An equation $p^a=q^b+r^c$ for powers of primes I´m preparing for math contests and found the following problem from this pdf.
Find all integers $a, b, c >1$ and all prime numbers $p, q, r$ which satisfy the equation 
$p^a=q^b+r^c$
($a, b, c$ and $p, q, r$ does not have to be distinct).
I guess we can solve this problem by examine possible divisors, from which we conclude that $p$ is a divisor in both $q$ and $r$, or none of them. But since $p, q, r$ are all primes, the first case is only possible if $p=q=r=2$. 
If $p=q=r=2$ we see that $a=k+1$ and $b=c=k$ for some integer $k$ satisfies the equation. Let now $p, q, r$ be distinct. Then, because of parity, one (and only one) of them must be 2.  But I´ve not come further than that. Any suggestions?
 A: This is only a very partial answer.  The problem strikes me as much more difficult than its origin (a "practice" problem in a section on congruences in a book on number theory for math contests) would suggest.
As the OP observes, one of the three primes must be $2$.  Setting aside the obvious family of solutions $2^{a+1}=2^a+2^a$ (with $a\ge2$), we're left looking for pairs of perfect powers of odd primes whose sum or difference is a power of $2$ (and not $2$ itself).  This suggests scanning OEIS sequence A244623:
$$9, 25, 27, 49, 81, 121, 125, 169, 243, 289, 343, 361,\ldots$$
Looking just for consecutive powers that differ by a power of $2$, I found
$$\begin{align}
25&=9+16\\
81&=49+32\\
125&=121+4\\
5041&=4913+128\\
\end{align}$$
(The last example is $71^2=17^3+2^7$.)  If there is a pattern here, I fail to see it.  Maybe one of the gifted 13-16-year-olds for whom the book was written can weigh in....
Added later:  A paragraph near the end of section D9 in Richard Guy's Unsolved Problems in Number Theory suggest this is, indeed, a very hard problem.  The paragraph considers the equation $p^m-q^n=2^h$, where the primes $p$ and $q$ and the positive integer $h$ are fixed, and the question is how many solutions $(m,n)$ there are.  According to UPiNT (third edition), it is apparently a fairly recent (2002) result that each triple $(p,q,h)$ admits at most one solution $(m,n)$.
