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-