Evaluate sqrt of (2003)(2005)(2007)(2009)+16 by using algebraic method.

Answer 1

Let a = 2006. Then you have
(a-3)(a-1)(a+1)(a+3) = (a-1)(a+1)*(a-3)(a+3) = (a^2-1)(a^2-9)
Let b = a^2+4 = 2006^2+4. Now you have
= (b-5)(b+5) = b^2-25

Hmmm. Stuck here so far.
Anyone see a way out?

Answer 2

Oops. My bad.
Let b = a^2-5 = 2006^2-5. Now you have
= (b+4)(b-4) = b^2-16

So,
(b^2-16 + 16 = b^2 = (2006^2-5)^2

So,
√((2003)(2005)(2007)(2009)+16) = 2006^2-5 = 4024031

Answer 3

heroic. I was doing similar scratching for about half an hour :)