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.

-Case

then

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-