Navigation 
 Home 
 Search 
 Site map 

 Contact Graeme 
 Home 
 Email 
 Twitter

 Skip Navigation LinksMath Help > Math Puzzles > Unsolved Problems > Catalan's Conjecture

Catalan's Conjecture

The conjecture made by Belgian mathematician Eug�nie Charles Catalan in 1844 that 8 and 9 (23 and 32) are the only consecutive powers (excluding 0 and 1). In other words,

32 - 23 = 1

is the only nontrivial solution to Catalan's Diophantine problem

xa - yb = ±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 2a = (2k+1)b±1, where a>1, b>1, k>0 is k=1, a=3, b=2, i.e. 23=32-1.

This special case can be phrased this way: "no power of 2 differs by 1 from any power other than 32".

Suppose 2a = (2k+1)b + 1.

If b is odd then 2a = (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 2a, so b can't be odd.

If b is even then (2k+1)b = 1 mod 4, so
2a = (2k+1)b+1 = 2 mod 4.
But 2a = 0 mod 4 (because a>1), a contradiction.

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

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

If b is odd then 2a = (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 2a = (2k+1)b-1 = (4k2+4k+1)b/2-1 =
      (4k2+4k)((4k2+4k+1)b/2-1 + (4k2+4k+1)b/2-2 + ... + 1)
The first factor, (4k2+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 2a = 3b-1 = 9b/2-1 = 8(9b/2-1 + 9b/2-2 + ... + 1)

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

If b/2 is even, then 2a = 3b-1 = 81b/4-1 = 80(81b/4-1 + 81b/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 x2 = y3±1, where x>1, y>1 is x=3, y=2, i.e. 23=32-1.

 . . . . . . . .

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.