A few weeks ago, [I asked on math.stackexchange.com](https://math.stackexchange.com/q/729107/9020) how many quadruples of non-negative integers $(a,b,c,d)$ satisfy the following equation:


$$2^a + 3^b = 2^c + 3^d \quad (a \neq c)$$


I found 5 quadruples: $5 = 2^2 + 3^0 = 2^1 + 3^1$, $11 = 2^3 + 3^1 = 2^1 + 3^2$, $17 = 2^4 + 3^0 = 2^3 + 3^2$, $35 = 2^5 + 3^1 = 2^3 + 3^3$, $259 = 2^8 + 3^1 = 2^4 + 3^5$


I didn't get an answer, but only a link to an [OEIS sequence](http://oeis.org/A085634) (no more quadruples below $10^{4000}$), so I'm asking the same question here. 



> 
> Is there a way to prove that they are [not] infinite?
> 
> 
> 


And, more generally, are there known tuples for which the following equation:
$$p\_{i\_1}^{a\_1} + p\_{i\_2}^{a\_2} + ... + p\_{i\_n}^{a\_n}=p\_{i\_1}^{b\_1} + p\_{i\_2}^{b\_2} + ... + p\_{i\_n}^{b\_n}$$
holds for infinitely many (or holds only for finitely many) $a\_i,b\_i$?



How many solutions to $2^a + 3^b = 2^c + 3^d$?
That these are all the solutions to $2^a+3^b=2^c+3^d$ was conjectured by Pillai in the 1930s and proved by Stroeker and Tijdeman in 1982, using lower bounds for linear forms in logarithms. Finiteness was proved by Pillai.


Oh, and I should have added that the answer is "no" to your second question via a result of Evertse, provided you have no "vanishing subsums". 


