**Hint:** First, try some positive integers.

Hopefully you will find some integer pairs that work? Now, can you use mathematics to discover more...?

**Solution Commentary:** Multiple approaches to this problem are possible. And some can lead you astray, as implied in by Kicab Castaneda-Mendez (in a comment to Alison's Prezi): "Curious: I have two advanced degrees in math and statistics and I've never had to explain that I don't just add bigger and bigger numbers. I do like the rectange puzzle: how many rectangles with integer sides have area equal to perimeter. The answer seems to imply there are an infinite, nonzero answers. However, I wonder if Alison has considered two other answers (that align with the idea of teaching math in nonlinear manner, i.e., exploratoy and inventive): 1) there is no such rectange because regardless of what units are used the area will be in units squared while the perimeter will be in units; 2) an infinite number of such rectangles because as soon as you find one, you can change the scale. What learnings come from accepting these two answers as also correct, besides that solution to 4s=s^2?"

Why is Kicab's reasoning incorrect in his concern?

As to possible ways to approach to solving the problem, M.J. (Bellingham) creatively used a visual model, viewing (2m+2n) as a plane (i.e. addition table) intersecting (mn) as a hyperbolic paraboloid (i.e. the multiplication table). Wow...non-linear mixing of the simple and the complex!

My solution was more mundane. If 2m+2n = mn, then m = (2n)/(n-2) = 2+4/(n-2). Now, since m is an integer, the term 4/(n-2) must be an integer for an integer n. The only possible values of n are n=3 and n=4. Thus, the only possible pairs for the rectangle's dimensions are (3,6) and (4,4)...ignoring the reality of the case (0,0).

T.R. (Bellingham) says: "What I did was pretty primitive--I graphed the equation 2(x + y) = xy in the first quadrant, noticed the points .....(4,4), (3,6),(6,3) and realized there are no solutions with the minimum dimension greater than 4 (from the graph), and none where the minimum is 1 or 2 (from the equation)."

And, M.N. (Norway) suggests: "Area vs. Perimeter problem: here's how I thought about it - match edge units of the rectangle 1 to 1 with square units inside the rectangle. Match the edge units with adjacent square units, and you'll find 4 edge units that don't have a mate. They'll have to match with square units inside the rectangle not touching the edges. So the inside portion of the rectangle must be rectangle with area 4: a 2x2 square or a 1x4 rectangle, which leads to two solutions, 4x4 and 3x6. (Maybe easier to see with a picture)."