An approach of incoherence of mathematical arguments

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An approach of incoherence of mathematical arguments

Jamel Ghanouchi

To cite this version:

An approach of incoherence of mathematical arguments

Jamel Ghanouchi

ECOLE SUPERIEURE DE SCIENCES ET TECHNIQUES DE TUNIS Marsa 2070, Tunisia

jamel.ghanouchi@topnet.tn

Abstract

Lemma 1 If U, X, Y verify U = X + Y , then u, x, y, z as defined verify (1) and (2) which follow

Because

∀x2, y2rational, ∃z2 rational verifyin

1 z2 = 1 x2 + 1 y2

Until infinity. For i

(xi+ yi) = (√xi+1+√yi+1)2 > xi+1+ yi+1 >0

And xi+ yi rational for i > 2

xi =√xi+1(√xi+1+√yi+1) > xi+1 >0

xi rational for i > 2

yi =√yi+1(√xi+1+√yi+1) > yi+1 >0

yirational for i > 2

zi =

xiyi

xi+ yi

=√xi+1yi+1 > zi+1 =

xi+1yi+1

x_i+1+ yi+1

>0

zi rational for i > 1 and

Proof of lemma 2 By induction x=√x₂(√x₂+√y₂) =√x₂(x + y)12 x₂ = x 2 x+ y Also y2 = y2 x+ y

The lemma is verified for i = 2. If it is true for i, from (3) and (4) xi =√xi+1(√xi+1+√yi+1) = √xi+1(xi+ yi)

And j=i−2 Y j=0 (x(2j)+ y(2j)) = (x (2i−1₎ − y(2i−1)) x_{− y} Lemma 3 Let x_{6= y} Which means xi 6= yi xi = x(2i−1) x(2i−1₎ − y(2i−1₎(x − y) = U X2i−1 X2i−1 − Y2i−1(X − Y ) And yi = y(2i−1) x(2i−1₎ − y(2i−1₎(x − y) = U Y2i−1 X2i−1 − Y2i−1(X − Y ) ui = xi+ yi = Uc X2i−1 + Y2i−1 X2i−1 − Y2i−1(X − Y ) Lemma 4 xi− yi = x − y As xi− yi = xi+1− yi+1 = x − y

Ghanouchi theorem We will prove that 1_z = 1_x + 1_y and u = x + y imply x _{= y. We have supposed x − y 6= 0. A priori nothing allows to say that} x is different or equal to y. Our reasoning leaded us to x = y, without any condition on x and y. Why this impossible result ? If kUn _{= k}

y= kUn_k jX nj j z = kjX nj j (kU n − kjX nj j ) It is lemma (1) u= x + y And 1 x + 1 y = 1 z

If there is an undecidability, the sequences must lead to an impossibility. It is xy(x − y) = 0 for all x, y. Let us prove that xy(x − y) = 0. As

xi = x2i−1 x2i−1 − y2i−1(x − y) And yi = y2i−1 x2i−1 − y2i−1(x − y) Thus x_{≥ y ⇒ lim}

i−→∞(xi) = x − y, i−→∞lim (yi) = 0

And

y_{≥ x ⇒ lim}

i−→∞(yi) = y − x, i−→∞lim (xi) = 0

The series

In all what follows, we suppose x ≥ y. As √_x

iyi = yi−1− yi= xi−1− xi

So

xi− xi+1 =√xi+1yi+1

If x ≤ y j_=∞ X j=2 (√xjyj) = lim i−→∞(x − xi+1) = x

The applications of the series

Hence

j=i

X

j=2

((−1)j√xjyj) = x − x2− (x2− x3) + (x3− x4) − ... + (−1)i(xi−1− xi)

= x − 2x2+ 2x3− ... + 2(−1)i−1xi−1+ (−1)i+1xi

= A − B + y + x − y − lim i−→∞((−1) i+1_{(x − y))} = A − B + A + B + (x − y) lim i−→∞((1 − (−1) i+1₎₎ = 2A + (x − y) lim i−→∞((1 − (−1) i+1₎₎ Also i = 2k + 1 U1 = U = 2A + 2(x − y); U2= V = 2A i= 2k U1 = V = 2A; U2 = U = 2A + 2(x − y) U3 = 2 lim i−→∞( j=i X j=1 ((−1)j+1yj)) = A − B + y = 2A U4+ 2y = 2 lim i−→∞( j=i−1 X j=1 ((−1)j+1yj)) = A − B + y = 2A Let U_{1− U2− 2x = 2(x − y) lim} i−→∞((−1) i+1₎ i= 2k + 1 U_{1− 2x = U4} U₂= U3− 2x

U₁+U2+2x = U3+U4+2x = 2(A−B)+2x = 2U1+2y−2x = 2U2+6x−2y = 4A+2(2x−y) = 2U3+2x−2y = 2U4+2

= 4A + 2x − 2y = 4A + 6x − 2y But U1(U2+ 2x) = 4A2+ 4A(x − y) U3(U4+ 2y) = 4A2 U1(U2+ 2x) = U1(U1− 2(x − y) lim i−→∞((−1) i+1₎₎

= (2U1+ 2y − 2(x − y) limi−→∞((−1)

i+1₎₎2

4 + (U1+ U2+ 2x + 2(y − x))(x − y)

= (2U1+ 2y − 2(x − y) limi−→∞((−1)

i+1₎₎2

4 +(2U1+2y−2(x−y) limi−→∞((−1)

i+1_))(x−y)

= (U1+y)2+(x−y)2−2(U1+y)(x−y) lim i−→∞((−1) i+1_)+(2U 1+2y−2(x−y) lim i−→∞((−1) i+1_))(x−y) = U₁2_{− 2U}1(x − y) lim i−→∞((−1) i+1₎₎

= U₁2+ y2+ 2yU1+ (x − y)2+ 2(U1+ y)(x − y)(1 − lim i−→∞((−1) i+1₎₎₊ −2(x − y)2 lim i−→∞((−1) i+1₎ y2+ 2U1(x − y) + 2yU1+ (x − y)2− 2(x − y)2 lim i−→∞((−1) i+1₎₊

+2y(x − y)(1 − lim

i−→∞((−1)

i= 2k + 1

y2+ 2xU1+ (x − y)2− 2(x − y)2 = 0

y2+ 2xU1− (x − y)2= 0 = 2xU1− x2+ 2xy

U1 = U = 2A + 2(x − y) = A − B + 2x − y = −2B + 2x

y2_{+ 2x(2A + 2(x − y)) − (x − y)}2= 0 y2_{+ 4xA + 4x(x − y) − (x − y)}2 = 0

= 4xA − 2xy + 3x2 = 0

= 4x(y − B) − 2xy + 3x2= 2xy + 3x2− 4xB = 0 2xU1 = 4xA + 4x(x − y) = 2xy − 3x2+ 4x2− 4xy = x2− 2xy

or 2U1 = x − 2y = 4A + 4(x − y) 4A = −3x + 2y = 4y − 4B 4B = 2y + 3x 2U2 = 4A = −3x + 2y 4U1+ 4U2 = −4x = 4U3+ 4U4 = 4A − 4B = −6x It means x= 0 As it is impossible, x = y. A= lim k−→∞( j=k X j=1 (√x2jy2j)) B = lim k−→∞( j=k X j=1 (√x2j+1y2j+1)) U1 = 2 lim k−→∞( j=i+1 X j=1 ((−1)j+1xj)) = A − B + x + (−1)i+1(x − y) U2 = 2 lim k−→∞( j=i X j=2

((−1)j+1xj)) = A−B−x−(−1)i+1(x−y) = U1−2x−2(−1)i+1(x−y)

But

U₁2+ U1(2(B − A) − 2x + 2y) − 2xU3+ (B − A)2+ y2= 0

U₂2+ U2(2(B − A) + 2x − 2y) + 2xU4+ (B − A)2+ y2= 0

U1 = A − B + x − y ±p(A − B + x − y)2+ 2xU3− (A − B)2− y2

U2 = A − B + y − x ±p(A − B + y − x)2− 2xU4− (A − B)2− y2

= A − B + y − x ±px2_{− 2xy + 2(A − B)(y − x) − 2xU}

4 = A − B + y − 2x

Hence

x2= x2− 2xy + 2(A − B)(y − x) − 2x(A − B − y) Thus

(A − B)(2y − 4x) = 0 Another proof :

U1 = A − B + x − y ±p(x − y)2+ 2(x − y)(A − B) + 2xU3− y2 = 2x − y

U2 = A − B + y − x ±p(x − y)2+ 2(y − x)(A − B) − 2xU4− y2 = y − 2x

U₁2_{− U2}2_{+ 2(B − A)(U}1− U2) − (2x − 2y)(U1+ U2) − 2x(U3+ U4)

= (U1− U2)(U1+ U2+ 2(B − A)) − 4(2x − y)(A − B) = 0 = 4(y − 2x)(A − B)

And

2A = 2B = y

U1 = 2x − y = y = −U2= U3= −U4

U1U2− U3U4 = (2x − y)(y − 2x) + y2 = −4x(x − y) = −x(U1− U2− 2y)

U1(U2+ 2x) + U2(U1− 2x) + 4xy + 2y2 = 0

= U1(U2+2x)+U2(U1−2x)+2y(2x+y) = 0 = (2x−y)y−y(y−2x)+2y(2x+y) = 0

= 2xy − y2− y2+ 2xy + 4xy + 2y2= 8xy = 0 As it is impossible, it means x − y = 0.

The series are convergent. We have proved lim

i−→∞(xi) = x − y = 0

Let anonther proof :

= lim m−→∞((e 1 √ 2m − 1)y 2p−1e− p−1 √ 2m + (e 1 √ 2m − 1) k=m X k=p+1 (y_2k−1e− 2_k−p−1 √ 2m )) = lim m−→∞((e 1 √ 2m − 1) k=m X k=p+1 (y_2k−1e− 2_k−p−1 √ 2m )) = lim m−→∞(A) = limm−→∞(S) For p_{+ 1 = m ⇒ lim}

m−→∞(A) = limm−→∞(S) = limm−→∞((e

= lim m−→∞((e 1 √ 2m − 1)x 2p−1e− p−1 √ 2m + (e 1 √ 2m − 1) k=m X k=p+1 (x_2k−1e− 2_k−p−1 √ 2m )) = lim m−→∞((e 1 √ 2m _{− 1)} k=m X k=p+1 (x_2k−1e− 2_k−p−1 √ 2m )) = lim m−→∞(A) = limm−→∞(S) For p_{+ 1 = m ⇒ lim}

m−→∞(A) = limm−→∞(S) = limm−→∞((e

1 √ 2m − 1)x 2m−1e− m √ 2m) = 0 Consequently lim m−→∞((e 1 √ 2m _{− 1)} k=m X k=1 (x_2k−1e− 2k √ 2m)) = 0 We deduce 0 < lim m−→∞( k=2m X k=1 ((−1)k+1xke− k √ 2m)) = lim m−→∞( k=2m X k=1 ((−1)k+1xke− k √ 2m)) = lim m−→∞( k=m X k=1 (√x_2ky_2ke−√2k 2m)) < lim m−→∞( k=m X k=1 (√x_2ky_2k)) < lim m−→∞( k=2m X k=1 (√xkyk)) = y Thus lim m−→∞( k=2m X k=1 ((−1)k+1(xk− yk)e− k √ 2m)) = 0 = lim m−→∞((x − y) k=2m X k=1 (e−√k 2m)) = lim m−→∞((x − y)e −√1 2m1 − e −√2m 1 + e− 1 √ 2m ) = x − y 2 = 0 And x − y = 0 Let us recapitulate xi> xi+1, yi > yi+1 ⇒ x = y

xi = xi+1 = xi+1+ √xi+1yi+1 ⇒ xy = 0

yi = yi+1 = yi+1+ √xi+1yi+1 ⇒ xy = 0

Another proof : let A= j=i−1 Y j=1 (xj+ yj xj ) = x xi B= j=i−1 Y j=1 (xj+ yj yj ) = y yi A − B = j_=i−1 Y j=1 (xj+ yj xj ) − j_=i−1 Y j=1 (xj+ yj yj ) = (x 2i−1 − y2i−1)x (x − y)x2i−1 − (x2i−1_{− y}2i−1)y (x − y)y2i−1 = (xy 2i−1 − yx2i−1)(x2i−1 − y2i−1) (x − y)x2i−1_y2i−1

= ((x − y)y

2i−1

+ y(y2i−1− x2i−1))(x2i−1 − y2i−1) (x − y)x2i−1_y2i−1

= 1

x2i−1 −

y(x2i−1_{− y}2i−1)2 (x − y)x2i−1_y2i−1

= 1 x2i−1 − x − y x AB A= 1 x2i−1 + B(1 − x − y x A) = x xi = 1 x2i−1 + y yi(1 − x − y xi ) xyi = xiyi x2i−1 + y(xi− (xi− yi)) xyi = xiyi x2i−1 + yyi x= xi x2i−1 + y lim

i−→∞(x) = x = limi−→∞(

A= xB B_{(x − y) + y}

B = yA

x − A(x − y)

A+ B = xB(x − A(x − y)) + yA(B(x − y) + y) (B(x − y) + y)(x − A(x − y)) = x

2_{B − ABx(x − y) + ABy(x − y) + y}2_A

(B(x − y) + y)(x − A(x − y)) (x2xiy+ y2xyi− xy(x − y)2)xi

(yxi(x − y) + yyixi)(xxi− x(x − y))

= xyi+ yxi xiyi

(xyi+ yxi)(yxi(x − y) + yyixi)xyi)) = xiyi(x2xiy+ y2xyi− xy(x − y)2)

(xyi+ yxi)(yxi(x − y) + yyixi)x = xi(x2xiy+ y2xyi− xy(x − y)2)

lim

i−→∞((xyi+ yxi)(yxi(x − y) + yyixi)x) = y 2

(x − y)3 = lim

i−→∞(xi(x 2_x

iy+ y2xyi− xy(x − y)2)) = (x−y)2(x2y −xy(x−y)) = xy2(x−y)2

Thus x − y = 0. Another : A+ B = j=i−1 Y j=1 (xj+ yj)( 1 Qj=i−1 j=1 (xj) + _Qj=i−11 j=1 (yj) ) = j=i−1 Y j=1 (xj+ yj)( x x2i−1 + y y2i−1) = j=i−1 Y j=1 (xj+ yj)( xyi+ yxi xiy2i−1 ) = j_=i−1 Y j=1 (xj+ yj)((x − 1)yi+ (y − 1)xi + xi+ yi xiy2i−1 ) = j=i Y j=1 (xj+ yj xj )( 1 y2i−1 + j=i−1 Y j=1 (xj+ yj)(x − 1 x2i−1 + y − 1 y2i−1) = x xi+1 ( 1 y2i−1 + x − 1 x ( x xi ) +y − 1 y ( y yi )) = x xi + y yi xxiyi( xiyixi+1 y2i−1 + ( x − 1 x )xxi+1yi+ ( y − 1

y )yxixi+1) − xyixi+1− yxixi+1 = 0 lim i−→∞(xxiyi( xiyixi+1 y2i−1 + ( x − 1 x )xxi+1yi+ ( y − 1

And y= x + u′ _{⇒ x = x} 3(3 + 3y − x x + (x − y)2 x2 ) x − y = 3(x − y) + (y − x)(3x_x3 + 3y3 y + (y − x)x3 x2 − (y − x)y3 y2 ) (y − x)(−2 +3x_x3 + 3y3 y + (y − x)x3 x2 + (x − y)y3 y2 ) = 0 (y − x)(−2(x + y)(x2+ y2) + 3x3+ 3y3_{+ (y − x)x}2_{+ (x − y)y}2) = 0 = (y − x)(−2x3− 2y3− 2x2y − 2xy2+ 2x3+ 2y3+ x2y+ xy2) = 0 = (y − x)(−x2y − xy2) = 0 Another proof, if x ≥ y let di= _xyi_i, for an n

And xi− 1 ≥ x − y = xi− yi ⇒ yi≥ 1 Or yi= yi xi− yi(x − y) ≥ 1 ⇒ yi(x − y) ≥ x − y ⇒ limi−→∞ (yi(x − y)) = 0 ≥ x − y ≥ 0

Then x = y. Another proof : as

y2_{i − x}2_{i = (y − x)(x}i+ yi) = (yi+1− xi+1)(xi+1+ yi+1+ 2√xi+1yi+1)

= y2_{i+1− x}2_i+1+ 2√xi+1yi+1(yi+1− xi+1)

√_y

i+1 = Z is solution of

Z4+ 2√xi+1Z3− 2

q

x3_{i+1Z − x}2_i+1+ x2_{i − y}2_i = 0 And √xi+1= Z′is solution of

Z′4_{+ 2}√_y i+1Z′3− 2 q y_i+13 Z′_{− y}2 i+1+ yi2− x2i = 0 And Z4+ 2Z′_Z3 − 2Z′3_{Z − Z}′4_{+ x}2 i − y2i = 0 Also Z′4_{+ 2ZZ}′3_{− 2Z}3_Z′_{− Z}4_{+ y}2 i − x2i = 0 x2_{i − yi}2_{= (x − y)(Z}′_{+ Z)}2 = (x − y)(Z2+ 2Z′_{Z+ Z}′2₎

Z4+ 2Z′_Z3_{+ (x − y)Z}2_{+ (−2Z}′3_{+ 2Z}′_{(x − y))Z − Z}′4_{+ (x − y)Z}′2_{= 0}

w3+(3 4z ′2₋1 2(x−y))w 2 +(−₁₆3 z′4₋3 4z ′2_{(x−y))w+(−} 97 512z ′6₋ 41 128z ′4_{(x−y)) = 0} 512L3_{− 160(x − y)L}2_{− 9728(x − y)}2_{L −}97(x − y) 3 64 − 41 4 (x − y) 3 _{= 0} 32768L3_{− 10240(x − y)L}2_{− 621992(x − y)}2_{L − 97(x − y)}3_{− 656(x − y)}3 = 0 w′3_{+ (}3 4z 2₋1 2(y −x))w ′2_{+ (−} 3 16z 4₋3 4z 2_{(y −x))w}′_{+ (−}97 512z 6₋ 41 128z 4_{(x −y)) = 0} 512L′3_{+ 640(x − y)L}′2_{− 4608(x − y)}2_L′_{− 261(x − y)}3 _{= 0} p_{= (−} 3 16z ′4₋ 3 4z ′2_{(x − y)) −} ( 3 4z′2−12(x − y))2 3 = −3 8z ′4_{− z}′2_{(x − y) −} 1 12(x − y) 2

L′′_{is the limit of p in the infinity.}

128L′′_{= −3(x − y)}2 − 32(x − y)2−32 3 (x − y) 2 384L′′_{= −137(x − y)}2 q= ( 3 4z′2−12(x − y)) 27 (2( 3 4z ′2₋1 2(x − y)) 2 − 9(− 3 16z ′4₋3 4z ′2_{(x − y)))+} +(−₅₁₂97 z′6₋ 41 128z ′4_{(x − y)}2₎ = 1 16z ′6₊10 24z ′4_{(x − y) −} 9 16z ′4_{(x − y) +} 1 24(x − y) 2_z′2₋1 4(x − y) 2_z′2₋ 1 108(x − y) 3 = 1 16z ′6₊ 1 16z ′4_{(x − y) −} 5 24(x − y) 2_z′2₋ 1 108(x − y) 3

L′′′_{is the limit of q in the infinity.}

= −(Z2− Z′2₎29

8 = 0 The solution is

Z2_{− Z}′2 _{= y}

i+1− xi+1 = y − x = 0

And lim i−→∞(x 1 22i−1 i ) = lim i−→∞(y 1 22i−1 i ) = 1 Thus 0 < lim i−→∞(x 1 2i−1 i − y 1 2i−1 i ) = 1 − y x < lim i−→∞( ( x−y x 1 2i i −y 1 2i i )(x2i−11 y 1 22i−1 i − y 1 2i−1x 1 22i−1 i ) x2i1y 1 2i ) = 0 Another proof : let

xi= aixi+1= xi+1+ √xi+1yi+1⇒ ai = 1 +

r yi+1

xi+1

= 1 + yi xi

And

yi= biyi+1 = yi+1+√xi+1yi+1⇒ bi = 1 +

r_x i+1 yi+1 = 1 +xi yi Or ai= 1 + ai bi = 1 + yi xi And bi= 1 + bi ai = 1 +xi yi With aibi = ai+ bi We have

xi= bi√xi+1yi+1, yi = ai√xi+1yi+1

Let ai= ciai+1, bi = dibi+1 Another proof : if x > y aixi = biyi = xi+ yi 2 − a bi− 2 = yi x, 2 − ai b − 2 = y xi xix yyi − 1 = (b − 2)(bi− 2) (2 − a)(2 − ai) − 1 = xix − yiy yiy = bib − 2bi− 2b − aia+ 2a + 2ai (2 − a)(2 − ai) = (x − y)xi+ y(xi− yi) yiy = (x − y)(xi+ y) yiy = (bi− ai)b + ai(b − a) + 2(a − b) + 2(ai− bi) (2 − a)(2 − ai) = (ai− bi)(2 − b) + (a − b)(2 − ai) (2 − a)(2 − ai) = √_x

i+1yi+1x2y2((ai− bi)(2 − b) + (a − b)(2 − ai))

√_x

= (y − x)(2 √_x

2y2− x) + (y − x)(2√xi+1yi+1− yi)

(2√x2y2− y)(2√xi+1yi+1− yi)

= (y − x)( √_x

2(√y2−√x2) + √yi+1(√xi+1−√yi+1))

√_y

2(√x2−√y2)√yi+1(√xi+1−√yi+1)

= (y − x)(_y√y −x

i+1(√xi+1−√yi+1)

+_√ 1 y2(√x2−√y2)) = (x − y)( xi yiy + 1 yi ) Or (x−y)(_y√y x

i+1(√xi+1−√yi+1)−

xi

y√y_i+1(√xi+1+ √yi+1)−

1 √_y

2(√x2−√y2)−

1 √_y

i+1(√xi+1+ √yi+1)

) = 0

= (x − y)( −bix

y(√yi−pbixi+1)√yi

− xi yyi +_√ b y_{(√y −}√bx2) − 1 yi ) = 0 = (x − y)(−bi √_y

i(√y −√bx2)x − xi(√yi−pbixi+1)(√y −√bx2)

(√y −√bx2)y(√yi−pbixi+1)yi

+b(√yi−pbixi+1)√yyi− ( √_y

i−pbixi+1)(√y −√bx2)y

(√y −√bx₂)y(√yi−pbixi+1)yi

) = 0 And (x − y)2( bix √ b√yi √_y 2+ √x2 + √ bixi √_x

i+1+ √yi+1

(√_{y −}pbx2)+

− b

√ bi

√_x

i+1+ √yi+1

√_yy

i+

√ bi

√_x

i+1+ √yi+1

(√_{y −}pbx2)y) = 0

(x−y)2(pbix

√

b√yi(√xi+1+√yi+1)+(√y2+√x2)(xi(√y−

p

bx2)−b√yyi+(√y−

p

bx2)y)) = 0

= (x−y)2(x√b(√x_i+1+√y_i+1)2+(√y₂+√x₂)(xi(√y−

p bx_2)−b√yyi+(√y− p bx₂)y)) = 0 = (x−y)2(x√b(xi+yi)+(√y2+√x2)(xi(√y − p bx2)−b√yyi+(√y − p bx2)y)) = 0 (x − y)2(pbx₂(xi+ yi) + xi(√y − p bx_{2) − b}√yyi+ (√y − p bx₂)y) = 0 = (x − y)2(pbx2(yi− y) +√y(xi− byi+ y)) = 0 = (x − y)2(pbx2(yi− y) + p by2(xi− byi+ y)) = 0 (x − y)2(√x₂(yi− y) +√y2(xi− byi+ y)) = 0 Thus x − y = 0 Another proof, let

x_{= c(x − y) = d(x + y)} c >1

y_{= (c − 1)(x − y) = (1 − d)(x + y)} xi = ci(x − y) = di(xi+ yi)

di = ci 2ci− 1 ci = di 2di− 1) (2ci− 1)(2di− 1) = 1 But 2ci− 1 = r 2ci− 1 2di− 1 = xi+ yi xi− yi √ 2ci− 1(xi− yi) =p2di− 1(xi+ yi) = xi r xi+ yi xi− yi − √_x iyi=√xiyi+ yi r xi− yi xi+ yi 2√xiyi = x2_i + y_i2 q x2_{i − y}2_i lim i−→∞(−2 √_x iyi+ x2_i + y_i2 q x2 i − yi2 ) = 0 = x − y Another way : x2_i = xi+1(xi+ yi) = cidi(x2i − y2i) = ci+1(x2i − y2i)

= d2_i(xi+ yi)2 = di+1(xi+1+ yi+1)(xi+ yi)

d_i+1 = 2d 2 i 2(2di− 1)2+ 1 = 2d 2 i 4d2 i − 4di+ 2 = 2d 2 i 8d2 i − 8di+ 3 4d2_i − 4di+ 1 = 0 = (2di− 1)2 2di= 1, thus 2xi = 2di(xi+ yi) = xi+ yi

And x − y = 0. Another proof :

4(2ci−1)(ci−1)2−4(2di−1)(di−1)2 = 8(ci−1)3−8(di−1)3+4(ci−1)2−4(di−1)2 = 4(2ci− 1)(ci− 1 2 − 1 2) 2 − 4(2di− 1)(di− 1 2 − 1 2) 2 = (2ci− 1)(2ci− 1 − 1)2− (2di− 1)(2di− 1 − 1)2 = (2ci− 1)3− (2di− 1)3− 2(2ci− 1)2+ 2(2di− 1)2+ 2ci− 2di = (ci− di)(8(ci− 1)2+ 8(ci− 1)(di− 1) + 8(di− 1)2+ 8cidi− 8) = (2ci− 2di)((2ci− 1)2+ 1 + (2di− 1)2− 2(2ci− 2di)(4cidi− 2) + 2(ci− di) = (ci− di)(2(2ci− 1)2+ 2 + 2(2di− 1)2− 16cidi+ 8 + 2) Or (ci−di)(2(2ci−1)2+2(2di−1)2−16cidi+12−8c2i−32cidi−8d2i−16+8cidi−8−8cidi+8) = 0 = (ci− di)(8c2i + 8d2i − 16cidi+ 4 − 48cidi− 8c2i − 8d2i + 4) = 0 = 8(ci− di)(−8cidi+ 1) = 0

Thus ci− di = 0 or y = 0, impossible : x − y = 0. Another proof :

= 24 r ((ci− di)2 4 (ci− 1)(1 − di) =p2(ci− di) 4 p(ci− 1)(di− 1) A=q4 c6_i(2ci− 1)3+ 4 q c4_i(2ci− 1)2d2i(2di− 1)+ 4 q c2_i(2ci− 1)d4i(2di− 1)2+ 4 q d6_i(2di− 1)3 = q4 c6_i(2ci− 1)3+ 4 q c4_i(2ci− 1)d2i + 4 q c2_id4_i(2di− 1) + 4 q d6_i(2di− 1)3 = q4 c6_i(2ci− 1)3+ 4 q c6_i(2di− 1) + 4 q d6_i(2ci− 1) + 4 q d6_i(2di− 1)3 =q4 c6_i(2ci− 1)( 4 p(2ci− 1)2+ 4 p(2di− 1)2)+ 4 q d6_i(2di− 1)( 4 p(2ci− 1)2+ 4 p(2di− 1)2) A(q4 c2_i(2ci− 1) − 4 q d2_i(2di− 1)) = c2i(2ci− 1) − d2i(2di− 1) = (ci− di)(2c2i + 2d2i) = Ap2(ci− di) 4 p(ci− 1)(di− 1) = (ci− di) A 4 √ cidi (ci− di)(2c2i + 2d2i) 4 p cidi= 2(ci− di)(4 q c6_i(2di− 1) + 4 q d6_i(2ci− 1)) Thus (ci− di)4 q c6_i(2ci− 1)(p(2c4 i− 1)2−p(2d4 i− 1)2) = (ci− di)4 q d6_i(2di− 1)(p(2c4 i− 1)2−p(2d4 i− 1)2)

It means ci− di= 0. As it is impossible, x = y. Another proof :

ci = c2i−1 c2i−1 − (c − 1)2i−1 ci− 1 = (c − 1) 2i−1 c2i−1 − (c − 1)2i−1

= 1 (c2i − (c − 1)2i )(2ci+1− 1) (2ci− 1)(2ci+1− 1) = 1 c22_i−1 (1 −yi+1 xi+1)(1 − yi xi) lim i−→∞((2ci− 1)(2ci+1− 1)) = 1 = 1 c22_i−1 = 0 Impossible ! It means x − y = 0. ci+ di− 1 di− ci+ 1 = x 2 i + yi2 x2 i − y2i − 2xiyi = 1 + y2i x2 i 1 −y2i x2 i − 2 yi xi = 1 + ci+1−1 ci+1 1 − ci+1−1 ci+1 − 2 ci−1 ci = 2ci+1− 1 1 − 2ci+1+ 2ci+1ci = −1 + 2 ci+1 ci 1 − 2ci+1+ 2ci+1_ci = −1 + 2_c ci+1(x − y) i(x − y) − 2ci+1ci(x − y) + 2ci+1(x − y) = −1 + 2_x xi+1 i− 2xi+1ci+ 2xi+1 = −1 + 2 √_x i+1 √_x

i+1+ √yi+1− 2ci√xi+1+ 2√xi+1

= −1 + 2

√_x

i+1

(3 − 2ci)√xi+1+ √yi+1

= (2ci− 1) √_x

i+1−√yi+1

(3 − 2ci)√xi+1+ √yi+1

lim

i−→∞(x 4

i − x3i − 3x3iyi+ xiyi3+ 3x2iyi− y3i − 2xiyi2+ 3x2iyi2)

= (x − y)3(x − y − 1) = 0

Thus x − y = 0. Another proof : if x 6= y, ∀x1 = p1, y1= q1, ∃r1verifying

(p1− q1) integer p1=√p2(√p2−√q2) p1integer q1=√q2(√p2−√q2) q1 integer r1= x1y1 x1− y1 =√p2q2 r2 rational And

∀p2, q2rational, ∃r2rational with

1 r₂ = 1 q₂ − 1 p₂

Until infinity. For i

(pi− qi) = (√pi+1−√qi+1)2

And pi− qirational for i > 2

pi =√pi+1(√pi+1−√qi+1)

pirational for i > 2

qi = √qi+1(√pi+1−√qi+1)

Lemma a We have (pi− qi) = (√pi+1−√qi+1)2 pi = x2 i−1 j=i−2 Y j=0 (x2j_{− y}2j)−1 ₍₅₎ qi= y2 i−1 j_=i−2 Y j=0 (x2j− y2j)−1 ₍₆₎

Proof of lemma a By induction

p=√p2(√p2−√q2) =√p2(x − y) 1 2 p2 = x2 x − y Also q2= y2 x − y True for i = 2. If it is true for i, from (5) and (6)

pi=√pi+1(√pi+1−√qi+1) =√pi+1(pi− qi)

Now let 1 ri = 1 qi − 1 pi

The same calculus will lead to p′ i = x (2i−1₎ i j_=i−2 Y j=0 (x(2_i j)_{− y}(2_i j))−1 = x(2i) j_=i−2 Y j=0 (x(2j+1)− y(2j+1))−1 = xipi Also q′ i = qiyi But p′ iyi = piyixi= qix2i q′ ixi= qixiyi= piy2i p′ iyi+ q′ixi= (p′i+1− q′i+1)xiyi

= (pi+1xi+1− qi+1yi+1)xiyi

= qix2i+piyi2= (pi+1−√pi+1qi+1)(yi+1+√xi+1yi+1)2+(−qi+1+√pi+1qi+1)(xi+1+√xi+1yi+1)2

= (pi+1xi+1− qi+1yi+1)√xi+1yi+1(xi+ yi)

But

(pi+1−√pi+1qi+1)√xi+1+ (−qi+1+√pi+1qi+1)√yi+1 =

pi+1xi+1− qi+1yi+1

√_x

i+ yi

Thus

((pi+1−√pi+1qi+1)yi+1+ (−qi + 1 −√pi+1qi+1)xi+1)(xi+ yi)

= (pi+1x√i+1− qi+1yi+1 xi+ yi

y_i+1 √_x

i+1

(xi+ yi)+

−(−qi+1+√pi+1qi+1)

s y3_i+1 xi+1

(xi+ yi)+

+(−qi+1+√pi+1qi+1)xi+1)(xi+ yi)

= (pi+1xi+1− qi+1yi+1)√xi+1yi+1(xi+ yi)

And

(pi+1xi+1− qi+1yi+1)(√xi+1yi+1−

yi+1

xi

)

= (−qi+1+√pi+1qi+1)(xi+1−

(xi+1− y_i+12 xi+1 )(√xi+1yi+1− y_i+1 xi ) = (−y_xi+1 i+1 + r_y i+1 xi+1 )(xi+1− s y3 i+1 xi+1 ) Or (xi+1− y_i+12 xi+1 )(√xi+1− √_y i+1 xi ) = (− √_y i+1 xi+1 + s 1 xi+1 )(xi+1− s y_i+13 xi+1 ) In the infinity p(x − y)3₌√_{x − y = 0} In the limits lim

i−→∞(pi) = L, i−→∞lim (qi) = L ′

From

pi = pi+1−√pi+1qi+1, qi = −qi+1+√pi+1qi+1

We deduce L_{= L −}√LL′_{, L}′ _{= −L}′₊√_LL′ _{⇒ L}′ _{= 0} But pi+ qi = pi+1− qi+1 p_i−1+ q_i−1= pi− qi ... p₂+ q2 = p3− q3 p+ q = x + y = p2− q2

qi+ qi−1+ ... + q2+ q + p = pi+1− qi+1− qi− qi−1− ... − q2

q_i+1+ 2qi+ 2qi−1+ ... + 2q2+ q = pi+1− p

2p_4k−1+ 2p_4k−2+ ... + 2p2 = p4k+p4k−1+p4k−2+p4k−3+...+p3+p2−p−q4k−q4k−1+q4k−2+q4k−3−q4k−4−...−q4−q3+q2+q Thus −p4k+p4k−1+p4k−2+...+p2−p = −(q4k+q4k−1)+(q4k−2+q4k−3)−...−(q4+q3)+q2+q And we pose −(q4k+ q4k−1) + (q4k−2+ q4k−3) + ... − (q4+ q3) + q2+ q = u4k = −p4k+ p4k−1+ ... + p2− p And v_{4k= −(q4k}+ q_4k−1) + q_4k−2+ q_4k−3 Hence u_4k= v4k+ v4(k−1)+ ... + v1 Or u4k− u_4(k−1) = v4k = −p4k+ p4k−1+ p4k−2+ p4k−3+ ... − p − (−p4k−4+ p4k−5+ ... − p) = −p4k+ p4k−1+ p4k−2+ p4k−3+ 2p4k−4 And lim k−→∞(v4k) = 0 = 4L We deduce lim i−→∞(pi− qi) = L − L ′_{= 0} = lim −→∞( x2i−1_{− y}2i−1 Qj=i−2 j=1 (x2 j − y2j )) = lim i−→∞((x − y) j=i−2 Y j=0 (x 2j + y2j x2j − y2j)) ≥ x − y ≥ 0 Or x − y = 0. Another proof : pi = x2i−1 Qj=i−2 j=0 (x2 j − y2j ) = x 2i−1 (x − y) (x − y)Qj=i−2 j=0 (x2 j − y2j ) i= 2k

(x − y)(x − y)(x2− y2)(x4_{− y}4)...(x2i−2 _{− y}2i−2) = (x2+ y2_{− 2xy)(x}2_{− y}2)(x4_{− y}4)...(x2i−2_{− y}2i−2) = (x4− y4− 2xy(x2− y2))(x4− y4)...(x2i−2− y2i−2)

= (x16_−y16_−2x4y4(x8_−y8_)−2xy(x2_−y2)(x4_−y4)(x8_−y8)(x16_−y16)...(x2i−2_−y2i−2) = (x25+ y25_{− 2x}24y24_{− 2x}22y22(x8_{− y}8)(x16_{− y}16_{) − 2xy(x}2_{− y}2)...(x16_{− y}16))... = x2i−1 + y2i−1_{− 2} j=i−2 X j=0 (x2jy2j k=i−2 Y k=j ((x2k_{− y}2k))) pi = x − y 1 +yi xi − 2 Pj=i−2 j=0 ( yj+1 xj+1 Qk=i−2 k=j (1 − yk+1 xk+1)) = x − y 1 + yi xi − 2 Pj=i−2 j=0 ( yj+1 xj+1 Qk=i−2 k=j (xx−yk+1)) x_k+1> x_i−1 x − y x_k+1 < x − y x_i−1 pi< x − y 1 +yi xi − 2 Pj=i−2 j=0 ( y(x−y)i−2−j xi−1−j i−1 ) pi< x − y 1 + yi xi − 2 y x_i−1( 1−(x−y xi−1)i−1 1−_xi−1x−y ) lim i−→∞(

1 − (xx−y_i−1)i−1

Another approach Let us suppose that for every complex number x ; x, for which, for every p1, p2, b. We can pose for p1and p2numbers verifying p1> p2.

2x = p1+ p2+ 2b

b depends of p1and p2. Then

x= p1+ p2

2 + b

And for all x, p1, p2, exists y whose expression is

y= p1− p2 2 + b We pose        x₁= p1+ 2b x2= p2− 2b x3= p2+ 2b x₄= p1− 2b ⇒            x= p1+p2 2 + b = p1+x2 2 + 2b = x1+p2 2 = x1+x2 2 + b = p1+x3 2 = x1+x3 2 − b = x4+x3 2 + b = x4+x2 2 + 3b y= p1−p2 2 + b = p1−x2 2 = x1−p2 2 = x1−x2 2 − b = p1−x3 2 + 2b = x1−x3 2 + b = x4−x3 2 + 3b = x4−x2 2 + b x1+ x2 = p1+ p2

LEMMA 1 The following formula            x= p1+p2 2 + b = p1+x2 2 + 2b = x1+p2 2 = x1+x2 2 + b = p1+x3 2 = x1+x3 2 − b = x4+x3 2 + b = x4+x2 2 + 3b y= p1−p2 2 + b = p1−x2 2 = x1−p2 2 = x1−x2 2 − b = p1−x3 2 + 2b = x1−x2 3 + b = x4−x2 3 + 3b = x4−x2 2 + b x₁+ x2 = p1+ p2

imply that ∃p1, p2prime numbers which verify

b= 0 .

if

kk′ _{= 0 ⇒ b = 0}

we will suppose

kk′ _{6= 0}

But ∀x, y, ∃φ verifying x = φy

Hence 4α2+ 4β2+ 16 + 8αβ + 16α + 16β = 4α2+ 4β2_{− 8αβ} Or αβ+ α + β + 1) = 0 = (α + 1)(β + 1) It means b = 0. And 2x = p1+ p2

Thus α = β = −1. b can not be different of zero. If

(x1− x2)(x1+ x3) = 0 ⇒ (x4+ x2)(x4− x3) 6= 0 let ₍ x4+x2 p1+p2 = p1+p2−4b p1+p2 = 1 − 4b p1+p2 p1+p2 x4+x2 = x4+x2+4b x4+x2 = 1 + 4b x4+x2 we pose ( 2k + 1 = 1 −p14b+p2 2k′_{+ 1 = 1 +} 4b x4+x2 let ₍ x4−x3 p1−p2 = p1−p2−4b p1−p2 = 1 − 4b p1−p2 p1−p2 x4−x3 = x4−x3+4b x4−x3 = 1 + 4b x4−x3 we pose ( 2m + 1 = 1 − p14b−p2 2m′_{+ 1 = 1 +} 4b x4−x3 ⇒ x = p1+3p2 4 = p1+p2 2 or p1+p2 2 = p1+p2 4 and x = p1 = p2 or p1 = p2 = 0 it is impossible, then (x1− x2)(x1+ x3) 6= 0 and b= 0

Conclusion Our approach of whatever equation allowed us to prove that for every x, y, then x − y = 0.

Références

[1] Darmon, H. and Granville, A. "On the Equations and ." Bull. London Math. Soc. , 27, 513-543 (1995).