NumberTheoryBasic PolynomialAndEquation UK Intermediate
2013


Problem - 2852
Calculate the value of $$\dfrac{2014^4+4 \times 2013^4}{2013^2+4027^2}-\dfrac{2012^4+4 \times 2013^4}{2013^2+4025^2}$$

By Sophie Germain's Identity (see # 3863), we have \begin{align} &\dfrac{2014^4+4 \times 2013^4}{2013^2+4027^2}-\dfrac{2012^4+4 \times 2013^4}{2013^2+4025^2}\\ &=\dfrac{(2014^2 + 2\times 2013^2-2\times 2014\times 2013)(2014^2 + 2\times 2013^2+2\times 2014\times 2013)}{2013^2+(2013+2014)^2}\\ &-\dfrac{(2012^2 + 2\times 2013^2-2\times 2012\times 2013)(2012^2 + 2\times 2013^2+2\times 2012\times 2013)}{2013^2+(2013+2012)^2}\\ &=(2014^2 + 2\times 2013^2-2\times 2014\times 2013)-(2012^2 + 2\times 2013^2-2\times 2012\times 2013)\\ &=2014^2-2012^2 -2\times2013\times(2014-2012)\\ &=(2014+2012)\times(2014-2012)-2\times 2013\times(2014-2012)\\ &=(2014-2012)\times(2014+2012-2\times2013)\\ &=\boxed{0} \end{align}

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