A proof of Catalan-Mihailescu theorem

Jamel Ghanouchi

jamel.ghanouchi@topnet.tn

Abstract
( MSC=11D04) More than one century after its formulation by the Belgian mathe-
matician Eugene Catalan, Preda Mihailescu has solved the open problem. But, is
it all ? Mihailescu’s solution utilizes computation on machines, we propose here
not really a proof as it is entended classically, but a resolution of an equation like
the resolution of the polynomial equations of third and fourth degrees. This solu-
tion is totally algebraic and does not utilize, of course, computers or any kind of
calculation.
(Keywords : Diophantine equations, Catalan equation ; Algebraic resolution)

Introduction
Catalan theorem has been proved in 2002 by Preda Mihailescu. In 2004, it became
officially Catalan-Mihailescu theorem. This theorem stipulates that there are not
consecutive pure powers. There do not exist integers stricly greater than 1, X > 1
and Y > 1, for which with exponants strictly greater than 1, p > 1 and q > 1,

Y p = Xq + 1

but for (X, Y, p, q) = (2, 3, 2, 3). We can verify that

32 = 23 + 1

Euler has proved that the equation X 3 +1 = Y 2 has this only solution. We propose
in this study a general solution. The particular cases already solved concern p = 2,
solved by Ko Chao in 1965, and q = 3 which has been solved in 2002. The case
q = 2 has been solved by Lebesgue in 1850. We solve here the equation for the
general case.

The first proof

1
Catalan equation is z c+2 = xa+2 + 1. With x and z coprime and greater than zero.
z is odd and x is even. We have

2(a1 S1 + a2 S2 ) = (a1 + a2 )(S1 + S2 ) + (a1 − a2 )(S1 − S2 ) (7)

But

z c+2 = xa+2 + 1 = xxa+1 + 1

And (7)

2(xa+2 + 1) = (x + 1)(xa+1 + 1) + (x − 1)(xa+1 − 1)

For a > 0, c > 0, with (7)

1
(x + 1)(xa+1 + 1) = ((x + 1)2 (xa + 1) + (x + 1)(x − 1)(xa − 1))
2

1 1
(x + 1)2 (xa + 1) = 2 ((x + 1)3 (xa−1 + 1) + (x + 1)2 (x − 1)(xa−1 − 1))
2 2

Until a
1 1
(x + 1)a (x2 + 1) = ((x + 1)a+1 (x + 1) + (x + 1)a (x − 1)(x − 1))
2a−1 2a

We add

2(xa+2 + 1)
m=a
X (x + 1)m
1 a+1
= (x + y) (x + 1) + (x − 1) ( (xa+1−m − 1)))
2a 2m
m=0

If we multiply by 2a

2a+1 (xa+2 + 1) = 2a+1 z c+2

m=a
X
a+1
= (x + 1) (x + 1) + (x − 1) (2a−m (x + y)m (xa+1−m − 1))
m=0

But x and z are coprimes, z is odd and x is even. Also

2(xa+2 − 1) = (x − 1)(xa+1 + 1) + (x + 1)(xa+1 − 1)

1
(x + 1)(xa+1 − 1) = ((x + 1)(x − 1)(xa + 1) + (x + 1)2 (xa − 1))
2

2
 1 1
(x + 1)2 (xa − 1) = 2 ((x + 1)2 (x − 1)(xa−1 + 1) + (x + 1)3 (xa−1 − 1))
2 2

Until a
1 1
(x + 1)a (x2 − 1) = ((x + 1)a (x − 1)(x + 1) + (x + 1)a+1 (x − 1))
2a−1 2a

We add

X (x + 1)mm=a
a+2 1
2(x − 1) = a (x + 1)a+1 (x − 1) + (x − 1) ( (1 + xa+1−m )))
2 2m
m=0

We multiply by 2a
m=a
X
a+1 a+2 a+1
2 (x − 1) = (x + 1) (x − 1) + (x − 1) (2a−m (x + 1)m (1 + xa+1−m ))
m=0

We have
m=a
X
a+1 a+2 a+1
2 (x + 1) = (x + 1) (x + 1) + (x − 1) (2a−m (x + 1)m (−1 + xa+1−m ))
m=0

We deduce
m=a
X
2a+2 xa+2 = 2x(x + 1)a+1 + 2(x − 1) (2a−m (x + 1)m xa+1−m )
m=0

And
m=a
X
2a+2 = 2(x + 1)a+1 − 2(x − 1) (2a−m (x + 1)m )
m=0

Thus
m=a
X
2a+1 xa+1 = (x + 1)a+1 + (x − 1) (2a−m (x + 1)m xa−m )
m=0
a+1 a a a−1
= (x + 1) + (x − 1)(2 x + 2 (x + 1)xa−1 + ... + 2(x + 1)a−1 x + (x + 1)a )
And
m=a
X
2a+1 = (x + 1)a+1 − (x − 1) (2a−m (x + 1)m )
m=0

= (x + 1)a+1 − (x − 1)(2a + 2a−1 (x + 1) + ... + 2(x + 1)a−1 + (x + 1)a )

We deduce
m=a
X
2a+1 (xa+1 − 1) = (x − 1) (2a−m (x + y)m (xa−m + 1))
m=0

3
Or
m=2a
X
22a+1 (x2a+1 − 1) = (x − 1) (22a−m (x + 1)m (x2a−m + 1))
m=0

And
m=a
X
2a+1 (αxa+1 − β) = (α − β)(x + 1)a+1 + (x − 1) (2a−m (x + y)m (αxa−m + β))
m=0

∀(α, β), particularly α = xa , β = 1, thus

2a+1 (x2a+1 − 1)
m=a
X
= (xa − 1)(x + 1)a+1 + (x − 1) (2a−m (x + 1)m (x2a−m + 1))
m=0

And
22a+1 (x2a+2 − 1)
m=a
X
= 2a (xa − 1)(x + 1)a+1 + (x − 1) (22a−m (x + 1)m (x2a−m + 1))
m=0
m=2a
X
= (x − 1) (22a−m (x + 1)m (x2a−m + 1))
m=0

And
2a (xa − 1)(x + 1)a+1
m=a
X
= (x − 1)( ((x + 1)m (22a−m )(x2a−m + 1 − (x2a−m + 1)))+
m=0
m=2a
X
+(x − 1) (22a−m (x + 1)m (x2a−m + 1))
m=a+1
m=2a
X
= (x − 1) (22a−m (x + 1)m (x2a−m + 1))
m=a+1

It means
m=2a
X
a a+1
(x − 1)(x + 1) = (x − 1) (2a−m (x + 1)m (x2a−m + 1))
m=a+1

1 1
= (x − 1)( (x + 1)a+1 (xa−1 + 1) + ... + a−2 (x + 1)2a−2 (x2 + 1)
2 2
1 1
+ a−1 (x + 1)2a−1 (x + 1) + a−1 (x + 1)2a )
2 2
1 1
= (x − 1)( (x + 1)a+1 (xa−1 + 1) + ... + a−2 (x + 1)2a−2 (x2 + 1)
2 2
1
+ a−2 (x + 1)2a )
2

4
Or
1
(x + 1)a+1 (xa − 1 − (x − 1)(xa−1 + 1))
2
1 1
= (x − 1)(x + 1)a+2 ( 2 (xa−2 + 1) + ... + a−1 (x + 1)a−2 )
2 2
1
= (x + 1)a+1 (−x + xa−1 )
2
And (x + 1) a+2 divises this expression, it means that x + 1 divises x and it is
impossible ! The equation is possible for a = 1 or a = 2. Those cases have been
studied in the past. The only soltion is (X, Y, a, c) = (2, ±3, 3, 2).

The second approach Let

Xp − 1 7 − Xp
c= p , c0 = p
Y 2 Y 2

If
c0 = 1 ⇒ (7 − X p )2 = X q + 1
If we suppose p > q
48 − 14X p + X 2p = X q
Thus
48
= 1 + 14X p−q − X 2p−q
Xq
is an integer, hence X q divises 48, three solutions

X q = 4 ⇒ X q + 1 = 5 6= Y p

And
X q = 16 ⇒ X q + 1 = 1 6= Y p
And
Xq = 8 ⇒ Xq + 1 = 9 = Y p ⇒ q > p
And the hypothesis p > q is impossible ! Now let q > p
48
= X q−p + 14X q−p − X p
Xp
is an integer, hence X p divises 48, three solutions

X p = 8 ⇒ (7 − X p )2 = 1 6= X q + 1

And
X p = 16 ⇒ (7 − X p )2 = 81 6= X q + 1
And

X p = 4 ⇒ X = 2, p = 2 ⇒ (7 − X p ) = 9 = X q + 1 = Y 2 ⇒ Y = ±3, q=3

Now, if c = 1, thus
(X p − 1)2 = X q + 1

5
Thus
X 2p − 2X p = X q
As X q > X p , thus q > p and

X q−p = X p − 2 ≥ 2

We deduce
q − p > 0, p>1
if we divise by X
2
X p−1 − X q−p−1 =
X
is an integer, thus
X=2
and
2p−1 − 1 = 2q−p−1
In the right, we have an even number and in the left, an odd one : this equality is
possible if p = 2. We have

2 − 1 = 1 = X q−3 ⇒ q = 3 ⇒ Y 2 = 23 + 1 = 9 ⇒ Y = ±3

If (c2 − 1)(c02 − 1) 6= 0. Thus let

p p
d = Xp − Y 2 , d0 = Y 2 + Xp

Or
d + d0 p d0 − d
X p − d = d0 − X p ⇒ X p = ⇒Y 2 = Xp − d =
2 2
But
p 7c + c0
cc0 Y 2 = c0 (X p − 1) = c(7 − X p ) ⇒ X p =
c + c0
p Xp − 1 6
⇒Y 2 = =
c c + c0
But
d + d0 7c + c0 d0 − d 6
= , =
2 c + c0 2 c + c0
Hence
7c + c0 − 6 0 7c + c0 + 6
d= , d =
c + c0 c + c0
We will give three proofs that (c − 1)(c0 − 1) = 0. Let

7c0 + c 7c0 + c − 6 7c0 + c + 6

Z= , e= , e0 =
c0 + c c + c0 c0 + c
We have

(c − 1)Z = (c − 1)(8 − X p ) = 8(c − 1) − cd + 1 = c(8 − d) − 7 = ce0 − 7

And

(c0 − 1)Z = (c0 − 1)(8 − X p ) = 8(c0 − 1) − c0 d0 + 7 = c0 (8 − d0 ) − 1 = c0 e − 1

6
But
(7c + c0 − 6)Z
d=
(c + c0 )Z
6(c − 1)Z + (c + c0 )Z
=
(c + c0 )Z
6ce0 − 42 + 7c0 + c 7c + c0 − 6
= =
7c0 + c c + c0
Thus

(6ce0 − 42)(c + c0 ) + (7c0 + c)(c + c0 ) = (7c + c0 )(7c0 + c) − 42c0 − 6c

And
(6ce0 − 42)(c + c0 ) + (7c0 + c)(c + c0 − 7c − c0 + 6) = 0
= 6(ce0 − 7)(c + c0 ) + 6(7c0 + c)(1 − c) = 0
Or
c2 e0 + cc0 e0 − 7c − 7c0 − 7cc0 + c − c2 + 7c0 = 0
= c2 e0 + cc0 e0 − 6c − 7cc0 − c2 = 0
= c2 (e0 − 1) + cc0 (e0 − 7) − 6c = 0
= c2 (e0 − 7) + 6c2 + cc0 (e0 − 7) − 6c = 0
Which means firstly
c(e0 − 7)(c + c0 ) + 6c(c − 1) = 0
Or
(e0 − 7)(c + c0 ) + 6(c − 1) = 0
And
(7c0 + c − 6)Z
e=
7c0 + c
6(c0 − 1)Z + (c + c0 )Z
=
7c0 + c
6(c0 e − 1) + 7c0 + c 7c0 + c − 6
= =
7c0 + c c + c0
Or
6(c0 e − 1)(c + c0 ) + (7c0 + c)(c + c0 ) = (7c0 + c)2 − 6(7c0 + c)
And
6(c0 e − 1)(c + c0 ) + (7c0 + c)(c + c0 − 7c0 − c + 6) = 0
= 6(c0 e − 1)(c + c0 ) + (7c0 + c)(6 − 6c0 ) = 0
Which means
(c0 e − 1)(c + c0 ) + (7c0 + c)(1 − c) = 0
= c02 e + cc0 e − c − c0 + 7c0 + c − 7cc0 − c2 = 0
= c02 e + cc0 e + 6c0 − 7cc0 − c2 = 0

7
We have also
c2 e0 + cc0 e0 − 6c − 7cc0 − c2 = 0
We sustract
c2 e0 − c02 e + cc0 (e0 − e) − 6(c + c0 ) = 0
= c2 (e0 − e) + e(c2 − c02 ) + cc0 (e0 − e) − 6(c + c0 ) = 0
Or
c(e0 − e)(c + c0 ) + e(c − c0 )(c + c0 ) − 6(c + c0 ) = 0
And
c(e0 − e) + e(c − c0 ) − 6 = 0
Or
c(e0 − e − 6) + (e − 1)(c − c0 ) + 6c + c − c0 − 6 = 0
= c(e0 − e − 6) + (e − 1)(c − c0 ) + 7c − c0 − 6 = 0
−6c + 6 − 6c0 + 6 6(c0 − 1)
= c( ) + ( )(c − c0 ) + 7(c − 1) − (c0 − 1) = 0
c + c0 c + c0
Hence

6c(1 − c) + 6c(1 − c0 ) + 6(c0 − 1) + 7(c − 1)(c + c0 ) − (c0 − 1)(c + c0 ) = 0

Or
(c − 1)(−6c + 7c + 7c0 ) + (c0 − 1)(−6c + 6 − c − c0 ) = 0
= (c − 1)(7c0 + c) + (c0 − 1)(−7c − c0 + 6) = 0
But
c(c − 1 − 7(c0 − 1)) + c0 (7(c − 1) − (c0 − 1)) + 6(c0 − 1) = 0
Or
(c − c0 )(c + c0 ) + 6(c − 1)c0 − 6(c0 − 1)c + 6(c0 − 1) = 0
= (c − c0 )(c + c0 ) + 6(c − c0 ) + 6(c0 − 1) = 0
We deduce
(c − c0 )(c + c0 ) + 6(c − 1) = 0
Or
(c − 1)(c + c0 + 6) − (c0 − 1)(c + c0 ) = 0
Thus
c−1 c + c0 7c + c0 − 6
= =
c0 − 1 c + c0 − 6 7c0 + c
8c + 2c0 − 6 8(c − 1) + 2(c0 − 1) + 4
= 0 =
8c + 2c − 6 8(c0 − 1) + 2(c − 1) + 4
2c0 + 2
=
2c + 2
It means
c2 − 1 = c02 − 1
Consequently
c2 = c02

8
Thus
(c2 − c02 )Y p = 12(X p − 4) = 0
(X, p) = (2, 2)
Y = Y 2 = 2q + 1
p

(Y, q) = (±3, 3)
Second proof : We have

7c0 + c − 6 7c0 + c + 6
e= , e0 =
c + c0 c + c0
And let
d0 = αd, e0 = βe
But
d0 − e0 = d − e = αd − βe
Thus
(β − 1)e = (α − 1)d
And
β−1 d d0 β
= = 0
α−1 e eα
Hence
αβ − α d0
= 0
αβ − β e
If we sustract 1
β−α d0 − e0 6(c − c0 )
= 0
= 0
αβ − β e 7c + c + 6
β−α d−e 6(c − c0 ) 6β(c − c0 )
= = 0 = 0
α−1 e 7c + c − 6 7c + c + 6
6(β − 1)(c − c0 )
=
12
2(β − α) = (α − 1)(β − 1)(c − c0 )
e0 d0 e0 d − ed0
= 2(− ) = 2( )
e d ed
(7c0 + c + 6)(7c + c0 − 6) − (7c0 + c − 6)(7c + c0 + 6)
= 2( )
ed
−24(7c0 + c) + 24(7c + c0 ) 144(c − c0 )
= =
ed ed
Or

144(c − c0 ) = (α − 1)(β − 1)de(c − c0 ) = (c − c0 )(d0 − d)(e0 − e) = 4Y p (c − c0 )

4Y p = 4(X q + 1) 6= 144 ⇒ c = c0
And
p p
(c − 1)Y 2 − (c0 − 1)Y 2 = d − 1 − (7 − d0 ) = 0 = d + d0 − 8 = 2X p − 8

9
And (X, p) = (2, 2) But
Y p − 1 = Y 2 − 1 = 2q
Y must be of the form Y = ±2γ (2k + 1) And

22γ (4k 2 + 4k + 1) − 1 = 2q ⇒ (γ, k) = (0, 1) ⇒ 8 = 2q ⇒ (Y, q) = (±3, 3)

Another proof : We have

p cd − 1
cY 2 = c(X p − d) = X p − 1 ⇒ X p =
c−1
And
(d − 1)2
Y p = (X p − d)2 =
(c − 1)2
If we pose
c2 Y 3p = b2 Y 2q
We have
c(d − 1)3
Yq =
b(c − 1)3
And
b2 − 1 c2 Y 3p−2q − 1
=
c2 − 1 b2 Y 2q−3p − 1
Y −2q c2 Y 3p − Y 2q b2 c2 Y 3p − Y 2q
=( )( ) = ( )( )
Y −3p b2 Y 2q − Y 3p c2 b2 Y 2q − Y 3p
We deduce
b2 c2 − c2 c2 Y 3p − Y 2q
=
b2 c2 − b2 b2 Y 2q − Y 3p
Or
b2 c2 − c2 b2 − c2 (c2 + 1)Y 3p − (b2 + 1)Y 2q
− 1 = =
b2 c2 − b2 b2 (c2 − 1) b2 Y 2q − Y 3p
If we develop

b2 (c2 − 1)((c2 + 1)Y 3p − (b2 + 1)Y 2q ) = (b2 − c2 )(b2 Y 2q − Y 3p )

Or
b2 Y 2q (b2 − c2 + (c2 − 1)(b2 + 1)) = Y 3p (b2 (c4 − 1) + b2 − c2 )
= c2 Y 3p (b2 − c2 + (c2 − 1)(b2 + 1)) = c2 Y 3p (b2 c2 − 1)
We deduce
b2 − c2 + (c2 − 1)(b2 + 1) = c2 b2 − 1
Also
b2 c2 − b2 c2 − b2 (b2 + 1)Y 2q − (c2 + 1)Y 3p
− 1 = =
b2 c2 − c2 c2 (b2 − 1) c2 Y 3p − Y 2q
If we develop

c2 Y 3p (c2 − b2 + (b2 − 1)(c2 + 1)) = Y 2q (c2 (b4 − 1) + c2 − b2 )

= b2 Y 2q (c2 − b2 + (b2 − 1)(c2 + 1)) = Y 2q b2 (c2 b2 − 1)

10
And we have
c2 − b2 + (b2 − 1)(c2 + 1) = b2 (c2 b2 − 1)
And
b2 − c2 + (c2 − 1)(b2 + 1) = c2 (c2 b2 − 1)
We add
(b2 − 1)(c2 + 1) + (c2 − 1)(b2 + 1) = (b2 + c2 )(c2 b2 − 1)
= 2b2 c2 − 2 = (b2 + c2 )(c2 b2 − 1)
Consequently
(b2 + c2 − 2)(c2 b2 − 1) = 0
If
b2 + c2 = 2
We have
c2 (b2 + c2 ) = 2c2
And
b2 (b2 + c2 ) = 2b2
Thus

b2 c2 − 1 = −c4 + 2c2 − 1 = −(c2 − 1)2 = −b4 + 2b2 − 1 = −(b2 − 1)2

But
b2 + c2 = 2
Or
2b2 = 2 + b2 − c2
And

2b2 c2 = c2 (2 + b2 − c2 ) = c2 (2 + b2 ) − c4 = (2 − b2 )(2 + b2 ) − c4 = 4 − b4 − c4

And
4b2 c2 − 4 = −b4 − c4 + 2b2 c2 = −(b2 − c2 )2
Consequently

4(b2 c2 − 1) = −4(c2 − 1)2 = −(b2 − c2 )2 = −4(b2 − 1)2

Or
4b2 c2 = 4 − (b2 − c2 )2 = (2 − b2 + c2 )(2 + b2 − c2 )
We deduce
4b2 c2 −4b2 c2
b2 − c2 = − 2 = +2
2 − b2 + c2 2 + b2 − c2
4b2 c2 − 4 + 2b2 − 2c2 −4b2 c2 + 4 + 2b2 − 2c2
= =
2 − b2 + c2 2 + b2 − c2
−4(c2 − 1)2 + 2b2 − 2c2 −4c4 + 6c2 − 4 + 2b2
= =
2 − b2 + c2 2c2

11
 4(c2 − 1)2 + 2b2 − 2c2 4c4 − 10c2 + 4 + 2b2
= =
2 + b2 − c2 2b2
Thus

2b4 −2b2 c2 = 4c4 −10c2 +4+2b2 = 4c4 −12c2 +8 = 4(c2 −1)(c2 −2) = −4b2 (c2 −1) = −4b2 c2 +4b2

Also

2c4 −2b2 c2 = −4b2 c2 +4c2 = −4c2 (b2 −1) = 4(b2 −1)(b2 −2) = 4b4 −12b2 +8 = 4b4 −10b2 +4+2c2

In consequence

2b4 − 4c4 − 2b2 c2 = −10c2 + 4 + 2b2 = 8 − 12c2

And
2c4 − 4b4 − 2b2 c2 = −10b2 + 4 + 2c2 = 8 − 12b2
Hence
4(b2 − c2 ) − 2c4 − 2b2 c2 = −12c2 + 8
And
4(c2 − b2 ) − 2b4 − 2b2 c2 = −12b2 + 8
It means
−2c4 − 2b2 c2 = −8c2 + 8 − 4b2 = −4c2 = 4b2 − 8
And
−2b4 − 2b2 c2 = −8b2 + 8 − 4c2 = −4b2 = 4c2 − 8
Or
−2c4 − 2b2 (2b2 + c2 ) = −8 = −2b4 − 2c2 (2c2 + b2 )
And
2b4 − 2c4 = 4(b2 − c2 ) = 2b2 (2b2 + c2 ) − 2c2 (2c2 + b2 )
And finally
2b2 (2b2 + c2 − 2) = 2c2 (2c2 + b2 − 2)
= 2b2 (2b2 + c2 − b2 − c2 ) = 2c2 (2c2 + b2 − b2 − c2 )
= 2b4 = 2c4
Thus
b2 + c2
b2 = c2 = =1
2
And if
c2 b2 = 1
Thus if we suppose (c2 − 1)(b2 − 1) 6= 0, we have

(b2 + 1)(c2 + 1) = (b2 + 1) + (c2 + 1) = 4 − (1 − b2 ) − (1 − c2 )

And
(b2 − 1)(c2 − 1) = (1 − b2 ) + (1 − c2 ) = 4 − (b2 + 1)(c2 + 1)

12
But
1 1 4 b2 − 1 + c2 − 1
+ = 1 = +
b2 + 1 c2 + 1 (b2 + 1)(c2 + 1) (b2 + 1)(c2 + 1)
And
−1 −1 4 b2 + 1 + c2 + 1
+ 2 =1= 2 − 2
b2 −1 c −1 (b − 1)(c − 1) (b − 1)(c2 − 1)
2

4 1 − b2 + 1 − c2
= −
(b2 + 1)(c2 + 1) (b2 + 1)(c2 + 1)
We deduce
4 4
− 2
(b2 + 1)(c + 1) (b − 1)(c2 − 1)
2

−b2 − 1 − c2 − 1 1 − b2 + 1 − c2
= +
(b2 − 1)(c2 − 1) (b2 + 1)(c2 + 1)
4(b2 − 1)(c2 − 1) − 4(b2 + 1)(c2 + 1)
=
(b4 − 1)(c4 − 1)
−(b2 + 1)2 (c2 + 1) − (c2 + 1)2 (b2 + 1) − (b2 − 1)2 (c2 − 1) − (c2 − 1)2 (b2 − 1)
=
(b4 − 1)(c4 − 1)
Thus
4(b2 − 1)(c2 − 1) − 4(b2 + 1)(c2 + 1)

= −(b2 + 1)2 (c2 + 1) − (c2 + 1)2 (b2 + 1) − (b2 − 1)2 (c2 − 1) − (c2 − 1)2 (b2 − 1)

(b2 − 1)(c2 − 1)(4 + b2 − 1 + c2 − 1) + (b2 + 1)(c2 + 1)(−4 + b2 + 1 + c2 + 1) = 0

= (b2 − 1)(c2 − 1)(b2 + c2 + 2) + (b2 + 1)(c2 + 1)(−2 + b2 + c2 ) = 0
= ((b2 − 1)(c2 − 1) − (c2 + 1)(b2 + 1))(b2 + c2 − 2) + 4(b2 − 1)(c2 − 1) = 0
= −2(b2 + c2 )(b2 + c2 − 2) + 4(2 − b2 − c2 ) = 0
= −2(b2 + c2 + 2)(b2 + c2 − 2) = 0
Consequently
b2 + c2 = 2
As
b2 c2 = 1 ⇒ b2 = c2 = 1
c2 − 1 can not be different of zero ! As we saw, the solution is

(c2 − 1)(c02 − 1) = 0

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