A nice Functional Equation

This post was originally posted by: Rijul Saini

I solved a cute functional equation today, and was also wondering what I was doing being a contributor here, so that resulted in this.
IMO Short List 2002 A01
Find all functions f: \mathbb{R} \rightarrow \mathbb{R} such that


for all real x,y


Let P(x,y) denote the assertion f(f(x)+y) = 2x + f(f(y)-x).
P(0,x) \implies f(x+f(0)) = f(f(x))
P(f(y),y) \implies f(f(f(y)) + y ) = 2f(y)+f(0) \implies f(f(y+f(0))+y) = 2f(y) + f(0)Taking y=-f(0) here, implies f(0) = 2f(-f(0)) +f(0) \implies f(-f(0)) = 0.

P(-\frac x2, -f(-\frac x2) - f(0)) \implies x = f(f(-f(-\frac x2) - f(0)) + \frac x2)

Therefore, f is onto i.e. covers all values of x.

Now,let f(0) = c, and define a function g : \mathbb{R} \rightarrow \mathbb{R} as g(x) = f(x - c).

Therefore, our original assertion P(x,y) for f changes into the equivalent assertion Q(x,y) for g, as being
g(g(x+c) + y + c) = 2x + g(g(y+c) - x +c)

As f is onto, therefore g is also onto. Also, g(0) = 0. And then finally,

Q(-c,x-c) \implies g(x) = -2c + g(g(x) + 2c) \implies g(g(x) + 2c) = g(x) + 2c

As g is onto, therefore, g(x) covers all values of x, and therefore, g(x) + 2c also covers all values of x,

Thus, g(x) = x for every x \in \mathhbb{R}.

And therefore, f(x) = x+f(0) for every x \in \mathbb{R} and this is indeed a solution.


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