by wu2481632, Jan 30, 2017, 3:59 AM 
such that 
for all 
(Romania 2009)
by
blue8931, Jan 30, 2017, 5:03 AM
is a solution of the functional equation, then 
for some constant 
must be a solution of the functional equation. Therefore there exists some function satisfying the given conditions such that 
, unless 
.
by
FlyingCucumber, Jan 30, 2017, 4:35 PM
by
FlyingCucumber, Jan 30, 2017, 4:37 PM
by
FlyingCucumber, Jan 30, 2017, 4:49 PM
by
aftermaths, Jan 30, 2017, 10:16 PM
by
shiningsunnyday, Jan 31, 2017, 4:52 AM
by
zephyrcrush78, Feb 1, 2017, 3:10 AM
To share with readers my favorite problem I came across today :) (Shout for contrib.)
May 2017
April 2017
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January 2017
December 2016
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,
;
so 
.
.
.
which reduces to
which is simply
After manipulations this reduces to
Now we put in 
into the above equation, giving us
The RHS reduces to
and the LHS reduces to
Equating these gives
Now we can put 
and 
together by subtracting them, implying 
.
for constant 
.
and the RHS reduces to 
.
Similarly to the above case, we can switch 
out for 
Putting the above two equations together gives 
,
and so if ![\[f((x+1)^3)=(x+1)f((x+1)^2)=(x+1)^2f(x+1) \Rightarrow f((x+1)^2)=(x+1)f(x+1) \Rightarrow f(x^2)=xf(x)\]](http://latex.artofproblemsolving.com/4/7/7/477180ddfd9f831b8a5e13c27efcca1840f285c4.png)
so by 
to
?
