Catalan's Conjecture

The conjecture made by Belgian mathematician Eugénie 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

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, Metsänkylä 2003), according to Mathworld. 

A Special Case of Catalan's Conjecture

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+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, so b can't be odd.

If b is even then 2^a = (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 2^a = 3^b-1 = 9^b/2-1 = 8(9^b/2-1 + 9^b/2-2 + ... + 1)

If b¹2 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.

Internet References

Catalan's Conjecture in Wolfram's Mathworld

Eugénie Charles Catalan in Wolfram's Scienceworld

George Baloglou's "thirteen" page

Related pages in this website

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