Find three squares such that each minus the product of the three gives a square.
Classical Rules: Let a "square" be any number that is the square of a rational number.
After making a program and entering a limit, the first result was the following set:
(1/9), (441/625), (1296/1369) are the first three squares that work.
The product is (571536/7700625) = (63504/855625).
(1/9) - (63504/855625) = (284089/7700625) = (533/2775)^2
(441/625) - (63504/855625) = (21609/34225) = (147/185)^2
(1296/1369) - (63504/855625) = (746496/855625) = (864/925)^2
So, since each of the differences is a square, these values will work.
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Posted by Justin
on 2006-05-27 23:30:23 |