Lemma 1:
I claim that for any n x n x n cube (n even or odd) the number of clockwise twists away from the solved orientation of all the corners summed together is congruent to 0 (mod 3).

Proof:
We will define two opposite colors of the cube to be our reference for orientation. Let the U and D face colors together represent our reference for defining the orientations of the corners. The number of clockwise twists on a corner is then defined to be the number of counter-clockwise rotations required to place the reference colored sticker on the corner onto the U face if the corner is in the U layer, or onto the D face of the corner is in the D layer.

The only turns that affect the corners on any n x n x n cube are the 18 outer face turns F, F', F2, B, B', B2, R, R', R2, L, L', L2, U, U', U2, D, D', D2. We can see that rotating the entire cube preserves the orientation of the corners based on our definition above, since the reference colors will change with all the pieces given any x or z rotation. Also it is easy to see that a y rotation does not affect the orientation of the corners either. This is also easily shown by rotating the entire solved cube, clearly the corners are still all correctly oriented.

Since the orientation is unaffected by cube rotations, then we only need to examine how each of the possible turns on one face in each pair of opposite faces affects the corners. This in effect examines for both faces, since we could simply rotate the opposite face into the position of the face we have examined, and the same results would still hold.

1) How a U turn affects the orientations of the corners. Starting the from the initial solved state where every corner has no clockwise twists on it, it is easy to see that no U turn will chance the orientation of any corners on the cube. Therefore a U turn keeps the orientation of the corners of the cube congruent to 0 (mod 3).

2) All of the following cases for this step will assume that we start from the solved state of the cube, where every piece is in the correct position and has the correct orientation. This is how the puzzle is scrambled and solved, and therefore it is valid to examine it this way.

a) How an F turn affects the orientations of the corners. Perform the move F. Now by examining where the reference colored stickers are each of the corners FUL, FUR, FDL, FDR, we see that FUL and DBR both have 2 clockwise twists on them, and FUR and DBL both have one clockwise twist on them. The orientation of the B layer corners has been unaffected. So the total number of clockwise twists on the corners is 6, which is congruent to 0 (mod 3).
b) How an F' turn affects the orientations of the corners. Perform the move F'. By the same process as above we now see that both FUL and DBR have 2 clockwise twists on them, and both FUR and DBL have one clockwise twist on them. The B layer corners are again unaffected, so the total number of clockwise twists is 6, which is congruent to 0 (mod 3).
c) How an F2 turn affects the orientations of the corner. It is easy to see that the orientations of the corners is unaffected here, and hence the total number of clockwise twists on all corners remains 0.

3) All of the following cases for this step will assume that we start from the solved state of the cube, where every piece is in the correct position and has the correct orientation. This is how the puzzle is scrambled and solved, and therefore it is valid to examine it this way.

a) How a R turn affects the orientations of the corners. Perform the move R. Again by the same process as above we see that RUF and DBR both have two clockwise twists on them, and both UBR and FDR have one clockwise twist on them. This totals to 6 and is congruent to 0 (mod 3).
b) How a R' turn affects the orientations of the corners. Perform the move R'. RUF and DBR both have two clockwise twists on them, and both UBR and FDR have one clockwise twist on them. This totals to 6 and is congruent to 0 (mod 3).
c) How a R2 turn affects the orientations of the corners. Again it is clear that an R2 move does not affect the orientations, and hence the total number of clockwise twists is still 0.

So having exhausted all the possible moves that affect the corners on any n x n x n cube, we can conclude that for any n x n x n cube (n even or odd) that the number of clockwise twists on the corners must be congruent to 0 (mod 3).

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