Let ( u 2 − v 2 ,2 u v , u 2 + v 2 ) ∈ T + ¯ . We then ask ourselves the question of knowing under what conditions a Pythagorician triplet ( u 1 2 − v 1 2 ,2 u 1 v 1 , u 1 2 + v 1 2 ) is P -equivalent (in T + ¯ ), to the chosen one?

Recall that if ( d , e ¯ , S ) , ( d 1 , e ¯ 1 , S 1 ) ∈ ( 2 ℕ + 1 × c o p 2 ℕ + 1 ) × ℕ ∗ are respectively, the Pythagorician parameters associated with the Pythagorician triplets ( u 2 − v 2 ,2 u v , u 2 + v 2 ) and ( u 1 2 − v 1 2 ,2 u 1 v 1 , u 1 2 + v 1 2 ) , then: { d = u − v ; e ¯ = 2 1 − S v . and { d 1 = u 1 − v 1 ; e ¯ 1 = 2 1 − S 1 v 1 . , and thus we get:

Proposition 1.5 Let ( u 2 − v 2 , 2 u v , u 2 + v 2 ) , ( u 1 2 − v 1 2 , 2 u 1 v 1 , u 1 2 + v 1 2 ) ∈ T + ¯ , two Pythagorician triplets, and ≅ P be the relation of P -equivalence in T + ¯ . Then, there is an equivalence between the following 2 propositions:

1) ( u 1 2 − v 1 2 ,2 u 1 v 1 , u 1 2 + v 1 2 )   ≅ P   ( u 2 − v 2 ,2 u v , u 2 + v 2 ) .

2) { u 1 = u + ( 2 S 1 − S − 1 ) v ; v 1 = 2 S 1 − S v . ⇔ ( u 1 v 1 ) = ( 1 2 S 1 − S − 1 0 2 S 1 − S ) ( 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 ∈ ℕ ∗ .

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

∃ k ∈ ℤ such that { u 1 = u + ( 2 k − 1 ) v ; v 1 = 2 k v .

Note that for such a k, we have 2 k v and 2 − k v 1 ∈ ℕ ∗ , and in case that v is odd, then necessarily k = 0.

(resp. the second proposition of (ii) by: ∃ k ∈ ℤ such that ( u 1 v 1 ) = ( 1 2 k − 1 0 2 k ) ( u v ) ).

3) Denote by M P = { ( 1 2 k − 1 0 2 k ) ,     k ∈ ℤ } , then ( M P , × ) is a commutative group.

This relation is indeed an equivalence relation because

( 1 2 S 1 − S − 1 0 2 S 1 − S ) − 1 = ( 1 2 S − S 1 − 1 0 2 S − S 1 ) , and

( 1 2 S 2 − S 1 − 1 0 2 S 2 − S 1 ) ( 1 2 S 1 − S − 1 0 2 S 1 − S ) = ( 1 2 S 2 − S − 1 0 2 S 2 − S ) .

➢ Suppose that u ≡ 0 ( mod 2 ) then S = 1 and v ≡ 1 ( mod 2 ) .

If u 1 ≡ 0 ( mod 2 ) ⇒ 2 S 1 − 1 − 1 ≡ 0 ( mod 2 ) ⇒ S 1 = 1 ; but then ( u 1 , v 1 ) = ( u , v ) .

So in an equivalence class of a Pythagorician triplet, which contains a Pythagorician triplet ( u 2 − v 2 ,2 u v , u 2 + v 2 ) , with u even then it’s the only triplet to have this property, i.e. all ( u 1 2 − v 1 2 ,2 u 1 v 1 , u 1 2 + v 1 2 ) which are equivalent to it, are such that u 1 ≡ 1 ( mod 2 ) . We can therefore choose ( u 2 − v 2 ,2 u v , u 2 + v 2 ) as an canonical exceptional representative element of its class.

➢ Now suppose that u ≡ 1 ( mod 2 ) then S ≥ 2 and v ≡ 0 ( mod 2 ) . We note that v = 2 S − 1 e ¯ .

Consider now the element ( u 1 2 − v 1 2 ,2 u 1 v 1 , u 1 2 + v 1 2 ) belonging to the class of ( u 2 − v 2 ,2 u v , u 2 + v 2 ) , and defined by: { u 1 = u + ( 2 1 − S − 1 ) v v 1 = 2 1 − S v , then v 1 = e ¯ ≡ 1 ( mod 2 ) , and ( u 1 2 + v 1 2 ) − ( u 1 2 − v 1 2 ) 2 = e ¯ 2 ≡ 1 ( mod 2 ) ⇒ λ 1 = 1 , hence u 1 ≡ 0 ( mod 2 ) , and it is the only Pythagorician triplet having this property, in the class of ( u 2 − v 2 ,2 u v , u 2 + v 2 ) , 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 ( u 1 2 − v 1 2 ,2 u 1 v 1 , u 1 2 + v 1 2 ) ∈ T + ¯ , then there exists a single ( u 2 − v 2 ,2 u v , u 2 + v 2 ) ∈ T + ¯ such that u ≡ 0 ( mod 2 ) , and ( u 1 2 − v 1 2 ,2 u 1 v 1 , u 1 2 + v 1 2 ) ∈ ( u 2 − v 2 ,2 u v , u 2 + v 2 ) ¯ .

So then: { u 1 = u + ( 2 S 1 − 1 − 1 ) v ; v 1 = 2 S 1 − 1 v .

And as a result:

( u 2 − v 2 , 2 u v , u 2 + v 2 ) ¯ = { ( u 1 2 − v 1 2 , 2 u 1 v 1 , u 1 2 + v 1 2 ) / { u 1 = u + ( 2 S 1 − 1 − 1 ) v ; v 1 = 2 S 1 − 1 v . ,   S 1 ∈ ℕ ∗ } , and those classes form a partition of T + ¯ , when u , v ∈ ℕ ∗ , u ≡ 0 ( mod 2 ) , u + v ≡ 1 ( mod 2 ) , g c d ( u , v ) = 1 .

Definition 1.5 1) For simplicity one can also note: ( u ,   v ) ¯ = ( u 2 − v 2 ,   2 u v ,   u 2 + v 2 ) ¯

2) When u ≡ 0 ( mod 2 ) , one notes:

( u , v ) ¯ ≤ l = { ( u + ( 2 S 1 − 1 − 1 ) v ,2 S 1 − 1 v )   where   S 1 ≤ l } ,

it is the class of ( u 2 − v 2 ,2 u v , u 2 + v 2 ) truncated in l .

We get the following proposals:

Proposition 1.7 Let ( a , b = 2 s b 1 , c ) ∈ T + ¯ , let’s recall cf. corollary 1.1. that S = s − λ ( s − 1 ) = 1 ⇔ λ = 1 .

1) The Complete set of representatives (called canonical complete set of representatives) of T + ¯ / ≅ P : Namely C 1 = { ( a , b , c ) ∈ T + ¯ / λ = 1 } , is the set of non-trivial Pythagorician solutions, positive, primitive “said evens (with reference to u even)”, which correspond to the Pythagorician solutions of T + ¯ whose adjacent side is ≡ − 1 ( mod 4 ) .

2) C a r d ( C 1 ) = C a r d ( 2 ℕ + 1 × c o p 2 ℕ + 1 ) = + ∞ , (because there is an infinity of prime numbers), However for ( d , e ¯ ) fixed, the element ( a , b , c ) ( d , e ¯ ,1 ) is the unique element of C 1 (i.e. even), which is present in ( a , b , c ) ( d , e ¯ ,1 ) ¯ . Thus: although they are in infinite number, we deduce that there is a relative rarefaction of the solutions ( a , b , c ) ∈ C 1 (i.e. even ones), compared to those ∉ C 1 , that is to say the set of “odd” solutions ( a , b , c ) , whose cardinality is equal to: C a r d ( ( 2 ℕ + 1 × c o p 2 ℕ + 1 ) × ( ℕ ∗ \ { 1 } ) ) = C a r d ( C 1 ) × C a r d ( ℕ ∗ \ { 1 } ) , which is much greater than C a r d ( C 1 ) .

Remark 1.3 So, in any class ( a , b , c ) ( d , e ¯ , S ) ¯ , there is exactly one element such as u ≡ 0 ( mod 2 ) , i.e. whose adjacent side is ≡ − 1 ( mod 4 ) . This one is ( a , b , c ) ( d , e ¯ ,1 ) = ( d 2 + 2 e ¯ d ,2 e ¯ 2 + 2 e ¯ d ,2 e ¯ 2 + 2 e ¯ d + d 2 ) , and then u = e ¯ + d ≡ 0 ( mod 2 ) , where d = g c d ( a , c − b ) , and e = g c d ( b , c − a ) .

And there are on the other hand in this same class: C a r d ( ℕ ∗ \ { 1 } ) other elements, all having u ≡ 1 ( mod 2 ) , they are: ( a , b , c ) ( d , e ¯ , S ) = ( d 2 + ( 2 S e ¯ ) d ,   ( 2 S e ¯ ) 2 2 + ( 2 S e ¯ ) d ,     ( 2 S e ¯ ) 2 2 + ( 2 S e ¯ ) d + d 2 ) where S ≥ 2 , then: u = 2 S − 1 e ¯ + d ≡ 1 ( mod 2 ) .

· It is difficult to observe this phenomenon, because if we fix a bound for the hypothenuse, and we use the parameterization in ( u , v ) 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 ( d , e ¯ , S ) 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 ≥ 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 ( u , v ) ¯ ) by 5, this corresponds to the application: i ↦ C a r d ( i ) 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 ≤ 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 = { ( 2 , 1 ) ¯   , ( 4 , 1 ) ¯   , ( 4 , 3 ) ¯   , ( 6 , 1 ) ¯   , ( 6 , 5 ) ¯ } and C a r d ( 1 ) 5 = 22 5 = 4.4 ;

5^th block = { ( 14 , 9 ) ¯   , ( 14 , 11 ) ¯   , ( 14 , 13 ) ¯   , ( 16 , 1 ) ¯   , ( 16 , 3 ) ¯ } and C a r d ( 5 ) 5 = 14 5 = 2.8 ;

22^th block = { ( 34 , 3 ) ¯   , ( 34 , 5 ) ¯   , ( 34 , 7 ) ¯   , ( 34 , 9 ) ¯   , ( 34 , 11 ) ¯ } and C a r d ( 22 ) 5 = 6 5 = 1.2 .

(The classes ( u , v ) ¯ cf. Definition 1.5., are easy to get from Table 3. The values use to draw the curve (with Excel software) are:

i th   block 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 C a r d ( i ) 5 4.4 3.8 3 3 2.8 2 2.6 2.4 1.4 2.6 1 2 2 1.4 1

i th   block 16 17 18 19 20 21 22 23 24 25 C a r d ( i ) 5 2 1 1.6 1.6 1.2 1.4 1.2 1.2 1 1

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

· In Table 1, we consider the truncated classes ( u , v ) ¯ ≤ l , when u , v ∈ ℕ ∗ , 1 ≤ v < u ≤ 10 , u ≡ 0 ( mod 2 ) , u + v ≡ 1 ( mod 2 ) , g c d ( u , v ) = 1 and l ≤ 15 .

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

· In this second table, which is double entry: We give the positive non-trivial primitive Pythagorician triplets ( a , b , c ) = ( u 2 − v 2 , 2 u v , u 2 + v 2 ) , where u > v ; u + v ≡ 1 ( mod 2 ) and g c d ( u , v ) = 1 ;

as well as the corresponding Pythagorician parameters

( d , e ¯ , S ) ∈ ( 2 ℕ + 1 × c o p 2 ℕ + 1 ) × ℕ ∗ .

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

( a , b , c ) = ( d 2 + ( 2 S e ¯ ) d ,   ( 2 S e ¯ ) 2 2 + ( 2 S e ¯ ) d ,   ( 2 S e ¯ ) 2 2 + ( 2 S e ¯ ) d + d 2 ) , where

( d , e ¯ , S ) ∈ ( 2 ℕ + 1 × c o p 2 ℕ + 1 ) × ℕ ∗ , note that:

d = g c d ( a , c − b ) , e ¯ = g c d ( b , c − a ) 2 S , b = 2 s b 1 and S = s − λ ( s − 1 ) , with λ ∈ { 0,1 } / c − a 2 ≡ λ ( mod 2 ) .

We give as well the corresponding usual Pythagorician parameters: u > v ; u + v ≡ 1 ( mod 2 ) and g c d ( u , v ) = 1 ; such that a = u 2 − v 2 ; b = 2 u v ; c = u 2 + v 2 .

➢ Consider the equation: x 2 + y 2 = 2 z 2 with x , y , z , positive, coprime in pairs and x > y (cf. [9] , for another type of resolution). We suppose that ( x , y , z ) ≠ ( 1,1,1 ) that is to say that it is non-trivial. It is clear that: x ≡ y = z ≡ 1 ( mod 2 ) . But then:

x 2 + y 2 = 2 z 2 ⇔ ( x − y 2 ) 2 + ( x + y 2 ) 2 = z 2 (1.6)

(with x − y 2 ,     x + y 2 ,     z coprime in pairs and x − y 2 ,     x + y 2 of opposite parity). Let β ∈ { 0,   1 } defined by:

x + ( − 1 ) β y 2 ≡ 0 ( mod 2 ) ⇔ x − ( − 1 ) β y 2 ≡ 1 ( mod 2 ) ; (1.7)

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

( x − ( − 1 ) β y 2 ) 2 +     ( x + ( − 1 ) β y 2 ) 2 = z 2 (1.8)

That is: ( x − ( − 1 ) β y 2 ,     x + ( − 1 ) β y 2 = 2 s b 1 ,   z ) ∈ T + ¯ , (where s ≥ 2 ; b 1 odd).

For such a triplet, we know (cf. Definition 1.2.) That there exists λ ∈ { 0,1 } , such that:

z − x − ( − 1 ) β y 2 2 ≡ λ ( mod 2 ) (1.9)

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

S = s − λ ( s − 1 )     and     S ′ = S − 1 = ( s − 1 ) ( 1 − λ ) . (1.10)

Then, ∃     e ¯ odd, such that we have the expression of the Pythagorician divisors associated with (1.8):

{ ( d , d ″ ) = ( g c d ( x − ( − 1 ) β y 2 ,     z − x + ( − 1 ) β y 2 ) ,     x − ( − 1 ) β y 2 d )     and     x − ( − 1 ) β y 2 = d d ″ ; { e = g c d ( x + ( − 1 ) β y 2 ,     z − x − ( − 1 ) β y 2 ) = 2 S e ¯ ; e ″ = x + ( − 1 ) β y 2 e = 2 s − S b 1 e ¯ .               And     x + ( − 1 ) β y 2 = e e ″ .

In addition we have: { d ″ = d + e = d + 2 S e ¯ ; e ″ = e 2 + d = 2 S − 1 e ¯ + d .

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 = 2 z 2 is solvable; (ii) { x = e ″ 2 − ( e 2 ) 2 + 2 ( e 2 ) e ″ ; y = ( − 1 ) β [ − e ″ 2 + ( e 2 ) 2 + 2 ( e 2 ) e ″ ] z = e ″ 2 + ( e 2 ) 2 . ;

(iii) { x = ( d ″ + d 2 ) 2 − ( d ″ − d 2 ) 2 + 2 ( d ″ − d 2 ) ( d ″ + d 2 ) ; y = ( − 1 ) β [ − ( d ″ + d 2 ) 2 + ( d ″ − d 2 ) 2 + 2 ( d ″ − d 2 ) ( d ″ + d 2 ) ] z = ( d ″ + d 2 ) 2 + ( d ″ − d 2 ) 2 . ;

(iv) { x = 2 × 2 2 S ′ e ¯ 2 + 4 × 2 S ′ e ¯ d + d 2 ; y = ( − 1 ) β [ 2 × 2 2 S ′ e ¯ 2 − d 2 ] ; z = 2 × 2 2 S ′ e ¯ 2 + 2 × 2 S ′ e ¯ d + d 2 .

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: β ∈ { 0,1 } such that: ( − 1 ) β = s i g n ( 2 × 2 2 S ′ e ¯ 2 − d ) 2 i.e. y ≥ 0 .

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

( x ,   y ,   z ) = ( u 2 − v 2 + 2 u v ,   u 2 − v 2 − 2 u v ,   u 2 + v 2 ) when β = 1 , and ( u 2 − v 2 + 2 u v ,   v 2 − u 2 + 2 u v ,   u 2 + v 2 ) when β = 0 , with ( u ,   v ) = ( e ″ ,   e 2 ) = ( d ″ + d 2 ,   d ″ − d 2 ) , such that: u > v , g c d ( u , v ) = 1 , u + v ≡ 1 ( mod 2 ) .

In other words by: ( x , y , z ) = ( u 2 − v 2 + 2 u v ,   | u 2 − v 2 − 2 u v | ,   u 2 + v 2 ) .

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 = 2 z 2 is in bijection with the set ( 2 ℕ + 1 × c o p 2 ℕ + 1 ) × ℕ , as follows:

( x y z ) ↦ f ( d = g c d ( x − ( − 1 ) β y 2 ,   z − x + ( − 1 ) β y 2 ) e ¯ = g c d ( x + ( − 1 ) β y 2 ,   z − x − ( − 1 ) β y 2 ) 2 × 2 S ′ S ′ ) , where β and S ′ are defined in (1.7), (1.9) (1.10).

Whose reciprocal bijection is:

( d e ¯ S ′ ) ↦ f − 1 ( x = 2 × 2 2 S ′ e ¯ 2 + 4 × 2 S ′ e ¯ d + d 2 y = ( − 1 ) β [ 2 × 2 2 S ′ e ¯ 2 − d 2 ] z = 2 × 2 2 S ′ e ¯ 2 + 2 × 2 S ′ e ¯ d + d 2 ) , where β ∈ { 0,1 } and ( − 1 ) β = s i g n ( 2 × 2 2 S ′ e ¯ 2 − d 2 ) .

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 Σ , the set of non-trivial, primitive and positive solutions of x 2 + y 2 = 2 z 2 , then ( x 1 , y 1 , z 1 ) ≅ ( x 2 , y 2 , z 2 ) ⇔ ( d 1 , e ¯ 1 ) = ( d 2 , e ¯ 2 ) define an equivalence relation on Σ , and each class ( x , y , z ) ¯ contains a unique solution ( x 0 , y 0 , z 0 ) for which S ′ 0 = ( s 0 − 1 ) ( 1 − λ 0 ) = 0 i.e. λ 0 = 1 (cf. (1.10)), and which will be its canonical representative. These solutions characterized by S ′ = 0 , are “rare”, they are of the type: ( x = 2 e ¯ 2 + 4 e ¯ d + d 2 y = ( − 1 ) β [ 2 e ¯ 2 − d 2 ] z = 2 e ¯ 2 + 2 e ¯ d + d 2 ) ; their cardinality is equal to: C a r d ( 2 ℕ + 1 × c o p 2 ℕ + 1 ) , while the solutions characterized by S ′ ≥ 1 , have for cardinality: C a r d ( 2 ℕ + 1 × c o p 2 ℕ + 1 ) × C a r d ( ℕ ∗ ) .

Exemples 1.1 Let’s take the minimum possible value ( d , e ¯ ) = ( 1 , 1 ) , we get the minimal solution of x 2 + y 2 = 2 z 2 , that is ( 7,1,5 ) = f − 1 ( 1,1,0 ) , whose class is { f − 1 ( 1,1, S ′ ) , S ′ ∈ ℕ } , i.e.:

( 7,1,5 ) ¯ = { ( 7 , 1 , 5 ) , ( 17,7,13 ) , ( 49,31,41 ) , ( 161,127,145 ) , ( 577,511,545 ) , ⋯ }

➢ We finish this paragraph, using methods similar to that of Pythagorician divisors, to solve the equation x 2 + 2 y 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 , y , z are positive and primitive, therefore necessarily x , z are odd, y ≠ 0 is even and x , y , z are pairwise coprime. Let’s define:

β ∈ { 0,1 }   by : z + ( − 1 ) β x 2 ≡ 0 ( mod 2 ) ,   and   therefore   z − ( − 1 ) β x 2 ≡ 1 ( mod 2 ) . (1.11)

Recall (cf. Reminder 1.1.) That g c d ( z + ( − 1 ) β x 2 ,   z − ( − 1 ) β x 2 ) = 1 .

Then: y 2 = 2 × z + ( − 1 ) β x 2 × z − ( − 1 ) β x 2 .

Hence ∃     a 1 even and a 2 odd, such that y = a 1 a 2 , g c d ( a 1 , a 2 ) = 1 and:

{ z + ( − 1 ) β x 2 = a 1 2 2 ; z − ( − 1 ) β x 2 = a 2 2 .   ⇒   g c d ( y ,     z + ( − 1 ) β x 2 ) = g c d ( a 1 a 2 ,     a 1 2 2 ) = a 1 ;

So, if we set:

e = g c d ( y ,   z + ( − 1 ) β x 2 ) = a 1     and     e ″ = y e = a 2 (1.12)

it follows that:

{ z + ( − 1 ) β x 2 = e 2 2 ; z − ( − 1 ) β x 2 = e ″ 2 .   ⇒   { x = ( − 1 ) β ( e 2 2 − e ″ 2 ) = { e 2 2 − e ″ 2 ,   if   β = 0 ; − e 2 2 + e ″ 2 ,   if   β = 1. y = e e ″ ; z = e ″ 2 + e 2 2 .

To sum up:

If β ∈ { 0,     1 } is defined by: z + ( − 1 ) β x 2 ≡ 0 ( mod 2 ) ⇔ z − ( − 1 ) β x 2 ≡ 1 ( mod 2 ) , and that: g c d ( y ,     z + ( − 1 ) β x 2 ) = e , and 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 + 2 y 2 = z 2 is solvable;

(ii) { x = ( − 1 ) β ( 2 ( e 2 ) 2 − e ″ 2 ) = { 2 ( e 2 ) 2 − e ″ 2 ,   if   β = 0 ; e ″ 2 − 2 ( e 2 ) 2 ,   if   β = 1. y = 2 e ″ ( e 2 ) ; z = e ″ 2 + 2 ( e 2 ) 2 .

Remark 1.6 We have ( x ,   y ,   z ) = ( | u 2 − 2 v 2 | ,     2 u v ,     u 2 + 2 v 2 ) where ( u , v ) = ( e ″ , e 2 ) .

Let ( x , y , z ) be a positive non-trivial and primitive solution x 2 + 2 y 2 = z 2 , then we have:

y = 2 s y 1     and     s ≥ 1, y 1     odd .     Let ’ s     put     S ″ = s − 1 ∈ ℕ . (1.13)

Then since e ″ is odd, there exists odd e ¯ such that: e = 2 s e ¯ with e ¯ e ″ = y 1 from where y = 2 s e ¯ e ″ .

We have the following Proposition:

Proposition 1.9 The set of non-trivial, primitive and positive solutions of the equation x 2 + 2 y 2 = z 2 is in bijection with the set ( 2 ℕ + 1 × c o p 2 ℕ + 1 ) × ℕ , as follows:

( x y z )     ↦   g   ( e ″ = y g c d ( y ,     z + ( − 1 ) β x 2 ) e ¯ = g c d ( y ,     z + ( − 1 ) β x 2 ) 2 × 2 S ″ S ″ ) , where β and S ″ are defined in (1.11) and (1.13).

Whose reciprocal bijection is:

( e ″ e ¯ S ″ )     ↦   g − 1   ( x = ( − 1 ) β ( 2 ( 2 S ″ e ¯ ) 2 − e ″ 2 ) y = 2 e ″ ( 2 S ″ e ¯ ) z = e ″ 2 + 2 ( 2 S ″ e ¯ ) 2 ) , where β ∈ { 0,1 } and ( − 1 ) β = s i g n ( 2 ( 2 S ″ e ¯ ) 2 − e ″ 2 ) .

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:

( x , y , z ) ≅ ( x 1 , y 1 , z 1 ) ⇔ { e ″ = y g c d ( y , z + ( − 1 ) β x 2 ) = y 1 g c d ( y 1 , z 1 + ( − 1 ) β 1 x 1 2 ) = e ″ 1 e ¯ = g c d ( y , z + ( − 1 ) β x 2 ) 2 × 2 S ″ = g c d ( y 1 , z 1 + ( − 1 ) β 1 x 1 2 ) 2 × 2 S ″ 1 = e ¯ 1

As an example, consider the smallest positive and primitive solution ( 1,2,3 ) of this equation, obtained for ( e ″ , e ¯ , S ″ ) = ( 1,1,0 ) , i.e. ( 1,2,3 ) = g − 1 ( 1,1,0 ) , then:

( 1,2,3 ) ¯ = { g − 1 ( 1,1, S ″ ) , S ″ ∈ ℕ } = { ( 1 , 2 , 3 ) , ( 7,4,9 ) , ( 31,8,33 ) , ( 127,16,129 ) , ⋯ }