Find all positive integers $a, b, c$ such that $21^a+ 28^b= 35^c$ . 
Find all positive integers $a, b, c$ such that $21^a+ 28^b= 35^c$.

It is clear that the equation can be rewritten as follows: 
$$
(3 \times 7)^a+(4 \times 7)^b=(5 \times 7)^c
$$
 If $a=b=c=2$ then this is the first possible answer to this issue. It is also obvious that the sum of $(3*7)^a+(4*7)^b$ must end and be divisible by $5$. Since $21^a$ always ends at $1$, then $28^b$ should end at $4$ . Defined $b$  as $b=2+4k$ -- even positive integer.

Note that $21=3\times7$, $28=4\times7$ and $35=5\times7$, and so by unique factorization the numbers $21^a$, $28^b$ and $35^c$ are all distinct for all positive integers $a$, $b$ and $c$.
By unique factorization we see that the left hand side of
$$21^a+28^b=35^c,$$
is divisible by $7^{\min\{a,b\}}$ and hence $c\geq\min\{a,b\}$. Moreover the right hand sides of
$$21^a=35^c-28^b
\qquad\text{ and }\qquad
28^b=35^c-21^a,$$
are divisible by $7^{\min\{b,c\}}$ and $7^{\min\{a,c\}}$, respectively, because $28^b\neq35^c\neq21^a$. This implies
$$a\geq\min\{b,c\}
\qquad\text{ and }\qquad
b\geq\min\{a,c\},$$
from which it follows that $a=b=c$. Dividing out the factor $7^a$ leaves us with
$$3^a+4^a=5^a,$$
which clearly has the unique solution $a=2$.
