Given that each of x, y and z is a positive integer, it is observed that the equation 5^x + 12^y = 13^z is satisfied for x=y=z=2.

Does there exist any further solution?

   10    for Tot=2 to 2400
   20     for X=1 to Tot-1
   30       Y=Tot-X
   40       Lhs=5^X+12^Y
   50       Z=log(Lhs)/log(13)
   60       Zr=int(Z+0.5)
   70       if Lhs=13^Zr then print X;Y;Zr,Lhs;13^Zr
   80     next X
   90    next Tot

finds only

 2  2  2         169  169

so that means there is no solution with x and y totalling under 2400, other than 2,2,2.

Posted by Charlie on 2012-05-26 13:49:24