Powers Differing By 1 (Catalan)

Catalan's Conjecture is that the equation a^n - b^m = 1  has no 
integer solutions with m,n > 1  other than a=3,n=2,b=2,m=3.  In 1976
Tijdeman that this conjecture is true except for possibly a finite 
number of exceptions, but the proof is not simple.

On the other hand, certain special cases of this conjecture can be
proven by very simple arguments.  For example, it's easy to show
that there do not exist powers 2^n and 3^m such that |2^n - 3^m| = 1
other than the two cases 2,3 and 8,9.  There are several ways of
proving this.  Consider the case 3^m - 1 = 2^n.  Notice that if m is 
composite of the form m=ab, then (3^a - 1) and (3^b - 1) must both 
be powers of 2.  It follows that for each prime p that divides m we 
must have (3^p - 1) = 2^u for some integer u.  However, no odd 
prime power of 3 is congruent to 1 (mod 8), so we can rule out any 
odd prime divisors of m (for u>2). This leaves only the case 
3^(2^k) - 1 = 2^u, which has the solutions (k=0,u=1) and (k=1,u=3). 
However, there can be no solution for k>1 because 3^(2^k)-1 is 
divisible by  (3^2 + 1) = 2*[5]  for all k>1.

A slightly modified approach is to observe that if 3^s = (2^r + 1) 
then clearly 3^s < 2^(r+1), so we have s < (r+1) ln(2)/ln(3).  
Also, by simple congruence arguments, if 3^s - 1 is divisible by 2^r, 
then s is divisible by 2^(r-2).  (This just extends the fact used 
above that if 3^s - 1 is divisible by 8, then s must be even.)  So 
we have
             2^(r-2)  <=   s   <   (0.6309)(r+1)

For r=3 this requires 2 <= s < 2.52, so s=2 is a possibility 
(fortunately), but for any r > 3 the value of 2(r-2) exceeds 
0.6309(r+1), so no other solutions are possible.

It's also easy to show that there are no positive integral solutions 
to  2^m - 3^n = 1  other than  m=2, n=1.  Written in the form  
2^m - 1 = 3^n  it's obvious that if m is composite of the form ab 
then (2^a - 1) and (2^b - 1) must both be powers of 3.  It follows 
that for each prime p that divides m we must have 2^p - 1 = 3^u for 
some integer u.  However, no odd power of 2 is congruent to 1 
(mod 3), so we can rule out any odd prime divisor of m.  This leaves 
only the case 2^(2^k) - 1 = 3^n, which has the solutions {k=0,n=0} 
and {k=1,n=1}, but there can be no solution for k>1 because 2^4 - 1 
is divisible by 5, and thus is not a pure power of 3.

Along these same lines, Ellison proved that  

                |2^n - 3^m| > (2^x) exp(-x/10)  

for n>27, and Tijdeman proved there exists a number c >=1  such that

                  |2^n - 3^m| > (2^n)/(n^c)

Results of this kind are relevant to a problem discussed in the note
Sequence Partitionable Into Powers of 2 or 3.