Given the square of (n-2008) + the square of (2007-n) = 2, find the value of (2007-n) (n-2008)
Let a = n-2008, B = 2007-n
Then a square + b square = 2
(a + b) square = a square + b square + 2 * a * b = 1. (2)
Bring (1) into (2)
Then a * b = - 1 / 2
So (2007-n) * (n-2008) = - 1 / 2
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