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



Diophantine equation $3^a+1=3^b+5^c$
I can't resist this: The young [Chris Skinner](https://www.sciencedirect.com/science/article/pii/0022314X90901125) 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](https://projecteuclid.org/download/pdf_1/euclid.pjm/1102724775) 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 $3^3+1 = 3 + 5^2$. 


