Algebra (Olympiad)hardFree

Sophie Germain Identity Evaluation | IOQM

JEE Maths question with a full step-by-step solution.

Question
Evaluate
(20144+4×2013420132+40272)−(20124+4×2013420132+40252).\left(\frac{2014^4 + 4 \times 2013^4}{2013^2 + 4027^2}\right) - \left(\frac{2012^4 + 4 \times 2013^4}{2013^2 + 4025^2}\right).
Solution
Answer: 0
Step 1: Sophie Germain identity. For all real a,ba, b,
a4+4b4=[(a−b)2+b2][(a+b)2+b2].a^4 + 4b^4 = \big[(a-b)^2 + b^2\big]\big[(a+b)^2 + b^2\big].
 Step 2: First fraction. Take a=2014a = 2014, b=2013b = 2013. Then a−b=1a - b = 1 and a+b=4027a + b = 4027, so
20144+4⋅20134=[12+20132][40272+20132].2014^4 + 4 \cdot 2013^4 = \big[1^2 + 2013^2\big]\big[4027^2 + 2013^2\big].
 Dividing by the denominator 20132+402722013^2 + 4027^2,
20144+4⋅2013420132+40272=1+20132.\frac{2014^4 + 4 \cdot 2013^4}{2013^2 + 4027^2} = 1 + 2013^2.
 Step 3: Second fraction. Take a=2012a = 2012, b=2013b = 2013. Then a−b=−1a - b = -1 and a+b=4025a + b = 4025, so
20124+4⋅20134=[(−1)2+20132][40252+20132]=[1+20132][40252+20132].2012^4 + 4 \cdot 2013^4 = \big[(-1)^2 + 2013^2\big]\big[4025^2 + 2013^2\big] = \big[1 + 2013^2\big]\big[4025^2 + 2013^2\big].
 Dividing by the denominator 20132+402522013^2 + 4025^2,
20124+4⋅2013420132+40252=1+20132.\frac{2012^4 + 4 \cdot 2013^4}{2013^2 + 4025^2} = 1 + 2013^2.
 Step 4: Subtract.
(1+20132)−(1+20132)=0.\big(1 + 2013^2\big) - \big(1 + 2013^2\big) = 0.
 Answer: 00.
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