Posted by Ben on .
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

math 
tchrwill,
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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:
1xy  2x  2y = 0 or
2x = 2y/(y2)
3Clearly, both x and 2y/(y2) must also be integers.
4Set x = k from which y then becomes 2k/(k2).
5Assume 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.