rwg>[...] Prettier still (except for taking x -> +or-1)

(fourth:)

(-(1/2))*x*Sum[(-q)^n/(1 + q^n*x), {n, 0, Infinity}] == (1/2)*Sum[(-x)^j/(1 + q^j), {j, 1, Infinity}]

These Lambert transformations are all the same easy trick, which I'm not stating because I've already asked Funster Neil (who owes us his sliding block report) to figure it out for tomorrow.

The trick is simply to expand the summands as infinite geometric series (which Mathematica steadfastly prohibits), and then interchange the orders of summation. As you should always do the same thing to both sides of an equation, this leaves you with what you started with, except with the sides switched. Thus you fail to prove anything.

This is similar to when you try to put your left shoe on your right foot. How do you know which is wrong--the shoe or the foot? The only safe procedure is to switch both, and place your right shoe on your left foot. When this too fails, you conclude as above that the problem has no solution. This is why Julian goes barefoot. --rwg PS: As Neil describes in his blog, the later "hardest" 4x4 puzzles are no harder, and sometimes easier, than in Nick's interactive collection<http://www.puzzleworld.org/SlidingBlockPuzzles/bickford.htm>, which includes puzzles where the goal piece is nonconvex, and technically fails to make a full diagonal traverse, due to misfitting a corner. This is a shame, because that four-piece 26-mover is a gem.