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?

Using the results from earlier, we know x and z must be even for there to be a solution with x,y,z > 1. But if that is so, then 4^y = 5^(2n) - 3^(2m) where 2n=z and 2m=x.

We can factor the RHS to (5^n+3^m)(5^n-3^m). Since this expression is equal to 4^y, both terms must be powers of 2. But both terms cannot be multiples of 4 -- if they were, their difference would be a multiple of 4 and that difference, 2*3^m is clearly not. So at least one term isn't a multiple of 4. But if both terms must be powers of two and both terms cannot be multiples of 4 then one term must be exactly 2 (and the other term must be twice a power of four so the product is exactly a power of 4).

Now, the sum term can't be exatly two (unless n=m=0 but that contradicts the assumption that x,y,z >1 and gives the solution 0,1,0 already known) so the difference term must be exactly two. That is, 5^n - 3^m = 2 for some n, m. (The given solution is n=m=1)

That's about as far as I can go tonight. But every solution to the equation above corresponds to a set of x,y,z that meets the problems requirements.  If there's only one integral solution then there are no new solutions; if not, we'll know the next largest solution.

Posted by Paul on 2005-09-05 10:09:06