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 ($2^{2^{n}}$). 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

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

with GCD(X,Y)=1

We pose

$u=U^{2n}$

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

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

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

then

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

and

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

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

LEMMA 1

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

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

Firstly

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

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

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

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

We pose

$x_{1}=x$

and

$y_{1}=y$

but

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

$\exists {z_{1}}$

verifying

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

and

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

then

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

or

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

we pose

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

and

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

also

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

and

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

which means

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

and

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

and

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

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

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

$x_{1}$ is an integer

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

$y_{1}$ is an integer

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

$z_{2}$ is rational, because

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

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

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

until infinity. For i

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

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

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

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

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

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

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

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

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

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

LEMMA 2

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

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

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

Proof of lemma 2

By induction

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

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

also

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

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

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

and

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

but (H) and (H'), then

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

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

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

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

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

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

$\forall {i>1}$

but, $\forall {a,b}$

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

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

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

the expressions of the sequences become, for

$x\neq {y}$

or

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

LEMMA 3

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

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

and

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

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

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

LEMMA 4

The equations (1) have (2) a constant

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

THE GHANOUCHI'S THEOREM

The only solution of equations

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

and

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

is

$xy(x-y)=0$

Proof of Ghanouchi's theorem

As

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

and

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

if

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

and

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

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

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

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

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

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

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

$xy(x-y)=0$

The proofs : we have

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

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

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

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

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

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

And

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

Also

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

Let

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

But

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

Hence

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

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

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

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

We deduce

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

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

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

Hence

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

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

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

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

We deduce

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

If we add

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

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

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

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

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

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

The solution is

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

Another proof : we have

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

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

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

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

Because

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

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

And

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

Because

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

And

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

We deduce

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

In the infinity

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

Therefore

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

Another proof : let

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

Also

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

We deduce

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

And

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

Thus

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

And

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

Also

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

We have

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

Let also

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

And

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

We deduce

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

Because

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

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

Or

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

And

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

But

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

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

And

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

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

Thus

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

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

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

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

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

Hence

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

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

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

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

Another proof : We have

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

Let

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

We have

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

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

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

And

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

It means

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

And

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

But

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

Thus

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

And

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

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

And

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

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

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

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

Or

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

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

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

$f_{i}=1$

Else

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

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

We deduce

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

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

Thus

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

The expressions between the parenthesis are not equal, therefore

$f_{i}=1$

Else

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

Hence

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

Or

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

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

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

And, else

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

And

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

Another proof :

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

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

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

Also

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

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

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

So, we have

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

And

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

But

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

Let

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

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

Thus

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

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

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

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

Or

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

And

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

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

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

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

$xy(x-y)=0$

And there are others (see the series).

The Ghanouchi's series

As seen

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

we deduce

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

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

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

then

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

and

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

if $x>y$

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

if $x<y$

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

The applications of the Ghanouchi's series

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

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

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

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

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

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

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

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

or

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

and

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

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

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

can be not convergent. But

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

is convergent.

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

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

is convergent. The limit of

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

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

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

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

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

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

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

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

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

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

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

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

Also

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

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

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

But

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

We have

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

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

Thus

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

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

We have

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

And

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

Or

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

Hance

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

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

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

And

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

Consequently

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

Also

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

We have

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

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

Thus

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

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

And

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

And

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

Or

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

Hence

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

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

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

Or

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

Consequently

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

We deduce

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

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

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

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

Thus

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

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

And

$x-y=0$

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

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

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

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

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

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

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

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

let

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

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

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

We Recapitulate

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

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

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

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

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 $n\leq {2}$. 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

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

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

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

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

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

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

and

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

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

and

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

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

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

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

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

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

and

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

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

and

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

It is the expression of ${u'}_{i-1}$, the $u_{i-1}$ 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

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

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

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

$GCD(X,Y)=1$

We pose

$u=U^{2c}$

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

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

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

and

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

and

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

it is lemma 1 and, with

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

the solutions are

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

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

Another application is the following equation.

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

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

We conjecture and prove that there is no solutions for n>i(i-1) and $n_{j}>i(i-1)$, $\forall {j\in \{1,2,...,i\}}$. 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 $\forall {k\geq {2}}$ is

$X_{k}=0$

Proof of lemma 6

We pose

$u=U^{2n}$

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

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

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

or

$u=x+y$

and

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

it is lemma 1. Its solution is

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

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

We will generalize the definition of the sequences.

We will define general sequences.

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

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

are positive integers, then

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

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

for the equation

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

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

there are solutions, for example :

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

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

and

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

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

etc...

We suppose (e) verified and that $GCD(X_{k})=1$

$k\in \{1,2,...,i\};i\geq {2}$, soit

$x_{k}=U^{(i-1)n}X_{k}^{n_{k}}$

$k\in {\{1,2,...,i\}}$

and

$u=U^{in}$

and

$v=X_{1}^{n_{1}}X_{2}^{n_{2}}...X_{i}^{n_{i}}$

LEMMA 7

$x_{k}$, $k=1,2,...i$, u, v verify

$x_{1}+x_{2}+...+x_{i}=U^{(i-1)n}(X_{1}^{n_{1}}+X_{2}^{n_{2}}+...+X_{i}^{n_{i}})=U^{in}=u\quad \quad {(3)}$

and

${\frac {1}{v}}={\frac {1}{X_{1}^{n_{1}}X_{2}^{n_{2}}...X_{i}^{n_{i}}}}={\frac {U^{{i(i-1)}n}}{U^{(i-1)n}X_{1}^{n_{1}}U^{(i-1)n}X_{2}^{n_{2}}...U^{(i-1)n}X_{i}^{n_{i}}}}={\frac {u^{(i-1)}}{x_{1}x_{2}...x_{i}}}\quad \quad {(4)}$

We pose

$x_{k,0}=x_{k}$

$u_{0}=u$

$v_{0}=v$

and

$x_{k,1}=x_{k}^{i}(x_{1}+x_{2}+...+x_{i})^{-(i-1)}$

$k=1,2,...,i$

it implies

$u=x_{1}+x_{2}+...+x_{i}=(x_{1,1}^{\frac {1}{i}}+x_{2,1}^{\frac {1}{i}}+...+x_{i,1}^{\frac {1}{i}})^{i}>u_{1}>1$

and

$x_{k,0}=x_{k}=x_{k,1}^{\frac {1}{i}}(x_{1}+x_{2}+...+x_{i})^{\frac {(i-1)}{i}}$

$=x_{k,1}^{\frac {1}{i}}(x_{1,1}^{\frac {1}{i}}+x_{2,1}^{\frac {1}{i}}+...+x_{i,1}^{\frac {1}{i}})^{(i-1)}>x_{k,1}>0$

and

$v={\frac {x_{1,0}x_{2,0}...x_{i,0}}{u^{(i-1)}}}=x_{1,1}^{\frac {1}{i}}x_{2,1}^{\frac {1}{i}}...x_{i,1}^{\frac {1}{i}}>v_{1}={\frac {x_{1,1}x_{2,1}...x_{i,1}}{u_{1}^{(i-1)}}}>0$

The reasoning is available until infinity. Then

$u_{j}=x_{1,j}+x_{2,j}+...+x_{i,j}=(x_{1,{j+1}}^{\frac {1}{i}}+x_{2,{j+1}}^{\frac {1}{i}}+...+x_{i,{j+1}}^{\frac {1}{i}})^{i}>u_{j+1}>0$

and

$x_{k,j}=x_{k,{j+1}}^{\frac {1}{i}}(x_{1,{j+1}}^{\frac {1}{i}}+x_{2,{j+1}}^{\frac {1}{i}}+...+x_{i,{j+1}}^{\frac {1}{i}})^{(i-1)}>x_{k,{j+1}}>0$

and

$k=1,2,...,i$

and

$v_{j}={\frac {x_{1,j}x_{2,j}...x_{i,j}}{u_{j}^{(i-1)}}}=x_{1,{j+1}}^{\frac {1}{i}}x_{2,{j+1}}^{\frac {1}{i}}...x_{i,{j+1}}^{\frac {1}{i}}>v_{j+1}={\frac {x_{1,{j+1}}x_{2,{j+1}}...x_{i,{j+1}}}{u_{j+1}^{(i-1)}}}>0$

$x_{k,j},v_{j},u_{j}$ are positive $\forall {j>1}$, $\forall {k=1,2,...,i}$.

LEMMA 8

(P) is the expression :

$x_{k,j}=x_{k}^{i^{j}}(\prod _{l=0}^{l={j-1}}{x_{1}^{i^{l}}+x_{2}^{i^{l}}+...+x_{i}^{i^{l}}})^{-(i-1)}$

Proof of lemma 8

for j=1, it is verified because of the definition of $x_{k,1},u_{1},v_{1}$,

we suppose (P) true and the expression of $u_{j}$ implies, with (P), that

$x_{k,j}=x_{k,{j+1}}^{\frac {1}{i}}(x_{1,{j+1}}^{\frac {1}{i}}+x_{2,{j+1}}^{\frac {1}{3}}+...+x_{i,{j+1}}^{\frac {1}{i}})^{(i-1)}$

or

$x_{k,{j+1}}^{\frac {1}{i}}=x_{k,j}(x_{1,{j+1}}^{\frac {1}{i}}+x_{2,{j+1}}^{\frac {1}{i}}+...+x_{i,{j+1}}^{\frac {1}{i}})^{-(i-1)}$

and

$x_{k,{j+1}}=x_{k,j}^{i}(x_{1,{j+1}}^{\frac {}{i}}+x_{2,{j+1}}^{\frac {1}{i}}+...+x_{i,{j+1}}^{\frac {1}{i}})^{-(i-1)i}=x_{k,j}^{i}(x_{1,j}+x_{2,j}+...+x_{i,j})^{-(i-1)}$

$=x_{k}^{i^{j+1}}\prod _{l=0}^{l={j-1}}{(x_{1}^{i^{l}}+x_{2}^{i^{l}}+...+x_{i}^{i^{l}})^{-{i(i-1)}}(x_{1}^{i^{j}}+x_{2}^{i^{j}}+...+x_{i}^{i^{j}})^{-(i-1)}}\prod _{l=0}^{l={j-1}}{(x_{1}^{i^{l}}+x_{2}^{i^{l}}+...+x_{i}^{i^{l}})^{(i-1)^{2}}}$

$=x_{k}^{i^{j+1}}\prod _{l=0}^{l=j}{(x_{1}^{i^{l}}+x_{2}^{i^{l}}+...+x_{i}^{i^{l}})^{-(i-1)}}$

then

$x_{k,j}=x_{k}^{i^{j}}\prod _{l=0}^{l={j-1}}{(x_{1}^{i^{l}}+x_{2}^{i^{l}}+...+x_{i}^{i^{l}})^{-(i-1)}}$

Why are not they solutions for $n>i(i-1),n_{k}>i(i-1)$ ?

We suppose

$n=n_{k}=i(i-1)$

the formula becomes

$x_{k,j}=x_{k}^{i^{j}}(\prod _{l=0}^{l={j-1}}{x_{1}^{i^{l}}+x_{2}^{i^{l}}+...+x_{i}^{i^{l}}})^{-(i-1)}$

$=U^{n{i^{j}}}X_{k}^{n_{k}{i^{j}}}(\prod _{l=0}^{l={j-1}}{(U^{n{i^{l}}}X_{1}^{n_{1}{i^{l}}}+U^{n{i^{l}}}X_{2}^{n_{2}{i^{l}}}+...+U^{n{i^{l}}}X_{i}^{n_{i}{i^{l}}})})^{-(i-1)}$

$=U^{i(i-1){i^{j}}}X_{k}^{i(i-1){i^{j}}}(\prod _{l=0}^{l={j-1}}{(U^{i(i-1){i^{l}}}X_{1}^{i(i-1){i^{l}}}+...+U^{i(i-1){i^{l}}}X_{i}^{i(i-1){i^{l}}})})^{-(i-1)}$

$=U^{(i-1)i^{j+1}}X_{k}^{(i-1)i^{j+1}}(\prod _{l=0}^{l={j-1}}{(U^{(i-1){i^{l+1}}}x_{1}^{(i-1){i^{l+1}}}+...+U^{(i-1){i^{l+1}}}X_{i}^{(i-1){i^{l+1}}})})^{-(i-1)}$

It is the expression of $x_{k,j+1}$ of the exponent i-1.

If we suppose that exist solutions for the exponent i-1,

there will exist solutions for an exposant not greater than i(i-1).

Some times, we must make attention to the initial change of the data. For example, let the following equationd

$kU^{n}=X^{n}+Y^{n}$

for some k integers like 7, there are solutions, for others like 2, there is no solution. It is too easy toi pose

$u=(kU^{n})^{2}$

$x=kU^{n}X^{n}$

$y=kU^{n}Y^{n}$

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

the lemma 1 is satisfied

$u=kU^{n}(X^{n}+Y^{n})=x+y$

${\frac {1}{z}}={\frac {k^{2}U^{2n}}{kU^{n}X^{n}kU^{n}Y^{n}}}={\frac {u}{xy}}={\frac {1}{x}}+{\frac {1}{y}}$

The correct solution is to pose

$u=U^{2n}$

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

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

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

Like this

$u=U^{2n}={\frac {kU^{n}(X^{n}+y^{n})}{k^{2}}}={\frac {x+y}{k}}$

and

${\frac {1}{z}}={\frac {U^{2n}}{U^{n}X^{n}U^{n}Y^{n}}}={\frac {u}{xy}}={\frac {x+y}{kxy}}={\frac {1}{kx}}+{\frac {1}{ky}}$

Conclusion

The conclusion is that Ghanouchi's sequences, series and numbers have several applications in all diophantine equations, we saw some of them and there are many others like Pilai equation, Smarandache equation, the Catalan equation, etc...