the solutions to 3^n - 2^m = 1 for all (n,m) is (2,3) and nothing else since we are talking about using only positive integers. Why is this the only solution?
CptBork, be generous and assume that is another solution and think about whether that is the set of natural number solutions (in short, whether such a set is finite or not).
Maybe (it's pretty late, but somehow this makes sense at the moment) : Look at \( 3^n - 1 = 2^m \) Factor the difference of two squares on the left, get \( 3^{\frac {n}{2}} - 1 \) for one of the factors. 2 must divide it (why?). It itself is the difference of two squares. Factor it, and keep going. Eventually, reach 3 - 1. Now this shows that if the original equation holds for any power of 3 greater than 2, one of that form holds for all in that cascade down to 2. Contradict, and conclude. And if that whole thing is stupid, a brief notice will do, thanks.
Ice, your method doesn't work for the solution n=m=1, because 2 is not a square. It would only work if m is even, so that the left hand side is a square. Both solutions have m is odd. I've never been very good at these but a line of thought which seems to give a useful representation is to write \(3^{n} = (2+1)^{n}\) and binomially expand : \(\sum_{k=0}^{n}\left( \begin{array}n \\ k \end{array}\right) \, 2^{k} = 2^{m} + 1\) \(\sum_{k=1}^{n}\left( \begin{array}n \\ k \end{array}\right) \, 2^{k} = 2^{m}\) \(\sum_{k=1}^{n-1}\left( \begin{array}n \\ k \end{array}\right) \, 2^{k} = 2^{m} - 2^{n}\) Though I've no idea if that's a blind alley or not...
Alpha, that's a dead end. I tried that last night but binomial coefficients are difficult to work with. At the least it shows that we cannot come up with a (simple) formula if the set of solutions is infinite. So that you all stop wasting your time... see here. Please Register or Log in to view the hidden image!
So wouldn't that make odd n impossible in the original equation ? - you don't get integers where they have to be. That doesn't matter - the idea is to prove impossibility of larger solutions, not find all the small ones. We have a special case of Catalan's {Conjecture} . That is not necessarily a waste of time here, eh? Just because someone has proven the general case - - -
Proceed to shame me. (The theorem was not proved by Catalan, by the way) http://www.ams.org/bull/2004-41-01/S0273-0979-03-00993-5/S0273-0979-03-00993-5.pdf
I think we can now all agree that a simply stated problem is not necessarily in fact a simple problem.
of course. this is the case with most problems in number theory. They are often very simple looking and easy to formulate, but require some very rigorous math to solve. Just look at Goldbach's conjecture, it sounds very simple but the solution is harder to find than black guy at a republican convention.
I actually some some black guys at the GOP convention. They were serving champagne to delegates while Palin bashed the "Washington out-of-touch elitists".
