# Diophantine equations

## Abstract

Our purpose in this article which has been published is to show how much diophantine equations are rich in analytic applications. Effectively, those equations allow to build amazing sequences, series and numbers. The question of the proof of some theorems remains of course, we will see it in this communication. We will make also an allusion to the very known Fermat numbers (${\displaystyle 2^{2^{n}}}$). We will see how this problem of the proof is actual and how it can be solved using amazing sequences and series.

## The Ghanouchi's sequences

Let us begin by Fermat equation (E), it is

${\displaystyle U^{n}=X^{n}+Y^{n}}$

with GCD(X,Y)=1

We pose

${\displaystyle u=U^{2n}}$

${\displaystyle x=U^{n}X^{n}}$

${\displaystyle y=U^{n}Y^{n}}$

${\displaystyle z=X^{n}Y^{n}}$

then

${\displaystyle u=U^{2n}=U^{n}(X^{n}+Y^{n})=x+y\quad \quad {(1)}}$

and

${\displaystyle {\frac {1}{z}}={\frac {1}{X^{n}Y^{n}}}={\frac {U^{2n}}{U^{n}X^{n}U^{n}Y^{n}}}={\frac {u}{xy}}={\frac {x+y}{xy}}={\frac {1}{x}}+{\frac {1}{y}}\quad \quad {(2)}}$

If U, X, Y are integers verifying equation (E), then u, x, y, z as defined verify

## LEMMA 1

${\displaystyle u=x+y\quad \quad {(1)}}$

${\displaystyle {\frac {1}{z}}={\frac {1}{x}}+{\frac {1}{y}}\quad \quad {(2)}}$

Firstly

${\displaystyle z=X^{n}Y^{n}}$

${\displaystyle u=(X^{n}+Y^{n})^{2}}$

${\displaystyle x=X^{n}(X^{n}+Y^{n})}$

${\displaystyle y=Y^{n}(X^{n}+Y^{n})}$

We pose

${\displaystyle x_{1}=x}$

and

${\displaystyle y_{1}=y}$

but

${\displaystyle \forall {x_{1},y_{1}}}$

${\displaystyle \exists {z_{1}}}$

verifying

${\displaystyle {\frac {1}{z_{1}}}={\frac {1}{x_{1}}}+{\frac {1}{y_{1}}}}$

and

${\displaystyle z_{1}={\frac {xy}{x+y}}=z}$

then

${\displaystyle (x_{1}+y_{1})z_{1}=x_{1}y_{1}}$

or

${\displaystyle x_{1}(y_{1}-z_{1})=z_{1}y_{1}}$

we pose

${\displaystyle y_{2}=y_{1}-z_{1}={\frac {z_{1}y_{1}}{x_{1}}}}$

and

${\displaystyle y_{1}(x_{1}-z_{1})=z_{1}x_{1}}$

also

${\displaystyle x_{2}=x_{1}-z_{1}={\frac {z_{1}x_{1}}{y_{1}}}}$

and

${\displaystyle x_{2}y_{2}=z_{1}^{2}}$

which means

${\displaystyle x_{1}=x_{2}+z_{1}=x_{2}+{\sqrt {x_{2}y_{2}}}}$

and

${\displaystyle y_{1}=y_{2}+z_{1}=y_{2}+{\sqrt {x_{2}y_{2}}}}$

and

${\displaystyle u_{1}=u=(x_{1}+y_{1})=({\sqrt {x_{2}}}+{\sqrt {y_{2}}})^{2}>x_{2}+y_{2}>0}$

${\displaystyle (x_{1}+y_{1})}$ is an integer

${\displaystyle x_{1}={\sqrt {x_{2}}}({\sqrt {x_{2}}}+{\sqrt {y_{2}}})>x_{2}>0}$

${\displaystyle x_{1}}$ is an integer

${\displaystyle y_{1}={\sqrt {y_{2}}}({\sqrt {x_{2}}}+{\sqrt {y_{2}}})>y_{2}>0}$

${\displaystyle y_{1}}$ is an integer

${\displaystyle z_{1}={\frac {x_{1}y_{1}}{x_{1}+y_{1}}}=X^{a}Y^{b}={\sqrt {x_{2}y_{2}}}>z_{2}={\frac {x_{2}y_{2}}{x_{2}+y_{2}}}>0}$

${\displaystyle z_{2}}$ is rational, because

${\displaystyle \forall {x_{2},y_{2}}}$ rationals

${\displaystyle \exists {z_{2}}}$ rational verifying

${\displaystyle {\frac {1}{z_{2}}}={\frac {1}{x_{2}}}+{\frac {1}{y_{2}}}}$

until infinity. For i

${\displaystyle (x_{i}+y_{i})=({\sqrt {x_{i+1}}}+{\sqrt {y_{i+1}}})^{2}>x_{i+1}+y_{i+1}>0}$

${\displaystyle x_{i}+y_{i}}$ is rational for i>1

${\displaystyle x_{i}={\sqrt {x_{i+1}}}({\sqrt {x_{i+1}}}+{\sqrt {y_{i+1}}})>x_{i+1}>0}$

${\displaystyle x_{i}}$ is rational for i>1

${\displaystyle y_{i}={\sqrt {y_{i+1}}}({\sqrt {x_{i+1}}}+{\sqrt {y_{i+1}}})>y_{i+1}>0}$

${\displaystyle y_{i}}$ is rational for i>1

${\displaystyle z_{i}={\frac {x_{i}y_{i}}{x_{i}+y_{i}}}={\sqrt {x_{i+1}y_{i+1}}}>z_{i+1}={\frac {x_{i+1}y_{i+1}}{x_{i+1}+y_{i+1}}}>0}$

${\displaystyle z_{i}}$ is rational for i>1, and

${\displaystyle {\frac {1}{z_{i+1}}}={\frac {1}{x_{i+1}}}+{\frac {1}{y_{i+1}}}}$

${\displaystyle (x_{i}+y_{i})=({\sqrt {x_{i+1}}}+{\sqrt {y_{i+1}}})^{2}}$

## LEMMA 2

${\displaystyle x_{i}}$ and ${\displaystyle y_{i}}$ have expressions

${\displaystyle x_{i}=x^{2^{i-1}}\prod _{j=0}^{j={i-2}}{(x^{2^{j}}+y^{2^{j}})^{-1}}\quad \quad {(H)}}$

${\displaystyle y_{i}=y^{2^{i-1}}\prod _{j=0}^{j={i-2}}{(x^{2^{j}}+y^{2^{j}})^{-1}}\quad \quad {(H')}}$

## Proof of lemma 2

By induction

${\displaystyle x={\sqrt {x_{2}}}({\sqrt {x_{2}}}+{\sqrt {y_{2}}})={\sqrt {x_{2}}}(x+y)^{\frac {1}{2}}}$

${\displaystyle x_{2}={\frac {x^{2}}{x+y}}}$

also

${\displaystyle y_{2}={\frac {y^{2}}{x+y}}}$

it is verified for i=2. We suppose (H) and (H') true for i, then

${\displaystyle x_{i}={\sqrt {x_{i+1}}}({\sqrt {x_{i+1}}}+{\sqrt {y_{i+1}}})={\sqrt {x_{i+1}}}(x_{i}+y_{i})^{\frac {1}{2}}}$

and

${\displaystyle x_{i+1}=x_{i}^{2}{(x_{i}+y_{i})}^{-1}}$

but (H) and (H'), then

${\displaystyle x_{i+1}=x^{2^{i}}\prod _{j=0}^{j=i-2}{(x^{2^{j}}+y^{2^{j}})^{-2}}(x^{2^{i-1}}+y^{2^{i-1}})^{-1}\prod _{j=0}^{j={i-2}}{(x^{2^{j}}+y^{2^{j}})}}$

${\displaystyle =x^{2^{i}}\prod _{j=0}^{j={i-1}}{(x^{2^{j}}+y^{2^{j}})^{-1}}}$

and it is true for i+1, also for ${\displaystyle y_{i}}$

${\displaystyle x_{i}}$ and ${\displaystyle y_{i}}$ can be written as it follows

${\displaystyle x_{i}=x^{2^{i-1}}\prod _{j=0}^{j=i-2}{(x^{2^{j}}+y^{2^{j}})^{-1}}}$

${\displaystyle y_{i}=y^{2^{i-1}}\prod _{j=0}^{j=i-2}{(x^{2^{j}}+y^{2^{j}})^{-1}}}$

${\displaystyle \forall {i>1}}$

but, ${\displaystyle \forall {a,b}}$

${\displaystyle a-b=(a^{2^{i-1}}-b^{2^{i-1}})\prod _{j=0}^{j={i-2}}{(a^{2^{j}}+b^{2^{j}})^{-1}}\quad \quad {(E)}}$

${\displaystyle x-y=(x^{2^{i-1}}-y^{2^{i-1}})\prod _{j=0}^{j={i-2}}{(x^{2^{j}}+y^{2^{j}})^{-1}}\quad \quad {(E')}}$

${\displaystyle \prod _{j=0}^{j={i-2}}{(x^{2^{j}}+y^{2^{j}})}={\frac {(x^{2^{i-1}}-y^{2^{i-1}})}{x-y}}}$

the expressions of the sequences become, for

${\displaystyle x\neq {y}}$

or

${\displaystyle x_{i}\neq {y_{i}}}$

## LEMMA 3

${\displaystyle x_{i}={\frac {x^{2^{i-1}}}{x^{2^{i-1}}-y^{2^{i-1}}}}(x-y)}$

${\displaystyle =U^{n}{\frac {X^{n{2^{i-1}}}}{X^{n{2^{i-1}}}-Y^{n{2^{i-1}}}}}(X^{n}-Y^{n})}$

and

${\displaystyle y_{i}={\frac {y^{2^{i-1}}}{x^{2^{i-1}}-y^{2^{i-1}}}}(x-y)}$

${\displaystyle =U^{n}{\frac {Y^{n{2^{i-1}}}}{X^{n{2^{i-1}}}-Y^{n{2^{i-1}}}}}(X^{n}-Y^{n})}$

${\displaystyle u_{i}=x_{i}+y_{i}=U^{n}{\frac {X^{n{2^{i-1}}}+Y^{n{2^{i-1}}}}{X^{n{2^{i-1}}}-Y^{n{2^{i-1}}}}}(X^{n}-Y^{n})}$

## LEMMA 4

The equations (1) have (2) a constant

${\displaystyle x_{i}-y_{i}=x-y}$

## THE GHANOUCHI'S THEOREM

The only solution of equations

${\displaystyle u=x+y,\quad ((1))}$

and

${\displaystyle {\frac {1}{z}}={\frac {1}{x}}+{\frac {1}{y}},\quad {(2)}}$

is

${\displaystyle xy(x-y)=0}$

## Proof of Ghanouchi's theorem

As

${\displaystyle x_{i}={\frac {x^{2^{i-1}}}{x^{2^{i-1}}-y^{2^{i-1}}}}(x-y)}$

and

${\displaystyle y_{i}={\frac {y^{2^{i-1}}}{x^{2^{i-1}}-y^{2^{i-1}}}}(x-y)}$

if

${\displaystyle x>y\Rightarrow {\lim _{i\longrightarrow {\infty }}{(x_{i})}=x-y,\quad \quad {\lim _{i\longrightarrow {\infty }}{(y_{i})}=0}}}$

and

${\displaystyle y>x\Rightarrow {\lim _{i\longrightarrow {\infty }}{(y_{i})}=y-x,\quad \quad {\lim _{i\longrightarrow {\infty }}{(x_{i})}=0}}}$

We will give several proofs that ${\displaystyle xy(x-y)=0}$ is the solution with the series. We recapitulate

${\displaystyle x_{i}>x_{i+1},\quad {y_{i}>y_{i+1}}\Rightarrow {x=y}}$

${\displaystyle x_{i}=x_{i+1}=x_{i+1}+{\sqrt {x_{i+1}y_{i+1}}}\Rightarrow {xy=0}}$

${\displaystyle y_{i}=y_{i+1}=y_{i+1}+{\sqrt {x_{i+1}y_{i+1}}}\Rightarrow {xy=0}}$

${\displaystyle \Rightarrow {xy(x-y)=0}}$

x and y are not différent, the initial hypothesis is false there the only solution is

${\displaystyle xy(x-y)=0}$

The proofs : we have

${\displaystyle y_{i}^{2}-x_{i}^{2}=(y-x)(x_{i}+y_{i})=(y_{i+1}-x_{i+1})(x_{i+1}+y_{i+1}+2{\sqrt {x_{i+1}y_{i+1}}})}$

${\displaystyle =y_{i+1}^{2}-x_{i+1}^{2}+2{\sqrt {x_{i+1}y_{i+1}}}(y_{i+1}-x_{i+1})}$

${\displaystyle {\sqrt {y_{i+1}}}=Z}$ is solution of the following equation

${\displaystyle Z^{4}+2{\sqrt {x_{i+1}}}Z^{3}-2{\sqrt {x_{i+1}^{3}}}Z-x_{i+1}^{2}+x_{i}^{2}-y_{i}^{2}=0}$

Also ${\displaystyle {\sqrt {x_{i+1}}}=Z'}$ is solution of

${\displaystyle {Z'}^{4}+2{\sqrt {y_{i+1}}}{Z'}^{3}-2{\sqrt {y_{i+1}^{3}}}Z'-y_{i+1}^{2}+y_{i}^{2}-x_{i}^{2}=0}$

And

${\displaystyle Z^{4}+2Z'Z^{3}-2{Z'}^{3}Z-{Z'}^{4}+x_{i}^{2}-y_{i}^{2}=0}$

Also

${\displaystyle {Z'}^{4}+2Z{Z'}^{3}-2Z^{3}Z'-Z^{4}+y_{i}^{2}-x_{i}^{2}=0}$

Let

${\displaystyle Z=u-{\frac {Z'}{2}}}$

But

${\displaystyle (u-{\frac {Z'}{2}})^{4}+2Z'(u-{\frac {Z'}{2}})^{3}-2{Z'}^{3}(u-{\frac {Z'}{2}})-{Z'}^{4}+x_{i}^{2}-y_{i}^{2}=0}$

Hence

${\displaystyle u^{4}+{\frac {{Z'}^{4}}{16}}-2u^{3}{Z'}+{\frac {3}{2}}u^{2}{Z'}^{2}+}$

${\displaystyle -{\frac {1}{2}}u{Z'}^{3}+2Z'u^{3}-3u^{2}{Z'}^{2}+{\frac {3}{2}}u{Z'}^{3}-{\frac {1}{4}}{Z'}^{4}+}$

${\displaystyle -2{Z'}^{3}u+{Z'}^{4}-{Z'}^{4}+x_{i}^{2}-y_{i}^{2}=0}$

${\displaystyle =u^{4}-({\frac {3}{2}}{Z'}^{2})u^{2}-{Z'}^{3}u-{\frac {3}{16}}{Z'}^{4}+x_{i}^{2}-y_{i}^{2}=0}$

We deduce

${\displaystyle (u^{2}-{\frac {3}{4}}{Z'}^{2})^{2}-{\frac {3}{4}}{Z'}^{4}-{Z'}^{3}u+x_{i}^{2}-y_{i}^{2}=0}$

Also ${\displaystyle Z'=v-{\frac {Z}{2}}}$ leads to

${\displaystyle (v-{\frac {Z}{2}})^{4}+2Z(v-{\frac {Z}{2}})^{3}-2{Z}^{3}(v-{\frac {Z}{2}})-{Z}^{4}+y_{i}^{2}-x_{i}^{2}=0}$

Hence

${\displaystyle v^{4}+{\frac {{Z}^{4}}{16}}-2v^{3}{Z}+{\frac {3}{2}}v^{2}{Z}^{2}+}$

${\displaystyle -{\frac {1}{2}}v{Z}^{3}+2Zv^{3}-3v^{2}{Z}^{2}+{\frac {3}{2}}v{Z}^{3}-{\frac {1}{4}}{Z}^{4}+}$

${\displaystyle -2{Z}^{3}v+{Z}^{4}-{Z}^{4}+y_{i}^{2}-x_{i}^{2}=0}$

${\displaystyle =v^{4}-({\frac {3}{2}}{Z}^{2})v^{2}-{Z}^{3}v-{\frac {3}{16}}{Z}^{4}+y_{i}^{2}-x_{i}^{2}=0}$

We deduce

${\displaystyle (v^{2}-{\frac {3}{4}}{Z}^{2})^{2}-{\frac {3}{4}}{Z}^{4}-{Z}^{3}v+y_{i}^{2}-x_{i}^{2}=0}$

${\displaystyle (u^{2}-{\frac {3}{4}}{Z'}^{2})^{2}+(v^{2}-{\frac {3}{4}}Z^{2})^{2}-{\frac {3}{4}}(Z^{4}+{Z'}^{4})-{Z'}^{3}u-Z^{3}v=0}$

${\displaystyle =(Z^{2}+{\frac {{Z'}^{2}}{4}}+ZZ'-{\frac {3}{4}}Z^{2})^{2}+({Z'}^{2}+{\frac {Z^{2}}{4}}+Z'Z-{\frac {3}{4}}{Z'}^{2})^{2}-{\frac {3}{4}}(Z^{4}+{Z'}^{4})-Z{Z'}^{3}-Z^{3}Z'-{\frac {{Z'}^{4}+Z^{4}}{2}}=0}$

${\displaystyle =({\frac {Z^{2}+{Z'}^{2}}{4}}+ZZ')^{2}+({\frac {Z^{2}+{Z'}^{2}}{4}}+Z'Z)^{2}-{\frac {5}{4}}(Z^{4}+{Z'}^{4})-ZZ'(Z^{2}+{Z'}^{2})=0}$

${\displaystyle ={\frac {Z^{4}+{Z'}^{4}+2Z^{2}{Z'}^{2}}{8}}+2Z^{2}{Z'}^{2}+ZZ'(Z^{2}+{Z'}^{2})-{\frac {5}{4}}(Z^{4}+{Z'}^{4})-ZZ'(Z^{2}+{Z'}^{2})=0}$

${\displaystyle =-{\frac {9}{8}}(Z^{4}+{Z'}^{4})+{\frac {9}{4}}Z^{2}{Z'}^{2}=0}$

${\displaystyle =-(Z^{2}-{Z'}^{2}){\frac {9}{8}}=0}$

The solution is

${\displaystyle Z^{2}-{Z'}^{2}=y-x=0}$

Another proof : we have

${\displaystyle {\sqrt {x_{i-1}}}+{\sqrt {x_{i}}}={\frac {x_{i-1}-x_{i}}{{\sqrt {x_{i-1}}}-{\sqrt {x_{i}}}}}={\frac {\sqrt {x_{i}y_{i}}}{{\sqrt {x_{i-1}}}-{\sqrt {x_{i}}}}}}$

${\displaystyle ={\frac {1}{{\sqrt {\frac {x_{i-1}}{x_{i}y_{i}}}}-{\sqrt {\frac {x_{i}}{x_{i}y_{i}}}}}}={\frac {1}{{\sqrt {\frac {\sqrt {x_{i-1}+y_{i-1}}}{{\sqrt {x_{i}}}y_{i}}}}-{\frac {1}{\sqrt {y_{i}}}}}}}$

${\displaystyle ={\frac {1}{\frac {{\sqrt[{4}]{x_{i-1}+y_{i-1}}}-{\sqrt[{4}]{x_{i}}}}{{\sqrt[{4}]{x_{i}}}{\sqrt {y_{i}}}}}}={\frac {{\sqrt[{4}]{x_{i}}}{\sqrt {y_{i}}}}{{\sqrt[{4}]{x_{i-1}+y_{i-1}}}-{\frac {\sqrt {x_{i-1}}}{\sqrt[{4}]{x_{i-1}+y_{i-1}}}}}}}$

${\displaystyle ={\frac {{\sqrt[{4}]{x_{i}}}{\sqrt {y_{i}}}({\sqrt[{4}]{x_{i-1}+y_{i-1}}})}{{\sqrt {x_{i-1}+y_{i-1}}}-{\sqrt {x_{i-1}}}}}\leq {\frac {{\sqrt[{4}]{x_{i}}}{\sqrt {y_{i}}}{\sqrt[{4}]{x_{i-1}+y_{i-1}}}}{\sqrt[{4}]{y_{i+1}}}}}$

Because

${\displaystyle {\sqrt {x_{i-1}+y_{i-1}}}-{\sqrt {x_{i-1}}}={\frac {y_{i-1}}{{\sqrt {x_{i-1}+y_{i-1}}}+{\sqrt {x_{i-1}}}}}\geq {\sqrt[{4}]{y_{i+1}}}}$

${\displaystyle y_{i-1}={\sqrt {y_{i}}}{\sqrt {x_{i-1}+y_{i-1}}}\geq {{\sqrt[{4}]{y_{i+1}}}({\sqrt {x_{i-1}+y_{i-1}}}+{\sqrt {x_{i-1}}})}={\frac {\sqrt {y_{i}}}{\sqrt[{4}]{x_{i}+y_{i}}}}({\sqrt {x_{i-1}+y_{i-1}}}+{\sqrt {x_{i-1}}})}$

And

${\displaystyle {\sqrt[{4}]{x_{i}+y_{i}}}{\sqrt {x_{i-1}+y_{i-1}}}\geq {{\sqrt {x_{i-1}+y_{i-1}}}+{\sqrt {x_{i-1}}}}}$

Because

${\displaystyle {\sqrt[{4}]{x_{i}+y_{i}}}={\frac {\sqrt[{4}]{x_{i-1}^{2}+y_{i-1}^{2}}}{\sqrt[{4}]{x_{i-1}+y_{i-1}}}}\geq {1+{\sqrt {\frac {x_{i-1}}{x_{i-1}+y_{i-1}}}}}}$

And

${\displaystyle {\sqrt[{4}]{x_{i-1}^{2}+y_{i-1}^{2}}}\geq {{\sqrt[{4}]{x_{i-1}+y_{i-1}}}+{\sqrt[{4}]{x_{i-1}+y_{i-1}}}{\frac {\sqrt {x_{i-1}}}{\sqrt {x_{i-1}+y_{i-1}}}}}}$

We deduce

${\displaystyle {\sqrt {x_{i-1}}}+{\sqrt {x_{i}}}\leq {\frac {{\sqrt[{4}]{x_{i}}}{\sqrt {y_{i}}}}{\sqrt[{4}]{y_{i+1}}}}={\frac {{\sqrt[{4}]{x_{i}}}{\sqrt[{4}]{y_{i+1}}}({\sqrt[{4}]{x_{i}+y_{i}}})}{\sqrt[{4}]{y_{i+1}}}}={\sqrt[{4}]{x_{i}}}({\sqrt[{4}]{x_{i}+y_{i}}})}$

In the infinity

${\displaystyle 0<{\sqrt {x-y}}\leq {\lim _{i\longrightarrow {\infty }}{({\sqrt {x_{i-1}}}+{\sqrt {x_{i}}})}=2{\sqrt {x-y}}}\leq {\lim _{i\longrightarrow {\infty }}{({\sqrt[{4}]{x_{i}}}{\sqrt[{4}]{x_{i}+y_{i}}})}={\sqrt {x-y}}}}$

Therefore

${\displaystyle {\sqrt {x-y}}=2{\sqrt {x-y}}=0}$

Another proof : let

${\displaystyle x_{i}=a_{i}x_{i+1}=x_{i+1}+{\sqrt {x_{i+1}y_{i+1}}}\Rightarrow {a_{i}=1+{\sqrt {\frac {y_{i+1}}{x_{i+1}}}}=1+{\frac {y_{i}}{x_{i}}}}}$

Also

${\displaystyle y_{i}=b_{i}y_{i+1}=y_{i+1}+{\sqrt {x_{i+1}y_{i+1}}}\Rightarrow {b_{i}=1+{\sqrt {\frac {x_{i+1}}{y_{i+1}}}}=1+{\frac {x_{i}}{y_{i}}}}}$

We deduce

${\displaystyle a_{i}=1+{\frac {a_{i}}{b_{i}}}=1+{\frac {y_{i}}{x_{i}}}}$

And

${\displaystyle b_{i}=1+{\frac {b_{i}}{a_{i}}}=1+{\frac {x_{i}}{y_{i}}}}$

Thus

${\displaystyle a_{i}b_{i}=a_{i}+b_{i}}$

And

${\displaystyle x_{i}=c_{i}y_{i+1}=x_{i+1}+{\sqrt {x_{i+1}y_{i+1}}}\Rightarrow {c_{i}=b_{i}(b_{i}-1)}}$

Also

${\displaystyle y_{i}=d_{i}x_{i+1}=y_{i+1}+{\sqrt {x_{i+1}y_{i+1}}}\Rightarrow {d_{i}=a_{i}(a_{i}-1)}}$

We have

${\displaystyle x_{i}=b_{i}{\sqrt {x_{i+1}y_{i+1}}},\quad {y_{i}=a_{i}{\sqrt {x_{i+1}y_{i+1}}}}}$

Let also

${\displaystyle x_{i}=i{\alpha }_{i},\quad {y_{i}=i{\beta }_{i}}}$

And

${\displaystyle {\alpha }_{i}=a'_{i}{\alpha }_{i+1},\quad {{\beta }_{i}=b'_{i}{\beta }_{i+1}}}$

We deduce

${\displaystyle {\frac {a_{i}}{b_{i}}}={\frac {y_{i}}{x_{i}}}={\frac {{\beta }_{i}}{{\alpha }_{i}}}={\frac {a'_{i}}{b'_{i}}}}$

Because

${\displaystyle i{\alpha }_{i}=x_{i}=x_{i+1}+{\sqrt {x_{i+1}y_{i+1}}}=(i+1)({\alpha }_{i+1}+{\sqrt {{\alpha }_{i+1}{\beta }_{i+1}}})}$

${\displaystyle i{\beta }_{i}=y_{i}=y_{i+1}+{\sqrt {x_{i+1}y_{i+1}}}=(i+1)({\beta }_{i+1}+{\sqrt {{\alpha }_{i+1}{\beta }_{i+1}}})}$

Or

${\displaystyle {\frac {{\alpha }_{i}}{{\beta }_{i}}}={\sqrt {\frac {{\alpha }_{i+1}}{{\beta }_{i+1}}}}={\frac {a'_{i}{\alpha }_{i+1}}{b'_{i}{\beta }_{i+1}}}}$

And

${\displaystyle b'_{i}={\frac {i+1}{i}}b_{i}\neq {2},\quad {a'_{i}={\frac {i+1}{i}}a_{i}\neq {2}}}$

But

${\displaystyle b'_{i}-2={\frac {{\alpha }_{i}}{\sqrt {{\alpha }_{i+1}{\beta }_{i+1}}}}-2=a'_{i}{\frac {{\alpha }_{i}}{{\beta }_{i}}}-2}$

${\displaystyle =(a'_{i}-2){\frac {x_{i}}{y_{i}}}+2({\frac {x_{i}}{y_{i}}}-1)}$

And

${\displaystyle a'_{i}-2={\frac {{\beta }_{i}}{\sqrt {{\alpha }_{i+1}{\beta }_{i+1}}}}-2=b'_{i}{\frac {{\beta }_{i}}{{\alpha }_{i}}}-2}$

${\displaystyle =(b'_{i}-2){\frac {y_{i}}{x_{i}}}+2({\frac {y_{i}}{x_{i}}}-1)}$

Thus

${\displaystyle (a'_{i}-2)^{2}{\frac {x_{i}}{y_{i}}}+2({\frac {x_{i}}{y_{i}}}-1)(a'_{i}-2)}$

${\displaystyle =(b'_{i}-2)^{2}{\frac {y_{i}}{x_{i}}}+2({\frac {y_{i}}{x_{i}}}-1)(b'_{i}-2)}$

${\displaystyle =((a'_{i}-2){\frac {x_{i}}{y_{i}}}+2({\frac {x_{i}}{y_{i}}}-1))^{2}{\frac {y_{i}}{x_{i}}}+2({\frac {y_{i}}{x_{i}}}-1)((a'_{i}-2){\frac {x_{i}}{y_{i}}}+2({\frac {y_{i}}{x_{i}}}-1))}$

${\displaystyle =(a'_{i}-2)^{2}{\frac {x_{i}}{y_{i}}}+4({\frac {x_{i}}{y_{i}}}-1)^{2}{\frac {y_{i}}{x_{i}}}+4({\frac {x_{i}}{y_{i}}}-1)(a'_{i}-2)+2(1-{\frac {x_{i}}{y_{i}}})(a'_{i}-2)+4({\frac {y_{i}}{x_{i}}}-1)^{2}}$

${\displaystyle =(a'_{i}-2)^{2}{\frac {x_{i}}{y_{i}}}+4({\frac {x_{i}}{y_{i}}}-1)^{2}{\frac {y_{i}}{x_{i}}}+2({\frac {x_{i}}{y_{i}}}-1)(a'_{i}-2)+4({\frac {y_{i}}{x_{i}}}-1)^{2}}$

Hence

${\displaystyle 4({\frac {x_{i}}{y_{i}}}-1)^{2}{\frac {y_{i}}{x_{i}}}+4({\frac {y_{i}}{x_{i}}}-1)^{2}=0}$

${\displaystyle =4({\frac {(x-y)^{2}}{y_{i}^{2}}}){\frac {y_{i}}{x_{i}}}+4({\frac {(x-y)^{2}}{x_{i}^{2}}})=0}$

${\displaystyle =4(x-y)^{2}({\frac {1}{x_{i}y_{i}}}+{\frac {1}{x_{i}^{2}}})=0}$

${\displaystyle \Rightarrow {(x-y)^{2}=0}}$

Another proof : We have

${\displaystyle c_{i}d_{i}=a_{i}b_{i}}$

Let

${\displaystyle a_{i-1}b_{i-1}=f_{i}a_{i}b_{i}}$

We have

${\displaystyle f_{i}={\frac {a_{i-1}b_{i-1}}{a_{i}b_{i}}}={\frac {x_{i-1}y_{i-1}x_{i+1}y_{i+1}}{x_{i}^{2}y_{i}^{2}}}}$

${\displaystyle ={\frac {{\sqrt {x_{i}y_{i}}}({\sqrt {x_{i}}}+{\sqrt {y_{i}}})^{2}x_{i+1}y_{i+1}}{x_{i+1}y_{i+1}(x_{i}+y_{i})^{2}}}\leq {1}}$

${\displaystyle a_{i}b_{i}={\frac {a_{i}}{b_{i}}}+{\frac {b_{i}}{a_{i}}}+2=({\sqrt {\frac {a_{i}}{b_{i}}}}+{\sqrt {\frac {b_{i}}{a_{i}}}})^{2}=(a_{i-1}b_{i-1}-2)^{2}}$

And

${\displaystyle a_{i}b_{i}=e_{i}=(f_{i}a_{i}b_{i}-2)^{2}}$

It means

${\displaystyle e_{i}^{2}f_{i}^{2}-(4f_{i}+1)e_{i}+4=0}$

And

${\displaystyle e_{i}=a_{i}b_{i}={\frac {4f_{i}+1+{\sqrt {8f_{i}+1}}}{2f_{i}^{2}}}}$

But

${\displaystyle {\sqrt {a_{i}b_{i}}}=f_{i}a_{i}b_{i}-2\Rightarrow {f_{i}a_{i}b_{i}-{\sqrt {a_{i}b_{i}}}-2=0}}$

Thus

${\displaystyle {\sqrt {e_{i}}}={\sqrt {a_{i}b_{i}}}={\frac {1+{\sqrt {1+8f_{i}}}}{2f_{i}}}}$

And

${\displaystyle {\sqrt {e_{i}}}-2={\frac {1-f_{i}+{\sqrt {1+8f_{i}}}-3f_{i}}{2f_{i}}}={\frac {1-f_{i}+{\frac {1-f_{i}^{2}+8f_{i}(1-f_{i})}{{\sqrt {1+8f_{i}}}+3f_{i}}}}{2f_{i}}}}$

${\displaystyle =(1-f_{i}){\frac {1+{\frac {1+9f_{i}}{{\sqrt {8f_{i}1}}+3f_{i}}}}{2f_{i}}}}$

And

${\displaystyle e_{i}-4=a_{i}b_{i}-4=({\sqrt {e_{i}}}-2)({\sqrt {e_{i}}}+2)=({\sqrt {e_{i}}}-2)f_{i}a_{i}b_{i}=(1-f_{i})f_{i}a_{i}b_{i}}$

${\displaystyle {\frac {1-f_{i}^{2}+4f_{i}(1-f_{i})+{\frac {1-f_{i}^{2}+8f_{i}(1-f_{i})}{{\sqrt {8f_{i}+1}}+3f_{i}^{2}}}}{2f_{i}^{2}}}}$

${\displaystyle =(1-f_{i}){\frac {1+5f_{i}+{\frac {1+9f_{i}}{{\sqrt {8f_{i}+1}}+3f_{i}^{2}}}}{2f_{i}^{2}}}}$

${\displaystyle (1-f_{i})(f_{i}a_{i}b_{i}-{\frac {1+5f_{i}+{\frac {1+9f_{i}}{{\sqrt {1+8f_{i}}}+3f_{i}^{2}}}}{2f_{i}^{2}}})=0}$

Or

${\displaystyle (1-f_{i})(a_{i}b_{i}-{\frac {1+5f_{i}+{\frac {1+9f_{i}}{{\sqrt {1+f_{i}}}+3f_{i}^{2}}}}{2f_{i}^{3}}})=0}$

${\displaystyle =(1-f_{i})({\frac {f_{i}(1+4f_{i})+f_{i}{\sqrt {8f_{i}+1}}-1-5f_{i}-{\frac {1+9f_{i}}{{\sqrt {1+8f_{i}}}+3f_{i}^{2}}}}{2f_{i}^{3}}})=0}$

The expression between the parenthesis is not equal to zero, we deduce

${\displaystyle f_{i}=1}$

Else

${\displaystyle {\sqrt {a_{i}b_{i}}}={\frac {2}{f_{i}{\sqrt {a_{i}b_{i}}}-1}}}$

${\displaystyle {\sqrt {a_{i}b_{i}}}-2=2({\frac {2-f_{i}{\sqrt {a_{i}b_{i}}}}{f_{i}{\sqrt {a_{i}b_{i}}}}})=2({\frac {2(1-f_{i})+f_{i}(2-{\sqrt {a_{i}b_{i}}})}{f_{i}{\sqrt {a_{i}b_{i}}}}})}$

We deduce

${\displaystyle ({\sqrt {a_{i}b_{i}}}-2)(1+{\frac {2}{\sqrt {a_{i}b_{i}}}})=({\sqrt {a_{i}b_{i}}}-2)({\frac {{\sqrt {a_{i}b_{i}}}+2}{\sqrt {a_{i}b_{i}}}})}$

${\displaystyle =({\sqrt {a_{i}b_{i}}}-2)({\frac {f_{i}a_{i}b_{i}}{\sqrt {a_{i}b_{i}}}})=({\sqrt {a_{i}b_{i}}}-2)f_{i}{\sqrt {a_{i}b_{i}}}=4{\frac {1-f_{i}}{f_{i}{\sqrt {a_{i}b_{i}}}}}}$

Thus

${\displaystyle ({\frac {1+{\sqrt {1+8f_{i}}}}{2f_{i}}}-2)f_{i}{\sqrt {a_{i}b_{i}}}=(1-f_{i})({\frac {1+{\frac {1+9f_{i}}{{\sqrt {1+8f_{i}}}+3f_{i}}}}{2f_{i}}})=(1-f_{i}){\frac {4}{f_{i}{\sqrt {a_{i}b_{i}}}}}}$

The expressions between the parenthesis are not equal, therefore

${\displaystyle f_{i}=1}$

Else

${\displaystyle a_{i}-4=({\sqrt {e_{i}}}-2)({\sqrt {e_{i}}}+2)=(1-f_{i})({\frac {1+{\frac {1+9f_{i}}{{\sqrt {1+8f_{i}}}+3f_{i}}}}{2f_{i}}})({\frac {1+8f_{i}+{\sqrt {1+8f_{i}}}}{2f_{i}}})}$

Hence

${\displaystyle (1-f_{i})({\frac {1+5f_{i}+{\frac {1+9f_{i}}{{\sqrt {1+8f_{i}}}+3f_{i}^{2}}}}{2f_{i}^{2}}}-({\frac {1+{\frac {1+9f_{i}}{{\sqrt {1+8f_{i}}}+3f_{i}}}}{2f_{i}}})({\frac {1+8f_{i}+{\sqrt {8f_{i}+1}}}{2f_{i}}}))=0}$

Or

${\displaystyle (1-f_{i})(2+10f_{i}+2{\frac {1+9f_{i}}{{\sqrt {1+8f_{i}}}+3f_{i}^{2}}}-(1+8f_{i}+{\frac {(1+9f_{i}){\sqrt {8f_{i}+1}}}{{\sqrt {8f_{i}+}}+3f_{i}}}+{\sqrt {8f_{i}+1}}+{\frac {(1+8f_{i})(1+9f_{i})}{{\sqrt {8f_{i}+1}}+3f_{i}}}))=0}$

The expression between the parenthesis is not equal to zero, we deduce

${\displaystyle f_{i}=1\Rightarrow {x_{i}y_{i}(a_{i}-b_{i}-4)=(x-y)^{2}=0}}$

And, else

${\displaystyle 2f_{i}^{2}e_{i}=4f_{i}+1+{\sqrt {8f_{i}+1}}}$

${\displaystyle (2f_{i}^{2}e_{i}-4f_{i}-1)^{2}=8f_{i}+1=4f_{i}^{4}e_{i}^{2}+16f_{i}^{2}-16f_{i}^{3}e_{i}-4f_{i}^{2}e_{i}+8f_{i}+1}$

${\displaystyle f_{i}^{2}e_{i}^{2}+4-4f_{i}e_{i}-e_{i}=0}$

${\displaystyle (f_{i}^{2}e_{i}-4f_{i}-1)e_{i}=-4}$

${\displaystyle e_{i}={\frac {4}{1+4f_{i}-f_{i}^{2}e_{i}}}}$

${\displaystyle e_{i}-4={\frac {-16f_{i}+4f_{i}^{2}e_{i}}{1+4f_{i}-f_{i}^{2}e_{i}}}}$

${\displaystyle ={\frac {-16f_{i}+16f_{i}^{2}+4f_{i}^{2}(e_{i}-4)}{1+4f_{i}-f_{i}^{2}e_{i}}}}$

${\displaystyle (e_{i}-4)(1-{\frac {4f_{i}^{2}}{1+4f_{i}-f_{i}^{2}e_{i}}})}$

${\displaystyle =(f_{i}-1){\frac {16f_{i}}{1+4f_{i}-f_{i}^{2}e_{i}}}=4f_{i}e_{i}(f_{i}-1)}$

${\displaystyle =(e_{i}-4)(1-f_{i}^{2}e_{i})}$

${\displaystyle e_{i}={\frac {4f_{i}+1+{\sqrt {8f_{i}+1}}}{2f_{i}^{2}}}}$

${\displaystyle e_{i}-4={\frac {4f_{i}(1-f_{i})+1-f_{i}^{2}+{\sqrt {8f_{i}+1}}-3f_{i}^{2}}{2f_{i}^{2}}}}$

${\displaystyle ={\frac {4f_{i}(1-f_{i})+(1-f_{i})(1+f_{i})+{\frac {8f_{i}-8f_{i}^{4}+1-f_{i}^{4}}{{\sqrt {8f_{i}+1}}+3f_{i}^{2}}}}{2f_{i}^{2}}}}$

${\displaystyle =(1-f_{i})({\frac {5f_{i}+1+{\frac {9f_{i}^{3}+9f_{i}^{2}+9f_{i}+1}{{\sqrt {8f_{i}+1}}+3f_{i}^{2}}}}{2f_{i}^{2}}})}$

${\displaystyle (f_{i}-1)(4f_{i}e_{i}+(1-f_{i}^{2}e_{i})({\frac {5f_{i}+1+{\frac {9f_{i}^{3}+9f_{i}^{2}+9f_{i}+1}{{\sqrt {8f_{i}+1}}+3f_{i}^{2}}}}{2f_{i}^{2}}}))=0}$

${\displaystyle =(f_{i}-1)({\frac {4(4f_{i}+1+{\sqrt {8f_{i}+1}})}{2f_{i}}}+(1-{\frac {4f_{i}+1+{\sqrt {8f_{i}+1}}}{2}})({\frac {5f_{i}+1+{\frac {9f_{i}^{3}+9f_{i}^{2}+9f_{i}+1}{{\sqrt {8f_{i}+1}}+3f_{i}^{2}}}}{2f_{i}^{2}}}))=0}$

${\displaystyle =(f_{i}-1)({\frac {5f_{i}+1+{\frac {9f_{i}^{3}+9f_{i}^{2}+9f_{i}+1}{{\sqrt {8f_{i}+1}}+3f_{i}^{2}}}}{2f_{i}^{2}}}-({\frac {4f_{i}+1+{\sqrt {8f_{i}+1}}}{2}})({\frac {-3f_{i}+1+{\frac {9f_{i}^{3}+9f_{i}^{2}+9f_{i}+1}{{\sqrt {8f_{i}+1}}+3f_{i}^{2}}}}{2f_{i}^{2}}}))=0}$

And the expression between the parenthesis is not equal to zero, it means that

${\displaystyle f_{i}=1\Rightarrow {a_{i}b_{i}=e_{i}=4}}$

${\displaystyle a_{i}^{2}b_{i}^{2}-4a_{i}b_{i}=(a_{i}-b_{i})^{2}=0=(y_{i}-x_{i})^{2}x_{i+1}y_{i+1}}$

And

${\displaystyle (a_{i}-b_{i}){\sqrt {x_{i+1}y_{i+1}}}=y-x=0}$

Another proof :

${\displaystyle x_{i}}$ and ${\displaystyle y_{i}}$ are particular cases of ${\displaystyle v_{i}}$ and ${\displaystyle w_{i}}$ which follow

${\displaystyle v_{i}={\frac {x^{i}}{x^{i}-y^{i}}}(x-y),\quad {w_{i}={\frac {y^{i}}{x^{i}-y^{i}}}(x-y)}}$

${\displaystyle x_{i}=v_{2^{i-1}},\quad {y_{i}=w_{2^{i-1}}}}$

Also

${\displaystyle e_{i}={\frac {v_{i}+w_{i}}{w_{i}}},\quad {f_{i}={\frac {v_{i}+w_{i}}{v_{i}}}}}$

${\displaystyle e_{i}={\frac {e_{i}+f_{i}}{f_{i}}},\quad {f_{i}={\frac {e_{i}+f_{i}}{e_{i}}}}}$

${\displaystyle b_{i}=e_{2^{i-1}},\quad {a_{i}=f_{2^{i-1}}}}$

So, we have

${\displaystyle v_{i}-w_{i}=x-y}$

And

${\displaystyle e_{i}f_{i}=e_{i}+f_{i}}$

But

${\displaystyle b_{1}b_{2}...b_{i}=1+{\frac {e_{1}}{f_{1}}}+...+{\frac {e_{2^{i}-1}}{f_{2^{i}-1}}}}$

Let

${\displaystyle 1+{\frac {e_{1}}{f_{1}}}+{\frac {e_{2}}{f_{2}}}+...+{\frac {e_{i}}{f_{i}}}=1+{\frac {v_{1}}{w_{1}}}+{\frac {v_{2}}{w_{2}}}+...+{\frac {v_{i}}{w_{i}}}}$

${\displaystyle =1+{\frac {x}{y}}+{\frac {x^{2}}{y^{2}}}+...+{\frac {x^{i}}{y^{i}}}={\frac {1-{\frac {x^{(i+1)}}{y^{(i+1)}}}}{1-{\frac {x}{y}}}}={\frac {2-e_{i+1}}{2-e_{1}}}}$

Thus

${\displaystyle e_{i+1}-2={\frac {e_{i+1}-f_{i+1}}{f_{i+1}}}={\frac {v_{i+1}-w_{i+1}}{w_{i+1}}}={\frac {x-y}{w_{i+1}}}}$

${\displaystyle =(e_{1}-2)(1+(e_{1}-1)+(e_{2}-1)+...+(e_{i}-1))=(e_{1}-2)(1+i+(e_{1}-2)+(e_{2}-2)+...+(e_{i}-2))}$

${\displaystyle =({\frac {e_{1}-f_{1}}{f_{1}}})(1+i+{\frac {e_{1}-f_{1}}{f_{1}}}+{\frac {e_{2}-f_{2}}{f_{2}}}+...+{\frac {e_{i}-f_{i}}{f_{i}}})}$

${\displaystyle =({\frac {v_{1}-w_{1}}{w_{1}}})(1+i+{\frac {v_{1}-w_{1}}{w_{1}}}+{\frac {v_{2}-w_{2}}{w_{2}}}+...+{\frac {v_{i}-w_{i}}{w_{i}}})}$

Or

${\displaystyle (x-y)({\frac {1}{w_{i+1}}}-{\frac {1}{w_{1}}}(1+i+{\frac {x-y}{w_{1}}}+{\frac {x-y}{w_{2}}}+...+{\frac {x-y}{w_{i}}}))=0}$

And

${\displaystyle {\frac {1}{w_{i+1}}}-(1+i){\frac {1}{w_{1}}}}$

is not equal to zero for ${\displaystyle x-y=0}$, and it is the case, of course, of

${\displaystyle -{\frac {1}{w_{1}}}(x-y)({\frac {1}{w_{1}}}+...+{\frac {1}{w_{i}}})}$

therefore the expression between the parenthesis is not equal to zero. We gave the fourth proof that the only solution is

${\displaystyle xy(x-y)=0}$

And there are others (see the series).

## The Ghanouchi's series

As seen

${\displaystyle {\sqrt {x_{i}y_{i}}}=y_{i-1}-y_{i}=x_{i-1}-x_{i}}$

we deduce

${\displaystyle x_{i}-x_{i+1}={\sqrt {x_{i+1}y_{i+1}}}}$

${\displaystyle x_{i-1}-x_{i}={\sqrt {x_{i}y_{i}}}}$

${\displaystyle x_{1}-x_{2}=x-x_{2}={\sqrt {x_{2}y_{2}}}}$

then

${\displaystyle \sum _{j=2}^{j=i+1}{({\sqrt {x_{j}y_{j}}})}=x-x_{2}+x_{2}-x_{3}+...+x_{i}-x_{i+1}=x-x_{i+1}}$

and

${\displaystyle \sum _{j=2}^{j=\infty }{({\sqrt {x_{j}y_{j}}})}=\lim _{i\longrightarrow {\infty }}{(x-x_{i+1})}}$

if ${\displaystyle x>y}$

${\displaystyle \sum _{j=2}^{j=\infty }{({\sqrt {x_{j}y_{j}}})}=\lim _{i\longrightarrow {\infty }}{(x-x_{i+1})}=x-(x-y)=y}$

if ${\displaystyle x

${\displaystyle \sum _{j=2}^{j=\infty }{({\sqrt {x_{j}y_{j}}})}=\lim _{i\longrightarrow {\infty }}{(x-x_{i+1})}=x}$

## The applications of the Ghanouchi's series

${\displaystyle \sum _{j=2}^{j=i}{((-1)^{j}{\sqrt {x_{j}y_{j}}})}=x-x_{2}-(x_{2}-x_{3})+(x_{3}-x_{4})-...+(-1)^{i}(x_{i-1}-x_{i})}$

${\displaystyle =x-2x_{2}+2x_{3}-...+2(-1)^{i-1}x_{i-1}+(-1)^{i+1}x_{i}}$

${\displaystyle =2\sum _{j=2}^{j=i-1}{((-1)^{j+1}x_{j})}+x+(-1)^{i+1}x_{i}}$

${\displaystyle =2\sum _{j=1}^{j=i}{((-1)^{j+1}x_{j})}-x-(-1)^{i+1}x_{i}}$

${\displaystyle =\sum _{j=2}^{j=i-1}{((-1)^{j+1}x_{j})}+\sum _{j=1}^{j=i}{((-1)^{j+1}x_{j})}}$

${\displaystyle =2\sum _{j=2}^{j=i-1}{((-1)^{j+1}y_{j})}+y+(-1)^{i+1}y_{i}}$

${\displaystyle =2\sum _{j=1}^{j=i}{((-1)^{j+1}y_{j})}-y-(-1)^{i+1}y_{i}}$

${\displaystyle =\sum _{j=2}^{j=i-1}{((-1)^{j+1}y_{j})}+\sum _{j=1}^{j=i}{((-1)^{j+1}y_{j})}}$

or

${\displaystyle 2\sum _{j=1}^{j=i}{((-1)^{j+1}x_{j})}=\sum _{j=2}^{j=i}{((-1)^{j}{\sqrt {x_{j}y_{j}}})}+x+(-1)^{i+1}x_{i}}$

and

${\displaystyle 2\sum _{j=1}^{j=i}{((-1)^{j+1}y_{j})}=\sum _{j=2}^{j=i}{((-1)^{j}{\sqrt {x_{j}y_{j}}})}+y+(-1)^{i+1}y_{i}}$

As we do not know the limit of ${\displaystyle (-1)^{i+1}x_{i}}$, then

${\displaystyle \sum _{j=1}^{j=\infty }{((-1)^{j}x_{j})}}$

can be not convergent. But

${\displaystyle \sum _{j=2}^{j=\infty }{((-1)^{j}{\sqrt {x_{j}y_{j}}})}}$

is convergent.

Also, knowing that ${\displaystyle y_{i}}$ tends to zero in the infinity, we can say

${\displaystyle \sum _{j=1}^{j=\infty }{((-1)^{j}y_{j})}}$

is convergent. The limit of

${\displaystyle \sum _{j=2}^{j=i}{((-1)^{j}{\sqrt {x_{j}y_{j}}})}=2\sum _{j=1}^{j=i}{((-1)^{j+1}y_{j})}-y-(-1)^{i+1}y_{i}}$

${\displaystyle =2\sum _{j=1}^{j=i}{((-1)^{j+1}x_{j})}-x-(-1)^{i+1}x_{i}}$

exists and the series are convergent. It means that for x and y integers and for conditions on the exponents like ${\displaystyle n\geq {3}}$ for Fermat equation :

${\displaystyle \lim _{i\longrightarrow {\infty }}{(x_{i})}=x-y=0}$

It is confirmed by the fact that the général term of the series tends to zero. Let us prove it. We give two proofs. We must remark that we prove firstly that the following series are convergent, we do not present the proof, here. Let

${\displaystyle \sum _{k=1}^{k=2m}{((-1)^{k+1}x_{k}e^{-{\frac {k}{\sqrt {2m}}}})}}$

${\displaystyle =xe^{-{\frac {1}{\sqrt {2m}}}}-x_{2}e^{-{\frac {2}{\sqrt {2m}}}}+x_{3}e^{-{\frac {3}{\sqrt {2m}}}}-...+(-1)^{2m+1}x_{2m}e^{-{\frac {2m}{\sqrt {2m}}}}}$

${\displaystyle =xe^{-{\frac {2}{\sqrt {2m}}}}+x(e^{-{\frac {1}{\sqrt {2m}}}}-e^{-{\frac {2}{\sqrt {2m}}}})-x_{2}e^{-{\frac {2}{\sqrt {2m}}}}+x_{3}e^{-{\frac {4}{\sqrt {2m}}}}+x_{3}(e^{-{\frac {3}{\sqrt {2m}}}}-e^{-{\frac {4}{\sqrt {2m}}}})-x_{4}e^{-{\frac {4}{\sqrt {2m}}}}+...-x_{2m}e^{-{\frac {2m}{\sqrt {2m}}}}}$

${\displaystyle =xe^{-{\frac {2}{\sqrt {2m}}}}(e^{\frac {1}{\sqrt {2m}}}-1)+x_{3}e^{-{\frac {4}{\sqrt {2m}}}}(e^{\frac {1}{\sqrt {2m}}}-1)+...+x_{2m-1}e^{-{\frac {2m}{\sqrt {2m}}}}(e^{\frac {1}{\sqrt {2m}}}-1)+}$

${\displaystyle +(x-x_{2})e^{-{\frac {2}{\sqrt {2m}}}}+(x_{3}-x_{4})e^{-{\frac {4}{\sqrt {2m}}}}+...+(x_{2m-1}-x_{2m})e^{-{\frac {2m}{\sqrt {2m}}}}}$

${\displaystyle =(e^{\frac {1}{\sqrt {2m}}}-1)(xe^{-{\frac {2}{\sqrt {2m}}}}+x_{3}e^{-{\frac {4}{\sqrt {2m}}}}+...+x_{2m-1}e^{-{\sqrt {2m}}})+({\sqrt {x_{2}y_{2}}}e^{-{\frac {2}{\sqrt {2m}}}}+{\sqrt {x_{4}y_{4}}}e^{-{\frac {4}{\sqrt {2m}}}}+...+{\sqrt {x_{2m}y_{2m}}}e^{-{\frac {2m}{\sqrt {2m}}}})}$

${\displaystyle =(e^{\frac {1}{\sqrt {2m}}}-1)\sum _{k=1}^{k=m}{(x_{2k-1}e^{-{\frac {2k}{\sqrt {2m}}}})}+\sum _{k=1}^{k=m}{({\sqrt {x_{2k}y_{2k}}}e^{-{\frac {2k}{\sqrt {2m}}}})}}$

Also

${\displaystyle \sum _{k=1}^{k=2m}{((-1)^{k+1}y_{k}e^{-{\frac {k}{\sqrt {2m}}}})}}$

${\displaystyle =ye^{-{\frac {1}{\sqrt {2m}}}}-y_{2}e^{-{\frac {2}{\sqrt {2m}}}}+y_{3}e^{-{\frac {3}{\sqrt {2m}}}}-...+(-1)^{2m+1}y_{2m}e^{-{\frac {2m}{\sqrt {2m}}}}}$

${\displaystyle =(e^{\frac {1}{\sqrt {2m}}}-1)\sum _{k=1}^{k=m}{(y_{2k-1}e^{-{\frac {2k}{\sqrt {2m}}}})}+\sum _{k=1}^{k=m}{({\sqrt {x_{2k}y_{2k}}}e^{-{\frac {2k}{\sqrt {2m}}}})}}$

But

${\displaystyle (e^{\frac {1}{\sqrt {2m}}}-1)\sum _{k=1}^{k=m}{(y_{2k-1}e^{-{\frac {2k}{\sqrt {2m}}}})}=S}$

We have

${\displaystyle (e^{\frac {1}{\sqrt {2m}}}-1)\sum _{k=1}^{k=m}{(y_{2k-1}e^{-{\frac {2k+1}{\sqrt {2m}}}})}

${\displaystyle (e^{\frac {1}{\sqrt {2m}}}-1)e^{-{\frac {3}{\sqrt {2m}}}}\sum _{k=1}^{k=m}{(y_{2k-1}e^{-{\frac {2k-2}{\sqrt {2m}}}})}

Thus

${\displaystyle \lim _{m\longrightarrow {\infty }}{((e^{\frac {1}{\sqrt {2m}}}-1)\sum _{k=1}^{k=m}{(y_{2k-1}e^{-{\frac {2k-2}{\sqrt {2m}}}})})}=\lim _{m\longrightarrow {\infty }}{((e^{\frac {1}{\sqrt {2m}}}-1)y+(e^{\frac {1}{\sqrt {2m}}}-1)\sum _{k=2}^{k=m}{(y_{2k-1}e^{-{\frac {2k-2}{\sqrt {2m}}}})})}}$

${\displaystyle =\lim _{m\longrightarrow {\infty }}{((e^{\frac {1}{\sqrt {2m}}}-1)\sum _{k=2}^{k=m}{(y_{2k-1}e^{-{\frac {2k-2}{\sqrt {2m}}}})})}=\lim _{m\longrightarrow {\infty }}{(S)}}$

We have

${\displaystyle (e^{\frac {1}{{\sqrt {2}}m}}-1)\sum _{k=p}^{k=m}{(y_{2k-1}e^{-{\frac {2k-p}{\sqrt {2m}}}})}=A}$

And

${\displaystyle (e^{\frac {1}{\sqrt {2m}}}-1)\sum _{k=p}^{k=m}{(y_{2k-1}e^{-{\frac {2k-p+1}{\sqrt {2m}}}})}

Or

${\displaystyle (e^{\frac {1}{\sqrt {2m}}}-1)e^{-{\frac {2}{\sqrt {2m}}}}\sum _{k=p}^{k=m}{(y_{2k-1}e^{-{\frac {2k-p-1}{\sqrt {2m}}}})}

Hance

${\displaystyle \lim _{m\longrightarrow {\infty }}{((e^{\frac {1}{\sqrt {2m}}}-1)\sum _{k=p}^{k=m}{(y_{2k-1}e^{-{\frac {2k-p-1}{\sqrt {2m}}}})})}}$

${\displaystyle =\lim _{m\longrightarrow {\infty }}{((e^{\frac {1}{\sqrt {2m}}}-1)y_{2p-1}e^{-{\frac {p-1}{\sqrt {2m}}}}+(e^{\frac {1}{\sqrt {2m}}}-1)\sum _{k=p+1}^{k=m}{(y_{2k-1}e^{-{\frac {2k-p-1}{\sqrt {2m}}}})})}}$

${\displaystyle =\lim _{m\longrightarrow {\infty }}{((e^{\frac {1}{\sqrt {2m}}}-1)\sum _{k=p+1}^{k=m}{(y_{2k-1}e^{-{\frac {2k-p-1}{\sqrt {2m}}}})})}=\lim _{m\longrightarrow {\infty }}{(A)}=\lim _{m\longrightarrow {\infty }}{(S)}}$

And

${\displaystyle p+1=m\Rightarrow {\lim _{m\longrightarrow {\infty }}{(A)}=\lim _{m\longrightarrow {\infty }}{(S)}=\lim _{m\longrightarrow {\infty }}{((e^{\frac {1}{\sqrt {2m}}}-1)y_{2m-1}e^{-{\frac {m}{\sqrt {2m}}}})}=0}}$

Consequently

${\displaystyle \lim _{m\longrightarrow {\infty }}{((e^{\frac {1}{\sqrt {2m}}}-1)\sum _{k=1}^{k=m}{(y_{2k-1}e^{-{\frac {2k}{\sqrt {2m}}}})})}=0}$

Also

${\displaystyle (e^{\frac {1}{\sqrt {2m}}}-1)\sum _{k=1}^{k=m}{(x_{2k-1}e^{-{\frac {2k}{\sqrt {2m}}}})}=S}$

We have

${\displaystyle (e^{\frac {1}{\sqrt {2m}}}-1)\sum _{k=1}^{k=m}{(x_{2k-1}e^{-{\frac {2k+1}{\sqrt {2m}}}})}

${\displaystyle (e^{\frac {1}{\sqrt {2m}}}-1)e^{-{\frac {3}{\sqrt {2m}}}}\sum _{k=1}^{k=m}{(x_{2k-1}e^{-{\frac {2k-2}{\sqrt {2m}}}})}

Thus

${\displaystyle \lim _{m\longrightarrow {\infty }}{((e^{\frac {1}{\sqrt {2m}}}-1)\sum _{k=1}^{k=m}{(x_{2k-1}e^{-{\frac {2k-2}{\sqrt {2m}}}})})}=\lim _{m\longrightarrow {\infty }}{((e^{\frac {1}{\sqrt {2m}}}-1)x+(e^{\frac {1}{\sqrt {2m}}}-1)\sum _{k=2}^{k=m}{(x_{2k-1}e^{-{\frac {2k-2}{\sqrt {2m}}}})})}}$

${\displaystyle =\lim _{m\longrightarrow {\infty }}{((e^{\frac {1}{\sqrt {2m}}}-1)\sum _{k=2}^{k=m}{(x_{2k-1}e^{-{\frac {2k-2}{\sqrt {2m}}}})})}=\lim _{m\longrightarrow {\infty }}{(S)}}$

And

${\displaystyle (e^{\frac {1}{{\sqrt {2}}m}}-1)\sum _{k=p}^{k=m}{(x_{2k-1}e^{-{\frac {2k-p}{\sqrt {2m}}}})}=A}$

And

${\displaystyle (e^{\frac {1}{\sqrt {2m}}}-1)\sum _{k=p}^{k=m}{(x_{2k-1}e^{-{\frac {2k-p+1}{\sqrt {2m}}}})}

Or

${\displaystyle (e^{\frac {1}{\sqrt {2m}}}-1)e^{-{\frac {2}{\sqrt {2m}}}}\sum _{k=p}^{k=m}{(x_{2k-1}e^{-{\frac {2k-p-1}{\sqrt {2m}}}})}

Hence

${\displaystyle \lim _{m\longrightarrow {\infty }}{((e^{\frac {1}{\sqrt {2m}}}-1)\sum _{k=p}^{k=m}{(x_{2k-1}e^{-{\frac {2k-p-1}{\sqrt {2m}}}})})}}$

${\displaystyle =\lim _{m\longrightarrow {\infty }}{((e^{\frac {1}{\sqrt {2m}}}-1)x_{2p-1}e^{-{\frac {p-1}{\sqrt {2m}}}}+(e^{\frac {1}{\sqrt {2m}}}-1)\sum _{k=p+1}^{k=m}{(x_{2k-1}e^{-{\frac {2k-p-1}{\sqrt {2m}}}})})}}$

${\displaystyle =\lim _{m\longrightarrow {\infty }}{((e^{\frac {1}{\sqrt {2m}}}-1)\sum _{k=p+1}^{k=m}{(x_{2k-1}e^{-{\frac {2k-p-1}{\sqrt {2m}}}})})}=\lim _{m\longrightarrow {\infty }}{(A)}=\lim _{m\longrightarrow {\infty }}{(S)}}$

Or

${\displaystyle p+1=m\Rightarrow {\lim _{m\longrightarrow {\infty }}{(A)}=\lim _{m\longrightarrow {\infty }}{(S)}=\lim _{m\longrightarrow {\infty }}{((e^{\frac {1}{\sqrt {2m}}}-1)x_{2m-1}e^{-{\frac {m}{\sqrt {2m}}}})}=0}}$

Consequently

${\displaystyle \lim _{m\longrightarrow {\infty }}{((e^{\frac {1}{\sqrt {2m}}}-1)\sum _{k=1}^{k=m}{(x_{2k-1}e^{-{\frac {2k}{\sqrt {2m}}}})})}=0}$

We deduce

${\displaystyle 0<\lim _{m\longrightarrow {\infty }}{(\sum _{k=1}^{k=2m}{((-1)^{k+1}x_{k}e^{-{\frac {k}{\sqrt {2m}}}})})}}$

${\displaystyle =\lim _{m\longrightarrow {\infty }}{(\sum _{k=1}^{k=2m}{((-1)^{k+1}x_{k}e^{-{\frac {k}{\sqrt {2m}}}})})}}$

${\displaystyle =\lim _{m\longrightarrow {\infty }}{(\sum _{k=1}^{k=m}{({\sqrt {x_{2k}y_{2k}}}e^{-{\frac {2k}{\sqrt {2m}}}})})}<\lim _{m\longrightarrow {\infty }}{(\sum _{k=1}^{k=m}{({\sqrt {x_{2k}y_{2k}}})})}}$

${\displaystyle <\lim _{m\longrightarrow {\infty }}{(\sum _{k=1}^{k=2m}{({\sqrt {x_{k}y_{k}}})})}=y}$

Thus

${\displaystyle \lim _{m\longrightarrow {\infty }}{(\sum _{k=1}^{k=2m}{((-1)^{k+1}(x_{k}-y_{k})e^{-{\frac {k}{\sqrt {2m}}}})})}=0}$

${\displaystyle =\lim _{m\longrightarrow {\infty }}{((x-y)\sum _{k=1}^{k=2m}{(e^{-{\frac {k}{\sqrt {2m}}}})})}=\lim _{m\longrightarrow {\infty }}{((x-y)e^{-{\frac {1}{\sqrt {2m}}}}{\frac {1-e^{-{\sqrt {2m}}}}{1+e^{-{\frac {1}{\sqrt {2m}}}}}})}={\frac {x-y}{2}}=0}$

And

${\displaystyle x-y=0}$

the only solution is ${\displaystyle xy(x-y)=0}$, ${\displaystyle xy=0}$ if at least one of the sequences ${\displaystyle x_{i}}$ or ${\displaystyle y_{i}}$ is constant. And the second proof. Let

${\displaystyle x_{i}(t)=\sum _{k=2}^{k=i-1}{(x_{k}e^{-kt})}=\sum _{k=2}^{k=i}{(x_{k}e^{-kt})}-x_{i}e^{-it}={x'}_{i}(t)-x_{i}e^{-it}}$

${\displaystyle y_{i}(t)=\sum _{k=2}^{k=i-1}{(y_{k}e^{-kt})}=\sum _{k=2}^{k=i}{(y_{k}e^{-kt})}-y_{i}e^{-it}={y'}_{i}(t)-y_{i}e^{-it}}$

${\displaystyle u_{i}(t)=\sum _{k=2}^{k=i}{({\sqrt {x_{k}y_{k}}}e^{-kt})}=\sum _{k=2}^{k=i-1}{((x_{k-1}-x_{k})e^{-kt})}}$

${\displaystyle =xe^{-2t}+x_{2}(-e^{-2t}+e^{-3t})+x_{3}(-e^{-3t}+e^{-4t})+...+x_{i-1}(-e^{-(i-1)t}+e^{-it})-x_{i}e^{-it}}$

${\displaystyle =xe^{-2t}+x_{2}e^{-2t}(-1+e^{-t})+x_{3}e^{-3t}(-1+e^{-t})+...+x_{i-1}e^{-(i-1)t}(-1+e^{-t})-x_{i}e^{-it}}$

${\displaystyle =xe^{-2t}-x_{i}e^{-it}+(-1+e^{-t})(x_{2}e^{-2t}+x_{3}e^{-3t}+...+x_{i}e^{-(i-1)t})}$

${\displaystyle =xe^{-2t}-x_{i}e^{-it}+(-1+e^{-t})x_{i}(t)=xe^{-2t}-x_{i}e^{-it}+(-1+e^{-t}){x'}_{i}(t)-(-1+e^{-t})x_{i}e^{-it}=xe^{-2t}-x_{i}e^{-(i+1)t}+(-1+e^{-t}){x'}_{i}(t)}$

let

${\displaystyle t={\frac {1}{\sqrt {i}}}\Rightarrow {\lim _{i\longrightarrow {\infty }}{(u_{i}({\frac {1}{\sqrt {i}}}))}=\lim _{i\longrightarrow {\infty }}{(\sum _{k=2}^{k=i}{({\sqrt {x_{k}y_{k}}}e^{-{\frac {k}{\sqrt {i}}}})})}}=\lim _{i\longrightarrow {\infty }}{(\sum _{k=2}^{k=i}{({\sqrt {x_{k}y_{k}}})})}=y}$

${\displaystyle =x+\lim _{i\longrightarrow {\infty }}{((-1+e^{-{\frac {1}{\sqrt {i}}}})\sum _{k=2}^{k=i-1}{(x_{k}e^{-{\frac {k}{\sqrt {i}}}})}-x_{i}e^{-{\sqrt {i}}})}=x}$

${\displaystyle =y+\lim _{i\longrightarrow {\infty }}{((-1+e^{-{\frac {1}{\sqrt {i}}}})\sum _{k=2}^{k=i-1}{(y_{k}e^{-{\frac {k}{\sqrt {i}}}})}-y_{i}e^{-{\sqrt {i}}})}=y}$

We Recapitulate

${\displaystyle x_{i}>x_{i+1},\quad {y_{i}>y_{i+1}}\Rightarrow {x=y}}$

${\displaystyle x_{i}=x_{i+1}=x_{i+1}+{\sqrt {x_{i+1}y_{i+1}}}\Rightarrow {xy=0}}$

${\displaystyle y_{i}=y_{i+1}=y_{i+1}+{\sqrt {x_{i+1}y_{i+1}}}\Rightarrow {xy=0}}$

${\displaystyle \Rightarrow {xy(x-y)=0}}$

is the only solution of (1) and (2). This result is paradoxal, we remark that we have not put any condition on n, because there are solutions for ${\displaystyle n\leq {2}}$. The answer is related to Matiasevic theorem which claims that dos not exist an algorithm to prove theorems related to diophantine equations and we gave one : The approach must conduct then to an impossibility. We confirm Matiasevic theorem and prove it because our algorithm is available for n=1. The approach is more important than it appears, it is an answer to problems more general than Fermat theorem or Beal or Fermat-Catalan conjectures. We will try to prove them. The series become

${\displaystyle \sum _{k=2}^{k={\infty }}{((-1)^{k}{\sqrt {x_{k}y_{k}}})}=\sum _{k=2}^{k={\infty }}{((-1)^{k}x_{k})}=\sum _{k=2}^{k={\infty }}{((-1)^{k}y_{k})}}$

${\displaystyle =x_{2}-x_{3}+x_{4}-x_{5}+...}$

${\displaystyle =2\sum _{k=1}^{k={\infty }}{((-1)^{k+1}x_{k})}-x=x-2x_{2}+2x_{3}-2x_{4}+...}$

${\displaystyle =2\sum _{k=1}^{k={\infty }}{((-1)^{k+1}y_{k})}-y=y-2y_{2}+2y_{3}-2y_{4}+...}$

${\displaystyle \Rightarrow {3(x_{2}-x_{3}+x_{4}-x_{5}+...)=3\sum _{k=2}^{k={\infty }}{((-1)^{k}x_{k})}=3\sum _{k=2}^{k={\infty }}{((-1)^{k}y_{k})}=x=y}}$

${\displaystyle \Rightarrow {x_{2}-x_{3}+x_{4}-x_{5}+...=\sum _{k=2}^{k={\infty }}{((-1)^{k}x_{k})}=\sum _{k=2}^{k={\infty }}{((-1)^{k}y_{k})}={\frac {x}{3}}={\frac {y}{3}}}}$

and

${\displaystyle \sum _{k=2}^{k={\infty }}{({\sqrt {x_{k}y_{k}}})}=\sum _{k=2}^{k={\infty }}{(x_{k})}=\sum _{k=2}^{k={i}}{(y_{k})}}$

${\displaystyle =x_{2}+x_{3}+x_{4}+x_{5}+...=y_{2}+y_{3}+y_{4}+y_{5}+...=y=x}$

and

${\displaystyle x_{i}=x^{2^{i-1}}\prod _{j=0}^{j=i-2}{(x^{2^{j}}+y^{2^{j}})^{-1}}=y_{i}}$

${\displaystyle =x^{2^{i-1}}\prod _{j=0}^{j=i-2}{(2x^{2^{j}})^{-1}}={\frac {x^{2^{i-1}}}{2^{i-1}x^{2^{i-1}-1}}}}$

${\displaystyle =x_{i}={\frac {x}{2^{i-1}}}=y_{i}={\frac {y}{2^{i-1}}}}$

This development is in fact a test of impossibility. The sequences and series are a consequence of Fermat equation and of other diophantine equations (as we will see). The question now is : why are there solutions for n=2 ? The answer is in the formulas, as seen. It is important to note that for n=1, there are trivial solutions. But, for n=1, lemma 3 allows to write

${\displaystyle x_{i}={\frac {x^{2^{i-1}}}{x^{2^{i-1}}-y^{2^{i-1}}}}(x-y)=U{\frac {X^{2^{i-1}}}{X^{2^{i-1}}-Y^{2^{i-1}}}}(X-Y)}$

${\displaystyle =U{\frac {X^{2{2^{i-2}}}}{X^{2{2^{i-2}}}-Y^{2{2^{i-2}}}}}(X-Y)}$

and

${\displaystyle y_{i}={\frac {y^{2^{i-1}}}{x^{2^{i-1}}-y^{2^{i-1}}}}(x-y)=U{\frac {Y^{2^{i-1}}}{X^{2^{i-1}}-Y^{2^{i-1}}}}(X-Y)}$

${\displaystyle =U{\frac {Y^{2{2^{i-2}}}}{X^{2{2^{i-2}}}-Y^{2{2^{i-2}}}}}(X-Y)}$

and

${\displaystyle u_{i}=x_{i}+y_{i}=U{\frac {X^{2{2^{i-2}}}+Y^{2{2^{i-2}}}}{X^{2{2^{i-2}}}-Y^{2{2^{i-2}}}}}(X-Y)={\frac {X^{2{2^{i-2}}}+Y^{2{2^{i-2}}}}{X^{2{2^{i-2}}}-Y^{2{2^{i-2}}}}}(X^{2}-Y^{2})}$

It is the expression of ${\displaystyle {u'}_{i-1}}$, the ${\displaystyle u_{i-1}}$ exponent 2.

The case n=1 conducts to the case n=2 and as there are solutions for n=1, it will be the same for n=2 !

For n=4

${\displaystyle u_{i}=x_{i}+y_{i}={\frac {X^{4{2^{i-3}}}+Y^{4{2^{i-3}}}}{X^{4{2^{i-3}}}-Y^{4{2^{i-3}}}}}U(X-Y)}$

the case n=4 is different, in this formula the exponent i-3 does not guarantee the existence of the sequences if

i=2.

Then, the case n=2 is the only exception. The only solution for n>2 is xy(x-y)=0, there is no solution.

Another application is Beal equation. It is

${\displaystyle U^{c}=X^{a}+Y^{b}}$

${\displaystyle GCD(X,Y)=1}$

We pose

${\displaystyle u=U^{2c}}$

${\displaystyle x=U^{c}X^{a}}$

${\displaystyle y=U^{c}Y^{b}}$

${\displaystyle z=X^{a}Y^{b}}$

and

${\displaystyle u=U^{2c}=U^{c}(X^{a}+Y^{b})=x+y}$

and

${\displaystyle {\frac {1}{z}}={\frac {1}{X^{a}Y^{b}}}={\frac {U^{2c}}{U^{c}X^{a}U^{c}Y^{b}}}={\frac {u}{xy}}={\frac {x+y}{xy}}={\frac {1}{x}}+{\frac {1}{y}}}$

it is lemma 1 and, with

${\displaystyle U^{c}=X^{a}+Y^{b}}$

the solutions are

${\displaystyle xy(x-y)=U^{2c}X^{a}Y^{b}(U^{c}X^{a}-U^{c}Y^{b})=0}$

or impossible solutions for a>2 et b>2 et c>2.

Another application is the following equation.

${\displaystyle U^{n}=X_{1}^{n_{1}}+X_{2}^{n_{2}}+...+X_{i}^{n_{i}}}$

${\displaystyle GCD(X_{j},X_{k})=1;\forall {j,k,j\neq {k}}}$

We conjecture and prove that there is no solutions for n>i(i-1) and ${\displaystyle n_{j}>i(i-1)}$, ${\displaystyle \forall {j\in \{1,2,...,i\}}}$. We can not know when there are solutions as proved by Matiasevic when one of the exponent is less or equal to i(i-1).

## LEMMA 6

The solution ${\displaystyle \forall {k\geq {2}}}$ is

${\displaystyle X_{k}=0}$

## Proof of lemma 6

We pose

${\displaystyle u=U^{2n}}$

${\displaystyle x=U^{n}X_{k}^{n_{k}}}$

${\displaystyle y=U^{n}(U^{n}-X_{k}^{n_{k}})}$

${\displaystyle z=X_{k}^{n_{k}}(U^{n}-X_{k}^{n_{k}})}$

or

${\displaystyle u=x+y}$

and

${\displaystyle {\frac {1}{z}}={\frac {1}{x}}+{\frac {1}{y}}}$

it is lemma 1. Its solution is

${\displaystyle U=X_{k}=0;\forall {k\in {\{1,2,...,i\}}}}$

But, why are not they solutions for n>i(i-1) and ${\displaystyle n_{j}>i(i-1)}$ ?

We will generalize the definition of the sequences.

We will define general sequences.

Our goal is to prove that if ${\displaystyle X_{i}}$,

(${\displaystyle i\geq {2}}$), ${\displaystyle n_{i}}$, ${\displaystyle U}$, ${\displaystyle n}$, ${\displaystyle PGCD(X_{1},X_{2},...,X_{i})=1}$

are positive integers, then

${\displaystyle X_{a}=X_{b}=0;\forall {a,b=1,2,...,i}}$

${\displaystyle n_{k}>{i(i-1)},\forall {k=1,2,...,i},n>i(i-1)}$

for the equation

${\displaystyle X_{1}^{n_{1}}+X_{2}^{n_{2}}+...+X_{i}^{n_{i}}=U^{n}\quad \quad {(e)}}$

When ${\displaystyle n\leq {i(i-1)},n_{k}\leq {i(i-1)},k=1,2,...,i}$

there are solutions, for example :

${\displaystyle i=2}$ has ${\displaystyle 3^{2}+4^{2}=5^{2}}$

and ${\displaystyle i=3}$ has ${\displaystyle 3^{3}+4^{3}+5^{3}=6^{3}}$

and

${\displaystyle 95800^{4}+217519^{4}+414560^{4}=422481^{4}}$

and ${\displaystyle i=4}$ has ${\displaystyle 27^{5}+84^{5}+110^{5}+133^{5}=144^{5}}$

etc...

We suppose (e) verified and that