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September 22, 2014

September 22, 2014

Posted by **Ben** on Monday, September 24, 2007 at 8:13pm.

- math -
**tchrwill**, Tuesday, September 25, 2007 at 1:51pmSubjects

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Homework Help Forum: math

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Posted by Ben on Monday, September 24, 2007 at 8:13pm.

The number of feared by Pythagoreans, since it lies halfway between the only two integers that can be both the perimeter and the area of the same rectange

Therefore, we can have rectangles with both area and perimeter of 16 and 18.

The approach to the solution of a problem of this type, where integer answers are sought, is by treating it as a Linear Indeterminate Algebra problem or a Diophantine Problem as follows: Letting x and y be the integer sides of our rectangles, then the area = xy and the perimeter = 2(x+y). Given that xy = 2(x+y) we derive:

1--xy - 2x - 2y = 0 or

2--x = 2y/(y-2)

3--Clearly, both x and 2y/(y-2) must also be integers.

4--Set x = k from which y then becomes 2k/(k-2).

5--Assume values of k and compute x and y.

k...0...1...2...3...4...5....6....7....8....9....10.....50.....100......

x...-....1...2...3...4...5....6....7....8....9....10.....50....100......

y...-...inf..inf..6...4...3+..3....2+..2+..2+...2+...2+.....2+......

Thus 3 and 6, (or 6 and 3), and 4 and 4 are the only possible answers as y never reaches 2 as an integer.

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