# James Tanton’s rectangle problem

Saw this neat question from James Tanton today and thought it would make for a nice last project of the year. We are still stuck in Omaha due to flight issues, so no videos . . . .

Their first thought was to try out a few rectangles. After a few tries they found one that worked.

Next they tried some larger numbers and found that large rectangles didn’t seem to be close to having the property Tanton was looking for. They speculated the 2×3 rectangle was the only one.

My younger son thought that the reason the 2×3 rectangle was the only one was that “one side stays the same and the other side doubles” when you add one to both sides.

That observation led to my older son writing down an algebraic equation describing the situation. We checked that the 2×3 rectangle satisfied the equation.

Now, how is this equation going to help us? It took a while for them to figure out that solving for one of the variables would be helpful. Part of the problem is that the result doesn’t look that helpful – but both x and y have to be integers, so it is helpful.

Next we tried a few small values of y to see when (y+1) / (y – 1) would be integers. We found 2, but those two values described the same rectangle.

My younger son also noticed that the value of (y+1)/(y-1) was decreasing as y increased.

Finally, I showed them a “simple” way to see that the expression (y + 1) / (y – 1) decreased as y increased. Unfortunately I wrote the inequality backwards, but the proof at least shows how the right proof would go 🙂

So, a fun little project to end the year. Can’t wait to see what math 2016 brings!

## 4 thoughts on “James Tanton’s rectangle problem”

1. Michael says:

I like your solution since it develops naturally and nothing is pulled out of a hat.

One could attack the expression x = (y + 1)/(y – 1) at the end using a factoring argument. Since x is an integer, the integer y – 1 must divide y + 1 exactly. In particular, y – 1 is a common factor of itself and y + 1. But any common factor of two numbers a and b must divide their difference a – b (or any other linear combination), so we see that y – 1 must divide (y+1) – (y-1) = 2. This implies that y – 1 = 1 or 2 (negative values can be ruled out) so y = 2 or 3 and then x = 3 or 2.

There is another way to solve this that makes use of “factoring” in a different way. Starting from xy = x + y + 1, you can subtract both x and y to get:

xy – x – y = 1.

The lefthand side is (x – 1)(y – 1) – 1 so, adding 1 to both sides, we see that:

(x – 1)(y – 1) = 1 + 1 = 2.

The factors must be integers so one again arrives at (x,y) = (2,3) and (3,2) as the only (positive) solutions.

2. Joshua says:

Did you see the follow-up tweet: how many triangles are there with integer sides such that the area doubles when the side lengths are increased by 1?

1. mjlawler says:

I didn’t see that one, but I did see the one about rectangular solids.. My initial reaction to this problem is that it looks like a mess to approach algebraically!

2. mjlawler says:

Ha – wait, isn’t it just a parity argument with Heron’s forumla that shows you can’t double the area in this case?