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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 $$ 1+3^a = 3^b+5^c. $$

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    $\begingroup$ 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
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    $\begingroup$ 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
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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 $3^3+1 = 3 + 5^2$.

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

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