This is not a research problem, but challenging enough that I've decided to post it in here:

Determine all triples $(a,b,c)$ of non-negative integers, satisfying $$ 1+3^a = 3^b+5^c. $$

  • 5
    $\begingroup$ Papers are still published on such questions (Lucia gave two examples, and there are others), so it's research-level (and thus fair game for Mathoverflow) even if it did not arise in your own research. $\endgroup$ – Noam D. Elkies Apr 22 at 2:47
  • 2
    $\begingroup$ But if it's a puzzle to which OP already knows the answer, then I'd say it's not appropriate for MO. $\endgroup$ – Gerry Myerson Apr 22 at 12:35
  • $\begingroup$ NoamD.Elkies: thanks for your comment. GerryMyerson: thanks. Had I known that these type of Diophantine equations are still an active area of research, would have likely phrased the problem that way. Still trying to digest the concept --- as there is a section (tag) with elementary proofs, that are contest problems, as opposed to research problems, but challenging enough that people still post. $\endgroup$ – kawa Apr 22 at 14:14

I can't resist this: The young Chris Skinner showed that if $a$, $b$, $c$, $d$ are fixed positive integers, and $p$ and $q$ are positive coprime integers then the equation $$ ap^x + bq^y = c+ dp^z q^w $$ has a bounded number of solutions in $(x,y,z,w)$ and that a bound on these could be computed (and the equation solved in practice). This solves (in principle) the more general equation $1+3^a 5^d = 3^b+ 5^c$. Anyway, there is a large literature around such exponential diophantine equations, and Skinner's paper will give some references.

Following a reference from Skinner's paper, Theorem 4.01 of Brenner and Foster gives an explicit treatment of the equation $$ 3^a + 7^b=3^c+5^d, $$ which completely resolves the problem in this question (take $b=0$). Their proof is elementary, and the only non-trivial solution to the equation in the question is $3^3+1 = 3 + 5^2$.

  • $\begingroup$ Lucia, many thanks for the paper. $\endgroup$ – kawa Apr 22 at 0:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.