The perimeter of a rectangle in units equals its area in units squared. (4,4), (3,6), and (6,3) are three possible pairs of lengths for this rectangle.

Give another pair of positive integral sides for this rectangle or prove why there isn't another pair.

(In reply to Solution (and spoiler) I think by SilverKnight)

I agree with the solution previously posted by SilverKnight. Here is a way to solve the problem algebraically.

As mentioned by SK, the given constraints boil down to the equation

xy = 2x + 2y

Solving for y, we get

y = 2x/(x-2)

Now modify this a little:

y = 2x/(x-2)
= (2x-4+4)/(x-2)
= [2(x-2)+4]/(x-2)
= 2 + 4/(x-2)

We know that we are looking for integer values of y; hence the 4/(x-2) term must take an integer value. Well, what values of x could possibly make this term an integer? Remember that x also must be an integer, hence x-2 is also an integer. So, 4 divided by some integer gives us another integer. The only positive integral factors of 4 are 1, 2 and 4; this means that x-2 must equal one of these 3 values. Therefore x must equal 3, 4 or 6, and these values for x correspond to the 3 rectangles given as examples in the problem. Hence indeed there are no other solutions to find.

It's always nice to have different ways of looking at the same problem. I hope this method is clear to the reader and that some of you will find it interesting.

Thanks for the puzzle Gamer, it was fun!

-John

Posted by John Reid on 2003-09-12 18:01:11