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

                                         2^P = 3^Q + 5

this is not fully rigorous but it is as close as I am going to get.

start with
2^p=3^q+5
2^p-2=3^q+3
2(2^(p-1)-1)=3(3^(q-1)+1)
2/3=(3^(q-1)+1)/(2^(p-1)-1)
ln(2/3)=ln((3^(q-1)+1)/(2^(p-1)-1))
ln(2)-ln(3)=ln(3^(q-1)+1)-ln(2^(p-1)-1)
now let ln(3^(q-1)+1)=ln(3^(q-1))+d1
and ln(2^(p-1)-1)=ln(2^(p-1))-d2
then we have
ln(2)-ln(3)=(q-1)ln(3)+d1-(p-1)ln(2)+d2
p*ln(2)-q*ln(3)=d1+d2 (1)
now via computer aided search I have verified that
the only solutions are (p,q)= (3,1) and (5,3)
for p,q<=10000.

now lets look at f(x)=ln(x)
f'(x)=1/x so while f(x) is increasing, the difference between f(x) and f(x+1) is rapidly decreasing

thus for p,q>10000 both d1,d2 will both be nearly zero thus we can simplify (1) to
p*ln(2)-q*ln(3)=0
p*ln(2)=q*ln(3)
p/q=ln(3)/ln(2)
now since ln(3)/ln(2) is irrational
we can't have p/q=ln(3)/ln(2) with p,q integers.

thus there are no further solutions other than
(3,1) and (5,3)

Posted by Daniel on 2008-10-05 00:46:33