations

François Emmanuel Tanoé¹, Prosper Kouadio Kimou²

¹UFR Mathématiques et Informatique. Université Félix Houphouët BOIGNY, Abidjan, Côte D’ivoire.
²Laboratoire d’Informatique et de Mathématiques Appliquées. Institut Polytechnique Félix Houphouët BOIGNY, Yamoussoukro, Côte D’ivoire.

DOI: 10.4236/apm.2023.132003                

Abstract

We consider the Pythagoras equation X² +Y² = Z², and for any solution of the type (a,b = 2^sb₁ ≠0,c) ∈ N^*3, s ≥ 2, b₁odd, (a,b,c) ≡ (±1,0,1)(mod 4), c > a , c > b, and gcd(a,b,c) = 1, we then prove the Pythagorician divisors Theorem, which results in the following: , where (d,d′′) (resp. (e,eⁿ)) are unique particular divisors of a and b, such that a = dd′′ (resp. b = ee′′ ), these divisors are called: Pythagorician divisors from a, (resp. from b). Let’s put λ ∈{0,1}, defined by: and S = s -λ (s -1). Then such that . Moreover the map is a bijection. We apply this new tool to obtain a new classification of the primitive, positive and non-trivial solutions of the Pythagoras equations: a² + b² = c² via the Pythagorician parameters (d,e,S ). We obtain for (d, e) fixed, the equivalence class of any Pythagorician solution (a,b,c), checking , namely: . We also update the solutions of some Diophantine equations of degree 2, already known, but very important for the resolution of other equations. With this tool of Pythagorean divisors, we have obtained (in another paper) new recurrent methods to solve Fermat’s equation: a⁴ + b⁴ = c⁴, other than usual infinite descent method; and to solve congruent numbers problem. We believe that this tool can bring new arguments, for Diophantine resolution, of the general equations of Fermat: a^2p + b^2p = c^2p and a^p + b^p = c^p. MSC2020-Mathematical Sciences Classification System: 11A05-11A51-11D25-11D41-11D72.

Keywords

Pythagoras Equation, Pythagorician Triplets, Diophantine Equations of Degree 2, Factorisation-Gcd-Fermat’s Equations

1. Introduction

A question that many mathematicians have asked themselves since the 17th century is the following: “Is there a Diophantine method, allowing to solve Fermat’s equation: $a^p+b^p=c^p$ , or $a^{2p}+b^{2p}=c^{2p}$ , where p is prime”. Very interesting results were obtained by Fermat himself, in the case: $a^4+b^4=c^4$ , with his famous method of the “infinite descent”, by Sophie Germain via his famous prime numbers, Ernst Kummer and by Guy Terjanian via Diophantine methods cf. [1] p. 110-123, p. 360, p. 209 and [2] , and this, in particular with the assumption that p/abc or not.

But in the end, the proof of Wiles for the general case (1995) (cf. [3] ) which is the culmination of the new methods developed in the 20th century, in particular those counting the number of integer points of the elliptic curves, has in a way closed the problem of solving , by non-diophantine means.

The goal of this paper is to prove Theorem 1.1. It’s why, we therefore return naturally to the basis of this problem, by focusing on the well-known problem of solving the Pythagoras equation: $a^2+b^2=c^2$ . This gives us, a new parametrization for the expression of Pythagorean triples (which is a new tool that we call: “Pythagorician divisors”).

This new parametrization is very important because it allows to solve many Diophantine equations: Thus, the results found can be directly applied, to the case: $a^4+b^4=c^4$ (cf. [1] [4] ) producing a new demonstration, and in some extent, to the equation $a^{2p}+b^{2p}=c^{2p}$ , where p is any odd prime number.

From a historical point of view, note that the Pythagorean triples have been known since ancient Egypt, long before Pythagoras, as evidenced by the problems of the Berlin papyrus 6619, found in Thebes in 1858 and dated to 1680-1620BC, cf. Clagett M., in [5] . There are also Pythagorean triples in the Kahun papyrus (1800BC): $6^2+8^2={10}^2$ ; ${12}^2+{16}^2={20}^2$ , and more surprisingly, rational Pythagorean triples, which are: ${\left(1+\frac{1}{2}\right)}^2+2^2={\left(2+\frac{1}{2}\right)}^2$ ; ${\left(\frac{3}{4}\right)}^2+1^2={\left(1+\frac{1}{4}\right)}^2$ .

We will follow the following plan for the article: We first make some reminders, followed by the definitions of the Pythagorean divisors, as well as the associated theorems. We express the Pythagorean triples with these new parameters in different tables, and we explain an equivalence relation allowing a new classification of these triples. Finally we apply these methods to the resolution of certain important equations of degree 2, before concluding.

1.1. Notations-Reminders

Let’s make some reminders:

Reminders 1.1 $\forall a\mathrm{,}c\in {\mathbb{N}}^{\ast }$ , $\forall n\in \mathbb{N}$ , $n\ge 2$ , we have:

1) $c^n-a^n=\left(c-a\right)\times T_n\left(c,a\right)$ , where $T_n\left(c,a\right)={\displaystyle {\sum }_{k=0}^{n-1}}{c}^{n-1-k}{a}^k$ .

2) $gcd\left(c-a\mathrm{,}{T}_n\left(c\mathrm{,}a\right)\right)=gcd\left(n\mathrm{,}c-a\right)$

· Therefore if $n=p$ is prime, then $gcd\left(c-a\mathrm{,}{T}_p\left(c\mathrm{,}a\right)\right)\in \left\{\mathrm{1,}p\right\}$ .

· In particular $gcd\left(c-a\mathrm{,}{T}_2\left(c\mathrm{,}a\right)\right)\in \left\{\mathrm{1,2}\right\}$ .

3) In addition, if $gcd\left(a\mathrm{,}c\right)=1$ , then: $gcd\left(c\mathrm{,}{T}_n\left(c\mathrm{,}a\right)\right)=1=gcd\left(a\mathrm{,}{T}_n\left(c\mathrm{,}a\right)\right)$ .

Convention 1.1 Let $\left(a\mathrm{,}b\mathrm{,}c\right)\ne \left(\mathrm{0,0,0}\right)$ be a solution of the Pythagoras equation: $a^2+b^2=c^2$ . We agree for the following, unless otherwise stated, that $\left(a\mathrm{,}b\mathrm{,}c\right)\equiv \left(\pm \mathrm{1,0,1}\right)\left(\mathrm{mod}4\right)$ .

This in no way restricts the expression of the generality of the solutions of said equation, because $\left(b\mathrm{,}a\mathrm{,}c\right)$ is also a solution called “associated with $\left(a\mathrm{,}b\mathrm{,}c\right)$ ”, such that $\left(b\mathrm{,}a\mathrm{,}c\right)\equiv \left(\mathrm{0,}\pm \mathrm{1,1}\right)\left(\mathrm{mod}4\right)$ .

Reminders 1.2 Let $\left(a\mathrm{,}b\mathrm{,}c\right)\ne \left(\mathrm{0,0,0}\right)$ be a solution of $a^2+b^2=c^2$ . So:

· $\left(a\mathrm{,}b\mathrm{,}c\right)$ is said to be a positive solution if $a>0$ , $b\ge 0$ , $c>0$ .

· $\left(a,b,c\right)=\left(1,0,1\right)$ is said to be the positive trivial solution.

· $\left(a>0,b=2^s{b}_1\ne 0,c>0\right)$ , is said to be a positive non-trivial solution (then $s\ge 2$ , $b_1$ odd).

· $\left(a\mathrm{,}b\mathrm{,}c\right)$ is called a primitive solution if and only if $gcd\left(a\mathrm{,}b\mathrm{,}c\right)=1$ .

· Let $\left(a\mathrm{,}b\mathrm{,}c\right)$ be a positive primitive solution, then $\left(\pm ta\mathrm{,}\pm tb\mathrm{,}\pm tc\right)$ $t\in \mathbb{N}$ , form the set of solutions known as generated by $\left(a\mathrm{,}b\mathrm{,}c\right)$ .

Proposition 1.1 The set of all the solutions of the Pythagoras equation $a^2+b^2=c^2$ , is formed from the solutions generated by all the positive primitive solutions, and their associates (notice that the null solution belong to).

It is well known cf. [1] , [6] or [7] (see also Table 3, at the end of paragraph 1.3), that:

Proposition 1.2 The set of positive primitive solutions of $a^2+b^2=c^2$ , is given by the following set:

$\left\{\left(u^2-v^2\mathrm{,}2uv\mathrm{,}{u}^2+v^2\right)\mathrm{<mo>\; ;\; </mo>}u\mathrm{,}v\in \mathbb{N}\mathrm{,}u>v\mathrm{,}u+v\equiv 1\left(\mathrm{mod}2\right)\text{and}gcd\left(u\mathrm{,}v\right)=1\right\}$ .

Note that in this article, we obtain in Corollary 1.2, our new parametrization, (in [8] there is another parametrization).

1.2. Pythagorician Divisors

We will describe and set the parameters we will need.

Definition 1.1 Let $\left(a,b=2^s{b}_1\ne 0,c\right)\in {\mathbb{N}}^{\ast 3}$ , $s\ge 2$ , $b_1$ odd, $\left(a\mathrm{,}b\mathrm{,}c\right)\equiv \left(\pm \mathrm{1,0,1}\right)\left(\mathrm{mod}4\right)$ , $c>a>0$ , $c>b>0$ , and $a\mathrm{,}b\mathrm{,}c$ pairwise relativly prime.

1) We call divisors of such a triplet, coming from a (resp. from b) the unique couple $\left(d\mathrm{,}{d}^{″}\right)$ (resp. $\left(e\mathrm{,}{e}^{″}\right)$ ) defined by:

$\{\begin{array}{l}\left(d\mathrm{,}{d}^{″}\right)=\left(gcd\left(a\mathrm{,}c-b\right)\mathrm{,}\frac{a}{d}\right)\mathrm{,}\text{note}\text{that}a=d{d}^{″}\mathrm{<mo>\; ;\; </mo>}\hfill \\ \left(e\mathrm{,}{e}^{″}\right)=\left(gcd\left(b\mathrm{,}c-a\right)\mathrm{,}\frac{b}{e}\right)\mathrm{,}\text{note}\text{that}b=e{e}^{″}\mathrm{.}\hfill \end{array}$

2) When in addition, we suppose that $\left(a\mathrm{,}b\mathrm{,}c\right)$ is a non-trivial, primitive and positive Pythagorician triplet, that is to say that: $a^2+b^2=c^2$ , then those just defined divisors are call Pythagorician divisors of such a triplet, coming from a (resp. From b) and are the unique couples $\left(d\mathrm{,}{d}^{″}\right)$ (resp. $\left(e\mathrm{,}{e}^{″}\right)$ ):

Definition 1.2 Let, as above, be a triplet: $\left(a,b=2^s{b}_1,c\right)\in {\mathbb{N}}^{\ast 3}$ , $s\ge 2$ , $b_1$ odd. So:

1) $\lambda \in \left\{\mathrm{0,1}\right\}$ is defined by: $\frac{c-a}{2}\equiv \lambda \left(\mathrm{mod}2\right)$ .

2) $S\in {\mathbb{N}}^{\ast }$ , is defined by: $S=s-\lambda \left(s-1\right)=\{\begin{array}{l}s\mathrm{,}\text{if}\lambda =\mathrm{0\; <mo>\; ;\; </mo>}\hfill \\ \mathrm{1,}\text{otherwise}\text{.}\hfill \end{array}$

And note that: $s-S=\lambda \left(s-1\right)=\{\begin{array}{l}\mathrm{0,}\text{if}\lambda =\mathrm{0\; <mo>\; ;\; </mo>}\hfill \\ s-1\text{,otherwise}\text{.}\hfill \end{array}$

Proposition 1.3 Let as above, a triplet: $\left(a,b=2^s{b}_1,c\right)$ . In addition, we suppose that $\left(a\mathrm{,}b\mathrm{,}c\right)$ is a non-trivial, primitive and positive Pythagorician triplet, that is to say that: $a^2+b^2=c^2$ . Then:

1) $d\mathrm{,}{d}^{″}$ are odd and $gcd\left(d\mathrm{,}{d}^{″}\right)=1$ .

2) e is always even and $gcd\left(e\mathrm{,}{e}^{″}\right)=2^{\lambda }=\{\begin{array}{l}\mathrm{1,}\text{if}\lambda =\mathrm{0\; <mo>\; ;\; </mo>}\hfill \\ \mathrm{2,}\text{otherwise}\text{.}\hfill \end{array}$

Specifically, $\exists \mathrm{<mo>\; !\; </mo>}$ $\bar{e}\equiv 1\left(\mathrm{mod}2\right)$ and $pgcd\left(\bar{e}\mathrm{,}\frac{b_1}{\bar{e}}\right)=1$ , such that:

$\{\begin{array}{l}e=2^S\bar{e}=\{\begin{array}{l}{2}^s\bar{e}\text{,}\text{if}\lambda =\mathrm{0\; <mo>\; ;\; </mo>}\hfill \\ 2\bar{e}\text{,}\text{otherwise;}\hfill \end{array}\hfill \\ e^{″}=2^{s-S}\frac{b_1}{\bar{e}}=\{\begin{array}{l}\frac{b_1}{\bar{e}}\mathrm{,}\text{if}\lambda =\mathrm{0\; <mo>\; ;\; </mo>}\hfill \\ 2^{s-1}\frac{b_1}{\bar{e}}\text{,}\text{otherwise;}\hfill \end{array}\hfill \end{array}$ Implies that:

$gcd\left(e\mathrm{,}{e}^{″}\right)=2^{\lambda }=\{\begin{array}{l}gcd\left(2^s\bar{e}\mathrm{,}\frac{b_1}{\bar{e}}\right)=\mathrm{1,}\text{if}\lambda =\mathrm{0\; <mo>\; ;\; </mo>}\hfill \\ gcd\left(2\bar{e}\mathrm{,}{2}^{s-1}\frac{b_1}{\bar{e}}\right)=2\text{otherwise}\text{.}\hfill \end{array}$

Note that: $b_1=\{\begin{array}{l}\frac{e}{2^s}\times e^{″}=\bar{e}{e}^{″}\mathrm{,}\text{if}\lambda =\mathrm{0\; <mo>\; ;\; </mo>}\hfill \\ \frac{e}{2}\times \frac{e^{″}}{2^{s-1}}=\bar{e}\times \frac{e^{″}}{2^{s-1}}\text{,}\text{otherwise}\text{.}\hfill \end{array}$

And that the factors are coprime, given the fact that $gcd\left(e\mathrm{,}{e}^{″}\right)=2^{\lambda }$ .

For the proof of this proposition, see the proof of the first case and second case, of the the following Theorem.

Theorem 1.1 (of Pythagorician divisors). Let $\left(a,b=2^s{b}_1,c\right)\in {\mathbb{N}}^{\ast 3}$ , $s\ge 2$ , $b_1$ odd and $\lambda ,S$ as defined in definition 1.1 & 1.2. and proposition 1.3. Denote by $\left(d\mathrm{,}{d}^{″}\right)$ and $\left(e,e^{″}\right)=\left(2^S\bar{e},2^{s-S}\frac{b_1}{\bar{e}}\right)$ the Pythagorician divisors of $\left(a\mathrm{,}b\mathrm{,}c\right)$ . So, there is an equivalence between the following propositions:

(i) $a^2+b^2=c^2$ ; (ii) $\{\begin{array}{l}c-b=d^2\hfill \\ c+b={d^{″}}^2\hfill \end{array}$ ; (iii) $\{\begin{array}{l}c-a=\frac{e^2}{2}=\frac{{\left(2^S\bar{e}\right)}^2}{2};\hfill \\ c+a=2{e^{″}}^2=2{\left(2^{s-S}\frac{b_1}{\bar{e}}\right)}^2.\hfill \end{array}$

Proof 1 Let us show that: (i) Û (ii) Û (iii).

· It is clear that each system (ii) and (iii) leads to the equality (i) $a^2+b^2=c^2$ .

· Conversely, let us show that:

$a^2+b^2=c^2$ realized $\Rightarrow$ each one of the systems (ii) and (iii) is solvable.

Note that: $a^2+b^2=c^2\Rightarrow a^2=\left(c-b\right)\left(c+b\right)$ and $b^2=\left(c-a\right)\left(c+a\right)$ .

Let us put: ${\Delta }_a=gcd\left(c-b\mathrm{,}c+b\right)$ and ${\Delta }_b=gcd\left(c-a\mathrm{,}c+a\right)$ .

Consequently (cf. Reminders 1.1.):

${\Delta }_a=gcd\left(c-b\mathrm{,}c+b\right)=gcd\left(c-b\mathrm{,}{T}_2\mathrm{<mo\; stretchy="false">\; (\; </mo>}c\mathrm{,}b\right)=gcd\left(\mathrm{2,}c-b\right)=1$

(because $c-b$ is odd),

While:

${\Delta }_b=gcd\left(c-a\mathrm{,}c+a\right)=gcd\left(c-a\mathrm{,}{T}_2\mathrm{<mo\; stretchy="false">\; (\; </mo>}c\mathrm{,}a\right)=gcd\left(\mathrm{2,}c-a\right)=2$

(because $c-a$ is even).

· 1^st case: ${\Delta }_a=1\Rightarrow c-b\equiv c+b\equiv 1\left(\mathrm{mod}2\right)$ .

But $\{\begin{array}{l}{a}^2=\left(c-b\right)\left(c+b\right)\text{and}\hfill \\ {\Delta }_a=gcd\left(c-b\mathrm{,}c+b\right)=1\hfill \end{array}\Rightarrow \exists \mathrm{<mo>\; !\; </mo>}\delta \mathrm{,}{\delta }^{″}\in 2\mathbb{N}+\mathrm{1\; <mo>\; /\; </mo>}gcd\left(\delta \mathrm{,}{\delta }^{″}\right)=1$ ,

Checking: $\{\begin{array}{l}c-b={\delta }^2\hfill \\ c+b={{\delta }^{″}}^2\hfill \end{array}\Rightarrow a^2={\left(\delta {\delta }^{″}\right)}^2\Rightarrow a=\delta {\delta }^{″}$ .

Let’s take a look at the quantities d and $d^{″}$ :

$gcd\left(a\mathrm{,}c-b\right)=gcd\left(\delta {\delta }^{″}\mathrm{,}{\delta }^2\right)=\delta =d\Rightarrow {\delta }^{″}=\frac{a}{d}=d^{″}\Rightarrow \{\begin{array}{l}c-b=d^2;\hfill \\ c+b={d^{″}}_2^2.\hfill \end{array}$

Hence as announced in the system (ii).

In conclusion the point ii) is checked by $\left(d\mathrm{,}{d}^{″}\right)=\left(gcd\left(a\mathrm{,}c-b\right)\mathrm{,}\frac{a}{d}\right)$ , the Pythagorician divisors coming from a, as defined in definition 1.1.; moreover, as stated in proposition 1.3., we have $gcd\left(d\mathrm{,}{d}^{″}\right)=gcd\left(\delta \mathrm{,}{\delta }^{″}\right)=1$ , as noted in the proof.

· 2^nd case: ${\Delta }_b=pgcd\left(c-a\mathrm{,}c+a\right)=2$ .

Then: $c-a\equiv c+a\equiv 0\left(\mathrm{mod}2\right)$ .

Thus: $b^2=\left(c-a\right)\left(c+a\right)=4\left(\frac{c-a}{2}\right)\left(\frac{c+a}{2}\right)$ with $gcd\left(\frac{c-a}{2}\mathrm{,}\frac{c+a}{2}\right)=1$ (cf. Reminders 1.1.2)). So we have:

${\left(\frac{b}{2}\right)}^2=\left(\frac{c-a}{2}\right)\left(\frac{c+a}{2}\right).$ (1.1)

We have 2 sub-cases:

α) 1^st sub-case: $\frac{c-a}{2}\equiv 0\left(\mathrm{mod}2\right)\iff \frac{c+a}{2}\equiv 1\left(\mathrm{mod}2\right)$ i.e. $\lambda =0$ , so in this case: $S=s$ .

Then $\exists \mathrm{<mo>\; !\; </mo>}E\mathrm{,}{\epsilon }^{″}\in {\mathbb{N}}^{\ast }$ such that: $\left(E\mathrm{,}{\epsilon }^{″}\right)\equiv \left(\mathrm{0,1}\right)\left(\mathrm{mod}2\right)$ and $gcd\left(E\mathrm{,}{\epsilon }^{″}\right)=1$ ,

Checking: $\{\begin{array}{l}\frac{c-a}{2}=E^2;\hfill \\ \frac{c+a}{2}={{\epsilon }^{″}}^2.\hfill \end{array}$

Hence (1.1) $\Rightarrow {\left(\frac{b}{2}\right)}^2={\left(E{\epsilon }^{″}\right)}^2\Rightarrow \frac{b}{2}=E{\epsilon }^{″}\Rightarrow b=2E{\epsilon }^{″}$ .

Consider then the Pythagorician divisors coming from $b:e\underset{def}{=}gcd\left(b\mathrm{,}c-a\right)$ .

Then: $e=gcd\left(2E{\epsilon }^{″}\mathrm{,2}{E}^2\right)=2gcd\left(E{\epsilon }^{″}\mathrm{,}{E}^2\right)=2E$ ;

From where: $\{\begin{array}{l}e=2E\hfill \\ b=2E{\epsilon }^{″}=e{\epsilon }^{″},\hfill \end{array}$ Thus, according to definition 1.1: $e^{″}=\frac{b}{e}={\epsilon }^{″}$ , and is indeed the second Pythagorician divisors coming from b and associated with e.

But: $e{e}^{″}=2^s{b}_1=b\Rightarrow \exists !\bar{e}\equiv 1\left(\mathrm{mod}2\right)$ and $\bar{e}=\frac{b_1}{e^{″}}$ , such that:

$e=2E=2^s\bar{e}=2^{s-\lambda \left(s-1\right)}\bar{e}=2^S\bar{e}$ , since we are in the case where $\lambda =0$ . (1.2)

Note that $E=2^{S-1}\bar{e}.$

But then: $\{\begin{array}{l}c-a=\frac{e^2}{2};\hfill \\ c+a=2{e^{″}}^2.\hfill \end{array}$ Hence exactly point (iii).

In addition:

As: $e^{″}\equiv 1\left(\mathrm{mod}2\right)$ and $b=e{e}^{″}=2^s{b}_1$ , we deduce that all the points of Proposition 1.3. are proved, since:

$gcd\left(E\mathrm{,}{e}^{″}\right)=1\iff pgcd\left(\frac{e}{2}\mathrm{,}{e}^{″}\right)=1\Rightarrow gcd\left(e\mathrm{,}{e}^{″}\right)=1=2^0=2^{\lambda }$ :

Let’s note that $gcd\left(\bar{e}\mathrm{,}{e}^{″}\right)=1$ (because $gcd\left(e\mathrm{,}{e}^{″}\right)=1$ ).

This demonstrates point 2) of proposition 1.3. in the case $\lambda =0$ .

β) 2^nd sub-case: $\frac{c-a}{2}\equiv 1\left(\mathrm{mod}2\right)\iff \frac{c+a}{2}\equiv 0\left(\mathrm{mod}2\right)$ i.e. $\lambda =1$ , in this case: $S=1$ .

This time in (1.1): $\exists \mathrm{<mo>\; !\; </mo>}\left(\bar{\epsilon }\mathrm{,}{\epsilon }^{″}\right)\equiv \left(\mathrm{1,0}\right)\left(\mathrm{mod}2\right)$ with $gcd\left(\bar{\epsilon }\mathrm{,}{\epsilon }^{″}\right)=1$ , verifying:

$\begin{array}{c}\{\begin{array}{l}\frac{c-a}{2}={\bar{\epsilon }}^2;\hfill \\ \frac{c+a}{2}={{\epsilon }^{″}}^2.\hfill \end{array}\Rightarrow {\left(\frac{b}{2}\right)}^2={\left(\bar{\epsilon }{\epsilon }^{″}\right)}^2.\\ \Rightarrow \frac{b}{2}=\bar{\epsilon }{\epsilon }^{″}\Rightarrow b=2^s{b}_1=2\bar{\epsilon }{\epsilon }^{″}.\end{array}$

We have: ${\epsilon }^{″}=2^{s-1}\frac{b_1}{\bar{\epsilon }}=2^{s-S}\frac{b_1}{\bar{\epsilon }}$ (remember that $s\ge 2$ ). Note that the points of proposition 1.3. are thus demonstrated, if $\bar{\epsilon }=\bar{e}$ and ${\epsilon }^{″}=e^{″}$ .

· Let us calculate the Pythagorician divisors coming from b:

$e\underset{def}{=}gcd\left(b\mathrm{,}c-a\right)=gcd\left(2\bar{\epsilon }{\epsilon }^{″}\mathrm{,2}{\bar{\epsilon }}^2\right)=2gcd\left(\bar{\epsilon }{\epsilon }^{″}\mathrm{,}{\bar{\epsilon }}^2\right)=2\bar{\epsilon }=2^S\bar{\epsilon }\underset{def}{=}{2}^S\bar{e}$ (1.3)

Then: $\bar{e}=\bar{\epsilon }$ , and from $e{\epsilon }^{″}=2\bar{\epsilon }{\epsilon }^{″}=b\Rightarrow {\epsilon }^{″}=e^{″}$ (cf. definition 1.1).

So, we have: $\{\begin{array}{l}c-a=\frac{e^2}{2};\hfill \\ c+a=2{e^{″}}^2.\hfill \end{array}$ hence point (iii), which completes the proof of the theorem.

Note that $\frac{e}{2}=\bar{e}$ is odd and $e^{″}$ is even. However $gcd\left(\bar{e}\mathrm{,}{e}^{″}\right)=1$ , hence:

$gcd\left(e\mathrm{,}{e}^{″}\right)=gcd\left(2\bar{e}\mathrm{,}{e}^{″}\right)=2gcd\left(\bar{e}\mathrm{,}\frac{e^{″}}{2}\right)=2=2^{\lambda }$ .

Which completes the proof of Proposition 1.3.

$■$

From which we deduce, the notations remaining the same, the following corollary, and which shows that the usual well-known parameters u, v (cf. Proposition 1.2.) are in fact intimately linked to the Pythagorician divisors, which is a remarkable fact in itself, and that the Pythagorician divisors are interdependent.

Corollary 1.1 Suppose that: $a^2+b^2=c^2$ is realized, with $\left(a\mathrm{,}b\mathrm{,}c\right)$ taken as a non-trivial, primitive and positive solution. So: the usual well-known parameters $u>v\in \mathbb{N}$ (cf. proposition 1.2), which express the solution $\left(a,b,c\right)=\left(u^2-v^2,2uv,u^2+v^2\right)$ , are unique and are related to the Pythagorician divisors as follows:

1) $\{\begin{array}{l}e=2v;\hfill \\ e^{″}=u\hfill \end{array}$ ; and $\{\begin{array}{l}d=u-v;\hfill \\ d^{″}=u+v.\hfill \end{array}$ ; in particular: $\bar{e}=2^{1-S}v$

2) From where we deduce:

$\{\begin{array}{l}u=\frac{d^{″}+d}{2}=e^{″}=\frac{e}{2}+d;\hfill \\ v=\frac{d^{″}-d}{2}=\frac{e}{2}=e^{″}-d.\hfill \end{array}$ (1.4)

Then:

$\{\begin{array}{l}d=-\frac{e}{2}+e^{″}=d^{″}-e=-d^{″}+2e^{″};\hfill \\ d^{″}=\frac{e}{2}+e^{″}=d+e=-d+2e^{″}.\hfill \end{array}$ And $\{\begin{array}{l}e=2\left(d^{″}-e^{″}\right);\hfill \\ e^{″}=\frac{e}{2}+d.\hfill \end{array}$ (1.5)

3) $\lambda =1\iff S=1\iff u\equiv 0\left(\mathrm{mod}2\right)\iff v=\bar{e}\equiv 1\left(\mathrm{mod}2\right)\iff a\equiv -1\left(\mathrm{mod}4\right)$ .

4) $\lambda =0\iff S=s\ge 2\iff u\equiv 1\left(\mathrm{mod}2\right)\iff v=2^{S-1}\bar{e}\equiv 0\left(\mathrm{mod}2\right)\iff a\equiv 1\left(\mathrm{mod}4\right)$ .

We have the following corollary, still under the same assumptions and notations as above:

Corollary 1.2 There is equivalencies between the following propositions:

(j) $a^2+b^2=c^2$ ; (jj) $\{\begin{array}{l}a=d{d}^{″}={\left(\frac{d^{″}+d}{2}\right)}^2-{\left(\frac{d^{″}-d}{2}\right)}^2;\hfill \\ b=2\left(\frac{d^{″}+d}{2}\right)\left(\frac{d^{″}-d}{2}\right);\hfill \\ c={\left(\frac{d^{″}+d}{2}\right)}^2+{\left(\frac{d^{″}-d}{2}\right)}^2.\hfill \end{array}$

(jjj) $\{\begin{array}{l}a={e^{″}}^2-{\left(\frac{e}{2}\right)}^2;\hfill \\ b=e{e}^{″}=2e^{″}\left(\frac{e}{2}\right);\hfill \\ c={e^{″}}^2+{\left(\frac{e}{2}\right)}^2.\hfill \end{array}$ (jv) $\{\begin{array}{l}a=d^2+\left(2^S\bar{e}\right)d;\hfill \\ b=\frac{{\left(2^S\bar{e}\right)}^2}{2}+\left(2^S\bar{e}\right)d;\hfill \\ c=\frac{{\left(2^S\bar{e}\right)}^2}{2}+\left(2^S\bar{e}\right)d+d^2.\hfill \end{array}$

Proof 2 As soon as Theorem 1.1. is verified, then Corollary 1.1. is trivially verified, and the same is true for Corollary 1.2., indeed:

- We have (j) Û (jj) Û (jjj) as consequencies of corollary 1.1.

- Concerning the result: (j) Û (vj) We can compare this result to the one found in [1] p. 7, but ours being an update and more precise, which additionally includes the correct identification of the odd parameters, and the value of s, this because we are using Pythagorician divisors and the parameter $\lambda \in \left\{\mathrm{0,1}\right\}$ , defined by $\frac{c-a}{2}\equiv \lambda \left(\mathrm{mod}2\right)$ .

Given that: $a=d{d}^{″}$ and $b=e{e}^{″}$ , this result is easily demonstrated from the value of e in proposition 1.3. 2) and the formulas (1.4) of Corollary 1.1. Then, for that of c, we use Theorem 1.1. (ii) or (iii). So from the hereafter formulas:

$\{\begin{array}{l}a=d\left(d+e\right)=d^2+ed;\\ b=e\left(\frac{e}{2}+d\right)=\frac{e^2}{2}+ed;\\ c=a+\frac{e^2}{2}=b+d^2=\frac{e^2}{2}+ed+d^2.\end{array}$ It suffices then to replace $e=2^S\bar{e}$ .

Definition 1.3 We now define the following sets:

1) $\overline{T^{+}}$ : The set of non-trivial, primitive and positive Pythagoras solutions of the type $\left(a,b=2^s{b}_1,c\right)\equiv \left(\pm 1,0,1\right)\left(\mathrm{mod}4\right)$ .

2) The set: $2\mathbb{N}+1\underset{cop}{\times }2\mathbb{N}+1=\left\{\left(x\mathrm{,}y\right)\in {\left(2\mathbb{N}+1\right)}^2\mathrm{<mo>\; /\; </mo>}gcd\left(x\mathrm{,}y\right)=1\right\}$ .

It is clear that $Card\left(2\mathbb{N}+1\underset{cop}{\times }2\mathbb{N}+1\right)=+\infty$ , because there is an infinity of pairs of distinct odd prime numbers i.e. $\left(p\mathrm{,}{p}^{\prime }\right)$ , and $p\ne p^{\prime }$ .

Corollary 1.3 $\overline{T^{+}}$ is in bijection with: $\left(2\mathbb{N}+1\underset{cop}{\times }2\mathbb{N}+1\right)\times {\mathbb{N}}^{\ast }$ , as follows:

$\begin{array}{l}\pi \mathrm{<mo>\; :\; </mo>}\overline{T^{+}}\to \left(2\mathbb{N}+1\underset{cop}{\times }2\mathbb{N}+1\right)\times {\mathbb{N}}^{\ast }\\ \left(\begin{array}{l}a\hfill \\ b\hfill \\ c\hfill \end{array}\right)\mapsto {\left(d\mathrm{,}\bar{e}\mathrm{,}S\right)}_{\left(a\mathrm{,}b\mathrm{,}c\right)}=\left(\begin{array}{c}d=gcd\left(a\mathrm{,}c-b\right)\\ \bar{e}=\frac{gcd\left(b\mathrm{,}c-a\right)}{2^S}\\ S\end{array}\right)\end{array}$

where $S=s-\lambda \left(s-1\right),$ with $s=v_2\left(b\right)$ and $\lambda$ defined in Definition 1.2;

Whose reciprocal bijection is:

$\begin{array}{l}{\pi }^{-1}\mathrm{<mo>\; :\; </mo>}\left(2\mathbb{N}+1\underset{cop}{\times }2\mathbb{N}+1\right)\times {\mathbb{N}}^{\ast }\to \overline{T^{+}}\\ \left(\begin{array}{c}d\\ \bar{e}\\ S\end{array}\right)\mapsto {\left(a\mathrm{,}b\mathrm{,}c\right)}_{\left(d\mathrm{,}\bar{e}\mathrm{,}S\right)}=\left(\begin{array}{l}a=d^2+\left(2^S\bar{e}\right)d\hfill \\ b=\frac{{\left(2^S\bar{e}\right)}^2}{2}+\left(2^S\bar{e}\right)d\hfill \\ c=\frac{{\left(2^S\bar{e}\right)}^2}{2}+\left(2^S\bar{e}\right)d+d^2\hfill \end{array}\right)\end{array}$

Remark 1.1 When $\left(a,b,c\right)=\left(u^2-v^2,2uv,u^2+v^2\right)\in \overline{T^{+}}$ , then

$\frac{c-a}{2}\mathrm{<mo>\; =\; </mo>}{v}^2\equiv \lambda \left(\mathrm{mod}2\right)$ , thus $\lambda$ is none other than, the parity indicator of v. Recall that: $2uv=2^s{b}_1$ with $b_1\equiv 1\left(\mathrm{mod}2\right)$ , but then $S=s-\lambda \left(s-1\right)$ , $d=u-v$ and $\bar{e}=2^{1-S}v$ (cf. corollary 1.1. 1). As a result, we also have:

$\begin{array}{l}\pi \mathrm{<mo>\; :\; </mo>}\overline{T^{+}}\to \left(2\mathbb{N}+1\underset{cop}{\times }2\mathbb{N}+1\right)\times {\mathbb{N}}^{\ast }\\ \left(\begin{array}{c}{u}^2-v^2\\ 2uv\\ u^2+v^2\end{array}\right)\mapsto {\left(d,\bar{e},S\right)}_{\left(a,b,c\right)}=\left(\begin{array}{l}d=u-v\hfill \\ \bar{e}=2^{1-S}v\hfill \\ S=s-\lambda \left(s-1\right)\hfill \end{array}\right)\end{array}$

Proof 3 1) Firstly, $\pi$ is well defined, because:

$gcd\left(d\mathrm{,}\bar{e}\right)=1$ , and $S\in {\mathbb{N}}^{\ast }$ . Let’s remind that (cf. corollary 1.1.):

$S=s-\lambda \left(s-1\right)=1\iff \lambda =1$ and $S=s-\lambda \left(s-1\right)=s\ge 2\iff \lambda =0$ .

2) $\pi$ is indeed surjective, because: $\forall \left(d\mathrm{,}\bar{e}\mathrm{,}S\right)\in \left(2\mathbb{N}+1\underset{cop}{\times }2\mathbb{N}+1\right)\times {\mathbb{N}}^{\ast }$ , its antecedent by $\pi$ is:

$\left(a=d^2+\left(2^S\bar{e}\right)d,b=\frac{{\left(2^S\bar{e}\right)}^2}{2}+\left(2^S\bar{e}\right)d,c=\frac{{\left(2^S\bar{e}\right)}^2}{2}+\left(2^S\bar{e}\right)d+d^2\right)$ which indeed belongs to $\overline{T^{+}}$ .

Since:

- $a^2+b^2=c^2$ and $\left(a\mathrm{,}b\mathrm{,}c\right)\ne \left(\mathrm{1,0,1}\right)$ is primitive and positive, therefore $\left(a\mathrm{,}b\mathrm{,}c\right)\in \overline{T^{+}}$ .

We distinguish the cases: $S=1$ and $S=s\ge 2$ . That is to say $\left(d\mathrm{,}\bar{e}\mathrm{,1}\right)$ and $\left(d,\bar{e},S=s\ge 2\right)$ . Denote by $v_2\left(x\right)$ the 2-adic valuation of the integer x. Then:

· For case $S=1$ :

$\{\begin{array}{l}\frac{c-a}{2}={\bar{e}}^2\equiv 1\left(\mathrm{mod}2\right)\Rightarrow \lambda =\mathrm{1\; <mo>\; ;\; </mo>}\hfill \\ b=2^{v_2\left(2\left({\bar{e}}^2+\bar{e}d\right)\right)}\left(\frac{b}{2^{v_2\left(2\left({\bar{e}}^2+\bar{e}d\right)\right)}}\right)\Rightarrow s=v_2\left(2\left({\bar{e}}^2+\bar{e}d\right)\right)\text{;}{b}_1=\frac{b}{2^{v_2\left(2\left({\bar{e}}^2+\bar{e}d\right)\right)}}\equiv 1\left(\mathrm{mod}2\right);\hfill \\ gcd\left(a\mathrm{,}c-b\right)=gcd\left(d\left(d+2\bar{e}\right)\mathrm{,}{d}^2\right)=d;\hfill \\ \frac{gcd\left(b\mathrm{,}c-a\right)}{2^S}=\frac{gcd\left(2\bar{e}\left(\bar{e}+d\right)\mathrm{,2}\bar{e}\times \bar{e}\right)}{2}=\bar{e}.\hfill \end{array}$

· And in the second case $S=s\ge 2$ :

$\{\begin{array}{l}\frac{c-a}{2}=2^{2\left(S-1\right)}{\bar{e}}^2\equiv 0\left(\mathrm{mod}2\right)\Rightarrow \lambda =\mathrm{0\; <mo>\; ;\; </mo>}\hfill \\ b=2^S\bar{e}\left(2^{S-1}\bar{e}+d\right)\Rightarrow s=S\text{and}{b}_1=\frac{b}{2^S\bar{e}}\equiv 1\left(\mathrm{mod}2\right);\hfill \\ gcd\left(a\mathrm{,}c-b\right)=gcd\left(d\left(d+2\bar{e}\right)\mathrm{,}{d}^2\right)=d;\hfill \\ \frac{gcd\left(b\mathrm{,}c-a\right)}{2^S}=\frac{gcd\left(2^S\bar{e}\times \left(2^{S-1}\bar{e}+d\right){\mathrm{,2}}^S\bar{e}\times \left(2^{S-1}\bar{e}\right)\right)}{2^S}=\bar{e}.\hfill \end{array}$

So in all cases: $\pi \left(a,b=2^s{b}_1,c\right)=\left(d,\bar{e},S\right)\Rightarrow \pi$ is surjective.

3) Elsewhere, $\pi$ is also injective, because if: $\left(a_1\mathrm{,}{b}_1\mathrm{,}{c}_1\right)$ , $\left(a_2\mathrm{,}{b}_2\mathrm{,}{c}_2\right)\in \overline{T^{+}}$ and that $\pi \left(a_1\mathrm{,}{b}_1\mathrm{,}{c}_1\right)=\pi \left(a_2\mathrm{,}{b}_2\mathrm{,}{c}_2\right)$ , then:

- $d_1=gcd\left(a_1\mathrm{,}{c}_1-b_1\right)=gcd\left(a_2\mathrm{,}{c}_2-b_2\right)=d_2\Rightarrow d_1=d_2$ .

- $S_1=s_1-{\lambda }_1\left(s_1-1\right)=s_2-{\lambda }_2\left(s_2-1\right)=S_2$ and

${\bar{e}}_1=\frac{gcd\left(b_1\mathrm{,}{c}_1-a_1\right)}{2^{S_1}}=\frac{gcd\left(b_2\mathrm{,}{c}_2-a_2\right)}{2^{S_2}}={\bar{e}}_2\Rightarrow e_1=2^{{}^{S_1}}{\bar{e}}_1=2^{S_2}{\bar{e}}_2=e_2$

But then:

$\left(d_1=d_2\text{and}{e}_1=e_2\right)\Rightarrow \{\begin{array}{l}{a}_1=d_1\left(d_1+e_1\right)=d_2\left(d_2+e_2\right)=a_2;\hfill \\ b_1=e_1\left(\frac{e_1}{2}+d_1\right)=e_2\left(\frac{e_2}{2}+d_2\right)=b_2;\hfill \\ c_1=b_1+d_1^2=b_2+d_2^2=c_2.\hfill \end{array}$ (cf. corollary 1.2. jv)).

Hence $\left(a_1\mathrm{,}{b}_1\mathrm{,}{c}_1\right)=\left(a_2\mathrm{,}{b}_2\mathrm{,}{c}_2\right)\Rightarrow \pi$ is injective.

Ultimately $\pi$ is bijective as stated.

We will retain the following proposition, (which is the counterpart of proposition 1.2.), the notations being unchanged:

Proposition 1.4

$\begin{array}{l}\overline{T^{+}}=\{\left(d^2+\left(2^S\bar{e}\right)d\mathrm{,}\frac{{\left(2^S\bar{e}\right)}^2}{2}+\left(2^S\bar{e}\right)d\mathrm{,}\frac{{\left(2^S\bar{e}\right)}^2}{2}+\left(2^S\bar{e}\right)d+d^2\right)\mathrm{<mo>\; /\; </mo>}\\ \begin{array}{c}\\ \\ \end{array}\left(d\mathrm{,}\bar{e}\right)\in 2\mathbb{N}+1\underset{cop}{\times }2\mathbb{N}+1\text{and}S\in {\mathbb{N}}^{\ast }\}.\end{array}$

The proof comes directly from the bijectivity of ${\pi }^{-1}$ .

1.3. Pythagorician Classes of Pythagoras Solutions and Tables

Let us keep the notations of the previous corollaries and propositions, then:

$\overline{T^{+}}=\left\{{\left(a\mathrm{,}b\mathrm{,}c\right)}_{\left(d\mathrm{,}\bar{e}\mathrm{,}S\right)}\mathrm{,}\text{such}\text{that}\left(d\mathrm{,}\bar{e}\right)\in 2\mathbb{N}+1\underset{cop}{\times }2\mathbb{N}+1\text{and}S\in {\mathbb{N}}^{\ast }\right\}\mathrm{.}$

This makes it possible to define the following concepts:

Definition 1.4 1) Let $\left(a\mathrm{,}b\mathrm{,}c\right)\in \overline{T^{+}}$ , we call Pythagorician parameters associated with $\left(a\mathrm{,}b\mathrm{,}c\right)$ , the elements forming the unique triplet $\left(d\mathrm{,}\bar{e}\mathrm{,}S\right)\in \left(2\mathbb{N}+1\underset{cop}{\times }2\mathbb{N}+1\right)\times {\mathbb{N}}^{\ast }$ , such that:

$d=gcd\left(a\mathrm{,}c-b\right)$ , $\bar{e}=\frac{gcd\left(b\mathrm{,}c-a\right)}{2^S}$ where $S=s-\lambda \left(s-1\right)$ .

It is clear that the Pythagorician parameters of ${\left(a\mathrm{,}b\mathrm{,}c\right)}_{\left(d\mathrm{,}\bar{e}\mathrm{,}S\right)}$ are worth $\left(d\mathrm{,}\bar{e}\mathrm{,}S\right)$ .

2) When $\left(d\mathrm{,}\bar{e}\right)\in 2\mathbb{N}+1\underset{cop}{\times }2\mathbb{N}+1$ is fixed, we call Pythagorician class crossing through the solution ${\left(a\mathrm{,}b\mathrm{,}c\right)}_{\left(d\mathrm{,}\bar{e}\mathrm{,1}\right)}$ , the set noted: $\overline{{\left(a\mathrm{,}b\mathrm{,}c\right)}_{\left(d\mathrm{,}\stackrel{¯}{e}\mathrm{,1}\right)}}=\left\{{\left(a\mathrm{,}b\mathrm{,}c\right)}_{\left(d\mathrm{,}\bar{e}\mathrm{,}S\right)}\mathrm{<mo>\; /\; </mo>}S\in {\mathbb{N}}^{\ast }\right\}$ .

It is clear that classes of 2 different solutions are pairwise disjoint, hence:

$\stackrel{¯}{{\left(a\mathrm{,}b\mathrm{,}c\right)}_{\left(d_1\mathrm{,}{\bar{e}}_1\mathrm{,1}\right)}}\cap \stackrel{¯}{{\left(a\mathrm{,}b\mathrm{,}c\right)}_{\left(d_2\mathrm{,}{\bar{e}}_2\mathrm{,1}\right)}}=\varnothing$ , $\forall \left(d_1\mathrm{,}{\bar{e}}_1\right)\ne \left(d_2\mathrm{,}{\bar{e}}_2\right)\in 2\mathbb{N}+1\underset{cop}{\times }2\mathbb{N}+1$ ; and that:

$\overline{T^{+}}={\displaystyle \underset{{\left(a\mathrm{,}b\mathrm{,}c\right)}_{\left(d\mathrm{,}\bar{e}\mathrm{,1}\right)}\in {\mathcal{C}}_1}{\cup }}\overline{{\left(a\mathrm{,}b\mathrm{,}c\right)}_{\left(d\mathrm{,}\stackrel{¯}{e}\mathrm{,1}\right)}}$ , where ${\mathcal{C}}_1=\left\{{\left(a\mathrm{,}b\mathrm{,}c\right)}_{\left(d\mathrm{,}\bar{e}\mathrm{,1}\right)}\mathrm{<mo>\; /\; </mo>}\left(d\mathrm{,}\bar{e}\right)\in 2\mathbb{N}+1\underset{cop}{\times }2\mathbb{N}+1\right\}$ .

3) We define an equivalence relation on $\overline{T^{+}}$ , noted $\underset{\mathcal{P}}{\cong }$ , as follows:

We say that $\left(a_1,b_1=2^{s_1}{b}_{1,1},c_1\right)$ , $\left(a_2,b_2=2^{s_2}{b}_{1,2},c_2\right)\in \overline{T^{+}}$ are $\mathcal{P}$ -equivalent in $\overline{T^{+}}$ , and we note:

$\left(a_1\mathrm{,}{b}_1\mathrm{,}{c}_1\right)\underset{\mathcal{P}}{\cong }\left(a_2\mathrm{,}{b}_2\mathrm{,}{c}_2\right)\iff \exists \left(d\mathrm{,}\bar{e}\right)\in 2\mathbb{N}+1\underset{cop}{\times }2\mathbb{N}+1/\left(a_1\mathrm{,}{b}_1\mathrm{,}{c}_1\right)$ and $\left(a_2\mathrm{,}{b}_2\mathrm{,}{c}_2\right)\in \overline{{\left(a\mathrm{,}b\mathrm{,}c\right)}_{\left(d\mathrm{,}\stackrel{¯}{e}\mathrm{,1}\right)}}$ $\iff$ $\left(gcd\left(a_1\mathrm{,}{c}_1-b_1\right)\mathrm{,}\frac{gcd\left(a_1\mathrm{,}{c}_1-b_1\right)}{2^{S_1}}\right)=\left(gcd\left(a_2\mathrm{,}{c}_2-b_2\right)\mathrm{,}\frac{gcd\left(a_2\mathrm{,}{c}_2-b_2\right)}{2^{S_2}}\right)$ , where: $\forall i\in \left\{\mathrm{1,2}\right\},{\lambda }_i\in \left\{\mathrm{0,1}\right\}/\frac{c_i-a_i}{2}\equiv {\lambda }_i\left(\mathrm{mod}2\right)$ , and $S_i=s_i-{\lambda }_i\left(s_i-1\right)$ where $s_i=v_2\left(b_i\right)$ .

- The equivalence class $\overline{\left(a\mathrm{,}b\mathrm{,}c\right)}$ containing the element $\left(a,b=2^s{b}_1,c\right)\in \overline{T^{+}}$ , is called the Pythagorician class of the solution $\left(a\mathrm{,}b\mathrm{,}c\right)$ , and if $\left(a\mathrm{,}b\mathrm{,}c\right)\in \overline{{\left(a\mathrm{,}b\mathrm{,}c\right)}_{\left(d\mathrm{,}\stackrel{¯}{e}\mathrm{,1}\right)}}$ , we get:

$\begin{array}{c}\overline{\left(a\mathrm{,}b\mathrm{,}c\right)}=\overline{{\left(a\mathrm{,}b\mathrm{,}c\right)}_{\left(d\mathrm{,}\stackrel{¯}{e}\mathrm{,1}\right)}}={\pi }^{-1}\left(\left\{d\right\}\times \left\{\bar{e}\right\}\times {\mathbb{N}}^{\ast }\right)\\ ={\pi }^{-1}\left(\left\{gcd\left(a\mathrm{,}c-b\right)\right\}\times \left\{\frac{gcd\left(b\mathrm{,}c-a\right)}{2^S}\right\}\times {\mathbb{N}}^{\ast }\right).\end{array}$

This unique element: ${\left(a\mathrm{,}b\mathrm{,}c\right)}_{\left(d\mathrm{,}\bar{e}\mathrm{,1}\right)}$ , is called the canonical exceptional representative of the class $\overline{\left(a\mathrm{,}b\mathrm{,}c\right)}$ .

3) We note: $\stackrel{¯}{{\left(a\mathrm{,}b\mathrm{,}c\right)}_{{\left(d\mathrm{,}\bar{e}\mathrm{,1}\right)}_{\le \mathcal{l}}}}=\left\{{\left(a\mathrm{,}b\mathrm{,}c\right)}_{\left(d\mathrm{,}\bar{e}\mathrm{,}S\right)}\mathrm{<mo>\; /\; </mo>1}\le S\le \mathcal{l}\right\}$ . This is the class of ${\left(a\mathrm{,}b\mathrm{,}c\right)}_{\left(d\mathrm{,}\bar{e}\mathrm{,1}\right)}$ truncated in $\mathcal{l}$ .

1.3.1. Equivalent Expression, for the Equivalence Relation $\underset{\mathcal{P}}{\cong }$ with Respect to the Parameterization $\left(u^2-v^2,2uv,u^2+v^2\right)$

Let $\left(u^2-v^2\mathrm{,2}uv\mathrm{,}{u}^2+v^2\right)\in \overline{T^{+}}$ . We then ask ourselves the question of knowing under what conditions a Pythagorician triplet $\left(u_1^2-v_1^2\mathrm{,2}{u}_1{v}_1\mathrm{,}{u}_1^2+v_1^2\right)$ is $\mathcal{P}$ -equivalent (in $\overline{T^{+}}$ ), to the chosen one?

Recall that if $\left(d\mathrm{,}\bar{e}\mathrm{,}S\right),\left(d_1\mathrm{,}{\bar{e}}_1\mathrm{,}{S}_1\right)\in \left(2\mathbb{N}+1\underset{cop}{\times }2\mathbb{N}+1\right)\times {\mathbb{N}}^{\ast }$ are respectively, the Pythagorician parameters associated with the Pythagorician triplets $\left(u^2-v^2\mathrm{,2}uv\mathrm{,}{u}^2+v^2\right)$ and $\left(u_1^2-v_1^2\mathrm{,2}{u}_1{v}_1\mathrm{,}{u}_1^2+v_1^2\right)$ , then: $\{\begin{array}{l}d=u-v;\hfill \\ \bar{e}=2^{1-S}v.\hfill \end{array}$ and $\{\begin{array}{l}{d}_1=u_1-v_1;\hfill \\ {\bar{e}}_1=2^{1-S_1}{v}_1.\hfill \end{array}$ , and thus we get:

Proposition 1.5 Let $\left(u^2-v^2,2uv,u^2+v^2\right),\left(u_1^2-v_1^2,2u_1{v}_1,u_1^2+v_1^2\right)\in \overline{T^{+}}$ , two Pythagorician triplets, and $\underset{\mathcal{P}}{\cong }$ be the relation of $\mathcal{P}$ -equivalence in $\overline{T^{+}}$ . Then, there is an equivalence between the following 2 propositions:

1) $\left(u_1^2-v_1^2\mathrm{,2}{u}_1{v}_1\mathrm{,}{u}_1^2+v_1^2\right)\underset{\mathcal{P}}{\cong }\left(u^2-v^2\mathrm{,2}uv\mathrm{,}{u}^2+v^2\right)$ .

2) $\{\begin{array}{l}{u}_1=u+\left(2^{S_1-S}-1\right)v;\hfill \\ v_1=2^{S_1-S}v.\hfill \end{array}\iff \left(\begin{array}{c}{u}_1\\ v_1\end{array}\right)=\left(\begin{array}{cc}1& 2^{S_1-S}-1\\ 0& 2^{S_1-S}\end{array}\right)(\; u\; v\; )$

Remark 1.2 1) Note that necessarily when the equivalence takes place, then simultaneously: $v_1=2^{S_1-S}v$ and $v=2^{S-S_1}{v}_1\in {\mathbb{N}}^{\ast }$ .

2) Let’s remark that we can replace in an equivalent way, the first proposition of (ii) by:

$\exists k\in \mathbb{Z}$ such that $\{\begin{array}{l}{u}_1=u+\left(2^k-1\right)v;\hfill \\ v_1=2^{k}v.\hfill \end{array}$

Note that for such a k, we have $2^{k}v$ and $2^{-k}{v}_1\in {\mathbb{N}}^{\ast }$ , and in case that v is odd, then necessarily k = 0.

(resp. the second proposition of (ii) by: $\exists k\in \mathbb{Z}$ such that $\left(\begin{array}{c}{u}_1\\ v_1\end{array}\right)=\left(\begin{array}{cc}1& 2^k-1\\ 0& 2^k\end{array}\right)\left(\begin{array}{c}u\\ v\end{array}\right)$ ).

3) Denote by ${\mathcal{M}}_P=\left\{\left(\begin{array}{cc}1& 2^k-1\\ 0& 2^k\end{array}\right)\mathrm{,}k\in \mathbb{Z}\right\}$ , then $\left({\mathcal{M}}_P\mathrm{,}\times \right)$ is a commutative group.

This relation is indeed an equivalence relation because

${\left(\begin{array}{cc}1& 2^{S_1-S}-1\\ 0& 2^{S_1-S}\end{array}\right)}^{-1}=\left(\begin{array}{cc}1& 2^{S-S_1}-1\\ 0& 2^{S-S_1}\end{array}\right)$ , and

$\left(\begin{array}{cc}1& 2^{S_2-S_1}-1\\ 0& 2^{S_2-S_1}\end{array}\right)\left(\begin{array}{cc}1& 2^{S_1-S}-1\\ 0& 2^{S_1-S}\end{array}\right)=\left(\begin{array}{cc}1& 2^{S_2-S}-1\\ 0& 2^{S_2-S}\end{array}\right)$ .

➢ Suppose that $u\equiv 0\left(\mathrm{mod}2\right)$ then S = 1 and $v\equiv 1\left(\mathrm{mod}2\right)$ .

If $u_1\equiv 0\left(\mathrm{mod}2\right)\Rightarrow 2^{S_1-1}-1\equiv 0\left(\mathrm{mod}2\right)\Rightarrow S_1=1$ ; but then $\left(u_1\mathrm{,}{v}_1\right)=\left(u\mathrm{,}v\right)$ .

So in an equivalence class of a Pythagorician triplet, which contains a Pythagorician triplet $\left(u^2-v^2\mathrm{,2}uv\mathrm{,}{u}^2+v^2\right)$ , with u even then it’s the only triplet to have this property, i.e. all $\left(u_1^2-v_1^2\mathrm{,2}{u}_1{v}_1\mathrm{,}{u}_1^2+v_1^2\right)$ which are equivalent to it, are such that $u_1\equiv 1\left(\mathrm{mod}2\right)$ . We can therefore choose $\left(u^2-v^2\mathrm{,2}uv\mathrm{,}{u}^2+v^2\right)$ as an canonical exceptional representative element of its class.

➢ Now suppose that $u\equiv 1\left(\mathrm{mod}2\right)$ then $S\ge 2$ and $v\equiv 0\left(\mathrm{mod}2\right)$ . We note that $v=2^{S-1}\bar{e}$ .

Consider now the element $\left(u_1^2-v_1^2\mathrm{,2}{u}_1{v}_1\mathrm{,}{u}_1^2+v_1^2\right)$ belonging to the class of $\left(u^2-v^2\mathrm{,2}uv\mathrm{,}{u}^2+v^2\right)$ , and defined by: $\{\begin{array}{l}{u}_1=u+\left(2^{1-S}-1\right)v\hfill \\ v_1=2^{1-S}v\hfill \end{array}$ , then $v_1=\bar{e}\equiv 1\left(\mathrm{mod}2\right)$ , and $\frac{\left(u_1^2+v_1^2\right)-\left(u_1^2-v_1^2\right)}{2}={\bar{e}}^2\equiv 1\left(\mathrm{mod}2\right)\Rightarrow {\lambda }_1=1$ , hence $u_1\equiv 0\left(\mathrm{mod}2\right)$ , and it is the only Pythagorician triplet having this property, in the class of $\left(u^2-v^2\mathrm{,2}uv\mathrm{,}{u}^2+v^2\right)$ , it’s why, we can therefore choose it as the canonical exceptional representative for this class.

We have just shown the following proposition:

Proposition 1.6 Let $\left(u_1^2-v_1^2\mathrm{,2}{u}_1{v}_1\mathrm{,}{u}_1^2+v_1^2\right)\in \overline{T^{+}}$ , then there exists a single $\left(u^2-v^2\mathrm{,2}uv\mathrm{,}{u}^2+v^2\right)\in \overline{T^{+}}$ such that $u\equiv 0\left(\mathrm{mod}2\right)$ , and $\left(u_1^2-v_1^2\mathrm{,2}{u}_1{v}_1\mathrm{,}{u}_1^2+v_1^2\right)\in \overline{\left(u^2-v^2\mathrm{,2}uv\mathrm{,}{u}^2+v^2\right)}$ .

So then: $\{\begin{array}{l}{u}_1=u+\left(2^{S_1-1}-1\right)v;\hfill \\ v_1=2^{S_1-1}v.\hfill \end{array}$

And as a result:

$\overline{\left(u^2-v^2,2uv,u^2+v^2\right)}=\left\{\left(u_1^2-v_1^2,2u_1{v}_1,u_1^2+v_1^2\right)/\{\begin{array}{l}{u}_1=u+\left(2^{S_1-1}-1\right)v;\hfill \\ v_1=2^{S_1-1}v.\hfill \end{array},S_1\in {\mathbb{N}}^{\ast }\right\}$ , and those classes form a partition of $\overline{T^{+}}$ , when $u\mathrm{,}v\in {\mathbb{N}}^{\ast }$ , $u\equiv 0\left(\mathrm{mod}2\right)$ , $u+v\equiv 1\left(\mathrm{mod}2\right)$ , $gcd\left(u\mathrm{,}v\right)=1$ .

Definition 1.5 1) For simplicity one can also note: $\overline{\left(u\mathrm{,}v\right)}=\overline{\left(u^2-v^2\mathrm{,}2uv\mathrm{,}{u}^2+v^2\right)}$

2) When $u\equiv 0\left(\mathrm{mod}2\right)$ , one notes:

${\overline{\left(u\mathrm{,}v\right)}}_{\le \mathcal{l}}=\left\{\left(u+\left(2^{S_1-1}-1\right)v{\mathrm{,2}}^{S_1-1}v\right)\text{where}{S}_1\le \mathcal{l}\right\}$ ,

it is the class of $\left(u^2-v^2\mathrm{,2}uv\mathrm{,}{u}^2+v^2\right)$ truncated in $\mathcal{l}$ .

1.3.2. Paradoxical Rarefaction of Pythagorician Triplets $\left(u^2-v^2,2uv,u^2+v^2\right)$ Such That $u\equiv 0\left(mod2\right)$ i.e. Whose Adjacent Side Is $\equiv -1\left(mod4\right)$ .

We get the following proposals:

Proposition 1.7 Let $\left(a\mathrm{,}b=2^s{b}_1\mathrm{,}c\right)\in \overline{T^{+}}$ , let’s recall cf. corollary 1.1. that $S=s-\lambda \left(s-1\right)=1\iff \lambda =1$ .

1) The Complete set of representatives (called canonical complete set of representatives) of $\overline{T^{+}}/\underset{\mathcal{P}}{\cong }$ : Namely ${\mathcal{C}}_1=\left\{\left(a\mathrm{,}b\mathrm{,}c\right)\in \overline{T^{+}}/\lambda =1\right\}$ , is the set of non-trivial Pythagorician solutions, positive, primitive “said evens (with reference to u even)”, which correspond to the Pythagorician solutions of $\overline{T^{+}}$ whose adjacent side is $\equiv -1\left(\mathrm{mod}4\right)$ .

2) $Card\left({\mathcal{C}}_1\right)=Card\left(2\mathbb{N}+1\underset{cop}{\times }2\mathbb{N}+1\right)=+\infty$ , (because there is an infinity of prime numbers), However for $\left(d\mathrm{,}\bar{e}\right)$ fixed, the element ${\left(a\mathrm{,}b\mathrm{,}c\right)}_{\left(d\mathrm{,}\bar{e}\mathrm{,1}\right)}$ is the unique element of ${\mathcal{C}}_1$ (i.e. even), which is present in $\overline{{\left(a\mathrm{,}b\mathrm{,}c\right)}_{\left(d\mathrm{,}\stackrel{¯}{e}\mathrm{,1}\right)}}$ . Thus: although they are in infinite number, we deduce that there is a relative rarefaction of the solutions $\left(a\mathrm{,}b\mathrm{,}c\right)\in {\mathcal{C}}_1$ (i.e. even ones), compared to those $\notin {\mathcal{C}}_1$ , that is to say the set of “odd” solutions $\left(a\mathrm{,}b\mathrm{,}c\right)$ , whose cardinality is equal to: $Card\left(\left(2\mathbb{N}+1\underset{cop}{\times }2\mathbb{N}+1\right)\times \left({\mathbb{N}}^{\ast }\backslash \left\{1\right\}\right)\right)=Card\left({\mathcal{C}}_1\right)\times Card\left({\mathbb{N}}^{\ast }\backslash \left\{1\right\}\right)$ , which is much greater than $Card\left({\mathcal{C}}_1\right)$ .

Remark 1.3 So, in any class $\overline{{\left(a\mathrm{,}b\mathrm{,}c\right)}_{\left(d\mathrm{,}\stackrel{¯}{e},S\right)}}$ , there is exactly one element such as $u\equiv 0\left(\mathrm{mod}2\right)$ , i.e. whose adjacent side is $\equiv -1\left(\mathrm{mod}4\right)$ . This one is ${\left(a\mathrm{,}b\mathrm{,}c\right)}_{\left(d\mathrm{,}\bar{e}\mathrm{,1}\right)}=\left(d^2+2\bar{e}d\mathrm{,2}{\bar{e}}^2+2\bar{e}d\mathrm{,2}{\bar{e}}^2+2\bar{e}d+d^2\right)$ , and then $u=\bar{e}+d\equiv 0\left(\mathrm{mod}2\right)$ , where $d=gcd\left(a\mathrm{,}c-b\right)$ , and $e=gcd\left(b\mathrm{,}c-a\right)$ .

And there are on the other hand in this same class: $Card\left({\mathbb{N}}^{\ast }\backslash \left\{1\right\}\right)$ other elements, all having $u\equiv 1\left(\mathrm{mod}2\right)$ , they are: ${\left(a\mathrm{,}b\mathrm{,}c\right)}_{\left(d\mathrm{,}\bar{e}\mathrm{,}S\right)}=\left(d^2+\left(2^S\bar{e}\right)d\mathrm{,}\frac{{\left(2^S\bar{e}\right)}^2}{2}+\left(2^S\bar{e}\right)d\mathrm{,}\frac{{\left(2^S\bar{e}\right)}^2}{2}+\left(2^S\bar{e}\right)d+d^2\right)$ where $S\ge 2$ , then: $u=2^{S-1}\bar{e}+d\equiv 1\left(\mathrm{mod}2\right).$

· It is difficult to observe this phenomenon, because if we fix a bound for the hypothenuse, and we use the parameterization in $\left(u\mathrm{,}v\right)$ the even and odd u will alternate, and in the end, after counting, there will be no significant difference (thus, Table 2 gives 121 quantities u which are even, and 118 quantities u which are odd). We get the same conclusion likewise if we use the parametrization $\left(d\mathrm{,}\bar{e}\mathrm{,}S\right)$ of Table 3, because given the fixed bound, the exponentiation $2^S$ , will be quickly out of control, and it will appear very quickly only $S=1$ , thus in Table 3 (which is a double entry one) we find the same conclusion, that is:121 triplets for which $S=1$ , and 118 other triplets for which $S\ge 2$ .

· For draw the below curve (cf. Figure 1), we consider the average of the cardinalities of the classes, taken per block of 5 classes, that is to say that we make the average obtained by taking the ratio of the cardinality of the i-th block of consecutive 5 classes (of type $\overline{\left(u\mathrm{,}v\right)}$ ) by 5, this corresponds to the application: $i\mapsto \frac{Card\left(i\right)}{5}$ ). The curve above reveals a hyperbolic behavior of the representatives of the classes. Where data were obtained from Table 3. The curve shows the average positioning of the classes, when $c\le 1500$ , most of the classes do not open, strictly speaking, but reveal their canonical complete set of representatives who is then isolated, and it’s why the curve tends towards the value 1.

As sample data:

1^st block $=\left\{\overline{\left(2,1\right)},\overline{\left(4,1\right)},\overline{\left(4,3\right)},\overline{\left(6,1\right)},\overline{\left(6,5\right)}\right\}$ and $\frac{Card\left(1\right)}{5}=\frac{22}{5}=4.4$ ;

5^th block $=\left\{\overline{\left(14,9\right)},\overline{\left(14,11\right)},\overline{\left(14,13\right)},\overline{\left(16,1\right)},\overline{\left(16,3\right)}\right\}$ and $\frac{Card\left(5\right)}{5}=\frac{14}{5}=2.8$ ;

22^th block $=\left\{\overline{\left(34,3\right)},\overline{\left(34,5\right)},\overline{\left(34,7\right)},\overline{\left(34,9\right)},\overline{\left(34,11\right)}\right\}$ and $\frac{Card\left(22\right)}{5}=\frac{6}{5}=$ $1.2$.

(The classes $\overline{\left(u\mathrm{,}v\right)}$ cf. Definition 1.5., are easy to get from Table 3. The values use to draw the curve (with Excel software) are:

$\begin{array}{cccccccccccccccc}{i}^{\text{th}}\text{block}& 1& 2& 3& 4& 5& 6& 7& 8& 9& 10& 11& 12& 13& 14& 15\\ \frac{Card\left(i\right)}{5}& 4.4& 3.8& 3& 3& 2.8& 2& 2.6& 2.4& 1.4& 2.6& 1& 2& 2& 1.4& 1\end{array}$

$\begin{array}{ccccccccccc}{i}^{\text{th}}\text{block}& 16& 17& 18& 19& 20& 21& 22& 23& 24& 25\\ \frac{Card\left(i\right)}{5}& 2& 1& 1.6& 1.6& 1.2& 1.4& 1.2& 1.2& 1& 1\end{array}$

Figure 1. Rarefaction of certain Pythagorician triples.

Now, we get Tables 1-3 hereafter, by using the formal calculus software: Maple.

· In Table 1, we consider the truncated classes ${\overline{\left(u\mathrm{,}v\right)}}_{\le \mathcal{l}}$ , when $u\mathrm{,}v\in {\mathbb{N}}^{\ast }$ , $1\le v<u\le 10$ , $u\equiv 0\left(\mathrm{mod}2\right)$ , $u+v\equiv 1\left(\mathrm{mod}2\right)$ , $gcd\left(u\mathrm{,}v\right)=1$ and $\mathcal{l}\le 15$ .

This classification shows as announced, the rarity of right-angled triangles, positive non-trivial primitives, with adjacent side congruent to $-1\left(\mathrm{mod}4\right)$ . In fact there are 13 pairs $\left(u\mathrm{,}v\right)$ canonical representatives, i.e. with even u, and 182 pairs $\left(u\mathrm{,}v\right)$ with odd u.

Table 1. Truncated classes ${\overline{\left(u,v\right)}}_{\le \mathcal{l}},u,v\in {\mathbb{N}}^{*},0\le v<u\le 10$ .

· In this second table, which is double entry: We give the positive non-trivial primitive Pythagorician triplets $\left(a,b,c\right)=\left(u^2-v^2,2uv,u^2+v^2\right)$ , where $u>v$ ; $u+v\equiv 1\left(\mathrm{mod}2\right)$ and $gcd\left(u\mathrm{,}v\right)=1$ ;

as well as the corresponding Pythagorician parameters

$\left(d\mathrm{,}\bar{e},S\right)\in \left(2\mathbb{N}+1\underset{cop}{\times }2\mathbb{N}+1\right)\times {\mathbb{N}}^{\ast }$ .

Table 2. Table of positive primitive Pythagorician triplets of type $\left(a\mathrm{,}b\mathrm{,}c\right)$ with $c<1500$ .

In Table 3, which is double entry too: This one gives the positive non-trivial primitive Pythagorician triplets:

$\left(a\mathrm{,}b\mathrm{,}c\right)=\left(d^2+\left(2^S\bar{e}\right)d\mathrm{,}\frac{{\left(2^S\bar{e}\right)}^2}{2}+\left(2^S\bar{e}\right)d\mathrm{,}\frac{{\left(2^S\bar{e}\right)}^2}{2}+\left(2^S\bar{e}\right)d+d^2\right)$ , where

$\left(d\mathrm{,}\bar{e},S\right)\in \left(2\mathbb{N}+1\underset{cop}{\times }2\mathbb{N}+1\right)\times {\mathbb{N}}^{\ast }$ , note that:

$d=gcd\left(a\mathrm{,}c-b\right)$ , $\bar{e}=\frac{gcd\left(b\mathrm{,}c-a\right)}{2^S}$ , $b=2^s{b}_1$ and $S=s-\lambda \left(s-1\right)$ , with $\lambda \in \left\{\mathrm{0,1}\right\}/\frac{c-a}{2}\equiv \lambda \left(\mathrm{mod}2\right)$ .

We give as well the corresponding usual Pythagorician parameters: $u>v$ ; $u+v\equiv 1\left(\mathrm{mod}2\right)$ and $gcd\left(u\mathrm{,}v\right)=1$ ; such that $a=u^2-v^2$ ; $b=2uv$ ; $c=u^2+v^2$ .

Table 3. Table of positive primitive Pythagorician triplets of type $\left(a\mathrm{,}b\mathrm{,}c\right)$ with $c<1500$ .

1.4. Application to the Resolution of the Equations $x^2+y^2=2z^2$ and $x^2+2y^2=z^2$

1.4.1. Resolution of the Diophantine Equation $x^2+y^2=2z^2$

➢ Consider the equation: $x^2+y^2=2z^2$ with $x,y,z$ , positive, coprime in pairs and $x>y$ (cf. [9] , for another type of resolution). We suppose that $\left(x\mathrm{,}y\mathrm{,}z\right)\ne \left(\mathrm{1,1,1}\right)$ that is to say that it is non-trivial. It is clear that: $x\equiv y=z\equiv 1\left(\mathrm{mod}2\right)$ . But then:

$x^2+y^2=2z^2\iff {\left(\frac{x-y}{2}\right)}^2+{\left(\frac{x+y}{2}\right)}^2=z^2$ (1.6)

(with $\frac{x-y}{2}\mathrm{,}\frac{x+y}{2}\mathrm{,}z$ coprime in pairs and $\frac{x-y}{2}\mathrm{,}\frac{x+y}{2}$ of opposite parity). Let $\beta \in \left\{\mathrm{0,}1\right\}$ defined by:

$\frac{x+{\left(-1\right)}^{\beta }y}{2}\equiv 0\left(\mathrm{mod}2\right)\iff \frac{x-{\left(-1\right)}^{\beta }y}{2}\equiv 1\left(\mathrm{mod}2\right)\mathrm{<mo>\; ;\; </mo>}$ (1.7)

Then, the Pythagoras Equation (1.6) is written as:

${\left(\frac{x-{\left(-1\right)}^{\beta }y}{2}\right)}^2+{\left(\frac{x+{\left(-1\right)}^{\beta }y}{2}\right)}^2=z^2$ (1.8)

That is: $\left(\frac{x-{\left(-1\right)}^{\beta }y}{2},\frac{x+{\left(-1\right)}^{\beta }y}{2}=2^s{b}_1,z\right)\in \overline{T^{+}}$ , (where $s\ge 2$ ; $b_1$ odd).

For such a triplet, we know (cf. Definition 1.2.) That there exists $\lambda \in \left\{\mathrm{0,1}\right\}$ , such that:

$\frac{z-\frac{x-{\left(-1\right)}^{\beta }y}{2}}{2}\equiv \lambda \left(\mathrm{mod}2\right)$ (1.9)

As in the previous section, and if necessary, we state:

$S=s-\lambda \left(s-1\right)\text{and}{S}^{\prime }=S-1=\left(s-1\right)\left(1-\lambda \right)\mathrm{.}$ (1.10)

Then, $\exists \bar{e}$ odd, such that we have the expression of the Pythagorician divisors associated with (1.8):

$\{\begin{array}{l}\left(d\mathrm{,}{d}^{″}\right)=\left(gcd\left(\frac{x-{\left(-1\right)}^{\beta }y}{2}\mathrm{,}z-\frac{x+{\left(-1\right)}^{\beta }y}{2}\right)\mathrm{,}\frac{\frac{x-{\left(-1\right)}^{\beta }y}{2}}{d}\right)\text{and}\frac{x-{\left(-1\right)}^{\beta }y}{2}=d{d}^{″}\mathrm{<mo>\; ;\; </mo>}\hfill \\ \{\begin{array}{l}e=gcd\left(\frac{x+{\left(-1\right)}^{\beta }y}{2}\mathrm{,}z-\frac{x-{\left(-1\right)}^{\beta }y}{2}\right)=2^S\bar{e};\hfill \\ e^{″}=\frac{\frac{x+{\left(-1\right)}^{\beta }y}{2}}{e}=2^{s-S}\frac{b_1}{\bar{e}}.\hfill \end{array}\text{And}\frac{x+{\left(-1\right)}^{\beta }y}{2}=e{e}^{″}.\hfill \end{array}$

In addition we have: $\{\begin{array}{l}{d}^{″}=d+e=d+2^S\bar{e};\hfill \\ e^{″}=\frac{e}{2}+d=2^{S-1}\bar{e}+d.\hfill \end{array}$

The notations being the same, we then apply the Pythagorician divisors theorem to (1.8), from which we deduce the following corollary:

Corollary 1.4 There is an equivalence between the following propositions (the solutions are supposed to be primitive and positive non-trivial).

(i) $x^2+y^2=2z^2$ is solvable; (ii) $\{\begin{array}{l}x={e^{″}}^2-{\left(\frac{e}{2}\right)}^2+2\left(\frac{e}{2}\right)e^{″};\hfill \\ y={\left(-1\right)}^{\beta }\left[-{e^{″}}^2+{\left(\frac{e}{2}\right)}^2+2\left(\frac{e}{2}\right)e^{″}\right]\hfill \\ z={e^{″}}^2+{\left(\frac{e}{2}\right)}^2.\hfill \end{array};$

(iii) $\{\begin{array}{l}x={\left(\frac{d^{″}+d}{2}\right)}^2-{\left(\frac{d^{″}-d}{2}\right)}^2+2\left(\frac{d^{″}-d}{2}\right)\left(\frac{d^{″}+d}{2}\right);\hfill \\ y={\left(-1\right)}^{\beta }\left[-{\left(\frac{d^{″}+d}{2}\right)}^2+{\left(\frac{d^{″}-d}{2}\right)}^2+2\left(\frac{d^{″}-d}{2}\right)\left(\frac{d^{″}+d}{2}\right)\right]\hfill \\ z={\left(\frac{d^{″}+d}{2}\right)}^2+{\left(\frac{d^{″}-d}{2}\right)}^2.\hfill \end{array};$

(iv) $\{\begin{array}{l}x=2\times 2^{2S^{\prime }}{\bar{e}}^2+4\times 2^{S^{\prime }}\bar{e}d+d^2;\hfill \\ y={\left(-1\right)}^{\beta }\left[2\times 2^{2S^{\prime }}{\bar{e}}^2-d^2\right];\hfill \\ z=2\times 2^{2S^{\prime }}{\bar{e}}^2+2\times 2^{S^{\prime }}\bar{e}d+d^2.\hfill \end{array}$

Proof 4 It suffices to apply Theorem 1.1., and Corollary 1.1. & 1.2., with respect to (1.8).

Remark 1.4 1) We have: $\beta \in \left\{\mathrm{0,1}\right\}$ such that: ${\left(-1\right)}^{\beta }=sign{\left(2\times 2^{2S^{\prime }}{\bar{e}}^2-d\right)}^2$ i.e. $y\ge 0$ .

2) Note that the non-trivial, primitive and positive solutions of the equation $x^2+y^2=2z^2$ are given by unique expressions of the type:

$\left(x,y,z\right)=\left(u^2-v^2+2uv,u^2-v^2-2uv,u^2+v^2\right)$ when $\beta =1$ , and $\left(u^2-v^2+2uv\mathrm{,}{v}^2-u^2+2uv\mathrm{,}{u}^2+v^2\right)$ when $\beta =0$ , with $\left(u,v\right)=\left(e^{″},\frac{e}{2}\right)=\left(\frac{d^{″}+d}{2},\frac{d^{″}-d}{2}\right)$ , such that: $u>v$ , $gcd\left(u\mathrm{,}v\right)=1$ , $u+v\equiv 1\left(\mathrm{mod}2\right)$ .

In other words by: $\left(x\mathrm{,}y\mathrm{,}z\right)=\left(u^2-v^2+2uv\mathrm{,}\left|u^2-v^2-2uv\right|\mathrm{,}{u}^2+v^2\right)$ .

Let us keep the same notations, then on the model of Corollary 1.3., We have:

Corollary 1.5 The set of non-trivial, primitive and positive solutions of the equation $x^2+y^2=2z^2$ is in bijection with the set $\left(2\mathbb{N}+1\underset{cop}{\times }2\mathbb{N}+1\right)\times \mathbb{N}$ , as follows:

$\left(\begin{array}{c}x\\ y\\ z\end{array}\right)\stackrel{f}{\mapsto }\left(\begin{array}{c}d=gcd\left(\frac{x-{\left(-1\right)}^{\beta }y}{2}\mathrm{,}z-\frac{x+{\left(-1\right)}^{\beta }y}{2}\right)\\ \bar{e}=\frac{gcd\left(\frac{x+{\left(-1\right)}^{\beta }y}{2}\mathrm{,}z-\frac{x-{\left(-1\right)}^{\beta }y}{2}\right)}{2\times 2^{S^{\prime }}}\\ S^{\prime }\end{array}\right)$ , where $\beta$ and $S^{\prime }$ are defined in (1.7), (1.9) (1.10).

Whose reciprocal bijection is:

$\left(\begin{array}{c}d\\ \bar{e}\\ S^{\prime }\end{array}\right)\stackrel{f^{-1}}{\mapsto }\left(\begin{array}{l}x=2\times 2^{2S^{\prime }}{\bar{e}}^2+4\times 2^{S^{\prime }}\bar{e}d+d^2\hfill \\ y={\left(-1\right)}^{\beta }\left[2\times 2^{2S^{\prime }}{\bar{e}}^2-d^2\right]\hfill \\ z=2\times 2^{2S^{\prime }}{\bar{e}}^2+2\times 2^{S^{\prime }}\bar{e}d+d^2\hfill \end{array}\right)$ , where $\beta \in \left\{\mathrm{0,1}\right\}$ and ${\left(-1\right)}^{\beta }=sign\left(2\times 2^{2S^{\prime }}{\bar{e}}^2-d^2\right)$ .

Proof 5 The proof is similar to that made in Corollary 1.3.

Remark 1.5 By reasoning similar to that already done in the paragraph 1.3, if we consider $\Sigma$ , the set of non-trivial, primitive and positive solutions of $x^2+y^2=2z^2$ , then $\left(x_1,y_1,z_1\right)\cong \left(x_2,y_2,z_2\right)\iff \left(d_1,{\bar{e}}_1\right)=\left(d_2,{\bar{e}}_2\right)$ define an equivalence relation on $\Sigma$ , and each class $\overline{\left(x\mathrm{,}y\mathrm{,}z\right)}$ contains a unique solution $\left(x_0\mathrm{,}{y}_0\mathrm{,}{z}_0\right)$ for which ${S^{\prime }}_0=\left(s_0-1\right)\left(1-{\lambda }_0\right)=0$ i.e. ${\lambda }_0=1$ (cf. (1.10)), and which will be its canonical representative. These solutions characterized by $S^{\prime }=0$ , are “rare”, they are of the type: $\left(\begin{array}{l}x=2{\bar{e}}^2+4\bar{e}d+d^2\hfill \\ y={\left(-1\right)}^{\beta }\left[2{\bar{e}}^2-d^2\right]\hfill \\ z=2{\bar{e}}^2+2\bar{e}d+d^2\hfill \end{array}\right)$ ; their cardinality is equal to: $Card\left(2\mathbb{N}+1\underset{cop}{\times }2\mathbb{N}+1\right)$ , while the solutions characterized by $S^{\prime }\ge 1$ , have for cardinality: $Card\left(2\mathbb{N}+1\underset{cop}{\times }2\mathbb{N}+1\right)\times Card\left({\mathbb{N}}^{\ast }\right)$ .

Exemples 1.1 Let’s take the minimum possible value $\left(d,\bar{e}\right)=\left(1,1\right)$ , we get the minimal solution of $x^2+y^2=2z^2$ , that is $\left(\mathrm{7,1,5}\right)=f^{-1}\left(\mathrm{1,1,0}\right)$ , whose class is $\left\{f^{-1}\left(\mathrm{1,1,}{S}^{\prime }\right)\mathrm{,}{S}^{\prime }\in \mathbb{N}\right\}$ , i.e.:

$\overline{\left(\mathrm{7,1,5}\right)}=\left\{\left(7,1,5\right)\mathrm{,}\left(\mathrm{17,7,13}\right)\mathrm{,}\left(\mathrm{49,31,41}\right)\mathrm{,}\left(\mathrm{161,127,145}\right)\mathrm{,}\left(\mathrm{577,511,545}\right)\mathrm{,}\cdots \right\}$

1.4.2. Resolution of the Diophantine Equation $x^2+2y^2=z^2$

➢ We finish this paragraph, using methods similar to that of Pythagorician divisors, to solve the equation $x^2+2y^2=z^2$ (One can compare our method with those applied to the resolution of similar equations cf. [10] and [11] ).

One can see also ref. [12] for details and applications to congruent numbers problem.

We suppose that $x\mathrm{,}y\mathrm{,}z$ are positive and primitive, therefore necessarily $x\mathrm{,}z$ are odd, $y\ne 0$ is even and $x\mathrm{,}y\mathrm{,}z$ are pairwise coprime. Let’s define:

$\begin{array}{l}\beta \in \left\{\mathrm{0,1}\right\}\text{by}\mathrm{<mo>\; :\; </mo>}\frac{z+{\left(-1\right)}^{\beta }x}{2}\equiv 0\left(\mathrm{mod}2\right)\mathrm{,}\\ \text{and}\text{therefore}\frac{z-{\left(-1\right)}^{\beta }x}{2}\equiv 1\left(\mathrm{mod}2\right)\mathrm{.}\end{array}$ (1.11)

Recall (cf. Reminder 1.1.) That $gcd\left(\frac{z+{\left(-1\right)}^{\beta }x}{2}\mathrm{,}\frac{z-{\left(-1\right)}^{\beta }x}{2}\right)=1$ .

Then: $y^2=2\times \frac{z+{\left(-1\right)}^{\beta }x}{2}\times \frac{z-{\left(-1\right)}^{\beta }x}{2}$ .

Hence $\exists a_1$ even and $a_2$ odd, such that $y=a_1{a}_2$ , $gcd\left(a_1\mathrm{,}{a}_2\right)=1$ and:

$\{\begin{array}{c}\frac{z+{\left(-1\right)}^{\beta }x}{2}=\frac{a_1^2}{2};\\ \frac{z-{\left(-1\right)}^{\beta }x}{2}=a_2^2.\end{array}\Rightarrow gcd\left(y\mathrm{,}\frac{z+{\left(-1\right)}^{\beta }x}{2}\right)=gcd\left(a_1{a}_2\mathrm{,}\frac{a_1^2}{2}\right)=a_1;$

So, if we set:

$e=gcd\left(y\mathrm{,}\frac{z+{\left(-1\right)}^{\beta }x}{2}\right)=a_1\text{and}{e}^{″}=\frac{y}{e}=a_2$ (1.12)

it follows that:

$\{\begin{array}{c}\frac{z+{\left(-1\right)}^{\beta }x}{2}=\frac{e^2}{2};\\ \frac{z-{\left(-1\right)}^{\beta }x}{2}={e^{″}}^2.\end{array}\Rightarrow \{\begin{array}{l}x={\left(-1\right)}^{\beta }\left(\frac{e^2}{2}-{e^{″}}^2\right)=\{\begin{array}{l}\frac{e^2}{2}-{e^{″}}^2\mathrm{,}\text{if}\beta =\mathrm{0\; <mo>\; ;\; </mo>}\hfill \\ -\frac{e^2}{2}+{e^{″}}^2\mathrm{,}\text{if}\beta =1.\hfill \end{array}\hfill \\ y=e{e}^{″};\hfill \\ z={e^{″}}^2+\frac{e^2}{2}.\hfill \end{array}$

To sum up:

If $\beta \in \left\{\mathrm{0,}1\right\}$ is defined by: $\frac{z+{\left(-1\right)}^{\beta }x}{2}\equiv 0\left(\mathrm{mod}2\right)\iff \frac{z-{\left(-1\right)}^{\beta }x}{2}\equiv 1\left(\mathrm{mod}2\right)$ , and that: $gcd\left(y\mathrm{,}\frac{z+{\left(-1\right)}^{\beta }x}{2}\right)=e$ , and $\frac{y}{e}=e^{″}$ .

We just come to show the following result:

Proposition 1.8 There is an equivalence between the following propositions (the solutions are supposed to be non-trivial, primitive and positive).

(i) $x^2+2y^2=z^2$ is solvable;

(ii) $\{\begin{array}{l}x={\left(-1\right)}^{\beta }\left(2{\left(\frac{e}{2}\right)}^2-{e^{″}}^2\right)=\{\begin{array}{l}2{\left(\frac{e}{2}\right)}^2-{e^{″}}^2\mathrm{,}\text{if}\beta =\mathrm{0\; <mo>\; ;\; </mo>}\hfill \\ {e^{″}}^2-2{\left(\frac{e}{2}\right)}^2\mathrm{,}\text{if}\beta =1.\hfill \end{array}\hfill \\ y=2e^{″}\left(\frac{e}{2}\right);\hfill \\ z={e^{″}}^2+2{\left(\frac{e}{2}\right)}^2.\hfill \end{array}$

Remark 1.6 We have $\left(x\mathrm{,}y\mathrm{,}z\right)=\left(\left|u^2-2v^2\right|\mathrm{,}2uv\mathrm{,}{u}^2+2v^2\right)$ where $\left(u\mathrm{,}v\right)=\left(e^{″}\mathrm{,}\frac{e}{2}\right)$ .

Let $\left(x\mathrm{,}y\mathrm{,}z\right)$ be a positive non-trivial and primitive solution $x^2+2y^2=z^2$ , then we have:

$y=2^s{y}_1\text{and}s\ge \mathrm{1,}{y}_1\text{odd}\text{.}\text{Let}’\text{s}\text{put}{S}^{″}=s-1\in \mathbb{N}\mathrm{.}$ (1.13)

Then since $e^{″}$ is odd, there exists odd $\bar{e}$ such that: $e=2^s\bar{e}$ with $\bar{e}{e}^{″}=y_1$ from where $y=2^s\bar{e}{e}^{″}$ .

We have the following Proposition:

Proposition 1.9 The set of non-trivial, primitive and positive solutions of the equation $x^2+2y^2=z^2$ is in bijection with the set $\left(2\mathbb{N}+1\underset{cop}{\times }2\mathbb{N}+1\right)\times \mathbb{N}$ , as follows:

$\left(\begin{array}{l}x\hfill \\ y\hfill \\ z\hfill \end{array}\right)\stackrel{g}{\mapsto }\left(\begin{array}{c}{e}^{″}=\frac{y}{gcd\left(y\mathrm{,}\frac{z+{\left(-1\right)}^{\beta }x}{2}\right)}\\ \bar{e}=\frac{gcd\left(y\mathrm{,}\frac{z+{\left(-1\right)}^{\beta }x}{2}\right)}{2\times 2^{S^{″}}}\\ S^{″}\end{array}\right)$ , where $\beta$ and $S^{″}$ are defined in (1.11) and (1.13).

Whose reciprocal bijection is:

$\left(\begin{array}{l}{e}^{″}\hfill \\ \bar{e}\hfill \\ S^{″}\hfill \end{array}\right)\stackrel{g^{-1}}{\mapsto }\left(\begin{array}{l}x={\left(-1\right)}^{\beta }\left(2{\left(2^{S^{″}}\bar{e}\right)}^2-{e^{″}}^2\right)\hfill \\ y=2e^{″}\left(2^{S^{″}}\bar{e}\right)\hfill \\ z={e^{″}}^2+2{\left(2^{S^{″}}\bar{e}\right)}^2\hfill \end{array}\right)$ , where $\beta \in \left\{\mathrm{0,1}\right\}$ and ${\left(-1\right)}^{\beta }=sign\left(2{\left(2^{S^{″}}\bar{e}\right)}^2-{e^{″}}^2\right)$ .

Remark 1.7 In the same way as the previous equations it is possible to classify the positive primitive solutions of this equation, in terms of equivalence classes:

$\left(x\mathrm{,}y\mathrm{,}z\right)\cong \left(x_1\mathrm{,}{y}_1\mathrm{,}{z}_1\right)\iff \{\begin{array}{l}{e}^{″}=\frac{y}{gcd\left(y\mathrm{,}\frac{z+{\left(-1\right)}^{\beta }x}{2}\right)}=\frac{y_1}{gcd\left(y_1\mathrm{,}\frac{z_1+{\left(-1\right)}^{{\beta }_1}{x}_1}{2}\right)}={e^{″}}_1\hfill \\ \bar{e}=\frac{gcd\left(y\mathrm{,}\frac{z+{\left(-1\right)}^{\beta }x}{2}\right)}{2\times 2^{S^{″}}}\mathrm{<mo>\; =\; </mo>}\frac{gcd\left(y_1\mathrm{,}\frac{z_1+{\left(-1\right)}^{{\beta }_1}{x}_1}{2}\right)}{2\times 2^{{S^{″}}_1}}={\bar{e}}_1\hfill \end{array}$

As an example, consider the smallest positive and primitive solution $\left(\mathrm{1,2,3}\right)$ of this equation, obtained for $\left(e^{″},\bar{e},S^{″}\right)=\left(\mathrm{1,1,0}\right)$ , i.e. $\left(\mathrm{1,2,3}\right)=g^{-1}\left(\mathrm{1,1,0}\right)$ , then:

$\overline{\left(\mathrm{1,2,3}\right)}=\left\{g^{-1}\left(\mathrm{1,1,}{S}^{″}\right)\mathrm{,}{S}^{″}\in \mathbb{N}\right\}=\left\{\left(1,2\mathrm{,}3\right)\mathrm{,}\left(\mathrm{7,4,9}\right)\mathrm{,}\left(\mathrm{31,8,33}\right)\mathrm{,}\left(\mathrm{127,16,129}\right),\cdots \right\}$

2. Conclusions and Perspectives

This new method exposed here, made it possible to solve the Pythagoras equation from its Pythagorician divisors, and to make a classification of the solutions of this one, highlighting the importance of the quantity $\lambda =0$ or 1, defined by $\frac{c-a}{2}\equiv \lambda \left(\mathrm{mod}2\right)$ , as well as other quantities coming directly from the Pythagorician divisors.

In addition, this method opens interesting prospects for resolution, concerning the Pythagoras equation and those which proceed from it, as we have exposed it in this paper, but probably also, concerning the quadratic equations, in particular those of Pell-Fermat of the type: $a^2-d{b}^2=1$ . This is how the well-known problem of the existence of the square area of the right-angled triangle could also be considered from this point of view, and the same applies to Fermat’s equations of the type $a^4+b^4=c^4$ , $a^{2p}+b^{2p}=c^{2p}$ or more generally those of the type $a^p+b^p=c^p$ .

Acknowledgements

It was during the “validation workshop of the quadrennial program of the status of ESATIC as an UIT center of excellence in cybersecurity”, on January 21, 2015 at ESATIC (Abidjan), that I met Dr. KIMOU, the co-author.

On this occasion we exchanged views on Pythagorician triplets, Fermat’s theorem and certain unconventional types of cryptographic models.

In 2017, I invited him to give a conference at the CIMPA-Abidjan 2017 Research School (April 10-22, 2017): Algebraic Number Theory and Applications & Cryptography, held at Félix University Houphbouët BOIGNY.

This is why, we are particularly pleased to thank Dr. Prosper Kouadio KIMOU, who believed in our research team, for having offered us, in a limited seminar at Félix Houphouët BOIGNY University (2018-2019), some results of his works, on the identified research theme, of which this article is one of the results, combined with our own results. We owe him the discovery and systematic use of the quantities $d=gcd\left(a\mathrm{,}c-b\right)$ and $e=gcd\left(b\mathrm{,}c-a\right)$ , for, among other things, the basic version of the Pythagorician divisors theorem, as well as, for the starting point for a new resolution of Fermat’s equations for $n=4$ , using these quantities.

Our thanks also go to Dr. François E. TANOE , for standardization and appellations, of the notion of “Pythagorician divisors”, and the Pythagorician divisors theorem in its final form, and its application to the resolutions of the equations $x^2+y^2=2z^2$ and $x^2+2y^2=z^2$ .

Our thanks also go to Dr. Kouassi Vincent KOUAKOU, for his availability, and who wrote the reports of the seminars.

Conflicts of Interest

The authors declare no conflicts of interest regarding the publication of this paper.

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