TITLE: The kissing number of a square, cube, hypercube?
QUESTION [35 upvotes]: How many nonoverlapping unit squares can (nonoverlappingly) touch one unit square?
By "nonoverlapping" I mean: not sharing an interior point.
By "touch" I mean: sharing a boundary point.
          

It seems the answer for a square in $\mathbb{R}^2$ should be $8$, 
and for a cube in $\mathbb{R}^3$, $26$.
But a 1999 paper by Larman and Zong, "On the Kissing Numbers of Some Special Convex Bodies."
Discrete Comput Geom 21:233–242 (1999). 
(Springer link) says

"In this note we determine the kissing numbers of octahedra, rhombic dodecahedra and
  elongated octahedra. In fact, besides balls and cylinders, they are the only convex bodies
  whose kissing numbers are exactly known."

In that paper, they were interested in the translative kissing number and the lattice kissing number,
whereas I want to consider arbitrary orientations of each square/cube.  Despite the quote above,
it seems this should be known...?

Update (30Dec12) The following explains (I believe) the $0.82$ in Henry Cohn's comment,
leading to his proof for $\le 9$ in $\mathbb{R}^2$:

REPLY [3 votes]: The problem for squares may have been settled first by J W T Youngs, A lemma on squares, American Math Monthly 46 (1939) 20-22. Youngs' proof can also be found in A M Gleason et al., The William Lowell Putnam Mathematical Competition – Problems and Solutions: 1938-1964, pp. 461-463, and in R Honsberger, Mathematical Morsels, pp. 82-89. A discussion of the history, and another proof, will be found in M S Klamkin et al., The kissing number of the square, Mathematics Magazine 68 (1995) 128-133. 
Likuan Zhao, The kissing number of the regular polygon, Discrete Mathematics 188 (1998) 293-296, lets $P_n$ be a regular polygon with $n$ sides, and $k(n)$ its kissing number. After citing earlier results $k(3)=12$, $k(4)=8$, and $k(6)=6$, the paper establishes $k(n)=6$ for $n>6$. Zhao and Junqin Xu, The kissing number of the regular pentagon, Discrete Math 252 (2002) 293-298, finish it off by proving $k(5)=6$.