Am I correct?
For example:
- The only time $3^a - 2^b = 1$ is $3^2 - 2^3$
- I believe that the only time $2^a - 3^b=1$ is $2^2 - 3^1 = 1$
This is not to say that all $p,q,c$ exist. Clearly $2^a - 3^b \ne 10$ for all $a,b$. I am thinking that if a solution exists, there is at most $1$ case where $p^a > q^b$ and at most $2$ if we accept either $q^b > p^a$ or $p^a > q^b$.
Here's my thinking:
- Let $p^a - q^b = c$
- Assume there is a second solution such that $p^d - q^e = p^a - q^b$ and $p^d > p^a$.
- $p^d - q^e = p^a - q^b$
- $p^a(p^{d-a}-1) = q^b(q^{e-b}-1)$
- Clearly, there exists $w$ where gcd$(w,p)=1$ and gcd$(w,q)=1$ and $p^aw = q^{e-b}-1$ and $q^bw = p^{d-a}-1$ and further $p^a \ne p^d \ne p^{d-a}$ and $q^b \ne q^e \ne q^{e-b}$
- So, it follows that $q^{e-b}-p^aw = p^{d-a} - q^bw$ and further that $q^{e-b} + q^bw = p^{d-a}+p^aw$
- Depending on whether $e-b > b$ or $b > e-b$ and whether $d-a > a$ or $a > d-a$, we could get one of the following:
$$q^b(q^{e-2b} + w) = p^a(p^{d-2a} + w)$$ $$q^{e-b}(1 + q^{2b-e}w) = p^{d-a}(1 + p^{2a-d}w)$$ $$\vdots$$
- Now, it seems to me (I haven't completed this yet) that it should be possible to show that $q^{e-2b} \ne q^b \ne q^e \ne q^{e-b} \ne \cdots$ and further that $p^{2a-d} \ne p^{p^a} \ne p^d \ne p^{d-a} \ne \cdots$
- And it seems to me that we can keep doing that ad infinitum where each $p^i < p^d$ and each new power of $p^i$ is distinct from the others.
- This then, if true, would lead to a contradiction since there are at most $d$ distinct powers between $p^d$ and $p$.
Does this make sense? Is this a well known result? It there a major flaw in my reasoning?
Thanks very much,
-Larry
Edit: Changed the question to make it clearer based on feedback from Tom Collinge.
There are certainly extra solutions if you allow $a =1 $ or $b = 1$ so I think you might want to exclude these. There is a pretty degenerate one in my comment, but here is a more substantial one: $2^{4} - 3^{1} = 2^{8} - 3^{5} = 13.$