**Catalan's Conjecture**

The conjecture made by Belgian mathematician
Eug�nie Charles Catalan in 1844 that 8 and 9 (2^{3}
and 3^{2}) are the only consecutive powers
(excluding 0 and 1). In other words,

3^{2} - 2^{3} = 1

is the only nontrivial solution to Catalan's Diophantine problem
x^{a} - y^{b} =
±1

The conjecture was finally proved by Preda Mihailescu in a manuscript
privately circulated on April 18, 2002. The proof has now appeared in
print (Mihailescu 2004) and is widely accepted as being correct and valid (Daems
2003, Mets�nkyl� 2003), according to Mathworld.

## A Special Case of Catalan's Conjecture

The only solution of 2^{a}
= (2k+1)^{b}±1, where a>1, b>1,
k>0 is k=1, a=3, b=2, *i.e.* 2^{3}=3^{2}-1.
This special case can be phrased this way: "no power of 2 differs by 1 from
any power other than 3^{2}".

Suppose 2^{a} = (2k+1)^{b} + 1.

If b is odd then 2^{a} = (2k+1)^{b}+1 = (2k+2)((2k+1)^{b-1}
- (2k+1)^{b-2} + ... + 1)

The second factor, (2k+1)^{b-1} - (2k+1)^{b-2} + ... + 1, is
an odd number greater than 1, which can't be a factor of 2^{a}, so b
can't be odd.

If b is even then (2k+1)^{b} = 1 mod 4, so

2^{a} = (2k+1)^{b}+1 = 2 mod 4.

But 2^{a} = 0 mod 4 (because a>1), a contradiction.

So there is no value of b such that 2^{a} = (2k+1)^{b} + 1.

Now suppose 2^{a} = (2k+1)^{b} - 1.

If b is odd then 2^{a} = (2k+1)^{b}-1 = (2k)((2k+1)^{b-1}
+ (2k+1)^{b-2} + ... + 1)

The second factor, (2k+1)^{b-1} + (2k+1)^{b-2} + ... + 1, is
an odd number greater than 1, so b can't be odd.

If b is even then 2^{a} = (2k+1)^{b}-1 = (4k^{2}+4k+1)^{b/2}-1
=

(4k^{2}+4k)((4k^{2}+4k+1)^{b/2-1}
+ (4k^{2}+4k+1)^{b/2-2} + ... + 1)

The first factor, (4k^{2}+4k), is a power of 2, so all its factors
must be powers of 2. In particular, k and k+1 must both be powers of 2,
which means k=1.

Since k=1, and b is even, we have 2^{a} = 3^{b}-1 = 9^{b/2}-1
= 8(9^{b/2-1} + 9^{b/2-2} + ... + 1)

If b≠2 and b/2 is odd, then 9^{b/2-1}+9^{b/2-2}+...+1 is
odd and larger than 1, and thus can't be a factor of 2^{a}, so b/2 is
even.

If b/2 is even, then 2^{a} = 3^{b}-1 = 81^{b/4}-1 =
80(81^{b/4-1} + 81^{b/4-2} + ... + 1), a multiple of 5, which
is a contradiction.

That leaves only b=2, giving our familiar solution.

## Another Special Case of Catalan's Conjecture

The only solution of x^{2}
= y^{3}±1, where x>1, y>1 is
x=3, y=2, *i.e.* 2^{3}=3^{2}-1.
. . . . . . . .

### Internet references

Catalan's
Conjecture in Wolfram's Mathworld

Eug�nie
Charles Catalan in Wolfram's Scienceworld

George
Baloglou's "thirteen" page

### Related pages in this website

The webmaster and author of this Math Help site is
Graeme McRae.