If x, y and z are integers, 3^x+4^y=5^z is satisfied for x=y=z=2.

Are there any other solutions?

(first, let's assume x,y,z are all greater than zero; we already know the solution where zero is allowed.)

consider the equation mod 3, 4, and 5 respectively.

mod 3: first term = 0 mod 3, second term = 1 mod 3 regardless of x or y. So third term must = 1 mod 3 which, since 5=-1mod 3 requires that z is even.

mod4: third term = 1 mod 4 and second term = 0 mod 4 regardless of values of y and z, so first term must = 1 mod 4. Since 3 = -1mod4, that requires x is also even.

mod 5: since x is even, first term is 9^p (x=2p) and is either -1 or 1 mod 5 according to whether p is odd or even. second term is also either -1 or 1 mod 5 according to whether y is odd or even; third term is 0 mod 5, so first and second terms must be negatives of each other mod 5. That means:

z is even
x is even
and EITHER
   x = 4a, y=2b+1
OR
   x=4a+2, y=2b

(for some a and b)
The known solution with x=y=z meets these requirements (a=0, b=1).

This doesn't reveal any other solutions and it doesn't prove there aren't any, but at least it limits where any other solutions might be hiding...

Posted by Paul on 2005-09-02 04:13:26