# Diophantine equation $3^a+1=3^b+5^c$

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

• 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. – Noam D. Elkies Apr 22 at 2:47
• But if it's a puzzle to which OP already knows the answer, then I'd say it's not appropriate for MO. – Gerry Myerson Apr 22 at 12:35
• 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. – kawa Apr 22 at 14:14

## 1 Answer

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

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