dissection of a $1\times 5$ rectangle to a square I've been thinking about the following problem:
We have a $1\times 5$ rectangle: how to cut it and reassemble it such that it forms a square?
Thanks a lot!
 A: 
A: You can do it with four pieces, and translations only (no rotations).


A: Here's a method that is surely far from optimal.
First, cut it into 5 $1\times1$ squares, and arrange them into a $2\times2$ square sitting next to a $1\times1$. Now there's a standard way to cut two squares into a total of four pieces that rearrange to form a single square. I wish I knew how to draw pictures. Anyway, let the small square be $ABCD$ With $C$ a vertex of the big square and $CD$ along the side $CEFG$ of the big square. Find $H$ on $CG$ such that $GH=AB$. Cut along $FH$ and along $AH$. Then the bits $FHG$, $ABH$, and $ADEFHA$ can be moved to form a square.

A: A standard approach to finding solutions to problems like this is to overlay two tilings. In the following image, the bright yellow rectangles give one 5-piece and two 4-piece solutions requiring only translations (one of which is the same as Robert Israel’s above):
$\hspace{1.15in}$ 
A: Cut a small square out of it and throw the rest away.  If you are not allowed to throw any part away, then crumple the rest into a ball and stack it on the square.  If you are not allowed to throw any part away and you are not allowed to overlap pieces, then I don't know the answer.
Edit:

Now there's a standard way to cut two squares into a totasl of four
  peices that rearrange to form a single square.
  I wish I knew how to draw pictures.


(source: cut-the-knot.org) 
A: We can consider this problem in terms of a tiling of the plane with squares. Or equivalently, we treat the large square as a torus, i.e., its top edge is glued to its bottom edge, and its left edge is glued to its right edge. Thus all the current solutions on this page are essentially equivalent.

Note that all of those right triangles are similar to the triangle with sides $1, 2, \sqrt5$.
