Learn · the 5×5
Bigger, and in one way simpler
Everything you learned on the 4×4 applies, with more pieces to sort. And one of the 4×4's two parity cases cannot happen here at all, and the reason is worth knowing.
Start with what has not changed. A 5×5 is solved by the same route as a 4×4: build the centres, gather the edge pieces into groups so each behaves as one, then finish the puzzle as a 3×3. Anyone who has worked through the 4×4 already has the method, and what follows is only what differs. Anyone who has not should start there.
The first difference is arithmetic. Each face now carries nine centre pieces instead of four, and each edge of the cube carries three pieces instead of two: a middle piece with a wing either side of it. That means 12 middle edge pieces, 24 wings, and 54 centre pieces in all. The 8 corners are unchanged.
The second difference is the one that matters, and it is a single piece. The middle of each face is a genuine fixed centre: it cannot go anywhere, exactly as on a 3×3. Only turning the middle slice moves it, and turning the middle slice is something you stop doing the moment you have decided which colour goes where. So on a 5×5 the colour scheme is not yours to choose. Every 5×5 is made with white opposite yellow, and you have to work with that.
6 pieces
Fixed centres
One per face, immovable. These are the anchor, and the reason a 5×5 behaves better than a 4×4.
12 pieces
Middle edges
One per edge of the cube, sitting dead centre. Each one behaves exactly like a 3×3 edge.
24 pieces
Wings
Two per edge, one either side of the middle. Gather each pair onto its middle and the edge is whole.
Two kinds of centre, one tool each
Around each fixed centre sit two rings of pieces: four that touch it edge to edge, which cubers call the + centres because with the fixed piece they make a plus sign, and four in the corners of the face. They are different pieces and they need different handling, so there are two sequences worth knowing rather than one.
2R2 2U2 2R2 2U2 moves 4 corner centres and nothing else on the entire puzzle. Run it 3 times and you are back where you started.
2R2 3U2 2R2 3U2 moves 4 + centres and nothing else. This one undoes itself if you run it twice.
Both leave every corner, every edge piece and every fixed centre untouched, so neither can cost you work you have already done. That is what makes them safe to try when a face will not come together.
The parity that cannot happen
On a 4×4 there are two positions the last layer can show that a 3×3 never could: one edge turned around, and two edges needing to change places with nothing else out of order. A 5×5 can show the first and never the second, and the fixed centre is the reason.
Here is the argument, which is short. Every arrangement of pieces is either even or odd, meaning it takes an even or an odd number of simple swaps to undo. Work out those labels for the corners, for the middle edges and for the fixed centres, and a fact falls out that holds for every possible sequence of moves: the corner label multiplied by the middle-edge label always equals the fixed-centre label. Turning an outer face flips the first two together and leaves the third alone. Turning the second layer alone leaves all three alone. Turning the middle slice flips the middle edges and the fixed centres together. There is no move anywhere that breaks the relation.
Now finish the centres. The fixed centres are home, so their label is even, and the relation forces the corner and middle-edge labels to match. Matching labels is precisely the 3×3 rule, and two edges wanting a swap with the corners already home would break it. So it cannot occur. A 4×4 has no fixed centre to pin the relation down, which is exactly why it suffers the case and a 5×5 does not.
The parity that does happen
The flipped edge survives, because it comes from somewhere else entirely: the wings. Turning an inner layer on its own moves four wings round in a single cycle, which is an odd rearrangement, and once the centres are built you have no move left that could make another. So an edge group can end up assembled the wrong way round.
The cure is the 4×4's algorithm, unaltered:
2R2 B2 U2 2L U2 2R' U2 2R U2 F2 2R F2 2L' B2 2R2
On a 5×5 it disturbs 2 wings, the two that need to trade, and leaves everything else in the puzzle exactly where it was: no corner, no middle edge, and none of the three kinds of centre. Run it twice and the puzzle returns to where it started.
Which means the whole of the 5×5, beyond the 4×4, is two centre sequences and one algorithm you already know.
The claim that a 5x5 cannot show the two-edges-to-swap case is proved rather than repeated: the build computes, for hundreds of random sequences drawn from every kind of move the puzzle has, that the sign of the corner arrangement multiplied by the sign of the middle-edge arrangement always equals the sign of the fixed-centre arrangement. Finish the centres and that last sign is fixed, so the first two cannot disagree. Every algorithm on the page is separately applied to a solved puzzle and measured piece by piece, and the animations run on an independent engine held to the same answers.