RETRACTED: On the Nature of Some Euler’s Double Equations Equivalent to Fermat’s Last Theorem

Abstract

In this work, I provide a new rephrasing of Fermat’s Last Theorem, based on an earlier work by Euler on the ternary quadratic forms. Effectively, Fermat’s Last Theorem can be derived from an appropriate use of the concordant forms of Euler and from an equivalent ternary quadratic homogeneous Diophantine equation able to accommodate a solution of Fermat’s extraordinary equation. Following a similar and almost identical approach to that of A. Wiles, I tried to translate the link between Euler’s double equations (concordant/discordant forms) and Fermat’s Last Theorem into a possible reformulation of the Fermat Theorem. More precisely, through the aid of a Diophantine equation of second degree, homogeneous and ternary, solved not directly, but as a consequence of the resolution of the double Euler equations that originated it, I was able to obtain the following result: the intersection of the infinite solutions of Euler’s double equations gives rise to an empty set and this only by exploiting a well-known Legendre Theorem, which concerns the properties of all the Diophantine equations of the second degree, homogeneous and ternary. The impossibility of solving the second degree Diophantine equation thus obtained is possible using well-known techniques at the end of 18th century (see Euler, Lagrange and Legendre) and perhaps present in Fermat’s brilliant mind.

1. Introduction

Fermat’s last theorem affirms: If n is an integer, greater than 2, there are not any positive integers $X,Y,Z$, so that it can be valid: $X^n+Y^n=Z^n.$

Fermat himself proved it for n = 4 ([1], pp. 108–112), ([2], II, Chap. XIII, § 202–209); it is consequent its validity also for n as a multiple of 4, because, if n is equal $4p$, for some positive integer p,

$$X^n+Y^n=Z^n\Rightarrow {\left(X^p\right)}^4+{\left(Y^p\right)}^4={\left(Z^p\right)}^4$$

and this is impossible.

In the same way, if we succeed in proving the theorem for a certain $k-$exponent, then it is valid for all the multiples of k.

As every positive integer greater than 2 is divisible either by a prime odd number (that is different from 2), or by 4, it will be then sufficient to prove the theorem for all those cases in which the exponent is a prime odd number ([3], pp. 203–207).

In this proof, we will discuss all those cases in which the exponent n is an odd number $>1$ and, from now on, we will indicate the Fermat Last Theorem with the acronym F.L.T.

2. Indeterminate Analysis of Second Degree

Our goal is to take care of the resolution, into integers, of quadratic equation with integer coefficients, depending on n unknowns ([4], Cap. I, pp. 60–69).

We will develop our considerations on the equation in three unknowns:

$$F\left(X,Y,Z\right)=a{X}^2+b{Y}^2+c{Z}^2+dXY+eXZ+fYZ=0$$

(1)

warning that, all what we will say, extends immediately to the case of n unknowns.

Since the (1) is a homogeneous equation, if ($A,B,C$) are the solutions, (mA, mB, mC) are also solutions.

Therefore we deem identical two solutions, such as ($A,B,C$) and (mA, mB, mC).

Such assumptions, will narrow the search to the only primitive solutions of Equation (1), that is, to those in which $X,Y$ and Z are pairwise relatively prime.

Let ($x,y,z$) be a solution in integers of the Equation (1) and then $F\left(x,y,z\right)=0$ and we put:

$$X=\rho ·x+\xi ,Y=\rho ·y+\eta ,Z=\rho ·z+\zeta$$

(2)

where $\xi ,\eta ,\zeta$ are arbitrary integer constants and $\rho$ an unknown to be determined, so that Equation (2) provides an integer solution for Equation (1).

It must be:

$F\left(X,Y,Z\right)={\rho }^2\left[a{x}^2+b{y}^2+c{z}^2+dxy+exz+fyz\right]+$

$$\rho ·\left[2a\xi ·x+2b\eta ·y+2c\zeta ·z+d\left(\xi ·y+\eta ·x\right)+e\left(\xi ·z+\zeta ·x\right)+f\left(\eta ·z+\zeta ·y\right)\right]+$$

$$\left[a\xi \xi +b\eta \eta +c\zeta \zeta +d\xi \eta +e\xi \zeta +f\eta \zeta \right]=0.$$

However, the coefficient of ${\rho }^2$, equal to $F\left(x,y,z\right)$, is null and the known term is $F\left(\xi ,\eta ,\zeta \right)$; so, set equal to M (with $M\ne 0$ due to the arbitrary of $\xi ,\eta ,\zeta$), the coefficient $\rho$ of the above equation is equal to $\rho =-\frac{F\left(\xi ,\eta ,\zeta \right)}{M}$.

Consequently, if an integer solution of Equation (1) is known, we have infinite other, by putting in Equation (2), in place of $\rho$, the value now found; then, except the divisor M, we have:

$$X=\xi ·M-xF\left(\xi ,\eta ,\zeta \right);Y=\eta ·M-yF\left(\xi ,\eta ,\zeta \right);$$

(3)

$$Z=\zeta ·M-zF\left(\xi ,\eta ,\zeta \right).$$

These are the general solutions of Equation (1).

To prove it, we will demonstrate, by appropriately selecting $\xi ,\eta ,\zeta$, the previous solutions provide a solution of Equation (1), given arbitrarily, and let this ($A,B,C$).

It is meanwhile F($A,B,C$) = 0; if now, in Equation (3), we write $\xi =A$, $\eta =B$, $\zeta =C$, we have the solution: $X=AM;Y=BM;Z=CM$, that, except the factor M, it is identified with the one already provided.

In conclusion:

Theorem 1.

Let(x, y, z) be an integer solution of Equation (1). All its integer solutions are given by Equation (3), except the integer divisor M.

Now, we solve the equation $F\left(X,Y,Z\right)=X^2+a{Y}^2-Z^2=0$ in integer numbers.

Keeping in mind that this equation is homogeneous, we know that we can consider identical the two solutions, as (1, 0, 1) and $\left(m,0,m\right)$.

Let us consider, at this point, the trivial solution (1, 0, 1) and we will have: $M=2\left(\xi -\zeta \right)$; $F\left(\xi ,\eta ,\zeta \right)={\xi }^2+a{\eta }^2-{\zeta }^2$ for which all the solutions, keeping in mind the Equation (3), are given by the relations:

$$X=2\xi \left(\xi -\zeta \right)-{\xi }^2-a{\eta }^2+{\zeta }^2={\left(\xi -\zeta \right)}^2-a{\eta }^2;Y=2\eta \left(\xi -\zeta \right)$$

$$Z=2\zeta \left(\xi -\zeta \right)-{\xi }^2-a{\eta }^2+{\zeta }^2=-{\left(\xi -\zeta \right)}^2-a{\eta }^2.$$

Therefore, assuming $\left(\xi -\zeta \right)=\theta$ and observing that from a solution $\left(x,y,z\right)$, we obtain the other changing sign to one, or two, or all $\left(x,y,z\right)$, we have:

$$X={\theta }^2-a{\eta }^2;Y=2\theta \eta ;Z={\theta }^2+a{\eta }^2$$

which provide us with all the primitive integer solutions of quadratic equation, except an appropriate integer divisor M.

In general, we have that all integer solutions for the equation $X^2+a{Y}^2=Z^2$ are:

$$X=k\left({\theta }^2-a{\eta }^2\right);Y=k\left(2\theta \eta \right);Z=k\left({\theta }^2+a{\eta }^2\right).$$

(4)

where $\theta , \eta$ are natural numbers and k a rational proportionality factor(see also [5], kap. V, §29, pp. 39–44).

3. On Homogeneous Ternary Quadratic Diophantine Equations ${\mathit{aX}}^{\mathbf{2}}+{\mathit{bY}}^{\mathbf{2}}-{\mathit{cZ}}^{\mathit{2}}=\mathbf{0}$

Theorem 2.

Let $x^n+y^n=z^n$, with $(x,y)=1$ and $n\ge 3$ has a solution, then there exists an equation $a{x}^2+b{y}^2=c{z}^2$, where $a,b,c$ are relatively prime and reduced to the minimum terms, whose a solution could be reduced to a solution of Fermat’s equation.

Proof. 

Let $X_1,Y_1,Z_1$ be three whole numbers pairwise relatively prime such as to satisfy the Fermat equation $x^n+y^n=z^n$.

Then, the following homogeneous ternary quadratic Diophantine equation, with $(V,T,P)=1$ exists:

$$X_1^n{V}^2+Y_1^n{T}^2=Z_1^n{P}^2.$$

(5)

We observe that with the following particular nontrivial solutions:

$V=1,T=1$ and $P=1$ or $V=T=P$ in Equation (5), we obtain the fundamental Hypothesis (Reductio ad Absurdum) of the F.L.T.:

$${X_1}^n+{Y_1}^n={Z_1}^n.$$

Now by the evident solutions, indicated above, we can derive an infinite number of solutions of Equation (5).

Let us remember that for Legendre’s Theorem if a ternary quadratic homogeneous Diophantine equation (assuming $a,b$ and c are fixed) has an integral solution, then the number of possible solutions is infinite.

Having said this, it is possible to transform the previous Diophantine Equation (5) into the following equivalent Diophantine equation, with $(V^{\prime },T^{\prime },P^{\prime })=1$:

$$X_1{V}^{\prime 2}+Y_1{T}^{\prime 2}=Z_1{P}^{\prime 2}.$$

(6)

It is sufficient to assume $V^{\prime }=X_1^{k}V, T^{\prime }=Y_1^{k}T, P^{\prime }=Z_1^{k}P$ where $k=\frac{n-1}{2}$ and $n>1$ are odd numbers.

Using the “fundamental theorem of Arithmetic” we can represent ([6], Theorem 19, p. 31):

$X_1=X_0{U}_1^2,Y_1=Y_0{U}_2^2,Z_1=Z_0{U}_3^2.$

In this case, it is possible to transform the previous Diophantine Equation (6) into the following equivalent Diophantine equation with the relative coefficients reduced to the minimum terms:

$$X_0{V}^{″2}+Y_0{T}^{″2}=Z_0{P}^{″2}.$$

In fact, assuming $V^{″}=U_1{V}^{\prime },T^{″}=U_2{T}^{\prime },P^{″}=U_3{P}^{\prime }$

We observe that $X_0,Y_0,Z_0$ are pairwise relatively prime and square-free numbers.

The proof ends here by properly also verifying the nature of exponent n.□

4. From the Concordant Forms of Euler to Fermat’s Last Theorem

Let $m,n\in \mathbf{Z}$$/\left\{0\right\}$ be integers with $m\ne n$. Following Euler (see [7]), the quadratic forms $X^2+m{Y}^2$ and $X^2+n{Y}^2$ (or the numbers m and n themselves) are called $concordant$ if there are integers $(X,Y,Z,T)$ with $Y\ne 0$ such that:

$$X^2+m{Y}^2=Z^2{X}^2+n{Y}^2=T^2.$$

(7)

In 1780, Euler seeks criteria for the treatment of the double Equation (7) and his interest and our own turns to proofs of impossibility for the cases m = 1, n = 3 or 4 and others, equivalent to these two ([8], Chap. III, §XVI, pp. 253–254).

In practice, Euler called $X^2+m{Y}^2$ and $X^2+n{Y}^2$ $concordant$ forms if they can both be made squares by choice of integers $X,Y$ each not zero; otherwise, $discordant$ forms. At this stage, let us introduce the following Euler double equations:

$$P^2+Y_1^n{Q}^2=V^2,P^2-X_1^n{Q}^2=T^2$$

(8)

with $X_1^n+Y_1^n=Z_1^n$ and $n>1$ odd number.

By multiplying the first two Equation (8) together, and multiplying by $\frac{P^2}{Q^6}$, with $P\ne 0$ and $Q\ne 0$, we get [9]:

$$\frac{P^2{V}^2{T}^2}{Q^6}=\frac{P^6}{Q^6}+\left(Y_1^n-X_1^n\right)\frac{P^4}{Q^4}-X_1^n{Y}_1^n\frac{P^2}{Q^2}.$$

(9)

If we then replace $\frac{P^2}{Q^2}$ by $X$ and also $\frac{PVT}{Q^3}$ by $Y$ we find that

$$Y^2=X\left(X-X_1^n\right)\left(X+Y_1^n\right).$$

This is known as Frey Elliptic curve ([10], pp. 154–156).

In Mathematics, a Frey curve or Frey–Hellegouarch curve is the elliptic curve:

$$Y^2=X\left(X-X_1^n\right)\left(X+Y_1^n\right)$$

(10)

or, equivalently:

$$Y^2=X\left[X^2+X\left(Y_1^n-X_1^n\right)-X_1^n{Y}_1^n\right]$$

(11)

associated with a (hypothetical) solution of Fermat’s equation: $X_1^n+Y_1^n=Z_1^n.$

In fact, the discriminant

$$\Delta =\sqrt{{\left(Y_1^n-X_1^n\right)}^2+4X_1^n{Y}_1^n}=X_1^n+Y_1^n=Z_1^n,$$

that determines the existence of the polynomial

$$\left(X-X_1^n\right)\left(X+Y_1^n\right)=X^2+X\left(Y_1^n-X_1^n\right)-X_1^n{Y}_1^n$$

is a perfect power of order n.

Frey suggested, in 1985, that the existence of a non-trivial solution to $X^n+Y^n=Z^n$ would imply the existence of a non-modular elliptic curve, viz. $Y^2=X(X-X^n)(X+Y^n)$.

This suggestion was proved by Ribet in 1986.

This curve is semi-stable and in 1993 Wiles announced a proof (subsequently found to need another key ingredient, furnished by Wiles and Taylor) that every semi-stable elliptic curve is modular, as in the semi-stable case of the Taniyama-Shimura-Weil conjecture [11,12].

Hence no non-trivial $X^n+Y^n=Z^n$ can exist.

Basically, thanks to does the spectacular work of A. Wiles, today we know that Frey’s elliptic curve does not exist and from this derives indirectly, as an absurd, the F.L.T.

Now, multiplying the first two Equation (8), respectively, by $X_1^n$ and by $Y_1^n$ and at the end adding together, we obtain the following homogeneous ternary quadratic equation (see Section 3):

$$X_1^n{V}^2+Y_1^n{T}^2=Z_1^n{P}^2$$

(12)

with the identity $X_1^n+Y_1^n=Z_1^n$ and $n>1$ odd number.

So, we can also enunciate the following conjecture:

Conjecture 1.

Fermat’s Last Theorem is true only if the homogeneous ternary quadratic Diophantine Equation (12) does not exist (in the sense that the Diophantine Equation (12) has no integer solutions).

Nobody prevents us from assuming the evident solution $V=T=P=1$ or $V=T=P$ in the Equation (12) and with this we obtain the solution of the Fermat equation: $X_1^n+Y_1^n=Z_1^n$.

Presently, from the Euler double Equation (8) by subtracting, we have:

$$V^2-T^2=Z_1^n{Q}^2.$$

This equation together with Equation (12) gives rise to a system perfectly equivalent to Euler’s double Equation (8) (see Section 5).

We have also with $V=T=1$ or $V=T$:

$$V^2-T^2=Z_1^n{Q}^2=0.$$

By definition, in Euler’s $concordant$ forms, Q is absolutely non-zero integer.

It follows that by $Z_1^n=0$ and the homogeneous ternary quadratic Diophantine Equation (12), it does not exist.

We observe that the same result can be achieved immediately if we assume $V=T=P=1$ or $V=T=P$ already in Equation (8), in fact with Q non-zero integer we even have $X_1^n=Y_1^n=0$ and therefore still $Z_1^n=0$.

Further verification of these conclusions is also possible in this way.

Let us introduce the following Euler double equations:

$$P^{\prime 2}+Y_1^n{Q}^2=V^2,P^{″2}-X_1^n{Q}^2=T^{″2}$$

(13)

with $X_1^n+Y_1^n=Z_1^n$ and $n>1$ odd number or

$$P^{\prime 2}+Y_1^n{Q}^2=V^2,P^{‴2}-X_1^n{Q}^{\prime 2}=T^{‴2}$$

(14)

with $X_1^n+Y_1^n=Z_1^n$ and $n>1$ odd number.

From Equation (4), we have the following solutions of first Euler equation of Equation (13):

$$P^{\prime }=k\left({\theta }^2-Y_1^n{\eta }^2\right),Q=k\left(2\theta \eta \right),V=k\left({\theta }^2+Y_1^n{\eta }^2\right)$$

(15)

and the following solutions of the second Euler equation of Equation (13):

$$P^{″}=k\left({\theta }^2+X_1^n{\eta }^2\right),Q=k\left(2\theta \eta \right),T^{″}=k\left({\theta }^2-X_1^n{\eta }^2\right)$$

(16)

or the following solutions of the second Euler equation of Equation (14):

$$P^{‴}=k^{\prime }\left({\theta }^{\prime 2}+X_1^n{\eta }^{\prime 2}\right),Q^{\prime }=k^{\prime }\left(2{\theta }^{\prime }{\eta }^{\prime }\right),T^{‴}=k^{\prime }\left({\theta }^{\prime 2}-X_1^n{\eta }^{\prime 2}\right).$$

(17)

Now, assuming $V=T=P$ with a Q non-zero integer, we have the following result due to Equations (15) and (16):

$$P=P^{\prime }=P^{″}\Rightarrow -Y_1^n=X_1^n\Rightarrow Z_1^n=0and$$

$$V=T^{″}\Rightarrow Y_1^n=-X_1^n\Rightarrow Z_1^n=0.$$

while, with Equations (15) and (17), we have:

$$P=P^{\prime }=V\Rightarrow -Y_1^n=Y_1^n\Rightarrow Y_1^n=0and$$

$$P=P^{‴}=T^{‴}\Rightarrow X_1^n=-X_1^n\Rightarrow X_1^n=0$$

and therefore still $Z_1^n=0$.

In conclusion, what has been described so far in relation to Conjecture 1 obviously does not have a demonstrative value, but allows us to state the following equivalent theorem:

Theorem 3

(Fundamental Theorem). Fermat’s Last Theorem is true if and only if a solution is not possible in integers of Equation (8) with the Q non-zero integer; that is, these are discordant forms.

In practice, this means that if the system of quadratic Equation (8) admits only the trivial solutions $(m,0,\pm m,\pm m)$, that include also (1,0,1,1), then the quadratic forms $P^2+Y_1^n{Q}^2$ and $P^2-X_1^n{Q}^2$ are a fortiori called $discordant$.

A complete and direct proof of this Theorem is formed in Section 6.

5. The Nature of Euler’s Double Equations through the Algebraic Geometry

In this section, we will concentrate on the following Euler’s concordant/discordant forms in Equation (8):

$$P^2+Y_1^n{Q}^2=V^2,P^2-X_1^n{Q}^2=T^2$$

with $X_1^n+Y_1^n=Z_1^n$ and $n\ge 3$.

In determining the nature of the Euler double equations and of an appropriate equivalent Diophantine system, we will make use of the description given by A. Weil ([8], Chap. II, App. IV, pp. 140–149) in order to provide some theoretical background to Fermat’s and Euler’s method of descent employed in the treatment of elliptic curves.

For simplicity, we consider the case where the roots of a cubic $\mathsf{\Gamma }$ are rational integers $\alpha ,\beta$ and $\gamma$.

$$y^2=f\left(x\right)=\left(x-\alpha \right)\left(x-\beta \right)\left(x-\gamma \right).$$

(18)

Weil considers an oblique quartic $\mathsf{\Omega }\left(A,B,C\right)$ in the space $\left(u,v,w\right)$

$$A{u}^2+\alpha =B{v}^2+\beta =C{w}^2+\gamma$$

(19)

with $u,v,w\in$Q and the following mapping of $\mathsf{\Omega }$ in $\mathsf{\Gamma }$

$$x=A{u}^2+\alpha ,y=\sqrt{ABC}uvw$$

(20)

where $A·B·C$ has to be a square.

In practice, Weil states that the determination of rational points of the curve $\mathsf{\Gamma }$ can be reduced to that of finding rational points of one or more appropriate quartics, such as (19), given a set of integers $A,B,C$ (positive or negative), considered squarefree, that is, not divisible by any square greater than 1, and such that the product $A·B·C$ is a square.

In homogeneous coordinates, $\mathsf{\Omega }\left(A,B,C\right)$ may be regarded as defined by the equation

$$A{U}^2+\alpha T^2=B{V}^2+\beta T^2=C{W}^2+\gamma T^2,$$

(21)

with integers $U,V,W,T$ without a common divisor.

Subsequently, after affirming that Equation (21) admits at least one solution, instead of defining $\mathsf{\Omega }=\mathsf{\Omega }\left(A,B,C\right)$ through (19), Weil writes it through the equation of two quadrics in $P^3$, that is: $\mathsf{\Phi }={\displaystyle \sum _{i,j=1}^4}{a}_{ij}{X}_i{Y}_j$ and $\mathsf{\Psi }={\displaystyle \sum _{i,j=1}^4}{b}_{ij}{X}_i{Y}_j$, with the condition $\mathsf{\Phi }=\mathsf{\Psi }=0$.

In detail, one has:

$$\mathsf{\Phi }\left(U,V,W,T\right)=\alpha \left(\beta -\gamma \right)\left(A{U}^2+\alpha T^2\right)+\beta \left(\gamma -\alpha \right)\left(B{V}^2+\beta T^2\right)+$$

$$\gamma \left(\alpha -\beta \right)\left(C{W}^2+\gamma T^2\right)=\alpha \left(\beta -\gamma \right)A{U}^2+\beta \left(\gamma -\alpha \right)B{V}^2+\gamma \left(\alpha -\beta \right)C{W}^2-\delta T^2$$

$$\mathsf{\Psi }\left(U,V,W,T\right)=\left(\beta -\gamma \right)A{U}^2+\left(\gamma -\alpha \right)B{V}^2+\left(\alpha -\beta \right)C{W}^2$$

where one has put $\delta =\left(\beta -\gamma \right)\left(\gamma -\alpha \right)\left(\alpha -\beta \right)$.

With this in mind, we consider the following assumptions

$$A=1,\alpha =0,B=1,\beta =X_1^n,C=1,\gamma =-Y_1^n.$$

(22)

In this case, Equation (18) would be reduced to the Frey elliptic curve:

$$Y^2=f\left(X\right)=\left(X\right)\left(X-X_1^n\right)\left(X+Y_1^n\right).$$

(23)

and the Euler double Equation (8) with the following assumptions, in order: $P=U,T=W,Q=T$ would be reduced to the oblique quartic $\mathsf{\Omega }\left(A,B,C\right)=\mathsf{\Omega }\left(1,1,1\right)$:

$$U^2=V^2-Y_1^n{T}^2=W^2+X_1^n{T}^2.$$

(24)

The product $ABC$ is, as required, a perfect square, and therefore it is certainly possible that the application (20) of the quartic $\mathsf{\Omega }$ on cubic $\mathsf{\Gamma }$.

The expressions of the two quadrics in $P^3$ become

$\mathsf{\Phi }\left(U,V,W,T\right)=-Y_1^n{X}_1^n{V}^2+X_1^n{Y}_1^n{W}^2+Z_1^n{X}_1^n{Y}_1^n{T}^2$ and

$$\mathsf{\Psi }\left(U,V,W,T\right)=-\left(Y_1^n+X_1^n\right)U^2+X_1^n{V}^2+Y_1^n{W}^2=-\left(Z_1^n\right)U^2+X_1^n{V}^2+Y_1^{n}W{.}^2$$

Finally, by $\mathsf{\Phi }=\mathsf{\Psi }=0$, they are translated into

$$\left(V^2-W^2\right)=\left(Z_1^n\right)T^2$$

(25)

and

$$X_1^n{V}^2+Y_1^n{W}^2=Z_1^n{U}^2.$$

(26)

Presently, Equations (25) and (26) with the following replacements:

$$T\Rightarrow W,Q\Rightarrow T,P\Rightarrow U$$

are none other than the equations of what we have described in the Section 4, that is:

$$\left(V^2-T^2\right)=\left(Z_1^n\right)Q^2$$

and

$$X_1^n{V}^2+Y_1^n{T}^2=Z_1^n{P}^2.$$

This alternative procedure confirms the validity of our conclusions: more precisely, I am referring to the fact that Euler’s double equations, as representatives of an evident oblique quartic of genus 1, can also be defined by means of a pair of equations of two quadrics in $P^3$, which establishes uniquely that the following Diophantine systems are perfectly equivalent:

$$\left\{\begin{array}{c}{P}^2+Y_1^n{Q}^2=V^2\hfill \\ P^2-X_1^n{Q}^2=T^2\hfill \end{array}\right.\left\{\begin{array}{c}{X}_1^n{V}^2+Y_1^n{T}^2=Z_1^n{P}^2\hfill \\ Z_1^n{Q}^2=V^2-T^2.\hfill \end{array}\right.$$

(27)

6. The Determination of the Parameter $\mathit{Q}$ in Euler’s Double Equations

Let us consider the first Diophantine equation of the second system (27):

$$X_1^n{V}^2+Y_1^n{T}^2=Z_1^n{P}^2$$

(28)

and we apply Theorem 1.

Now, we solve the equation $F\left(X,Y,Z\right)=a{X}^2+b{Y}^2-c{Z}^2=0$.

Keep in mind that this equation is homogeneous we known that we can consider identical the two solutions, as $(1,1,1)$ and $(m,m,m)$.

Let us consider, at this point, the solutions $(1,1,1)$ and we will have:

$$M=2\left(a\xi +b\eta -c\zeta \right);F\left(\xi ,\eta ,\zeta \right)=a{\xi }^2+b{\eta }^2-c{\zeta }^2$$

for which all the solutions, except the integer divisor M, keeping in mind Equation (3), are given by the relations:

$$X=a{\xi }^2-b{\eta }^2+2b\xi \eta +c\zeta \left(\zeta -2\xi \right);Y=-a{\xi }^2+b{\eta }^2+2a\xi \eta +c\zeta \left(\zeta -2\eta \right)$$

$$Z=-a{\xi }^2-b{\eta }^2-\zeta \left[c\zeta -2\left(a\xi +b\eta \right)\right].$$

Without the loss of generality, we assume that $\zeta =0$, therefore we reduce the intervention of the three integers $\xi , \eta$ and $\zeta$ and to only two of them.

In practice, we use the following equations instead of Equation (2):

$$X=\rho ·x+\xi ,Y=\rho ·y+\eta ,Z=\rho ·z$$

and eliminates the parameter $\rho$ to obtain the following parametric solutions of Equation (28):

$$V=\lambda \left(X_1^n{\xi }^2-Y_1^n{\eta }^2+2Y_1^n\xi \eta \right);T=\lambda \left(-X_1^n{\xi }^2+Y_1^n{\eta }^2+2X_1^n\xi \eta \right);$$

$$P=\lambda \left(X_1^n{\xi }^2+Y_1^n{\eta }^2\right).$$

(29)

where $\xi$ and $\eta$ are coprime integers and $\lambda$ is a rational proportionality factor.

Moreover $\xi$, $\eta$ and $\lambda$ are uniquely determinated, up to a simultaneous change of the sign of $\xi$ and $\eta$.

One standard method of obtaining the above parametrization can be found also in ([13], §6.3.2, pp. 343–346).

Now, from the second equation of the second system (27) with the Equation (29) and $\left(V,T\right)=1$, we have with $\lambda =\frac{1}{M}$:

$$Z_1^n{Q}^2=V^2-T^2=\frac{1}{M^2}\left[4\xi \eta \left(\xi -\eta \right)\left(X_1^n\xi +Y_1^n\eta \right)\left(X_1^n+Y_1^n\right)\right]\Rightarrow$$

$$Q^2=\frac{1}{M^2}4\xi \eta \left(\xi -\eta \right)\left(X_1^n\xi +Y_1^n\eta \right).$$

(30)

For the last factor $\left(X_1^n\xi +Y_1^n\eta \right)$, we can consider the following linear equation:

$$\left(X_1^n\xi +Y_1^n\eta \right)=h{Z}_1^n$$

(31)

which certainly, admitting the obvious solution $\xi =\eta =h$, provides us with all the solutions, and also with $\xi \ne \eta$, that is:

$$\xi =h+Y_1^n\theta ;\eta =h-X_1^n\theta .$$

(32)

Besides, we have:

$$\left(\xi -\eta \right)=Z_1^n\theta .$$

(33)

Therefore, bearing in mind that $\left(X_1,Y_1,Z_1\right)=1$, $\left(V,T,P\right)=1$ and $\left(\xi ,\eta \right)=1$, we also have that $\left(h,\theta \right)=1$.

Now, Equation (30) with Equations (31) and (33) and in addition with $M=2\left(a\xi +b\eta \right)=2\left(X_1^n\xi +Y_1^n\eta \right)$ provides:

$$Q^2=\frac{1}{4{\left(X_1^n\xi +Y_1^n\eta \right)}^2}4\xi \eta \left(\xi -\eta \right)\left(X_1^n\xi +Y_1^n\eta \right)=\frac{\xi \eta \theta Z_1^n}{h{Z}_1^n}=\xi \eta \frac{\theta }{h}.$$

(34)

Now, we will resort to the Corollary 6.3.8 ([13], p. 346).

In the case of $\left(V,T,P\right)=1$ we have that the rational proportionality parameter

in the Equation (29) is $\lambda =\frac{1}{r}$ with $r|2Y_1^n{Z}_1^n$.

Now, $\lambda =\frac{1}{M}\Rightarrow h=\frac{Y_1^n}{m}$ with $m\left|Y_1^n\right.$.

Without the loss of generality, we can verify only the following extreme case $m=1$ and $m=Y_1^n$ (see Appendix A).

In fact, thanks to the solutions (32), a single and appropriate value of h is sufficient for these equations to constitute the general solution of the linear Equation (31).

It follows that for $\theta =0,\pm 1,\pm 2,\dots$ Formula (32) give all the integral solutions of Equation (31).

The necessary condition is that h is an exact divisor of $Y_1^n$ and consequently $h=Y_1^n$ or $h=1$ both satisfy this condition.

In the first case with $h=Y_1^n$, we have from Equation (34): $Q^2=\left(1+\theta \right)\left(\theta \right)\left(Y_1^n-X_1^n\theta \right)$ with the three positive factors in brackets that are pairwise relatively prime.

By the uniqueness of the prime decomposition we have $\left(1+\theta \right)$ and $\theta$ should be equal to squares and this is absurd.

In the second case with $h=1,\theta >0$ and $X_1^n<0$ we have from Equation (34): $Q^2=\left(1+Y_1^n\theta \right)\left(\theta \right)\left(1-X_1^n\theta \right)$ with the three positive factors in brackets that are pairwise relatively prime.

By the uniqueness of the prime decomposition we have that:

$$\begin{array}{c}\xi =\left(1+Y_1^n\theta \right)=V_1^2;\eta =\left(1-X_1^n\theta \right)=T_1^2;P_1^2=1;\theta =Q_1^2.\hfill \end{array}$$

In conclusion, we have the further double Euler equations:

$$P_1^2+Y_1^n{Q}_1^2=V_1^2;P_1^2-X_1^n{Q}_1^2=T_1^2$$

with $Q>Q_1$, if compared with the double Euler equations of the first Diophantine system (27).

Repeating the argument indefinitely would give a sequence of positive integer $Q>Q_1>Q_2>Q_3>\dots$, which decreased indefinitely.

This is impossible, because it implies an “infinite descent” for parameter Q.

The determination of the parameter Q, as the rational integer is not equal to zero and ends here, but we must remember that the Equation (34) was determined only thanks to assuming the obvious solution $\xi =\eta =h$ of the linear Equation (31).

In this case, due to Equation (33), assuming $Z_1^n>0$, we have $\theta =0$ and this results in the zeroing of the parameter Q.

The double equations of Euler are discordant forms and so the F.L.T. turns out to be true, just as honestly announced by Fermat himself.

7. Conclusions

In this paper, we have tried a new rephrasing of F.L.T. making use of elementary techniques, maybe present in Fermat’s brilliant mind.

We show that making use of the concordant forms of Euler and a ternary quadratic homogeneous Diophantine equation, it is possible to derive a proof of the F.L.T. without recurring the modern techniques, but exploiting the important criterion of Legendre for determining the solutions of the ternary quadratic homogeneous equation.

The proof, here presented, is valid in the case of all odd exponents greater than one (see the proof of the Theorem 2).

We observe, however, that also in the case of exponent $n=4$ the double equations of Euler are discordant: in this case, in the double equations of Euler, defined by the expressions (7), it is sufficient assuming that $m=-n=1$.

More precisely, we have the following system of equations:

$$\left\{\begin{array}{c}{X}^2+Y^2=Z^2\hfill \\ X^2-Y^2=T^2\hfill \end{array}\right.$$

that has no solutions in the natural numbers.

This theorem of a "congruent number" was anticipated by Fibonacci in his book “The Book of squares” ([14], Chap. III, § VI-2, pp. 310–311), but with a not complete demonstration (the first complete proof was provided by Fermat with the equivalent Theorem: No Pythagorean triangle has square area) ([6], Chap. II, pp. 50–56).

In this work, we have not used the proof of non-existence of the Frey elliptic curve, but we have limited ourselves to proof of non-existence of the single homogeneous ternary quadratic equation Equation (5), defined in the proof of the Theorem 2, but whose origin (see Equation (12)) is implicit in the nature of Euler’s double equations.

The double equations of Euler gave rise in different ways to the elliptic curve of Frey and to a particular homogeneous ternary quadratic equation: both characterized by the presence of ${X_1}^n$, ${Y_1}^n$ and ${Z_1}^n$ in their coefficients.

For this, it was possible to use a similar strategy to build a reformulation of the F.L.T.