## May 1999

parallel to the sides of the rectangles.
Consider the function .
When integrated over each smaller rectangle it gives 0:
y
which is zero precisely when at least one of (y is an integer -- that is,
when the rectangle has at least one integer-length side.
So the integral over the large rectangle, being the sum of the integrals
over the small rectangles, is also zero, whence the large rectangle
has at least one integer-length side.
_{1} — y_{0})
.
Consider lattice points of the form (p,j/p) where k/p,j are integers.
Move the coordinates so that no lattice points are on the boundaries
of any rectangles.
For each small rectangle, the number of lattice points inside the
rectangle is divisible by k:
it's the product of two integers (the number of points along the bottom
edge, times the number along the left edge), at least one of which
is divisible by p (by integrality).
Therefore the number of lattice points in the large rectangle is
divisible by p.
Again this is the product of two integers, and since p is prime,
one of the integers is divisible by p,
whence the length of the
corresponding edge is an integer (within an error of p).
Since 1/p was arbitrary, that error must in fact be 0.p |

*If you have any problems you think we might enjoy, please send them in. All replies should be sent to:* ponder@il.ibm.com