A proof of Beal’s conjecture
Jamel Ghanouchi
RIME Dpt Mathematics
jamel.ghanouchi@topnet.tn
Abstract
More than one century after its formulation by the Belgian mathematician 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 of Catalan theorem as it is
entended classically, but a resolution of an equation like the resolution of the polynomial
equations of third and fourth degrees. This solution is totally algebraic and does not utilize, of
course, computers or any kind of calculation. We generalize our approach to Beal equation
and discuss the solutions. (Keywords : Diophantine equations, Catalan, Fermat-Catalan,
Conjectures, Proofs, 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  X q  1 but for (X,Y,p,q)=(2,3,2,3). We
can verify that 32  23  1 . Euler has proved that the equation Y 2  X 3  1 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 and prove that Beal
equation is related to this problem.
The proof
Catalan equation is Y p  X q  1 we pose c 
X p 1
p
2
and c ' 
7 X p
p
2
p
thus Y 2 
6
and
c c'
Y
Y
2
36  (c  c ')
7c  c '
X p  cY  1 
and X q  Y p  1 
but X p  7  c '  0 and c  c ' and
2
c c'
(c  c ')
p
2
p
Y 2  3  0  c  c '  2 . We will prove now that q  1  2 p and will discuss two cases :
c 2  1 and c 2  1 .
1) c 2  1
36  (c  c ')2  (7c  c ')2
1  c2
We have X q  X 2 p 
 36
 0 we deduce q  1  2 p
(c  c ')2
(c  c ')2
72c 2  2c 2 (c  c ')2  (7c  c ')2 64c 2  (7c  c ')2

0
(c  c ')2
(c  c ')2
36c 2  c 2 (c  c ')2  (7c  c ')2 36c 2  c 2 (c  c ')2  36c 2


0
(c  c ')2
(c  c ')2
2c 2 X q  X 2 p 
c2 X q  X 2 p
1
But 2c 2 X 2 p 1  2c 2 X q  X 2 p  2c 2  X
Lemma 1 :
u 2c 2 X 2 p 1  u 2 X 2 p 1  u 2c 2 X q  X 2 p  u 2c 2  X and
1 2 2
u c  X 2 p q  u 2c 2
4
n2
m2
and X 
n 1
m 1
we have q+1=2p or q+2=2p and (respectively) X  u 2c 2 or X 2  u 2 c 2
Proof of the lemma 1 :
We will give three proofs :
1
1
Let v verifying X 2 p  2a X p  1a1a  (  vc 2 ) X q  c 2Y p  (  vc 2 ) X q
u
u
1
1
1
 (c 2  vc 2  ) X q  c 2  0  c 2  (c 2  ) X q  vc 2 X q  c 2  (c 2  ) X q  vc 2 X q
u
u
u
with
1
c 2  (c 2  ) X q
Xq
2
q
u
c
(
X

1)

0


2v

0
c2 X q
u and
because
1
 X 2  wvc 2  vc 2 X q  c 2  c 2 X q  X q  2  wvc 2 X q
u
 (1  w)vc 2 X q  c 2  (c 2  X 2 ) X q  0 because if
c 2  X 2  X q 1  X 2 p  u 2c 2 X q  u 2 X q  2  X q 3 then q  1  2 p  q  2 , we have
1
1
(  vc 2 ) X q  2 p  1  2a X  p  1a1a X 2 p  (  vc 2 ) X q 1 2 p  X  2a X 1 p  1a1a X 1 2 p  1
u
u
2p
2p
p 1
because X ( X  1)  2 X  2a X  1a1a X and
u 2c 2 
X q 1 2 p 
1
 X q  4 2 p 
X3

X3
 1 with w verifying
X 2  (v  w)c 2
1
1
 vc 2
 vc 2
u
u
2
2
X ( X  1)  (v  w)c thus we have X 3  X q 42 p  1  2  q  3  2 p  0 it is proved !
Another proof :
m
 X  2  0 ). There exists always k for
We have n  0 and m  0 ( 2 X  1  X  1 
m 1
1
1
1
which k 1  2 2  k
2 X uc
2 X
2 k
q 1 2 p
X  2v (m  1)
2 X
 2 k
Thus
and with
v>2, (for example m+1=X/4 and v=2) we
k
2  2 k 1  X
2 p q 1
k 1
k 2
 2 k 1 1  2
v (2 p  q 1) 1 k
k 1
(m  1)
have
(m  1)
q 12 p
k 1
2
v (2 p q 1)  k 1
k 1
2 p  q 1
k 1
3
 2 k 1 if v(2p-q-1)>k+1 then
 1 and q+1-2p>0 else if v(2p-q-1)>k then
k
k
(m  1)q 12 p  2 k v (2 p q 1)  1 and q+1-2p>0 else q  1  2 p 
22k  X q 12 p  X v and
but
v
k
k
 2 and q  1  2 p  2 or
4  2k X v we can always choose w as k<2v then q  1  2 p 
v
1
1
1
 X q 2 p  2 2   X  u 2c 2
q  2  2 p  0 if q+1=2p then
X
uc
X
and if q+2=2p then
2
u 2 c 2 X q  u 2c 2 X 2 p 2  X 2 p  u 2c 2  X 2
and
22k  X q 22 p  2 k leads to
1
1
1
 X q 2 p  2 2  2  X 2  u 2c 2
2
X
uc
X
1
, but in this expression n  1
n 1
for all m, thus m | n  1  10 m  1  k  kc 2  1  10m particularly
Another proof : we have kc 2  1, k  1 we pose kc 2  1 
can not be less than 10  m
for k 
u 2 (1  10m )
 u 2c 2  X  u 2c 2  X  u 2c 2
X
Here
 X q 1  2c2 X q  X 2 p  X q 1  q  1  2 p
 2c 2Y p  2( X p  1)2  X q 1  X
2
  X q  1  2 X 2 p 1  4 X p 1  N  X  2
X
 2 p 1  22 p 1  0  p  2  q  2 p  1  3  Y  3
Now let
2) c 2  1
72  2(c  c ')2  (7c  c ')2 64  (7c  c ')2
2X q  X 2 p 

0
(c  c ')2
(c  c ')2
2
 X 2 p  q 1   1  2 p  q  1
X
72c 2  2c 2 (c  c ')2  (7c  c ')2 64c 2  (7c  c ')2
We have 2c 2 X q  X 2 p 

0
(c  c ')2
(c  c ')2
c X X
2
q
2p
36c 2  c 2 (c  c ')2  (7c  c ')2 36c 2  c 2 (c  c ')2  36c 2


0
(c  c ')2
(c  c ')2
Lemma 2 :
1
With u 2 c 2  X 2 p  q  u 2c 2 and q+1>2p then 2 p  q  2 p  1
4
Proof of the lemma 2 :
First proof : let v verifying
1
1
X 2 p  2a X p  1a1a  (  vc 2 ) X q  c 2Y p  (  vc 2 ) X q
u
u
1
 (c 2  vc 2  ) X q  c 2  0
u
1
 c 2  (c 2  ) X q  vc 2 X q
u
1
c 2  (c 2  ) X q
u
2v
0
2
q
c X
1
 X 2  wvc 2  (1  w)vc 2 X q  c 2  (c 2  X 2 ) X q  0
u
With
3
1
(  vc 2 ) X q  2 p  1  2a X  p  1a1a X 2 p  2
u
and
1
 vc 2
1
X 2  (v  w)c 2
2
2 p q
2 p q
2 p 1 q
 (  vc )  2 X
X
 2X
u

1
u
X
X
We have then 1>2p-q>0 another proof :
There exists always k for which
X
1
X
1
1
 u 2 c 2  k hence k  2  X 2 p 1 q  k and
k 1
2
2
2
2
2
k
X  2v (m  1)  2v ; v  2 (for example m+1=X/4 and v=2) then 2  2 k 1  X
1 2
v ( q 1 2 p )  k 1
k 1
(m  1)
q 1 2 p
k 1
q 1 2 p
k 1
2
k 2
k 1
or
3
 2 k 1 and if
2 p q 1
k 1
v ( q12 p )k 1
v(q  1  2 p)  k  1  0  (m  1)
 2 k 1  1  2 p  q  1  0 else if
v(q  1  2 p)  k  0  (m  1)2 p q 1  2v ( q 12 p )k  1  2 p  q  1  0 else
k
k
k
2 k
2 p  q 1
k
v
2 p  q 1 
2  X
 X but 4  2 X v we can always choose v as k<2v then
v
k
 2 p  q 1 
 2  2 p  1  q  0  2 p  q  2 p  1
v
There is another proof similar than higher.
With the same calculus than for c 2  1 , we prove that (in virtue of the lemma 2)
X  2c 2  c 2 X  2  X 2 p  c 2 X q  2 X q 1
X 2 p 1
 X q  X 2 p 1
2
For Catalan equation, q and 2p do not have the same parity thus q+1=2p. But for Catalan
0  (c 2  1)Y p  X 2 p  2 X p  X q

 X 2 p  2 X p  X 2 p 1  X 2 p 1  2 X p  0
 (c 2  1)Y p  X 2 p  2 X p  X q  0
 X 2 p  2 X p  X 2 p 1  0 
2
 X p 1  X p  2  N
X
 ( X , Y , p, q)( (2, 3, 2,3)
Generalization to Beal or Fermat-Catlan equation
We have Fermat-Catalan equation Y p  X q  Z c with X q  Z c  vX q we pose Y p  X q  1a1a
and c 
c' 
X p  1a
p
Y2
7a  X p
Y
p
2
p
, Y2 
p
6a
7 c  1a c '
36  (c  c ')2
, X p  cY 2  1a  a
, X q  Y p  1a1a  1a1a
,
c c'
c c'
(c  c ')2
p
2
with almost the same equalities than for Catalan but Y 
discuss two cases
1) c 2  1
4
6a
6
 2a 
 c  c ' , we
c c'
2
36a1a  1a1a (c  c ')2  (7a c  1a c ') 2
1  c2
X X 
 36a1a
0
(c  c ')2
(c  c ')2
64
 q  1  2 p and with u 2 
18 we have
u 2 36a1a c 2  u 21a1a c 2 (c  c ')2  (7a c  1a c ') 2 64a1a c 2  (7 a c  1a c ') 2
u 2c 2 X q  X 2 p 

0
(c  c ')2
(c  c ')2
36 1 c 2  1a1a c 2 (c  c ')2  (7a c  1a c ')2 36a1a c 2  (7 a c  1a c ') 2
c2 X q  X 2 p  a a

0
(c  c ')2
(c  c ')2
1
u 2 9a1a c 2  u 21a1a c 2 (c  c ')2  (7 a c  1a c ')2
9a1a u 2c 2  36a1a c 2
1 2 2 q
2p
2
u c X X 

0
4
(c  c ')2
(c  c ')2
By the same way u 2 c 2 X 2 p 1  u 2 c 2 X q  X 2 p  u 2 c 2  X and we prove then that in virtue of
the lemma 1, X  u 2c 2  q  1  2 p
 u 2 c 2Y p  u 2 ( X p  1a ) 2  X q 1  1a1a X
q
2p
u 21a1a  1a1a X

 X q  p 1  u 2 X p  u 2 2a  N  X p  ( X  u 2 )1a1a
p
X
And for Fermat-Catlan the solutions are q  1  2 p
2) c 2  1
The proof is the same than for Catalan, we prove that
u 2 36a1a  u 21a1a (c  c ')2  (7a c  1a c ')2 64a1a  (7a c  1a c ') 2
u2 X q  X 2 p 

0
(c  c ')2
(c  c ')2
Thus q  1  2 p and
u 2 36a1a c 2  u 21a1a c 2 (c  c ')2  (7a c  1a c ') 2 64a1a c 2  (7 a c  1a c ') 2

0
(c  c ')2
(c  c ')2
36 1 c 2  1a1a c 2 (c  c ')2  (7a c  1a c ')2 36a1a c 2  (7 a c  1a c ') 2
c2 X q  X 2 p  a a

0
(c  c ')2
(c  c ')2
1
u 2 9a1a c 2  u 21a1a c 2 (c  c ')2  (7 a c  1a c ')2
9a1a u 2c 2  36a1a c 2
1 2 2 q
2p
2
u c X X 

0
4
(c  c ')2
(c  c ')2
u 2c 2 X q  X 2 p 
We prove by the same calculus than higher in virtue of the lemma 2, then that
u 2 c 2  X  u 2 X q 1  c 2 X q  X 2 p  X q  2 thus 2 p  q  2; q  1  2 p
 (c 2  1)Y p  X 2 p  2a X p  X q  0
2a
 X p 1  X p  2  N
X
Thus Beal equation implies q+1=2p or q+2=2p , which means that if a=b=c=n then n=1 or
Z p  1a
n=2. It is Fermat theorem. We have also Y p  Z c  1a1a and d 
p
2
Y
p
p
p
7 Z
6a
7 d  1a d '
6
d' a p
; Y2 
 1a 1  v 
 d  d ' ; Z p  dY 2  1a  a
d d'
d d'
1 v
Y2
 X 2 p  2a X p  X 2 p 1  0 
5
36  (d  d ')2
Z  Y  1a1a  1a1a
(d  d ')2
we discuss two cases
3) d 2  1
64
u '2 
36(1  v)
u '2 36a1a d 2  u '2 1a1a d 2 (d  d ')2  (7 a d  1a d ') 2 64a1a d 2  (7 a d  1a d ') 2
u '2 d 2 Z c  Z 2 p 

0
(d  d ')2
(d  d ')2
36a1a  1a1a (d  d ') 2  (7 a d  1a d ') 2
1 d 2
c
2p
Z Z 

36
1
0
a a
(d  d ')2
(d  d ')2
c
p
d 2Z c  Z 2 p 
36a1a d 2  1a1a d 2 (d  d ') 2  (7 a d  1a d ') 2 36a1a d 2  (7 a d  1a d ') 2

0
(d  d ')2
(d  d ')2
Thus c<2p-1
2
2 2 p 1
 u '2 d 2 Z c  Z 2 p  u '2 d 2  Z and we can not prove in virtue of the lemma 1
But u ' d Z
2 2
because v<1, that Z  u ' d  c  1  2 p
And for Fermat-Catlan the solutions are not c+1=2p
4) d 2  1
The proof is the same than for Catalan, we prove that
u '2 36a1a  u '2 1a1a (d  d ') 2  (7a d  1a d ') 2 64a1a  (7a d  1a d ') 2

0
(d  d ')2
(d  d ')2
Thus c  2  2 p and
u '2 Z c  Z 2 p 
u '2 36a1a d 2  u '2 1a1a d 2 (d  d ')2  (7 a d  1a d ') 2 64a1a d 2  (7 a d  1a d ') 2

0
(d  d ')2
(d  d ')2
36 1 d 2  1a1a d 2 (d  d ')2  (7a d  1a d ')2 36a1a d 2  (7a d  1a d ')2
 aa

0
(d  d ')2
(d  d ')2
u '2 d 2 Z c  Z 2 p 
d 2Z c  Z 2 p
We can not prove by the same calculus than higher in virtue of the lemma 2 because v is
u '2 d 2  Z  u '2 Z c 1  d 2 Z c  Z 2 p  Z c  2
unknown and v<1, that
Thus Beal equation implies q+1=2p or q+2=2p or q=2p ! With Fermat equation there are
solutions for q=p=n=1 or q=p=n=2.
Conclusion
Catalan equation implies two other equations, and an original solution of Catalan equation
exists. By the same method we have proved that Beal equation does not have solutions for the
exponents greater than 2.
The bibliography
P. MIHAILESCU, A class number free criterion for catalan's conjecture, Journal of Number
theory, 99 (2003).
6
7
8