Definition 2.1 Let n be a positive integer, we say that n is a congruent number if there exists a rational number V such that V 2 − n and V 2 + n are simultaneously rational squares.

See the examples below for concrete cases.

Proposition 2.1 Let n ∈ ℕ ∗ , a square-free natural integer. There is equivalence between the following propositions:

(i) n is congruent.

(ii) ∃ U , V , W ∈ ℚ ∗ such that: { U 2 + n = V 2 ; V 2 + n = W 2 .

(iii) ∃ E , F , G ∈ ℚ ∗ such that: { E 2 + F 2 = G 2 ; 1 2 E F = n .

Proof 1 From Definition 2.1., it is clear that (i) ⇔ (ii).

Let us show that (ii) ⇔ (iii).

Ÿ If (ii) holds, then (iii) holds too with { E = W − U ; F = W + U ; G = 2 V .

Ÿ Conversely if (iii) holds, then (ii) holds too, with { U = F − E 2 ; V = G 2 ; W = F + E 2 .

Concerning equation (iii), we have the following additioning precisions:

Lemma 2.1 Let n ∈ ℕ ⋆ , a square-free natural integer, which is a congruent

number, i.e. such that there exists ( A α , B β , C γ ) a rational triplet, with A α , B β , C γ irreducibles such that: { ( A α ) 2 + ( B β ) 2 = ( C γ ) 2 ; 1 2 A α B β = n .

Then D = g c d ( A , B ) = 1 and δ = g c d ( α , β ) = 1 .

Proof 2 Note that we have: 1 2 A α B β = n , from which we deduce that: ( A β ) ( B α ) = 2 n ( α β ) 2 .

Ÿ Let us show first that D = p g c d ( A , B ) ≠ 2 :

Suppose that the converse holds: D = g c d ( A , B ) = 2 :

Then α and β are odd because A α and B β are irreducibles.

Moreover note that γ is odd too, because C = 2 C ′ is even (with C ′ odd), because otherwise we would have: ( A β γ ) 2 + ( B α γ ) 2 = ( C α β ) 2 odd, with A β γ , B α γ even, which is absurd. So as a consequence C is even and necessarly γ is odd.

Consequently we have:

( A β γ ) 2 + ( B α γ ) 2 = 4 ( C ′ α β ) 2 ⇔ ( A 2 β γ ) 2 + ( B 2 α γ ) 2 = ( C ′ α β ) 2 .

But then: γ ∕ C ′ α β ⇒ ( A 2 β ) 2 + ( B 2 α ) 2 = ( C ′ α β γ ) 2 .

Note that A 2 and B 2 can’t be both even, because g c d ( A , B ) = 2 , and that also A 2 and B 2 can’t be both odd, because if it was the case, we would have 2 ≡ 0 mod 4 . Which is absurd.

So looking to the left side, as exactly one term between A 2 and B 2 is even, then in particular C ′ is odd.

But this would be contradictory since:

1 2 A α B β = n ⇒ 1 2 A B = n α β ⇒ 2 A 2 B 2 = n α β = 2 n ′ α β , where n = 2 n ′ , and n ′

odd without square factors.

But then A 2 B 2 = n ′ α β remains odd ⇒ A 2 and B 2 are both odd, which contradicts the previous fact that exactly one between A 2 and B 2 is necessarily even.

So necessarily D = g c d ( A , B ) ≠ 2 .

Ÿ So D is odd. Suppose that D ≠ 1 , then from ( A β ) ( B α ) = 2 n ( α β ) 2 it comes D 2 / 2 n ( α β ) 2 .

But as g c d ( D , α β ) = 1 ⇒ D 2 / 2 n ⇒ D 2 / n which is absurd, because n is square free.

In conclusion: D = g c d ( A , B ) = 1 .

Otherwise:

D 2 ( A d ) ( B d ) = 2 δ 2 ( α δ ) ( β δ ) n ⇒ A B = 2 δ 2 ( α δ ) ( β δ ) n ⇒ δ / A B ⇒ δ = 1.

Because p g c d ( δ , A B ) = 1 .

We deduce the following theorem:

Theorem 2.1 Let n ∈ ℕ ⋆ a square-free natural integer, there is equivalence between the following propositions:

(i) n is a congruent number.

(ii) There exists f ∈ ℕ ⋆ and ( a , b , c ) ∈ T + ¯ (cf. Definition 2.2.), whose area Δ = 1 2 a b = n f 2 .

(iii) There exists f ∈ ℕ ⋆ and ( a , b , c ) ∈ ℕ ∗ prime in pairs, such that:

{ ( a − b 2 f ) 2 + n = ( c 2 f ) 2 ; ( c 2 f ) 2 + n = ( a + b 2 f ) 2 .

(i.e. ( a − b 2 f ) 2 , ( c 2 f ) 2 , ( a + b 2 f ) 2 are in an arithmetic progression of reason n).

Proof 3

Ÿ Let’s show that (i) ⇒ (ii). i.e. Assume n is congruent number; then there exists ( A α , B β , C γ ) a triple, as in Lemma 2.1., such that:

( ( A α ) 2 + ( B β ) 2 = ( C γ ) 2 ; 1 2 A α B β = n .

Let f ′ = α β γ , and multiply the two equations by f ′ 2 , we obtain:

( ( β γ A ) 2 + ( α γ B ) 2 = ( α β C ) 2 ; 1 2 α β γ 2 A B = n ( α β γ ) 2 . ⇒ ( ( β γ A ) 2 + ( α γ B ) 2 = ( α β C ) 2 ; 1 2 ( β γ A ) ( α γ B ) = n f ′ 2 .

By setting a ′ = β γ A ;   b ′ = α γ B and c ′ = α β C , we have:

( a ′ 2 + b ′ 2 = c ′ 2 ; 1 2 a ′ b ′ = n f ′ 2 .

Let g c d ( a ′ , b ′ , c ′ ) = D ′ , and simply take ( a , b , c ) = ( a ′ D ′ , b ′ D ′ , c ′ D ′ ) ∈ T + ¯ , and f = f ′ D ′ , then the condition (ii) of the Theorem is demonstrated.

Ÿ Let’s show that (ii) ⇒ (iii). Assume (ii) realized, i.e. a 2 + b 2 = c 2 , and 1 2 a b = n f 2 , with a , b , c prime in pair, then:

{ ( a − b 2 f ) 2 + n = a 2 + b 2 − 2 a b + 4 f 2 n 4 f 2 = ( c 2 f ) 2 ; ( c 2 f ) 2 + n = a 2 + b 2 + 4 f 2 n 4 f 2 = a 2 + b 2 + 2 a b 4 f 2 = ( a + b 2 f ) 2 . i.e. (iii) is checked.

Ÿ Let’s show that (iii) ⇒ (i).

If (iii) holds, then Proposition 2.1., (ii) holds too:

With U = a − b 2 f , V = c 2 f and W = a + b 2 f ; so point (i) of Theorem is true

because point (i) of the Proposition 2.1., is true.

In particular, as reminded in the introduction, we obtain formula (1.1), by applying Theorem 2.1., and taking into account the Pythagorician parameterization Theorem 2.5., and thus:

n is congruent ⇔ ∃ ( d , e ¯ , S ) where ( d , e ¯ ) ∈ 2 ℕ + 1 × c o p 2 ℕ + 1 , S ∈ ℕ ∗ such that:

Δ = 2 S − 1 d e ¯ ( d + 2 S − 1 e ¯ ) ( d + 2 S e ¯ ) = n f 2 .

However, for a given n, we do not know (in general) if such a rational triplet exists, and if we do know it, we do not know from which primitive integer Pythagorician triplet it comes.

Exemples 2.1

Ÿ Let us now consider these primitive integer Pythagorean triples as they vary and calculate the areas of the integer right-angled triangles a 2 + b 2 = c 2 such that Δ = 1 2 a b = n f 2 where n is a squareless integer, which means that: n is a congruent number:

We therefore see that the number 5 is congruent: Indeed very quickly, we obtain:

9 2 + 40 2 = 41 2 with: 1 2 × 9 × 40 = 180 = 5 × 6 2 .

In this example, the first right-angled integer triplet whose area is 5 × f 2 , is quite close in the list, and the corresponding rational right-angled triangle is:

( 3 2 ) 2 + ( 20 3 ) 2 = ( 41 6 ) 2 , with: 1 2 × 3 2 × 20 3 = 5 .

Compared to the Pythagoras equation with integer solutions: 9 2 + 40 2 = 41 2 , One find that

The squares ( ( b − a 2 ) 2 , ( c 2 ) 2 , ( b + a 2 ) 2 ) = ( ( 31 2 ) 2 , ( 49 2 ) 2 , ( 41 2 ) 2 ) , are in arithmetic progression of reason 180, We deduce from that with respect to the following Pythagorician rational equation: ( 3 2 ) 2 + ( 20 3 ) 2 = ( 41 6 ) 2 , that the squares ( 31 12 ) 2 ; ( 49 12 ) 2 and ( 41 12 ) 2 , are in arithmetic progression with reason 5 = 180 36 .

Ÿ There are, however, complicated cases, because, although existing, the primitive right triangles we are looking for could be very far down the list: As a specific example, the number 23 turns out to be a congruent one:

And we find a primitive integer right-angled triangle ( a , b , c ) checking:

1 2 a b = 23 f 2 with f ∈ ℕ ⋆ .

But this time, this one is far down the list. Indeed, we find:

279340175 2 + 860959008 2 = 905141617 2 ; with Δ 1 = 1 2 a b = 23 × 72306780 2 .

And the corresponding rational right-angled triangle is:

( 80155 20748 ) 2 + ( 41496 3485 ) 2 = ( 905141617 72306780 ) 2 , with: Δ 2 = 1 2 × 80155 20748 × 41496 3485 = 23 .

Which is also a caracterization cf. Proposition 2.1. (iii), so that 23 is congruent.

Thus (cf. Definition 2.1. & Theorem 2.1., (iii)) the squares in arithmetic progression of reason 23, are:

( ( b − a 2 f ) 2 , ( c 2 f ) 2 , ( b + a 2 f ) 2 ) = ( ( 581618833 144613560 ) 2 , ( 905141617 144613560 ) 2 , ( 1140299183 144613560 ) 2 )

Thus, constructing with large calculators, tables of congruent integers n without squares and checking (with usual Pythagorician triplets parameterization):

u v ( u 2 − v 2 ) = n f 2 , where u + v ≡ 1 ( mod 2 ) and g c d ( u , v ) = 1 ,

is a priori feasible; but show that a given square-free integer n, is congruent or not, is a very difficult problem, and which remains open to this day, despite progress due to numerous works, based, among others, on elliptic curves, and the BSD conjecture cf. [8] [9] [15] [16] [17] .

Remark 2.1 The problem of determining congruent numbers, or properties concerning them, remains open. For example, let n and m be two congruent numbers, we can ask ourselves if nm can also be congruent ? In [18] , pp 44 and 65, this problem is stated, and certain families of congruent numbers n and m such that nm are congruents, are proposed.

Consider n, a congruent number, then cf. Proposition 2.1.: ∃ E , F , G ∈ ℚ ∗ , such that: { E 2 + F 2 = G 2 1 2 E F = n

But then by setting: x = ( G 2 ) 2 et y = ( E 2 − F 2 ) G 8 , the following elliptic curve:

y 2 = x 3 − n 2 x , admits rational points. We recall Tunnel’s theorem (1983) cf. [8] :

Theorem 2.2 Let n ∈ ℕ , and square free. Consider then the following cardinals:

A n = C a r d ( { x , y , z ∈ ℤ   such   that : n = 2 x 2 + y 2 + 32 z 2 } ) ;

B n = C a r d ( { x , y , z ∈ ℤ   such   that : n = 2 x 2 + y 2 + 8 z 2 } ) ;

C n = C a r d ( { x , y , z ∈ ℤ   such   that : n = 4 x 2 + 2 y 2 + 6 z 2 } ) ;

D n = C a r d ( { x , y , z ∈ ℤ   such   that : n = 8 x 2 + 2 y 2 + 16 z 2 } ) .

Ÿ If we assume that n is a congruent number then necessarily:

A n = B n if n is even and 2 C n = D n if n is odd.

Ÿ Conversely if these equalities between cardinals hold, as well as the conjecture of Birch and Swinnerton-Dyer (cf. [9] and [10] ), for the elliptic curve: y 2 = x 3 − n 2 x , then n is a congruent number. See also [17] for recent results using these methods.

Proposition 2.2 There is equivalence between the following two propositions:

(i) ∀ ( a , b , c ) ∈ ℕ 3 , a b c ≠ 0 , g c d ( a , b ) = 1 and a 2 + b 2 = c 2 then 1 2 a b ≠ f 2 , ∀ f ∈ ℕ ⋆ .

(ii) ∀ ( a , b , c ) ∈ T + ¯ then 1 2 a b ≠ f 2 , ∀ f ∈ ℕ ⋆ .

Consequently, to solve the problem of Fermat’s right-angled triangle theorem, it suffices to prove it only for right-angled triangles of T + ¯ .

Let us now show the following theorem, which is one of the aims of the article: To say that 1 is not a congruent number, amounts to solving equation (1.1) with n = 1 .

Theorem 2.3 Let ( a , b , c ) ∈ T + ¯ then 1 2 a b ≠ f 2 , ∀ f ∈ ℕ ⋆ .

Proof 4 We have cf. Theorem 2.5. ∃ ! ( ( d , e ¯ ) , S ) ∈ ( 2 ℕ + 1 × c o p 2 ℕ + 1 ) × ℕ ⋆ such that:

( 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 ) .

We then show that equation (1.1) for n = 1 , is not solvable:

Δ = 2 S − 1 d e ¯ ( d + 2 S − 1 e ¯ ) ( d + 2 S e ¯ ) = f 2 .

We distinguish 2 cases according to the parity of S:

The 1^st case: S = 2 k , k ≥ 1 (which is trivial);

And the 2^th case S = 2 k + 1 , k ≥ 0 .

In this latter case, we will distinguish S = 1 odd, (which is also trivial); from case S = 2 k + 1 ≥ 3 odd, which is more complex. Let’s demonstrate.

Ÿ 1^st case: S = 2 k is even, k ≥ 1 . Then:

( a , b , c ) = ( d 2 + ( 2 2 k e ¯ ) d , ( 2 2 k e ¯ ) 2 2 + ( 2 2 k e ¯ ) d , ( 2 2 k e ¯ ) 2 2 + ( 2 2 k e ¯ ) d + d 2 ) .

Suppose there exists f ∈ ℕ ⋆ ∕ 1 2 a b = f 2 , then cf. (1.1) we get:

2 2 k − 1 d e ¯ ( d + 2 2 k − 1 e ¯ ) ( d + 2 2 k e ¯ ) = f 2 ⇒

2 ( e ¯ d + 2 2 k − 1 e ¯ 2 ) ( d 2 + ( 2 2 k e ¯ ) d ) = ( f 2 k − 1 ) 2 ⇒ 2 ≡ 0 ( mod 4 ) ,

which is absurd.

So 1 2 a b ≠ f 2 , ∀ f ∈ ℕ ⋆ , in this first case .

Ÿ 2^th case

- 1^er Subcase: S is odd, and S = 1 (i.e. k = 0).

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

Suppose there exists f ∈ ℕ ⋆ ∕ 1 2 a b = f 2 , then cf. (1.1) we have:

d e ¯ ( d + e ¯ ) ( d + 2 e ¯ ) = f 2 .

Note that f is necessarily even, and that the four factors are pairwise prime.

Then ∃ D 1 , E 1 , f ′ 1 ∈ 2 ℕ + 1 and f 1 ∈ ℕ ⋆ , Such that:

{ d = D 1 2 ; e ¯ = E 1 2 ; D 1 2 + E 1 2 = ( 2 f 1 ) 2 ; D 1 2 + 2 E 1 2 = f ′ 1 2 . (2.1)

But the 3^th equation ⇒ 2 ≡ 0 ( mod 4 ) , which is absurd.

Thus 1 2 a b ≠ f 2 , ∀ f ∈ ℕ ⋆ .

- 2^th Subcase: S is odd, and S = 2 k + 1 ≥ 3 .

( a , b , c ) = ( d 2 + ( 2 2 k + 1 e ¯ ) d , ( 2 2 k + 1 e ¯ ) 2 2 + ( 2 2 k + 1 e ¯ ) d , ( 2 2 k + 1 e ¯ ) 2 2 + ( 2 2 k + 1 e ¯ ) d + d 2 ) .

Suppose, for absurdity, that as above there exists: f ∈ ℕ ⋆ ∕ 1 2 a b = f 2 .

Then: d e ¯ ( d + 2 2 k e ¯ ) ( d + 2 2 k + 1 e ¯ ) = ( f 2 k ) 2 ≡ 1 ( mod 2 ) .

⇒ ∃ pairwise prime integers D k , E k , f k , f ′ k ∈ 2 ℕ + 1 such that:

{ d = D k 2 ; e ¯ = E k 2 ; D k 2 + ( 2 k E k ) 2 = f k 2 ; D k 2 + 2 ( 2 k E k ) 2 = f ′ k 2 . (2.2)

We then recognize the non-solvable system ( Ξ 0, k ) of Lemma 2.2. (with here k ≥ 1 ), which is found here, to be solvable, which is absurd.

Thus 1 is not congruent.

We now use similar approach to show that 2 is not congruent.

That is, to prove the following theorem:

Theorem 2.4 Let ( a , b , c ) ∈ T + ¯ Then 1 2 a b ≠ 2 f 2 , ∀ f ∈ ℕ ⋆ .

Proof 5 Cf. Theorem 2.5., ∃ ! ( ( d , e ¯ ) , S ) ∈ ( 2 ℕ + 1 × c o p 2 ℕ + 1 ) × ℕ ⋆ such that:

( 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 ) .

We then seek to solve (1.1) with n = 2 ; that is:

Δ = 2 S − 1 d e ¯ ( d + 2 S − 1 e ¯ ) ( d + 2 S e ¯ ) = 2 f 2 .

Let us show that this is impossible, by reasoning taking into account the parity of S: The 1^st case, wich indeed trivial, corresponds to S = 1 and S = 2 k + 1 ≥ 3 , odd, and of is the same type as in the previous 1^st case.

Consequently, we can therefore solve the remaining case which corresponds to:

2^th case: S = 2 k , even, with k ≥ 1 .

∃ ! ( ( d , e ¯ ) , S ) ∈ ( 2 ℕ + 1 × c o p 2 ℕ + 1 ) × 2 ℕ ⋆ such that:

( a , b , c ) = ( d 2 + ( 2 2 k e ¯ ) d , ( 2 2 k e ¯ ) 2 2 + ( 2 2 k e ¯ ) d , ( 2 2 k e ¯ ) 2 2 + ( 2 2 k e ¯ ) d + d 2 ) .

Suppose by absurd that there exists f ∈ ℕ ⋆ such that 1 2 a b = 2 f 2 .

Then cf. (1.1) : 2 2 k − 1 d e ¯ ( d + 2 2 k − 1 e ¯ ) ( d + 2 2 k e ¯ ) = 2 f 2 ⇒

d e ¯ ( d + 2 2 k − 1 e ¯ ) ( d + 2 2 k e ¯ ) = ( f 2 k − 1 ) 2 ≡ 1 ( mod 2 ) .

From where:

∃ D k − 1 , E k − 1 , f k − 1 , f ′ k − 1 ∈ 2 ℕ + 1 such that:

{ d = D k − 1 2 ; e ¯ = E k − 1 2 ; D k − 1 2 + 2 ( 2 k − 1 E k − 1 ) 2 = f k − 1 2 ; D k − 1 2 + 4 ( 2 k − 1 E k − 1 ) 2 = f ′ k − 1 2 . (2.3)

But here we recognize the non-solvable system ( Ξ 1, k − 1 ) of Lemma 2.2., (with here k ≥ 1 ), which is absurd. This shows that equation (1.1) is unsolvable in this case.

So as we had claimed: 1 2 a b ≠ 2 f 2 , ∀ f ∈ ℕ ⋆ .

That is to say that the number 2 is not congruent.

Let us now state and establish the following Lemmas, which allowed our two previous proofs.

Lemma 2.2 Let t ∈ { 0,1 } , and ∀ k ∈ ℕ , then the system of equations ( Ξ t , k ) { X 2 + 2 t ( 2 k Y ) 2 = Z 2 , X 2 + 2 t + 1 ( 2 k Y ) 2 = T 2 ; does not admit any pairwise prime solution

( X , Y , Z , T ) = ( D k , E k , f k , f ′ k ) ∈ ( 2 ℕ + 1 ) 4 .

Remark 2.2 a) When t = 0 : It is clear that ( Ξ 0 , k ) ⇔ { T 2 + ( 2 k Y ) 2 = Z 2 , T 2 − ( 2 k Y ) 2 = X 2 ;

which is not solvable, because it is well known that the sum and the difference of two squares cannot both be squares.

b) When t = 1 : It is clear that ( Ξ 1 , k ) ⇔ { T 2 + X 2 = 2 Z 2 , T 2 − X 2 = ( 2 k + 1 Y ) 2 ; is not solvable,

because we have:

T = u 2 + v 2 , X = u 2 − v 2 , 2 k + 1 Y = 2 u v ⇒ T 2 + X 2 = 2 Z 2 ⇒ u 4 + v 4 = Z 2 which is not resolvable ( [19] ).

However, we are going to propose another proof of this lemma.

Proof 6 CASE 1 : t = 0 .

we will demonstrate by induction on k, the proposition :

{ X 2 + ( 2 k Y ) 2 = Z 2 , X 2 + 2 ( 2 k Y ) 2 = T 2 ; does not admit any pairwise prime solution ( X , Y , Z , T ) = ( D k , E k , f k , f ′ k ) ∈ ( 2 ℕ + 1 ) 4 .

Ÿ Note first that the proposition is true for k = 0 and k = 1 .

Indeed, if k = 0 (resp. k = 1 ), then ( Ξ 0,0 ) , ( Ξ 0,1 ) does not admit solutions in ( 2 ℕ + 1 ) 4 , because otherwise we would have: D 0 2 + E 0 2 = f 0 2 (resp. D 1 2 + ( 2 E 1 ) 2 = f 1 2 ) which in both cases is impossible.

Ÿ Assume now that k ≥ 2 , and let’s suppose that this proposition is true until rank k − 1 , What forms our recurrence hypothesis:

( Ξ 0, l ) does not admit, up to rank k − 1 , any pairwise prime solution in

( 2 ℕ + 1 ) 4 (2.4)

Ÿ Let’s show that ( Ξ 0, k ) does not admit too, any pairwise prime solution in ( 2 ℕ + 1 ) 4 .

Suppose the converse, that is: ∃ ( D k , E k , f k , f ′ k ) ∈ ( 2 ℕ + 1 ) 4 , such that:

( Ξ 0, k ) { X 2 + ( 2 k Y ) 2 = Z 2 , X 2 + 2 ( 2 k Y ) 2 = T 2 ; is solvable in ( X , Y , Z , T ) = ( D k , E k , f k , f ′ k ) ∈ ( 2 ℕ + 1 ) 4 , and pairwise prime.

As a consequence: ( D k ,2 k E k , f ′ k ) is a non-trivial and positive solution of x 2 + 2 y 2 = z 2 , then (cf. Proposition 2.5.):

∃ ! ( e ¯ k − 1 , e ″ k − 1 ) ∈ ( 2 ℕ + 1 ) × c o p ( 2 ℕ + 1 ) ; and S 0 ∈ ℕ , such that:

{ D k = ( − 1 ) β ( 2 ( 2 S 0 e ¯ k − 1 ) 2 − e ″ k − 1 2 ) = ( − 1 ) β ( 2 2 k − 1 e ¯ k − 1 2 − e ″ k − 1 2 ) ; 2 k E k = 2 e ″ k − 1 ( 2 S 0 e ¯ k − 1 ) = e ″ k − 1 ( 2 k e ¯ k − 1 ) ; f ′ k = e ″ k − 1 2 + 2 ( 2 S 0 e ¯ k − 1 ) 2 = e ″ k − 1 2 + 2 2 k − 1 e ¯ k − 1 2 .

Because 2 k E k = 2 e ″ k − 1 ( 2 S 0 e ¯ k − 1 ) ⇒ S 0 = k − 1 ,

which implies:

( D k ,2 k E k , f ′ k ) = ( ( − 1 ) β × ( 2 2 k − 1 e ¯ k − 1 2 − e ″ k − 1 2 ) ,2 k e ″ k − 1 e ¯ k − 1 ,2 2 k − 1 e ¯ k − 1 2 + e ″ k − 1 2 ) .

But then:

D k 2 = ( 2 2 k − 1 e ¯ k − 1 2 − e ″ k − 1 2 ) 2

= e ″ k − 1 4 − 2 2 k e ¯ k − 1 2 e ″ k − 1 2 + ( 2 2 k − 1 e ¯ k − 1 2 ) 2 ; but as ( 2 k E k ) 2 = 2 2 k e ¯ k − 1 2 e ″ k − 1 2 .

= e ″ k − 1 4 − ( 2 k E k ) 2 + ( 2 2 k − 1 e ¯ k − 1 2 ) 2

⇒ D k 2 + ( 2 k E k ) 2 = e ″ k − 1 4 + ( 2 2 k − 1 e ¯ k − 1 2 ) 2 .

But ( Ξ 0, k ) solvable ⇒ D k 2 + ( 2 k E k ) 2 = f k 2 ⇒ e ″ k − 1 4 + ( 2 2 k − 1 e ¯ k − 1 2 ) 2 = f k 2 .

Thereby:

e ″ k − 1 4 = f k 2 − ( 2 2 k − 1 e ¯ k − 1 2 ) 2 ⇒ e ″ k − 1 4 = ( f k − 2 2 k − 1 e ¯ k − 1 2 ) ( f k + 2 2 k − 1 e ¯ k − 1 2 ) .

These two factors being coprimes, there exists odd coprimes numbers f ′ k − 1 , D k − 1 , such that: { D k − 1 4 = f k − 2 2 k − 1 e ¯ k − 1 2 ; f ′ k − 1 4 = f k + 2 2 k − 1 e ¯ k − 1 2 .

As a result:

f k = f ′ k − 1 4 + D k − 1 4 2 and 2 2 k − 1 e ¯ k − 1 2 = f ′ k − 1 4 − D k − 1 4 2 , from where : 2 2 k e ¯ k − 1 2 = f ′ k − 1 4 − D k − 1 4

Thus: D k − 1 4 + 2 2 k e ¯ k − 1 2 = f ′ k − 1 4 ;

Let’s rewrite that like this: ( D k − 1 2 ) 2 + ( 2 k e ¯ k − 1 ) 2 = ( f ′ k − 1 2 ) 2 , that is a Pythagoras equation:

Thus ( D k − 1 2 ,2 k e ¯ k − 1 , f ′ k − 1 2 ) is a Pythagorician triplet of T + ¯ . (Note that then necessarily k ≥ 2 ).

If we denote ( 2 k E k − 1 , f k − 1 ) as the Pythagorician divisors coming from 2 k e ¯ k − 1 (cf. Definition 2.3.), with respect that here λ = 0 , then:

2 k E k − 1 = g c d ( 2 k e ¯ k − 1 , f k − 1 2 − D k − 1 2 ) , and 2 k E k − 1 f k − 1 = 2 k e ¯ k − 1 .

And the Pythagorician divisors theorem (cf. Theorem 2.5.) implies that:

{ f ′ k − 1 2 − D k − 1 2 = ( 2 k E k − 1 ) 2 2 = 2 ( 2 k − 1 E k − 1 ) 2 ; f ′ k − 1 2 + D k − 1 2 = 2 f k − 1 2 .

We therefore obtain on the one hand:

D k − 1 2 + 2 ( 2 k − 1 E k − 1 ) 2 = f ′ k − 1 2 .

And on the other hand:

( D k − 1 2 + 2 ( 2 k − 1 E k − 1 ) 2 ) + D k − 1 2 = 2 f k − 1 2 ⇒ D k − 1 2 + ( 2 k − 1 E k − 1 ) 2 = f k − 1 2 .

And as a result we have:

{ D k − 1 2 + ( 2 k − 1 E k − 1 ) 2 = f k − 1 2 ; D k − 1 2 + 2 ( 2 k − 1 E k − 1 ) 2 = f ′ k − 1 2 . ⇒ ( Ξ 0, k − 1 ) is solvable, which contradicts the assumption of recurrence assumed to be true up to rank k − 1 .

So the Lemma when t = 0 , is proved. ■

CASE 2 : t = 1 .

In this case too, we will demonstrate by induction on k, the proposition:

{ X 2 + 2 ( 2 k Y ) 2 = Z 2 , X 2 + 4 ( 2 k Y ) 2 = T 2 ; does not admit any pairwise prime solution ( X , Y , Z , T ) = ( D k , E k , f k , f ′ k ) ∈ ( 2 ℕ + 1 ) 4 .

Ÿ This proposition is true for k = 0 . Because if not, then ( Ξ 1,0 ) should admit solutions in ( 2 ℕ + 1 ) 4 , consequently we would have:

D 0 2 + 2 E 0 2 = f 0 2 ⇒ 3 ≡ 1 ( mod 4 ) , which is absurd.

Ÿ Assume now that k ≥ 1 , and let’s suppose that this proposition is true until rank k − 1 , What forms our recurrence hypothesis:

( Ξ 1, l ) does not admit, up to rank k − 1 , any pairwise prime solution in

( 2 ℕ + 1 ) 4 (2.5)

Ÿ Let’s show that ( Ξ 1, k ) does not admit too, any pairwise prime solution in ( 2 ℕ + 1 ) 4 .

Suppose the converse, that is: ∃ ( D k , E k , f k , f ′ k ) ∈ ( 2 ℕ + 1 ) 4 , such that:

{ D k 2 + 2 ( 2 k E k ) 2 = f k 2 ; D k 2 + 4 ( 2 k E k ) 2 = f ′ k 2 . (2.6)

Let’s solve by factorization the equation D k 2 + 4 ( 2 k E k ) 2 = f ′ k 2 .

Define β ∈ { 0,1 } such that f ′ k + ( − 1 ) β D k 2 ≡ 0 ( mod 2 ) .

Then: ∃ ( e ¯ k − 1 , e ″ k − 1 ) ∈ ( 2 ℕ + 1 ) × c o p ( 2 ℕ + 1 ) such that:

f ′ k + ( − 1 ) β D k 2 × f ′ k − ( − 1 ) β D k 2 = ( 2 k E k ) 2 ⇒ { f ′ k + ( − 1 ) β D k 2 = 2 2 k e ¯ k − 1 2 ; 2 k E k = 2 k e ¯ k − 1 e ″ k − 1 ; f ′ k − ( − 1 ) β D k 2 = e ″ k − 1 2 .

Then:

{ D k = ( − 1 ) β ( 2 2 k e ¯ k − 1 2 − e ″ k − 1 2 ) ; 2 k E k = 2 k e ¯ k − 1 e ″ k − 1 ; f ′ k = 2 2 k e ¯ k − 1 2 + e ″ k − 1 2 .

⇒ ( D k ) 2 = ( 2 2 k e ¯ k − 1 2 ) 2 + ( e ″ k − 1 2 ) 2 − 2 ( 2 k e ¯ k − 1 e ″ k − 1 ) 2                               = ( 2 2 k e ¯ k − 1 2 ) 2 + ( e ″ k − 1 2 ) 2 − 2 ( 2 k E k ) 2 ;

⇒ ( D k ) 2 + 2 ( 2 k E k ) 2 = recurrencehypothesis f k 2 = ( 2 2 k e ¯ k − 1 2 ) 2 + ( e ″ k − 1 2 ) 2 ;

⇒ f k 2 − ( 2 2 k e ¯ k − 1 2 ) 2 = ( e ″ k − 1 2 ) 2 ;

⇒ ( f k − 2 2 k e ¯ k − 1 2 ) ( f k + 2 2 k e ¯ k − 1 2 ) = e ′ k − 1 4 ;

Then ∃ ( D k − 1 , f ′ k − 1 ) ∈ ( 2 ℕ + 1 ) × c o p ( 2 ℕ + 1 ) such that:

{ D k − 1 4 = f k − 2 2 k e ¯ k − 1 2 , f ′ k − 1 4 = f k + 2 2 k e ¯ k − 1 2 ; ⇒ ( f ′ k − 1 2 ) 2 − ( D k − 1 2 ) 2 = 2 ( 2 2 k e ¯ k − 1 2 ) ;

And ∃ ( E k − 1 , f k − 1 ) ∈ ( 2 ℕ + 1 ) × c o p ( 2 ℕ + 1 ) such that:

⇒ ( f ′ k − 1 2 − D k − 1 2 ) f ′ k − 1 2 + D k − 1 2 2 = 2 2 k e ¯ k − 1 2 ⇒ { f ′ k − 1 2 − D k − 1 2 = 2 2 k E k − 1 2 ; f ′ k − 1 2 + D k − 1 2 = 2 f k − 1 2 . ⇒

D k − 1 2 + ( 2 k E k − 1 ) 2 = f ′ k − 1 2 which can be rewritten in one hand: D k − 1 2 + 4 ( 2 k − 1 E k − 1 ) 2 = f ′ k − 1 2 ;

and, on the other hand:

f ′ k − 1 2 + D k − 1 2 = 2 f k − 1 2 ⇒ ( D k − 1 2 + 4 ( 2 k − 1 E k − 1 ) 2 ) + D k − 1 2 = 2 f k − 1 2 ⇒

D k − 1 2 + 2 ( 2 k − 1 E k − 1 ) 2 = f k − 1 2 .

So as a result:

⇒ { D k − 1 2 + 2 ( 2 k − 1 E k − 1 ) 2 = f k − 1 2 ; D k − 1 2 + 4 ( 2 k − 1 E k − 1 ) 2 = f ′ k − 1 2 . ⇒ ( Ξ 1, k − 1 ) is solvable, which contradicts the

assumption of recurrence assumed to be true up to rank k − 1 .

So the Lemma is proved. ■

Note that these two Lemmas induce the following Corollary:

Corollary 2.1 ∀ k , t ∈ ℕ , the systems of equations ( Ξ t , 0 ) { X 2 + 2 t ( 2 k Y ) 2 = Z 2 , X 2 + 2 t + 1 ( 2 k Y ) 2 = T 2 ; don’t admit any solutions in ( X , Y , Z , T ) = ( D k , E k , f k , f ′ k ) ∈ ( 2 ℕ + 1 ) 4 , and pairwise prime.

Proof 7 It suffices to consider the cases t even (resp. t odd), which lead in one, or the other of the cases, of the previous lemmas.

Let us recall here some results and definitions used in various proofs above.

All these concepts, definitions and proofs are found in [13] or [14] .

Convention 2.1 Let ( a , b , c ) ≠ ( 0,0,0 ) be a solution of the Pythagoras equation: a 2 + b 2 = c 2 . We agree for the following, unless otherwise stated, that ( a , b , c ) ≡ ( ± 1,0,1 ) ( mod 4 ) .

This in no way restricts the expression of the generality of the solutions of said equation, because ( b , a , c ) is also a solution called “associated with ( a , b , c ) ”, such that ( b , a , c ) ≡ ( 0, ± 1,1 ) ( mod 4 ) .

Definition 2.2 T + ¯ : is the set of non-trivial, primitive and positive Pythagoras solutions of the type ( a , b = 2 s b 1 , c ) ≡ ( ± 1 , 0 , 1 ) ( mod 4 ) .

It is well known that:

Proposition 2.3

T + ¯ = { ( u 2 − v 2 ,2 u v , u 2 + v 2 ) ; u , v ∈ ℕ , u > v > 0, u + v ≡ 1 ( mod 2 )   and   g c d ( u , v ) = 1 } .

Definition 2.3 Let ( a , b = 2 s b 1 , c ) ∈ T + ¯ , s ≥ 2 , b 1 odd.

-Denote by ( d , d ″ ) = ( g c d ( a , c − b ) , a d ) and ( e , e ″ ) = ( g c d ( b , c − a ) , b e ) = ( 2 S e ¯ ,2 s − S b 1 e ¯ ) , this quantities are the Pythagorician

divisors of ( a , b , c ) . The first ones are said coming from a, and the second coming from b.

-The number S ∈ ℕ ∗ , is defined by: S = s − λ ( s − 1 ) , where λ ∈ { 0,1 } is such that: c − a 2 ≡ λ ( mod 2 ) , and e ¯ is the suitable odd integer.

Let’s recall cf. [13] [14] .

Theorem 2.5 (Of Pythagorician divisors). Let ( a , b = 2 s b 1 , c ) ∈ T + ¯ , s ≥ 2 , b 1 odd.

There is equivalences between the following propositions:

(i) a 2 + b 2 = c 2 .

(ii) { c − b = d 2 ; c + b = d ″ 2 .

(iii) { c − a = e 2 2 = ( 2 S e ¯ ) 2 2 ; c + a = 2 e ″ 2 = 2 ( 2 s − S b 1 e ¯ ) 2 .

(iv) { a = d 2 + ( 2 S e ¯ ) d ; b = ( 2 S e ¯ ) 2 2 + ( 2 S e ¯ ) d ; c = ( 2 S e ¯ ) 2 2 + ( 2 S e ¯ ) d + d 2 .

Remark 2.3 About the Pythagirican divisors’s Theorem one can says that ( a , b = 2 s b 1 ≠ 0 , c ) ∈ T + ¯ , is a Pythagorician triplet if and only if ∃ d odd divisor of a, ∃ e even divisor of b such that: { a d = d + e ; b e = d + e 2 ; c = a + e 2 2 = b + d 2 .

Let’s put:

Definition 2.4 2 ℕ + 1 × c o p 2 ℕ + 1 = { ( x , y ) ∈ ( 2 ℕ + 1 ) 2 / g c d ( x , y ) = 1 } .

We have the following corollary:

Corollary 2.2 T + ¯ is in bijection with: ( 2 ℕ + 1 × c o p 2 ℕ + 1 ) × ℕ ∗ , as follows:

π : T + ¯ → ( 2 ℕ + 1 × c o p 2 ℕ + 1 ) × ℕ ∗         ( a b c ) ↦ ( d , e ¯ , S ) ( a , b , c ) = ( d = g c d ( a , c − b ) e ¯ = g c d ( b , c − a ) 2 S S )

Where S = s − λ ( s − 1 ) with s = v 2 ( b ) and λ defined in Definition 2.3.;

Whose reciprocal bijection is:

π − 1 : ( 2 ℕ + 1 × c o p 2 ℕ + 1 ) × ℕ ∗ → T + ¯               ( d e ¯ S ) ↦ ( a , b , c ) ( d , e ¯ , S ) = ( a = d 2 + ( 2 S e ¯ ) d b = ( 2 S e ¯ ) 2 2 + ( 2 S e ¯ ) d c = ( 2 S e ¯ ) 2 2 + ( 2 S e ¯ ) d + d 2 )

In particular we get the proposition:

Proposition 2.4

T + ¯ = { ( 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 ) / ( d , e ¯ )                 ∈ 2 ℕ + 1 × c o p 2 ℕ + 1   and   S ∈ ℕ ∗ } .

From all the above, we have:

Theorem 2.6 Let ( a , b , c ) ∈ T + ¯ , whose area is Δ = 1 2 a b . Then ∃ ! ( d , e ¯ , S ) ∈ ( 2 ℕ + 1 × c o p 2 ℕ + 1 ) × ℕ ∗ , such that:

1) { a = d 2 + ( 2 S e ¯ ) d ; b = ( 2 S e ¯ ) 2 2 + ( 2 S e ¯ ) d ; c = ( 2 S e ¯ ) 2 2 + ( 2 S e ¯ ) d + d 2 .

2) Δ = 1 2 a b = 2 S − 1 d e ¯ ( d + 2 S e ¯ ) ( 2 S − 1 e ¯ + d ) .

With: 2 S − 1 , d , e ¯ , ( d + 2 S e ¯ ) , ( 2 S − 1 e ¯ + d ) , pairwise prime.

Remark 2.4 Other interesting parametrizations of Pythagorician triplets can be found in [20] , [21] or [22] .

We now giving the positive, non-null and non-trivial primitive solutions (i.e. ≠ ( 1,0,1 ) ), of the equation x 2 + 2 y 2 = z 2 , cf. [14] . These solutions are necessary to solve the Fermat’s Theorem of the right-angled triangle, and that of the non-congruence of 2.

We get the following theorem cf. [13] [14] :

Proposition 2.5 There is an equivalence between the following propositions (the solutions are supposed to be non-null, non-trivial, primitive and positive).

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

(ii) ∃ e ″ , e ¯ coprimes, such that:

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

Remark 2.5 1) β ∈ { 0,1 } is defined by: z + ( − 1 ) β x 2 ≡ 0 ( mod 2 ) .

2) e ″ is an odd integer, as well as e ¯ , and are such that: e = 2 s e ¯ , and so y = 2 s e ¯ e ″ .

{ 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 ″ = s − 1 ∈ ℕ .

The Pythagorician divisors theorem, gives a new, simple original proof of Fermat’s right-angled triangle theorem, and that number 2 is not a congruent number. We do believe, that such a method seems to be usable for the general problem of congruent numbers, namely for other values of square-free integers n, and in particular for those where: n = p prime. With this method, we have been able to show that any prime number p ≡ 3 ( mod 8 ) , is not congruent; and that any prime number p ≡ 5 , 7 ( mod 8 ) are congruents. We seek to show, with our method, the resolution of the problem of congruent numbers for the prime numbers p ≡ 1 ( mod 8 ) . Note that for this last case, using the theory of elliptic curves, Evink, Top and Top, J., D. cf. [17] , showed on the one hand, that there exists an infinity of prime numbers p ≡ 1 ( mod 8 ) , which are not congruent (these being of

density > 1 8 ), and on the other hand, that the infinity of the prime numbers of

f ( ℤ ) are congruents, where f ( x ) = 8 x 4 + 16 x 3 + 12 x 2 + 4 x + 1 , is a polynomial verifying the conjecture of Bouniakowsky, see also [23] [24] , for related results.