# Integer solution to exponential diophantine equation

1. Sep 28, 2011

Hey everyone!

I was recently scribbling on paper, and after a series of ideas, I got stuck with a problem. That is, can I find out if there exists some integers A and B such that

$C=2^{A}3^{B}$

For some integer C?

For an arbitrary C, how do I know whether some $A, B \in \textbf{Z}$ exist?

2. Sep 28, 2011

### dodo

it shouldn't be harder than testing if the number is divisible by 2 or by 3; and, in that case, if you are interested in the actual values of A and B, just divide and iterate. Unless you mean really big numbers.

3. Sep 28, 2011

### RamaWolf

Code (Text):

Solving with a computer:

Factor C  givin a list of it's prime factors and their occurences;
if there i9s a factor > 3 then 'No, C is not of the required form'
else 'Yes, A and B are the number, the factor 2 (resp 3) occurs

Set A and B to zero
Loop2:
If C is even replace C by C / 2 andd add 1 to A
loop until C is odd
Loop3:
If C is multiple of 3 (add the digits modulo 3)
replace C by C / 3 andd add 1 to B
loop until C is not a multiple of 3
Test: it the remaining C is one, then 'Yes' else 'No'