2 0 1 3 2 + 4 0 2 7 2 2 0 1 4 4 + 4 ⋅ 2 0 1 3 4 − 2 0 1 3 2 + 4 0 2 5 2 2 0 1 2 4 + 4 ⋅ 2 0 1 3 4
Evaluate the above expression
( BMO 2014/1 )
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same way! @Cody Johnson Thanks for sharing this nice problem!
@敬全 钟 I think there should be an extra "2" before the "b^2" in the second line of your solution :) Good working!
After solving using Sophie-Germain, I found another interesting solution:
If f ( x ) = x 2 + ( 2 x + 1 ) 2 ( x + 1 ) 4 + 4 x 4 − x 2 + ( 2 x − 1 ) 2 ( x − 1 ) 4 + 4 x 4 , then we can compute f ′ ( x ) = 0 . Therefore, f is constant so f ( 2 0 1 3 ) = f ( 0 ) = 1 − 1 = 0 .
Let x=2013, Write out the equation in form of x, Use long division or equate coefficients to divide both sides, find that you get (x^2 +1 ) - (x^2+1), Move on with life
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Indeed, SG isn't necessary mean Singapore, but it is Sophie Germain's Identity which says a 4 + 4 b 4 = ( a 2 + 2 a b + 2 b 2 ) ( a 2 − 2 a b + b 2 ) , and we are going to use it to tackle this problem.
So, we can rewrite the numerator of the fraction 2 0 1 3 2 + 4 0 2 7 2 2 0 1 4 4 + 4 ( 2 0 1 3 4 ) into 2 0 1 4 4 + 4 ( 2 0 1 3 4 ) = ( 2 0 1 4 2 + 2 ( 2 0 1 4 ) ( 2 0 1 3 ) + 2 ( 2 0 1 3 2 ) ) ( 2 0 1 4 2 − 2 ( 2 0 1 4 ) ( 2 0 1 3 ) + 2 ( 2 0 1 3 2 ) ) ⟹ [ ( 2 0 1 4 + 2 0 1 3 ) 2 + 2 0 1 3 2 ] [ ( 2 0 1 4 − 2 0 1 3 ) 2 + 2 0 1 3 2 ] ⟹ ( 4 0 2 7 2 + 2 0 1 3 2 ) ( 1 2 + 2 0 1 3 2 ) Great! Now we can simplify the fraction into 2 0 1 3 2 + 4 0 2 7 2 2 0 1 4 4 + 4 ( 2 0 1 3 4 ) = 2 0 1 3 2 + 4 0 2 7 2 ( 4 0 2 7 2 + 2 0 1 3 2 ) ( 1 2 + 2 0 1 3 2 ) = 1 + 2 0 1 3 2 By using the same technique for the other fraction, we have 2 0 1 3 2 + 4 0 2 5 2 2 0 1 2 4 + 4 ( 2 0 1 3 4 ) = 2 0 1 3 2 + 4 0 2 5 2 ( 4 0 2 5 2 + 2 0 1 3 2 ) ( ( − 1 ) 2 + 2 0 1 3 2 ) = 1 + 2 0 1 3 2 Therefore, our final answer is 1 + 2 0 1 3 2 − ( 1 + 2 0 1 3 2 ) = 0 .