This has been posted many times before, for example here.

thank you, I used the search function for possible reposts with the word <<czech>> but not result came through

Very nice Functional Equation problem Let P(a,b) be the insertion of x=a and y=b into the functional equation in the problem P(x , (x^2 - f(x))/2 ) will get a result as follow : f( (x^2+ f(x))/2 ) = f( (x^2+f(x))/2) + 4f(x).(x^2-f(x))/2 0 = 2 f(x) (x^2- f(x)) from there we will get f(x) = 0 or f(x) = x^2 as the solution It's easy to check that both of them satisfy the functional equation given in the problem So, all functions f:R->R that satisfy the condition are f(x) = 0 and f(x)= x^2