Generalize This BMO Question

Algebra Level 3

201 4 4 + 4 201 3 4 201 3 2 + 402 7 2 201 2 4 + 4 201 3 4 201 3 2 + 402 5 2 \frac{2014^4+4\cdot2013^4}{2013^2+4027^2}-\frac{2012^4+4\cdot2013^4}{2013^2+4025^2}

Evaluate the above expression

( BMO 2014/1 )


The answer is 0.

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3 solutions

敬全 钟
Aug 11, 2014

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 ) a^4+4b^4=(a^2+2ab+2b^2)(a^2-2ab+b^2) , and we are going to use it to tackle this problem.

So, we can rewrite the numerator of the fraction 201 4 4 + 4 ( 201 3 4 ) 201 3 2 + 402 7 2 \frac{2014^4+4(2013^4)}{2013^2+4027^2} into 201 4 4 + 4 ( 201 3 4 ) = ( 201 4 2 + 2 ( 2014 ) ( 2013 ) + 2 ( 201 3 2 ) ) ( 201 4 2 2 ( 2014 ) ( 2013 ) + 2 ( 201 3 2 ) ) 2014^4+4(2013^4)=(2014^2+2(2014)(2013)+2(2013^2))(2014^2-2(2014)(2013)+2(2013^2)) [ ( 2014 + 2013 ) 2 + 201 3 2 ] [ ( 2014 2013 ) 2 + 201 3 2 ] \implies [(2014+2013)^2+2013^2][(2014-2013)^2+2013^2] ( 402 7 2 + 201 3 2 ) ( 1 2 + 201 3 2 ) \implies (4027^2+2013^2)(1^2+2013^2) Great! Now we can simplify the fraction into 201 4 4 + 4 ( 201 3 4 ) 201 3 2 + 402 7 2 = ( 402 7 2 + 201 3 2 ) ( 1 2 + 201 3 2 ) 201 3 2 + 402 7 2 = 1 + 201 3 2 \frac{2014^4+4(2013^4)}{2013^2+4027^2}=\frac{(4027^2+2013^2)(1^2+2013^2)}{2013^2+4027^2}=1+2013^2 By using the same technique for the other fraction, we have 201 2 4 + 4 ( 201 3 4 ) 201 3 2 + 402 5 2 = ( 402 5 2 + 201 3 2 ) ( ( 1 ) 2 + 201 3 2 ) 201 3 2 + 402 5 2 = 1 + 201 3 2 \frac{2012^4+4(2013^4)}{2013^2+4025^2}=\frac{(4025^2+2013^2)((-1)^2+2013^2)}{2013^2+4025^2}=1+2013^2 Therefore, our final answer is 1 + 201 3 2 ( 1 + 201 3 2 ) = 0 1+2013^2-(1+2013^2)=\boxed0 .

same way! @Cody Johnson Thanks for sharing this nice problem!

Kartik Sharma - 6 years, 10 months ago

@敬全 钟 I think there should be an extra "2" before the "b^2" in the second line of your solution :) Good working!

Jessica Wang - 5 years, 7 months ago
Cody Johnson
Aug 28, 2014

After solving using Sophie-Germain, I found another interesting solution:

If f ( x ) = ( x + 1 ) 4 + 4 x 4 x 2 + ( 2 x + 1 ) 2 ( x 1 ) 4 + 4 x 4 x 2 + ( 2 x 1 ) 2 f(x)=\frac{(x+1)^4+4x^4}{x^2+(2x+1)^2}-\frac{(x-1)^4+4x^4}{x^2+(2x-1)^2} , then we can compute f ( x ) = 0 f'(x)=0 . Therefore, f f is constant so f ( 2013 ) = f ( 0 ) = 1 1 = 0 f(2013)=f(0)=1-1=0 .

Naheem Ahmed
Aug 21, 2014

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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