Turkey TST 2014

[Nirjhor]

Find all functions $f:\mathbb R\mapsto \mathbb R$ satisfying \[f\left(f(y)+x^2+1\right)+2x=f(x+1)^2+y\] for all $(x,y)\in\mathbb R^2$.

[Phlembac Adib Hasan]

Suppose $P(x,y)$ denotes the original equation and $f(1)=c$.
$P(0,y)\Longrightarrow f(f(y)+1)=c^2+y\quad (1)$
$\Longrightarrow f$ is bijective.
Also $ c^2+f(y)+1=f(f(f(y)+1)+1)=f(c^2+y+1)$
Hence $P(c,y)\Longrightarrow f(f(y))=\text { constant }+y\quad (2)$
From (1) & (2), we see $f(f(y)+1)-f(f(y))=d$ for some constant $d$. And since $f$ is surjective, we further derive $f(x+1)-f(x)=d\forall x\in \mathbb R$.

Comparing $P(x-1,y)$ and $P(-x+1,y)$, we get \[4(x-1)=f(x)^2-f(-x+2)^2=f(x)^2-f(-x)^2-4f(-x)d-4d^2\quad (3)\]Substituting $x$ by $-x$ we get \[4(-x-1)=f(-x)^2-f(x)^2-4f(x)d-4d^2\quad (4)\]Adding (3) & (4) will give $f(x)+f(-x)=e$ for some constant $e$. But substituting $f(-x)$ by $e-f(x)$ in (3) reveals that $f$ is a linear function. So $f(x)=ax+b$. Now checking in the main equation shows that $a=1,b=0$. In other words, $f(x)=x\forall x$

[Fm Jakaria]

Main equation is denoted $(M)$. $f(1) = k$.
$(M)x = 0 \rightarrow f(f(y)+1) = k^2+y…(ii)$. So f is bijective.
$(ii) y-k^2 \rightarrow f(f(y-k^2)+1) = y….(ii-)$
$(M)(x,y)$ vs. $(-x,y) \rightarrow f(x+1)^2 – f(1-x)^2 = 4x ….(i)$
$(ii-) y = k^2+1 \rightarrow f(k+1) = k^2+1….(vii)$
$(ii)y = k+1 \rightarrow f(k^2+2) = k^2+k+1….(ix)$
$(i) x = k^2+1 \rightarrow f(-k^2)^2 \rightarrow (ix) \rightarrow (k^2+k+1)^2 –4(k^2+1)
…(x)$
Choose $x_0 : f(x_0) = -x^2.$
$(M) y = x_0 \rightarrow k+2x – f(x+1)^2 = x_0 \rightarrow (f) \rightarrow
f(k+2x – f(x+1)^2) = -x^2 ….(1)$
$(ii_) y = -x^2 \rightarrow f(f(-(x^2+k^2))+1) = -x^2 \rightarrow (1) \rightarrow
f(k+2x – f(x+1)^2)$
$\Rightarrow f(-(x^2+k^2)) = k+2x -1 –f(x+1)^2……(2)$
$(2)x=0 \rightarrow f(-k^2) = k-k^2-1…(iv)$
$(iv),(x) \rightarrow (k-k^2-1)^2 = (k^2+k+1)^ 2-4(k^2+1) \rightarrow k = 1$, as $k^2+1>0$.
$(ii_) \rightarrow f(f(y-1)+1) = y….(ii_)$
$(1) \rightarrow f(2x+1-f(x+1)^2) = -x^2….(1)$
$(1) x = 0 \rightarrow f(0) = 0$
$(2) \rightarrow f(-(x^2+1)) = 2x – f(x+1)^2…(2)$
Also $(M) y = 0 \rightarrow 2x-f(x+1)^2 = -f(x^2+1)$. In particular,
f(-x) = -f(x) for all $x \geqslant 1$ and hence also for all $x \leqslant -1$.
Now choose $x_1 \leqslant -2$ or $\geqslant 0.$ Using the last fact, and that f is injective;
$(M) (x_1,y) $ vs $ (-x_1-2. -4x_1+y-4)
\rightarrow f(y)+x_1^2+1 = f(-4x_1+y-4)+(x_1+2)^2+1$, for all $y$.
Set $y = 4x_1+4 \rightarrow f(4(x_1+1)) = 4(x_1+1)$.
So for all $x.\leqslant -4$ and $x \geqslant 4, f(x) = x$.
If $f(y-1) = y-1, (ii_)\rightarrow f(y) = y$. So analyzing the formula $f(x)=x$; it’s truth extends in this way:
$[-5,-4] \rightarrow [-4,-3] \rightarrow [-3, -2] \rightarrow ……….. \rightarrow [3,4]$.
So $f(x) = x$; $\forall x \in \mathbb{R}$.