Are there any other solutions?

First consider the degenerate cases (one of x,y,z is 0)

Consider x=0.  Then the equation becomes 1 + 4^y = 5^z, which means that 4^y and 5^z are consecutive if there is a solution.

Consider y=0.  Then the equation becomes 3^x + 1 = 5^z, which has no integral solutions for x,z.

Consider z=0.  Then the equation becomes 3^x + 4^y = 1, which has no integral solutions for x,z.

Now consider the non degenerate cases (all of x,y,z are at least 1)

Taking the equation mod 3 yields 1 = 2^z mod 3, which means z must be even.  Taking the equation mod 4 yields (-1)^x = 1, which means x must also be even.

If the substitution 2u=x and 2w=z is made, then the equation can be expressed as 2^(2y) = (5^w + 3^u) * (5^w - 3^u).  The sum of the factors is 2*5^w and the difference is 2*3^u.

Both the sum and difference of factors are multiples of 2 but not 4.  The only way for that to happen is if 2^(2y) factors as 2^(2y-1) * 2.  Then 5^w - 3^u = 2 and 5^w + 3^u = 2^(2y-1).

Substituting back into the expressions for the sum and difference of factors yields 5^w = 2^(2y-2)+1 = 4^(y-1)+1 and 3^u = 2^(2y-2) - 1 = 4^(y-1)-1.  That means 3^u, 4^(y-1), 5^w are three consecutive integers if there is a solution.

A solution must fall under one of two conditions.  1: 4^y and 5^z are consecutive or 2: 3^u, 4^(y-1), 5^w are consecutive.

If y=1,z=1 then a solution is 3^0 + 4^1 = 5^1.  If u=1,y=2,w=1 then a solution is 3^2 + 4^2 = 5^2.  If the exponents are larger (all at least 2) then Catalan's conjecture implies that there are no more solutions.

Edited on September 18, 2005, 3:19 am

Posted by Brian Smith on 2005-09-18 03:18:11