Catalan's Conjecture

Date: 06/02/2007 at 21:31:49
From: Thomas
Subject: Solve a^b = b^a + 1

Solve a^b = b^a + 1.  How many integer solutions are there?  Are there
an infinite number?  How are the solutions distributed?

I need some clues as to how to analyze this equation.  I noticed that
8 is 2^3 while 9 is 3^2. I started thinking about pairs of consecutive 
numbers where one is of the form a^b and the other is of the form b^a.
  
I expressed the relationship between a and b in the above equation and
tried tried taking the log of both sides, but didn't know how to
expand log(b^a + 1) or if this was the best approach.

Is there a way to analyze this problem other than random guessing?

Date: 06/03/2007 at 08:54:13
From: Doctor Vogler
Subject: Re: Solve a^b = b^a + 1

Hi Thomas,

Thanks for writing to Dr. Math.  The only integer solution with a>1 
and b>1 is the one you already found.  In fact, more is true; see

  "Catalan's Conjecture" From MathWorld - A Wolfram Web Resource.
    http://mathworld.wolfram.com/CatalansConjecture.html 

Of course, if b = 1, then a = 2, and if a = 1, then b = 0.  So these
would be the other integer solutions.

On the other hand, proving Catalan's Conjecture takes a lot of work,
so you could probably come up with a more accessible analysis of your
equation, sort of along the lines of

  Solving the Diophantine Equation x^y - y^x = x + y
    http://mathforum.org/library/drmath/view/66640.html 

I guess the way I would attack this problem is by using the function
f(x) defined in

  Solving the Equation x^y = y^x
    http://mathforum.org/library/drmath/view/66166.html 

A solution to your equation has

  a^b = b^a + 1

and therefore

  a^b > b^a

which means that

  b log a > a log b

and therefore

  (log a)/a > (log b)/b

or

  f(a) > f(b).

That means that either a or b is 1 or 2, or both are bigger than e and
therefore

  a < b

which means (since a and b are both integers) that

  a <= b - 1.

Next, I would try to prove that for fixed a >= 3, the function

  g(x) = a^x - x^a

is increasing on x > a.  (Hint: x > a > e implies a^x > x^a.)  That
means that

  g(b) >= g(a+1).

Next, I would try to prove that for a >= 3,

  g(a+1) > 1,

perhaps by defining another function

  h(x) = x^(x+1) - (x+1)^x - 1

and looking at its derivative....

Of course, you also have to check the special cases when a or b is 1
or 2, but these are much easier than the general case I just outlined.

If you have any questions about this or need more help, please write
back and show me what you have been able to do, and I will try to
offer further suggestions.

- Doctor Vogler, The Math Forum
  http://mathforum.org/dr.math/