KamalDoni wrote: Do there exist? functions $f \colon \mathbb{R} \to \mathbb{R}$ such that a) $f$ is a surjective function; b) $f(f(x))=(x-1)f(x)+2$ for all $x$ real. Assume that there exists such function. It follows from (b) that $f\big( f(1)\big)=2.$ Since $f$ is surjective, there exists $a$ such that $f(a)=0.$ From this, we get \[f(0)=f\big(f(a)\big)=(a-1)\cdot f(a)+2=2\] and \[f(2)=f\big(f(0)\big)=(0-1)\cdot f(0)+2=0.\] Since $f\big(f(1)\big)=2$ and $f(2)=0,$ we have \[0=f(2)=f\left(f\big(f(1)\big)\right) =\big(f(1)-1\big)\cdot f\big(f(1)\big)+2=2\big(f(1)-1\big)+2.\] Therefore, we have $f(1)=0.$ Also, we see that there exists $b$ such that $f(b)=1,$ so \[0=f(1)=f\big(f(b)\big)=(b-1)\cdot f(b)+2=b-1+2.\] From this, we obtain $b=-1$ and $f({-1})=1.$ Finally, we see that there exists $c$ such that $f(c)=-1.$ Therefore, \[1=f({-1})=f\big(f(c)\big)=(c-1)\cdot f(c)+2=3-c,\] so $c=2$ and $f(2)=-1.$ But we already have that $f(2)=0,$ a contradiction. So, there is no function satisfying the given conditions.

..................

What if we remove the condition a) of surjectivity? The trivial solution $f(0)=2$ and $f(x)=0$ for $x\neq 0$ exists. Are there other? Can one characterize all of them?

I think this problem proposed by IGOR VORONOVICH

Only because his propension for functional equations is well-known, or do you have any insider info?

yes-yes! But, Igor Voronovich in the Olympiad.

a little bit similar to the equation: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2558779&sid=363b0935d5d9e8b40d8a697d290e3905#p2558779

@mavropnevma Another function that was mentioned in the official solution is $f(x)=\frac{x-2}{x-1},x \ne 1$ and $f(1)=0$. There were no mentions of the general form.

amatysten wrote: @mavropnevma Another function that was mentioned in the official solution is $f(x)=\frac{x-2}{x-1},x \ne 1$ and $f(1)=0$. There were no mentions of the general form. Yes, I've seen that, since I was able to eventually get the official solutions.

Is this correct?

Wait... So there is actually a solution! However, the proposed function has some problems with the second condition, I think. I believe that the function might be a careless trap for many students. Is my solution correct? Thanks! Nice problem, though

There is no trap with the second condition. No solution exists. We were discussing what happens if we remove the first condition, when at least two examples were given above, but no-one exhibited their general form.

Wait, but the function $f(x)=\frac{x-2}{x-1}$ works for the first condition. There is just a "trap" when we combine BOTH conditions together.

No, the function $f(x)=\frac{x-2}{x-1}$ for $x\neq 1$ and $f(1)=0$ does not work for the first condition (it works for the second condition). For $y=1$ there is no $x$ such that $f(x)=y$.

oh yes... never noticed that. but why was that function mentioned in the official solution?

Because, as I already said it three times, both the official solution and my own were interested in what happened if we gave up on the first condition (surjectivity). It has nothing to do with the problem as asked, but it is an interesting thing to ask oneself. Maybe this can put an end to this discussion.

Oh, sorry. I did not notice from the thread that the official solution was actually interested in finding the functions when we disregarded the first condition. OOPS

Is this correct? Suppose $f(x)=f(y)\neq 0$ for some $x,y$, then $f(f(x))-(x-1)f(x)=f(f(y))-(y-1)f(y) \Rightarrow (x-1)f(x)=(y-1)f(y) \Rightarrow x=y$. Take $x=1$, then $ff(1)=2$, and surjectivity implies there exist $a$ such that $f(a)=0$, then $f(0)=f(f(a))=(a-1)f(a)+2=2$, then $f(f(1))=f(0)\neq 0\Rightarrow f(1)=0$. Take $b$ such that $f(b)=1$, then $0=f(1)=f(f(b))=(b-1)f(b)+2=b+1 \Rightarrow b=-1$, then $f(-1)=1$. Take $c$ such that $f(c)=-1$, then $f(-1)=f(f(c))=(c-1)f(c)+2=3-c\Rightarrow c=2$, so $f(2)=-1$. Finally, since $f(0)=2$, we have $f(2)=f(f(0))=(0-1)f(0)+2=0 \Rightarrow 0=f(2)=-1$, contradiction. No function exist.

@Garfield I believe your proof of bijectivity fails when $s=0$. My solution: We will first prove a Lame Lemma: $\textbf{Lame Lemma}$: For any two reals $x_1,x_2$, $f(x_1)=f(x_2)\neq 0\implies x_1=x_2$. (Note that this means $f$ is "almost" injective except when it's zero, making it "lamely" injective; hence the name of the result.) $\textbf{Proof}$: Say $f(x_1)=f(x_2)=y\ne 0$, then $f(f(x_1))=f(f(x_2))\implies (x_1-1)y+2=(x_2-1)y+2\implies x_1=x_2$, as claimed. $\square$ Now it's showtime. As always, let $P(x)$ stand for "$P$lug in $x$ in the given equation". Let $s_0\in \mathbb{R}$ be such that $f(s_0)=0$ (it exists by (i)). Then $P(s_0)\implies f(0)=2$, and $P(0)\implies f(f(0))=-f(0)+2\implies f(2)=0$. Now $P(1)\implies f(f(1))=2=f(0)$, so by Lame injectivity, $f(1)=0$. Now suppose $s_1$ is such that $f(s_1)=1$ (which exists again by (i)), and note that $$P(s_1)\implies 0=f(1)=f(f(s_1))=(s_1-1)f(s_1)+2=(s_1-1)+2=s_1+1.$$So $s_1=-1$. Thus $f(-1)=1$. Finally take $s_{-1}$ so that $f(s_{-1})=-1$; I won't bother repeating why this exists. $P(s_{-1})$ readily gives $1=(s_{-1}-1)(-1)+2\implies s_{-1}=2\implies f(2)=-1$. But as we discovered before, $f(2)=0$ already, so $0=-1$, which is blatantly wrong. This contradiction proves that no such $f$ exists. $\blacksquare$

Garfield wrote:

Yes you have a right version, this function is almost bijection, but u have not noticed that $s$ could be 0, so your solution is false.

Wait a minute..... is this really from the Zhautykov?? . This is so very easy.

cool Pick $r$ such that $f(r)=0$. Take $x=r\implies f(0)=2$. Now take $x=0\implies f(2)=0$. Consider $s$ such that $f(s)=1$. Then $x=s\implies f(1)=s+1$. Now let $x=1$, to get $f(f(1))=2$. Thus, $$0=f(2)=f(f(f(1)))=(f(1)-1)f(f(1))+2=2s+2,$$which implies $f(-1)=1$. Finally consider $t$ such that $f(t)=-1$. Then, $$1=f(-1)=f(f(t))=(t-1)f(t)+2=3-t,$$which gives $f(2)=-1$, contradiction since we already showed $f(2)=0$. It is interesting to note that $$f(x)=\frac{x-2}{x-1}$$almost works; just that it misses the value of $1$ in its range.

Here's a cool one which makes the full use of surjection. Assume $\exists$ such a function. It is easy to note that $f(f(1))=2$. Let $P(x)$ denote assertion Now pick $j \in \mathbb{R}$ with $f(j)=1$. Thus $f(f(j))=f(1)=P(j)=j+1$, thus, $f(j+1)=2$ implying $f(2)=P(j+1)=2j+2=2f(1)$. Now pick $k \in \mathbb R$ with $f(k)=0$, and $P(k) \implies f(0)=2$. Thus, $f(f(0))=0=f(2)=2f(1)$ implying $f(1)=f(2)=0$, contradiction! Oops its same as @above

f(2)=0 and f(2)=-1 Contradiction Very easy problem for IZHO p2

We hav e $f: surjective $ (1) if we show $f(a)=f(b)$ and $a$ is not equal to $b$ this function is not exist $x=1$ $\implies$ $f(f(1))=2$ $x=f(1)$ $\implies$ $f(2)=2f(1)$ (2) from (1) $f(a)=0$ and $x=a$ $\implies$ $f(0)=2$ $x=0$ $implies$ $f(2)=0$ (3) $\implies$ from (2) and (3) $f(1)=f(2)=0$ contradiction this problem is not conform to P2 $O.Y.SH.$

First of All first condition implies that $f$ Surjective Let $P(x)$ be assertion of $ff(x)=(x-1)f(x)+2$ 1.$P(0)$ $ff(0)=-f(0)+2$ Assume that there is $q$ which $f(q)=0$ Take $q$ to $x$ $ff(q)=(q-1)f(q)+2$ $\implies$ $f(0)=2$ And it implies $f(2)=0$ 2.$P(1)$ $ff(1)=2$ Assume that there is $p$ such that $f(p)=2$ $P(p)$ $ff(p)=(p-1)f(p)+2$ $0=2p$ So $f(p)=2$ is unique and $p=0$ Then, $ff(1)=f(0)$ so $f(1)=0$ So $f(1)=f(2)=0$ Take $b$ such that $f(b)=1$ $P(b)$ $\implies$ $f(-1)=-1$ Take $a$ such that $f(a)=-1$ $P(a)$ $\implies$ $f(2)=-1$ But it is contradicition so we are done. Very intresting+easy question .

Let $P(x)$ denote the assertion. Let $k$ be a real such that $f(k)=0$ $P(k): f(0)=(k-1)f(k)+2=2 \implies f(0)=2$ $f$ is injective at all points $\neq 0$. If $f(a)=f(b)$, $P(a)$ vs $P(b)$, $(a-1)f(a)=(b-1)f(a) \implies a=b$. $P(1): f(f(1))=2$, so $f(1)=0$. Also $P(0): f(2)=-2+2=0$. Now since $f$ is bijective at all points $\neq 0$, $f^{-1}(x)$ is defined for all nonzero $x$. $P(f^{-1}(1)): f(1)=(f^{-1}(1)-1)+2$. So $f^{-1}(1)=-1$. Now $P(f^{-1}(-1)): f(-1)=-(f^{-1}(-1)-1)+2 \implies 1=-f^{-1}(-1)+3 \implies f^{-1}(-1)=2$. So $f(2)=-1$, but $f(2)=0$ so contradiction and there is no such function.