This is a puzzle I bought. It has 5 coloured pegs, R(ed), Y(ellow),
G(reen), B(lue) and P(purple). Initially on each pegs, there are 4
discs (of SAME size, unlike Hanoi) but of different colours.
On the R(ed) peg, in order from top to bottom, the discs are Y, G, B, P.
On the Y peg, the discs are G, B, P, R. 
On the G peg, the discs are B, P, R, Y. 
On the B peg, the discs are P, R, Y, G. 
On the P peg, the discs are R, Y, G, B. 
The aim is to end the game with the 4 discs of each colour on their
matching peg subject to the rules:
1. Discs may be moved singly or in pairs.
2. A disc may be moved onto another disc of the same colour or onto an
empty peg of that colour.
3. When 2 discs are moved together, the bottom one is the indicator of
the colour and can be placed onto a disc or empty peg of the same
colour.
4. No more than 7 discs can be stacked on a peg at one time.
I've tried this puzzle many times but find it impossible to go far.
Can any mathematician confirm that this cannot be solved (I don't need
any proofs, just simple "No")? Or if you can provide the solution, I
will give you a good tip. Please try this puzzle out and give me your
comments!

No, the problem appears unsolvable. Even a simplified version of it
with only 3 different coloured pegs with 2 different colured discs and
similar rules cannot be solved.

Thank you for your comments. That's why I find this puzzle very
puzzling to solve with the given rules. Interestingly, the puzzle came
from a box with a "Marks and Spencer" label! Could it be that the
inventor simply wrote the instructions without even trying out the
puzzle himself, or even bothering whether it could be solved or not?

Problems of this kind are sometimes easier to work out going in
reverse, ie. start with the disks segregated on the pegs of matching
colors and see if the "mixed up" state (where disks are taboo on the
peg of the same color) can be attained.

regards, mathtalk-ga

I've been working on this puzzle, and I have a couple of observations to add.

The final state is homogeneous, with each disk resting on a peg or
other disk of the same color, while the initial state is free of such
matches.  Since a move can result in increasing the number of matches
by one and never decreases the number of coloring matches, at least
twenty moves are required, possibly (likely) more.

The other note is wondering whether as originally worded the move of a
pair of disks is allowed to invert the order of the two disks as they
are transferred.  In working with some smaller versions of the puzzle
I find that it cannot be solved without allowing one such move.

I wrote a Prolog program to search for solutions, but the depth of the
puzzle (20+ moves, some branching) makes it a computational challenge
for my laptop!

regards, mathtalk-ga