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 such that
for all real
denote the assertion
Therefore, is onto i.e. covers all values of .
Now,let , and define a function as .
Therefore, our original assertion for changes into the equivalent assertion for , as being
As is onto, therefore is also onto. Also, . And then finally,
As is onto, therefore, covers all values of , and therefore, also covers all values of ,
Thus, for every .
And therefore, for every and this is indeed a solution.