China 2011 tst quiz-1 problem 1

[SANZEED]

Let $n\geq 2$ be a given integer. Find all functions $f:\mathbb{R}\rightarrow \mathbb{R}$ such that
$f(x-f(y))=f(x+y^n)+f(f(y)+y^n), \qquad \forall x,y \in \mathbb {R}$.

[Fm Jakaria]

The given equation is denoted $(M)$. Let $f(0) = c$.
$(M) f(y),y \rightarrow f(y^n + f(y)) = \frac {c}{2} ….(i)$
$(i), (M) \rightarrow f(x – f(y)) = f(x + y^n) + \frac{c}{2} ….(M_1)$
$(M_1) x + f(a), a(\forall a \in \mathbb{R})$
$\rightarrow f(x + (f(a) + a^n)) = f(x) - \frac{c}{2}…(1)$
$(M_1) x – b^n , b (\forall b \in \mathbb{R})$
$\rightarrow f(x – (b^n + f(b)) = f(x) + \frac{c}{2}….(2)$
$(2) x + (a^n + f(a)) , b$
$\rightarrow f(x + (a^n + f(a)) – (b^n + f(b))) \rightarrow (1) \rightarrow f(x) ….(3)$

We consider two cases.

Case-1:
Suppose $x^n + f(x)$ is not constant over $x \in \mathbb{R}$.
Choose $a_0, b_0 \in \mathbb{R} : a_0^n + f(a_0) \neq b_0^n + f(b_0)$.
Now set $k = | (a_0 ^n + f(a_0)) – (b_0^n + f(b_0)) |$.
Then $(3) \rightarrow f(x + k) = f(x); \forall x \in \mathbb{R}$.
Now $(3) x, a+k, a (\forall (x,a) \in \mathbb{R}^2)$
$\rightarrow f (x + ((a+k)^n – a^n)) = f(x)$

Since $k > 0$, here the function $h(a) = (a+k)^n – a^n$ is a nonconstant polynomial of positive degree $n-1$ with positive leading coefficient; so it is continuous over
$a \in \mathbb{R}$, and
when $a \rightarrow \infty, h(a) \rightarrow \infty$.
So $h$ takes all sufficiently large value of $R$.
Let $S = range(h)$.
So for $\forall y \in \mathbb{S}$ and $\forall x \in \mathbb{R}$;
$f(x + y) = f(x)$.
Then $f$ is constant. This follows for $\forall (p,q) \in \mathbb{R}^2$ by taking
$s = (p + q + r)$, for $r \in \mathbb{R}$ so that both $(p+r)$ and $(q+r)$ are sufficiently large to stay in $S$. By the previous relation, $f(p) = f(q) = f(s)$.
$(M) \rightarrow c = 2c \rightarrow c = 0$. $f = 0$ works,

Case-2:
Suppose $f(x) = -x^n + d$ for some constant $d \in \mathbb{R}$.
$x = 0 \rightarrow d = c$. Now
$(1) \rightarrow (x+c)^n – x^n = \frac{c}{2}; \forall x \in \mathbb{R}$.
The LHS, as a polynomial function of $x$; is of positive degree iff $c \neq 0$. Because of RHS is constant; we must conclude that $c = 0$ and hence $f(x) = -x^n$ for all
real $x$.

The final solutions are
A) $f(x) = 0$; all real $x$.
B) $f(x) = -x^n$; all real $x$.