Catalan's Conjecture
a**p=b**q +/- 1
a==0 [mod 2] or b==0 [mod 2]
=> WLOG b==0 [mod 2]
Therefore, by similar to FLT, p=2
a**2=b**q +/- 1
b==0 [mod 2] => b**q==0 [mod 4]
a==1 [mod 2] => a**2==1 [mod 4]
Therefore,
b**q=a**2 - 1=(a+1)(a-1)
hcf (a+1,a-1)=2
=> either
a+1=b1**q, a-1=b2**q (clearly not possible) or
a+1=2b1**q, a-1=2**(q-1)b2**q => 2**(q-2)b2**q=b1**q - 1 or
a+1=2**(q-1)b2**q, a-1=2b1**q => 2**(q-2)b2**q=b1**q + 1
Therefore, by similar to FLT, b1=1
=> b2=1, q=3
i.e.
a=3, p=2, b=2, q=3
Copyright 1997 James Wanless