Determine all possible pair(s) of nonnegative integers (P, Q), that satisfy this equation:

                                         2^P = 3^Q + 5

The only two solutions appear to be (3,1) and (5,3). The left and right sides of the equation are 8 in the former case, and 32 in the latter.

list
   10      for I=1 to 1000
   20       P2=int(2^I+0.5)
   30       P3=P2-5
   40       if P3>0 then
   50        :P3=int(log(P3)/log(3)+0.5)
   60        :if 2^I=3^P3+5 then print I;P3,P2;3^P3+5
   80      next
OK

run
 3  1    8  8
 5  3    32  32
OK

Posted by Charlie on 2008-10-04 15:48:05