An algebra problem by I Gede Arya Raditya Parameswara

Algebra Level 4

Let f f be a functional equation that satisfy f ( − x ) − x f ( 1 x ) = − 2 x f(-x) - x f \left( \frac1x\right) = -\frac2x for x ≠ 0 x\ne0 .

If f ( 2017 ) = m n f(2017) = \frac mn , where m m and n n are coprime positive integers, find m + n m+n .


The answer is 8205740931.

I Gede Arya Raditya Parameswara by I Gede Arya Raditya Para… , 17

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1 solution

Tom Engelsman
Mar 4, 2017

Let us substitute − 1 x -\frac{1}{x} into the original functional equation to obtain the set:

f ( − x ) − x f ( 1 x ) = − 2 x ; f(-x) - xf(\frac{1}{x}) = -\frac{2}{x}; (i)

f ( 1 x ) + 1 x ⋅ f ( − x ) = 2 x . f(\frac{1}{x}) + \frac{1}{x} \cdot f(-x) = 2x. (ii)

Solving (ii) for f ( 1 x ) f(\frac{1}{x}) produces f ( 1 x ) = 2 x − 1 x ⋅ f ( − x ) f(\frac{1}{x}) = 2x - \frac{1}{x} \cdot f(-x) , and substituting this value into (i) yields:

f ( − x ) − x [ 2 x − 1 x ⋅ f ( − x ) ] = − 2 x ⇒ f ( − x ) = x 2 − 1 x ⇒ f ( x ) = x 2 + 1 x = x 3 + 1 x f(-x) - x[2x - \frac{1}{x} \cdot f(-x)] = -\frac{2}{x} \Rightarrow f(-x) = x^2 - \frac{1}{x} \Rightarrow f(x) = x^2 + \frac{1}{x} = \frac{x^3 + 1}{x} .

Hence, f ( 2017 ) = 201 7 3 + 1 2017 = 8 , 205 , 740 , 931 . f(2017) = \frac{2017^3 + 1}{2017} = \boxed{8,205,740,931}.

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