Let $f:\mathbb{R}\to\mathbb{R}$ satisfy $f\big(x-f(y)\big)=f\big(f(y)\big)+x\cdot f(y)+f(x)-1$ for all $x,y\in\mathbb{R}$. 
Let $f:\mathbb{R}\rightarrow \mathbb{R}$ be such that 
  $$f\big(x-f(y)\big)=f\big(f(y)\big)+x\cdot f(y)+f(x)-1$$
  for all $x,y \in\mathbb{R}$.  Then, what is the value of $|f(16)|-25$?

what i try
put $f(y)=z.$ then $f(x-z)=f(f(x))+xz+f(x)-1$
put $f(x)=t.$ then $f(x-z)=f(t)+zx+t-1$
How do i solve it Help me please 
 A: Let $c:=f(0)$.  Plugging in $f(y)$ for $x$ in the original functional equation yields
$$f\big(f(y)\big)=\frac{1+c}{2}-\frac{\big(f(y)\big)^2}{2}\tag{*}$$
for all real numbers $y$.    Then (*) implies that $c\neq 0$.
Taking $y:=0$ in the original functional equation, we obtain
$$f(x-c)-f(x)=f(c)-1+cx$$
for every real number $x$.  Using $c\neq 0$, this proves that, for each real number $t$, there exist real numbers $p_t$ and $q_t$ such that
$$t=f(p_t)-f(q_t)\,.$$
Replacing $x$ by $f(x)$ in the original functional equation, we get
$$f\big(f(x)-f(y)\big)=f\big(f(y)\big)+f(x)\,f(y)+f\big(f(x)\big)-1$$
for all real numbers $x$ and $y$.  By (*), the previous equation becomes
$$f\big(f(x)-f(y)\big)=c-\frac{\big(f(x)-f(y)\big)^2}{2}$$
for all $x,y\in\Bbb R$.  Replacing $x$ and $y$ in the equation above by $p_t$ and $q_t$, respectively, we have $$f(t)=c-\frac{t^2}{2}$$
for all $t\in \Bbb R$.  Comparing the previous equation with (*), we deduce that $c=1$.  That is, $$f(t)=1-\frac{t^2}{2}$$ for every real number $t$, and it can be easily seen that this function is indeed a solution to the given functional equation.
