Y by
f:R->R
f(xf(y)+y)=yf(x)+f(f(y)) and there are finite number of t such that f(t)= - t
f(xf(y)+y)=yf(x)+f(f(y)) and there are finite number of t such that f(t)= - t
This post has been edited 1 time. Last edited by jjkim0336, Apr 28, 2025, 11:29 PM
Reason: title issue
Reason: title issue