Since the RHS is a power of 2, we can't have at the same time, because the LHS would be a multiple of 3.
Without loss of generality -becuase of the symmetry on a,b,c- suppose , then or
- Case then or and .
Note that 2 doesn't divide the RHS, and so and can't be greater than 1 at the same time.
WLOG consider or then since the RHS is not divisible by 4, c is at most 3., and no less than 2 -since the RHS is even-.
For we get and for we have no solution since should be equal to 3.
Thus we have ; ; and all the permutations of the values of a,b,c there.
Then , again, since 4 doesn't divide the RHS, b and c cannot be both greater than 3.
WLOG assume . If -0 or 1- then and , so assume , then -by looking back at the equation- and so
If then (1), so actually 4 divides , and then we are forced to have , but that would imply that 8 divides , which doesn't happen in (1) !!
If then (2), so actually 2 divides , but 8 doesn't, so , here note that is impossible since this would imply which is a power of 2! -and can't be divisible by 3 -, so and we must have - in (2)-, and so i.e. which is impossible!.
So we get , , and basically we find that these are the same solutions we found on the other part! -considering the permutations-