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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,
is the only nontrivial solution to Catalan's Diophantine problem32 - 23 = 1
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.
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.
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Catalan's Conjecture in Wolfram's Mathworld
Eug�nie Charles Catalan in Wolfram's Scienceworld
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