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Source: India TST 2018 D1 P3

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#1 h Jul 18, 2018, 2:04 PM • 3 Y

Y by Amir Hossein, megarnie, Adventure10

Find all functions $f: \mathbb{R} \mapsto \mathbb{R}$ such that $f(x)f\left(yf(x)-1\right)=x^2f(y)-f(x),$ for all $x,y \in \mathbb{R}$ .

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#2 h Jul 18, 2018, 2:11 PM • 7 Y

Y by Amir Hossein, sa2001, AopsUser101, YC1math, ashrith9sagar_1, Commander_Anta78, Adventure10

The only functions which work are the identity and the zero-function. It is clear that both satisfy the equation. Now we show any valid function that is not zero is the identity. Suppose $f(x_0) \ne 0$ for some $x_0 \in \mathbb{R}$ .

If $f(0) \ne 0$ then $f(0)f(yf(0)-1)=-f(0)$ but $f \equiv -1$ is not a solution. So $f(0)=0$ .

Put $x=x_0, y=0$ so $f(-1)=-1$ . Put $y=x$ so $f(xf(x)-1)=x^2-1$ for all $x$ ; hence $f$ is surjective over $(-1, \infty)$ . Suppose $f(a)=0$ then $x=a, y=x_0$ implies $a=0$ . So $f$ is injective at $0$ .

Now $f(f(1)-1)=0$ so $f(1)=1$ . Put $x=1$ to get $f(y-1)=f(y)-1$ for all $y$ so $f(y-N)=f(y)-N$ for all integers $N \ge 0$ . Now $f(y)=f(y+N)-N$ so $f(y+N)=f(y)+N$ for all integers $N \ge 0$ . Consequently, $f$ is surjective over all of $\mathbb{R}$ .

Plug $f(yf(x)-1)=f(yf(x))-1$ to conclude $f(x)f(yf(x))=x^2f(y)$ . Put $y=1$ so $x^2=f(x)f(f(x))$ hence $f(yf(x))=f(y)f(f(x))$ for all $x \ne 0$ . For $x=0$ , last claim is obvious. Now surjectivity shows $f(ty)=f(t)f(y)$ for all $t,y \in \mathbb{R}$ so $f$ is multiplicative.

Thus, $f(z^2)=f(z)^2 \ge 0$ and $f(-z^2)=f(-1)f(z)^2=-f(z)^2 \le 0$ hence $f$ preserves the sign of the argument. Now pick $x>1$ and so $1 \ge x-\lfloor x \rfloor \ge 0$ hence $f(x-\lfloor x \rfloor) \ge 0$ and $f(x-\lfloor x \rfloor-1) \le 0$ hence $\lfloor x \rfloor \le f(x) \le \lfloor x \rfloor+1$ so $x-1 \le f(x) \le x+1$ for all $x>1$ . Now suppose $f(t_0) \ne t_0$ for some $t_0>1$ .

1. If $f(t_0)>t_0$ .

Now for $y>1$ sufficiently large, $yf(t_0)^2-2f(t_0) \le f(t_0)f(yf(t_0)-1)=t_0^2f(y)-f(t_0) \le yt_0^2+t_0^2-f(t_0)$ hence $y \le \frac{t_0^2+f(t_0)}{f(t_0)^2-t_0^2}$ which fails as $y \rightarrow \infty$ .

2. If $f(t_0)<t_0$ .

Again for $y>1$ sufficiently large, $yf(t_0)^2 \ge f(t_0)f(yf(t_0)-1)=t_0^2f(y)-f(t_0) \ge yt_0^2-t_0^2-f(t_0)$ hence $y \le \frac{t_0^2+f(t_0)}{-f(t_0)^2+t_0^2}$ which fails as $y \rightarrow \infty$ .

Finally, shifting down by large integers $N$ , we obtain $f(x)=x$ for all $x \in \mathbb{R}$ as desired. $\blacksquare$

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#3 h Jul 18, 2018, 9:01 PM • 3 Y

Y by Amir Hossein, ImbecileMathImbaTation, Adventure10

Easy for P3 (if problems order in difficult)and simple solution.
Let $P(x,y)$ be the assertion of $f(x)f(yf(x)-1)=x^2f(y)-f(x)$
$P(0,0)\to f(0)\cdot f(-1)=-f(0),$ then $1) f(0)=0,$ or $f(-1)=-1.$
$1)$ $P(x,0)\to f(x)\cdot f(-1)=-f(x).$
$1.1)$ For all $x,$ $f(x)\equiv 0,$ Indeed this solution work.
$1.2)$ $\exists a\in\mathbb{R} ,$ such that $f(a)\not= 0,$ them from $P(a,0)\to f(-1)=-1.$
$2)$ $P(-1,0)\to f(0)=0.$
Then from both condition we can get $f(0)=0,f(-1)=-1.$
Lemma: $f(a)\equiv 0 \iff a\equiv 0.$
Proof: we know $f(0)=0,$ let show $f(a)=0\to a=0.$
$P(a,-1)\to -a^2=0\to a=0.$ As desired.
From $P(1,1)\to f(1)\cdot f(f(1)-1)=0.$
$2.1)$ $f(1)=0.$
From $P(1,x)\to$ for all $x,$ $f(x)=0,$ but $f(-1)=-1.$ contradiction.
$2.2)$ $f(1)\not= 0\to f(f(1)-1)=0.$ From lemma we get $f(1)=1.$
Then $P(1,x+1)\to f(x+1)=f(x)+1,$ (or $f(x-1)=f(x)-1.$ )
Then our equation equivalent to $Q(x,y):f(x)\cdot f(yf(x))=x^2f(y).$
From $Q(x,1)\to f(x)\cdot f(f(x))=x^2.$
Also from $Q(x+1,1)\to (f(x)+1)\cdot f(f(f(x))+1)=f(x)\cdot f(f(x))+f(x)+f(f(x))+1=x^2+2x+1,$ then we get $f(f(x))=2x-f(x).$
From $f(x)\cdot f(f(x))=2x,$ use $f(f(x))=2x-f(x),$ we get $(x-f(x))^2=0.$
Then $f(x)\equiv x,$ for all $x.$ Indeed this solution work.

This post has been edited 2 times. Last edited by falantrng, Dec 28, 2018, 8:50 AM
Reason: Typo

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#4 h Dec 20, 2018, 11:08 AM • 4 Y

Y by rocketscience, mijail, Adventure10, Mango247

Nice problem. Here's my solution: Let $P(x,y)$ denote the given assertion. Note that the only constant function which satisfies the given equation is the zero function. So from now on we assume that $f$ is non-constant.

CLAIM For $c \in \mathbb{R}$ , we have $f(c)=0 \Leftrightarrow c=0$ .

Proof: Note that we have $P(-1,0) \Rightarrow f(-1)^2=f(0)-f(-1) \Rightarrow f(-1)(f(-1)+1)=f(0)$ Also, $P(0,0) \Rightarrow f(0)f(-1)=-f(0) \Rightarrow f(0)=0 \text{ OR } f(-1)=-1$ If $f(0)=0$ , then using the first relation, we get that either $f(-1)=0$ or $f(-1)=-1$ . But, when $f(-1)=0$ , we have $P(x,0) \Rightarrow f(x)=0$ , which contradicts the fact that $f$ is non-constant. That means $f(0)=0$ implies $f(-1)=-1$ . And, when $f(-1)=-1$ , then (again from the first relation), we get $f(0)=0$ . Summarizing the above, we can say that $f(0)=0$ and $f(-1)=-1$ are both true simultaneously.

Now, suppose that $f(c)=0$ . Then $P(c,-1) \Rightarrow c^2f(-1)=0 \Rightarrow c=0$ . $\Box$

Return to the problem at hand. Then we get $P(x,-1) \Rightarrow f(x) \cdot f(-f(x)-1)=-x^2-f(x)$

$P(-1,f(x)) \Rightarrow -f(-f(x)-1)=f(f(x))+1 \Rightarrow f(x) \cdot f(-f(x)-1)=-f(x) \cdot (f(f(x))+1)$ $\Rightarrow x^2+f(x)=f(x) \cdot (f(f(x))+1) \Rightarrow f(f(x)) \cdot f(x)=x^2 \text{ } (*)$
Now, $P(1,1) \Rightarrow f(1) \cdot f(f(1)-1)=0$ . Using our Claim, we get that $f(1)-1=0 \Rightarrow f(1)=1$ $($ as $f(1) \neq 0)$ . Then $P(1,y) \Rightarrow f(y-1)=f(y)-1 \Rightarrow \text{ By an easy induction, }f(y+n)=f(y)+n \text{ } \forall n \in \mathbb{N}$
Finally, Putting $x \Longrightarrow x+n$ in $(*)$ , and using the above relation, we get that $(f(f(x))+n)(f(x)+n)=x^2+2nx+n^2=f(f(x)) \cdot f(x)+2nx+n^2$ $\Rightarrow n(f(f(x))+f(x))=2nx \Rightarrow f(f(x))+f(x)=2x \text{ } (**)$
Using $(*)$ and $(**)$ , we can easily find that $f(f(x))=f(x)=x$ . Thus, our final answers are $\boxed{f \equiv 0 \text{ AND } f(x)=x \text{ } \forall x \in \mathbb{R}}$ Now one can easily verify that these solutions actually work. Hence, done. $\blacksquare$

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#5 h Dec 20, 2018, 12:44 PM • 2 Y

Y by Adventure10, Mango247

I think this solution is more elementary than the others, although many parts are similar to the given above.
Solution

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#6 h Apr 4, 2019, 2:21 PM • 2 Y

Y by Adventure10, Mango247

Sorry for double posting, but I have a different ending after proving the claim in post #4. Here's my solution: As proved in post #4, we have $f(0)=0,f(-1)=-1$ and that $f$ is non-constant. Now, $P(-1,y) \Rightarrow -f(-y-1)=f(y)+1 \overset{y \rightarrow -y}{\Longrightarrow} f(y-1)=-f(-y)-1$ Again, as shown in post #4, one easily gets that $f(y-1)=f(y)-1$ . Thus, we have $f(-y)=-f(y)$ , i.e. $f$ is odd. We will show that $f$ is injective also. Suppose we have $f(a)=f(b)$ for some $a,b \in \mathbb{R}$ . Then $P(a,-1)-P(b,-1) \Rightarrow a^2=b^2 \Rightarrow b=a \text{ OR } b=-a$ However, we cannot have $f(-a)=f(a)$ , as $f$ is odd. Hence, $f$ must be injective.

Now, as $f(y-1)=f(y)-1$ , so we can rewrite the problem condition as $P(x,y):= f(x)(f(yf(x))-1)=x^2f(y)-f(x) \Rightarrow f(x)f(yf(x))=x^2f(y)$ Then, $P(x,x)$ gives that $f(xf(x))=x^2$ . This means that $f$ is surjective for $x>0$ . But, as $f(-x)=-f(x)$ , so we get that $f$ is in fact always surjective. Now, using injectivity, one can easily prove that $f(1)=1$ , and so we have $P(x,1) \Rightarrow x^2=f(x)f(f(x)) \Longrightarrow f(x)f(yf(x))=x^2f(y)=f(x)f(f(x))f(y) \Rightarrow f(yf(x))=f(f(x))f(y)$ However, as $f$ is surjective, so we can take $f(x)=z$ , giving that $f$ is multiplicative also. Let $w=\frac{1}{f(x)}$ (As $f$ is surjective, so $\frac{1}{f}$ is also surjective). Then, using multiplicity, we have $P \left(y+\frac{1}{f(x)},x \right) \Rightarrow x^2f(y+w)=f(x)f(yf(x)+1)=f(x)(f(yf(x))+1)= x^2f(y)+f(x)$ $\Rightarrow \text{ As } x^2=f(x)f(f(x)) \text{, we get that } f(f(x))(f(y+w)-f(y))=1$ But, $f(x)=\frac{1}{w}$ , and so we get $f(y+w)-f(y)=\frac{1}{f \left(\frac{1}{w} \right)}=f(w) \Rightarrow f(y+w)=f(y)+f(w)$ where we use that $f \left(\frac{1}{w} \right) f(w)=f(1)=1$ . This means that $f$ is both additive and multiplicative, in which case it is well known that $f$ must be the identity function. Hence, done. $\blacksquare$

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#8 h May 22, 2019, 8:02 AM • 2 Y

Y by Adventure10, Mango247

Nice problem. Here is my solution:
Answer
Solution
Edit: Just realised that my solution is similar to @2above.

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#9 h Oct 28, 2019, 2:22 AM • 1 Y

Y by Adventure10

Good problem. Nice solutions. Learned much.

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#13 h Nov 11, 2019, 6:24 AM • 1 Y

Y by Adventure10

Nice problem

Solution
$f (x)\equiv 0$ if $f (x)\neq 0$ for some $x$ .

If $f (x)=f (y)$ then $x=y$ or $x=-y$
$P (0,y)\implies f (0)=0$ or $f (yf (0)-1)=-1\implies f (0)=0$
$P (x,0)\implies f (x)f (-1)=-f (x)\implies f (-1)=-1$
$P (-1,y)\implies -f(-y-1)=f (y)+1$

$\text {Restate the original equation as}$

$f (x)f (-yf (x))=x^2f (y)\implies f (x)f (f (x))=x^2$
$P (x,x): f (-xf (x))=x^2=-f (xf (-x))f (x)+f (-x)=0$
$f (y+1)=f (y)+1 f (x+1)+f (f (x+1))=(x+1)^2$
$f (x)+f (f (x))=2x$ , and $f (x)=x$
Hence, $f(x)=0$ $\forall x\in\mathbb R$ and $f(x)=x$ . Easy to verify.

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#14 h Jan 9, 2020, 5:43 PM • 2 Y

Y by AlastorMoody, Adventure10

Bonus for a TST P3!

anantmudgal09 wrote:

Find all functions $f: \mathbb{R} \mapsto \mathbb{R}$ such that $f(x)f\left(yf(x)-1\right)=x^2f(y)-f(x),$ for all $x,y \in \mathbb{R}$ .

Answers: $f\equiv 0$ and $f \equiv x \ \forall x \in \mathbb{R}$ .
Clearly both these functions satisfy, so we continue by proving them to be the only ones. Denote by $(x,y)$ the assertion $f(x)\cdot f(yf(x)-1) = x^2f(y)-f(x)$ .
$(x,x) \implies f(x)f(xf(x)-1) = f(x)(x^2-1) ... (1)$ . Now, $(0,y) \implies f \equiv 0$ (which is a solution) or $f(0)=0$ . Assuming $f$ to be non-constant,
take $f(0)=0$ . Now suppose there exists $x_0 \neq 0 \in \mathbb{R}$ such that $f(x_0) =0$ . Then, $(x_0,y) \implies x_0^2 \cdot f(y) = 0 \iff x_0 = 0$ , contradiction and thus $f$ is injective at $0 \implies f(xf(x) -1) = x^2-1 \implies f$ is surjective over $[-1,\infty)$ .
Now, $(x,1) \implies f(x) f(f(x)-1) = x^2 - f(x)$ . We now show that $f$ is injective. Indeed, if $f(a) =f(b)$ , then $a^2 = b^2$ . So suppose $a+b=0$ . Then we get that $f(af(x)-1) = f(-af(x)-1) \forall x \in \mathbb{R}$ . Using surjectivity, pick $\alpha$ such that $f(\alpha) = \frac{1}{a}$ . Then we have $f(0)=f(2)$ , contradiction $\implies f$ is injective. Now using similar arguments get that $f(x) = -f(-x)$ which implies $f$ is a bijection on $\mathbb{R}$ . Then $(x,f(y)) \implies x^2 \cdot f(f(y)) \cdot f(y) = y^2 \cdot f(x) \cdot f(f(x))$ and using $f(1)=1$ (obtainable from $(1,1)$ ) we establish $f(x) f(f(x)) = x^2$ which in turn implies $f(x+1) = f(x) + 1 \forall x$ and moreover, $f(x) f(yf(x)) = x^2 f(y) = f(x)f(f(x)) f(y) \implies f$ is multiplicative $\implies f(y)f(x+1) = f(y)f(x) + f(y)f(1) \iff f(xy+y) = f(xy) + f(y) \implies f$ is Cauchy and multiplicative and thus $f$ is identity.

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#15 h Jan 13, 2020, 10:02 PM • 1 Y

Y by Adventure10

Not that hard for P3

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#16 h Mar 21, 2021, 5:24 AM • 1 Y

Y by megarnie

$\clubsuit \color{magenta}{\textit{\textbf{ANS:}}}$ $f(x)=0 \quad \textrm{and} \quad f(x)=x \quad \forall x\in \mathbb{R}$ .
$\blacklozenge \color{blue}{\textit{\textbf{Proof:}}}$ It's easy to see that these are indeed solutions to the given FE. Let $P(x,y)$ be the given assertion, we have $P(x,0): f(x)(f(-1)+1)=x^2f(0)$ if $f(-1)\ne -1$ , we will have $f(x)=cx^2$ for some real number $c$ , plugging it into our FE, $c^4x^6y^2+\ldots =c^2x^2(cx^2y-1)^2=cx^2y^2-cx^2$ which means $c=0$ and $f(x)=0$ for all real $x$ . If $f(-1)=-1$ , then $f(0)=0$ . If there exists a real number $u$ , such that $f(u)=0$ , then $P(u,x): u^2f(x)=0 \quad \forall x\in \mathbb{R}$ which if $f(x)\ne 0$ for all real $x$ , then $u=0$ . So, $f$ is injective at $0$ and $f$ is not the zero function. Then, $P(x,x): f(xf(x)-1)=x^2-1 \quad \forall x\ne 0$ but since $f(0)=0$ , $f(xf(x)-1)=x^2-1 \quad \forall x\in \mathbb{R}$ . Now, by comparing $P(x, 1)$ and $P(1,f(x))$ , we have $\frac{x^2-f(x)}{f(x)}=f(f(x))-1 \implies f(x)f(f(x))=x^2 \quad \forall x\ne 0$ but again $f(0)=0$ , so this also holds for all real $x$ . Plugging this back to our original FE, we have $P(x,y):f(yf(x)-1)=f(f(x))f(y)-1$ and $P(1,x+1): f(x+1)=f(x)+1.$ Most importantly, $f$ is odd since $P(-1,x): f(-x)-1=f(-x-1)=-f(x)-1 \implies f(-x)=-f(x) \quad \forall x\in \mathbb{R}.$ Therefore, since $f$ is odd and $f(xf(x)-1)=x^2-1$ , $f$ is surjective over $\mathbb{R}$ . This means $P(x,f(y)): f(f(x)f(y))=f(f(x))f((y)) \implies f(xy)=f(x)f(y)$ $f$ is multiplicative and in particular, $f(x^2)=f(x)^2$ . Finally, as $\begin{cases} f(x+1)=f(x)+1 \\ f(x^2)=f(x)^2, \end{cases}$ this implies $f(x)=x$ (it's a well-known FE). $\quad \blacksquare$

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#17 h Apr 25, 2021, 4:54 PM

Y by

anantmudgal09 wrote:

Find all functions $f: \mathbb{R} \mapsto \mathbb{R}$ such that $f(x)f\left(yf(x)-1\right)=x^2f(y)-f(x),$ for all $x,y \in \mathbb{R}$ .

Easy, but nice.
Let $P(x,y)$ denote the given assertion. $f \equiv 0$ is obviously a solution. Henceforth, assume that $f \not \equiv 0$ . Now $P(1,1): f(1)f(f(1)-1)=0 \implies \exists c \quad \text{such that} f(c)=0$ $P(c,y): c^2f(y)=0$ since $f \not \equiv 0$ , we conclude that $f(t)=0 \iff t=0$ . So $f(1)-1=0 \implies f(1)=1$ . Observe that $P(x,1): f(f(x)-1)= \frac{x^2}{f(x)}-1 \quad (i)$ and $P(1,f(x)): f(f(x)-1)=f(f(x))-1 \quad (ii)$ On comparing $(i)$ and $(ii)$ we have that $f(x)f(f(x))=x^2 \forall x \in \mathbb{R} \quad (iii)$ . Now note that for all $x\not = 1,-1$ $P(x,x): f(xf(x)-1)=x^2-1 \implies f(x^2-1)=f(f(xf(x)-1))=\frac{(xf(x)-1)^2}{f(xf(x)-1)}=\frac{(xf(x)-1)^2}{x^2-1} (iv)$ $P(f(x),f(x)): f(f(x)f(f(x))-1)= f(x)^2-1 \overset{\text{using} (iii)}{\implies} f(x)^2-1=f(x^2-1)= \frac{(xf(x)-1)^2}{x^2-1}$ Upon simplification, the above result is equivalent to $(f(x)-x)^2=0 \implies f(x)=x$ . Also $P(x,0): f(-1)=-1$ . Hence we have two solutions, i.e.
$f(x)=x \quad \forall x \in \mathbb{R}$ $f(x)=0 \quad \forall x \in \mathbb{R}$ Hence, we are done.
Note

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#18 h Apr 25, 2021, 5:52 PM • 2 Y

Y by megarnie, kido2006

$\boxed{f(x)=0}$ works. Assume now that $\exists j:f(j)\ne0$ .

$P(1,1)\Rightarrow f(1)f(f(1)-1)=0\Rightarrow\exists k:f(k)=0$
$P(k,j)\Rightarrow k^2f(j)=0\Rightarrow k=0$ (injectivity at $0$ )

Either $f(1)=0$ or $f(f(1)-1)=0$ . If $f(1)=0$ then $1=0$ , absurd, thus $f(1)-1=0\Rightarrow f(1)=1$ .

$P(1,x+1)\Rightarrow f(x+1)=f(x)+1$
The assertion becomes $Q(x,y):f(x)f(yf(x))=x^2f(y)$ .
$Q(x,1)\Rightarrow f(x)f(f(x))=x^2$
$Q(x+1,1)\Rightarrow f(x)+f(f(x))=2x$
So $f(f(x))=2x-f(x)$ , thus $f(x)(2x-f(x))=x^2$ , which factors as $(x-f(x))^2=0$ , hence $\boxed{f(x)=x}$ , which is the only remaining solution.

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Project_Donkey_into_M4

#19 h Mar 3, 2022, 8:42 AM • 1 Y

Y by HoRI_DA_GRe8

Solution

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#20 h Mar 29, 2022, 1:35 PM

Y by

Hello, the equation in hand is f (x) f (y f(x) - 1) = x ^ 2 f (y) - f (x)
Assume there is some other function since f (x) =0
x = y = 0 => f (0) f (-1) = -f (0) => f(0) = 0 /f(-1) = -1
Taking f (0) = 0 and substitute y = 0
=> f (-1) = -1
Taking f (-1) = -1 and substitute y = 0
=> f (0) = 0
This proves that both statements provided before are equivalent
Considering that there exists an a which is not equal to 0 such that f (a) = 0,
Substituting x = a and y = -1. This will eliminate the case.
Substituting P (x, y)
P (-1, y) => -f (-y -1) = f(y) + 1 => f (y-1) = -f (-y) -1

Now substitute this in the original equation available to get
F (x) f (f (x)) = x^2
P(1, 1)
f(1) f (f(1) -1) = 0
f(1) = 1
Then, P (1,y) = f; This is an odd function

Use f (x +1) = f(x) +1 to get rid of the 1

Put x = x + 1, y = 1 to get
F (f(x)) = 2 x - f (x)

Now substitute f (x) f (f(x)) = x ^2 with f (f (x)) = 2x - f (x)
(f(x) - x) ^ 2 = 0
f(x) = x

Working functions: f (x) = 0 and f (x) =x

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#21 h Apr 8, 2022, 5:30 PM

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We will prove that $f\equiv0$ and $f(x)=x$ are the only solutions.
Let $P(x,y)$ be the assertion in $f(x)f\left(yf(x)-1\right)=x^2f(y)-f(x)$ .
It can be easily seen that the only constant function that works is $\boxed{f \equiv 0}$ , which is our first solution.
Let $f$ be non-constant.

Claim 1 : $f(0)=0$ .
Proof :
$P(0,y) : f(0)f(yf(0)-1)=-f(0)$
Assume possible $f(0)\neq0$ .
Then, $f(yf(0)-1)=-1$ .
As $f(0)\neq0$ , as we vary $y$ over $\mathbb{R}$ , $yf(0)-1$ also varies over $\mathbb{R}$ .
Hence, $f(x)=-1 \forall x$ .But $f\equiv-1$ doesn't satisfy given equation.
Contradiction!
Hence proved claim 1!

Claim 2 : $f$ injective.
Proof :
Let $f(a)=f(b)$ .
Let $c$ be such that $f(c)\neq0$ .
$P(a,c)-P(b,c) : a^2=b^2 \implies a = \pm b$ .
Hence, $f(a)=f(b) \implies a= \pm b ...(1)$ .
Assume that $a=-b$ ( $a,b \neq 0$ ) is possible.
$P(c,a)-P(c,b) : f(af(c)-1)=f(bf(c)-1) \overset{(1)}{\implies} af(c)-1= \pm (bf(c)-1) \overset{a=-b}{\implies} af(c)-1= \pm (-af(c)-1)$ .
If $af(c)-1=-af(c)-1$ , we get $a=0$ . Contradiction.
If $af(c)-1=-(-af(c)-1)$ , we get $-1=1$ . Contradiction!
Hence, if $f(a)=f(b)$ , $a=-b$ is not possible.
Combining with $(1)$ , we get $f(a)=f(b)\implies a=b$ .
Hence proved claim 2!

Hence note that , $f(x) \neq 0 \forall x\neq 0$ .

Claim 3 : $f(1)=1$ .
Proof :
$P(1,1) : f(1)\cdot f(f(1)-1)=0 \implies f(f(1)-1)=0 \implies f(1)-1=0 \implies f(1)=1$ .
(Here we used claim 1 and claim 2)
Hence Proved claim 3!

Claim 4 : $f(-1)=-1$ .
Proof :
$P(x,0) : f(x)f(-1)=-f(x) \implies f(-1)=-1$ . (Here we used claim 1 and claim 2)
Hence proved claim 4!

Claim 5 : The given equation reduces to $f(x)f(yf(x))=x^2f(y)$ .
Proof :
$P(1,y) : f(y-1)=f(y)-1...(2)$ . (Here we used claim 3)
$y\rightarrow yf(x) : f(yf(x)-1)=f(yf(x))-1$ .
Substituting this in given equation, we get $f(x)f(yf(x))=x^2f(y)$ .
Hence proved claim 4!

Let $R(x,y)$ be the assertion in $f(x)f(yf(x))=x^2f(y)$ .

Claim 6 : $f(x+1)=f(x)+1$ and $f(x^2)=f(x)^2$ .
Proof :
$(2) : f(y)=f(y-1)+1 \implies f(y+1)=f(y)+1$ .

$R(f(x),f(x)) : f(f(x)f(f(x)))=f(x)^2$ .
$R(x,1) : f(x)f(f(x))=x^2 \implies f(f(x)f(f(x))=f(x^2)$ .
Hence, $f(x)^2=f(x^2)$ (From last two equations).
Hence proved claim 6!

From claim 6, its well-known that ${f(x)=x}$ is the solution.
$\boxed{f(x)=x}$ indeed works and is our second solution!
Hence, we are done

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#22 h Apr 8, 2022, 9:40 PM

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Let $P(x,y)$ denote the given assertion.

$P(0,x): f(0)f(xf(0)-1)=-f(0)$ .

So either $f(0)=0$ or $f(xf(0)-1)=-1$ for any $x$ . If $f(0)\ne 0$ , then $xf(0)-1$ can take on any real value, which implies $f\equiv -1$ , which is not a solution. Thus, we have $f(0)=0$ .

Now, noting $\boxed{f\equiv 0}$ works, we can assume $f$ is non-constant.

Claim: $f$ is injective.
Proof: Suppose $f(a)=f(b)$ with $a\ne b$ .

$P(a,x): f(a)f(xf(a)-1)=a^2f(x)-f(a)$ .

$P(b,x): f(b)f(xf(b)-1)=b^2f(x)-f(b)$ .

This implies $a^2f(x)=b^2f(x)$ . If we set $x$ such that $f(x)\ne 0$ , then $a^2=b^2\implies a=\pm b\implies a=-b$ , since $a\ne b$ . Then $f(a)=f(-a)$ . In fact, this implies $f$ is injective at $0$ .

$P(a,-a): f(a)f(-af(a)-1)=a^2f(a)-f(a)$ .

$P(-a,a): f(a)f(af(a)-1)=a^2f(a)-f(a)$ .

If $f(a)\ne 0$ , then we have $f(-af(a)-1)=f(af(a)-1)$ . However, this implies either $af(a)+1=af(a)-1$ , or $-af(a)-1=af(a)-1$ , both are absurd. So $f(a)\ne f(-a)$ .

If $f(a)=0$ , then $a=0$ .

Since $f$ is injective at $0$ , $f$ is injective. $\blacksquare$

$P(x,0): f(x)f(-1)=-f(x)$ . If we set $x\ne 0$ , then we get $f(-1)=-1$ .

$P(1,1): f(1)f(f(1)-1)=0$ , so $f(1)=1$ .

$P(1,x): f(x-1)=f(x)-1\implies f(x+1)=f(x)+1$ .

Now we rearrange the FE.

We have $f(x)(f(yf(x))-1)=x^2f(y)-f(x)$ , so $f(x)f(yf(x))=x^2f(y)$
Let $Q(x,y)$ be the assertion here.

$Q(x,1): f(x)f(f(x))=x^2\implies f(f(x))=\frac{x^2}{f(x)}$ .

$P(x+1,1): (f(x)+1)f(f(x))=(x+1)^2-f(x)-1=x^2+2x-f(x)\implies f(f(x))=\frac{x^2+2x-f(x)}{f(x)+1}$ .

This implies $\begin{align*} \frac{x^2}{f(x)}=\frac{x^2+2x-f(x)}{f(x)+1} \\ \implies x^2(f(x)+1)=f(x)(x^2+2x-f(x)) \\ \implies x^2f(x)+x^2=x^2f(x)+2xf(x)-f(x)^2 \\ \implies x^2=2xf(x)-f(x)^2 \\ \implies f(x)^2-2xf(x)+x^2=0 \\ \implies (f(x)-x)^2=0 \\ \implies \boxed{f(x)=x} \\ \end{align*}$ which clearly works.

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#23 h Jun 19, 2022, 6:05 PM

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Also Singapore 2015 aops.com/community/c6h1618668p25462682

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#24 h Nov 9, 2023, 5:13 PM

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The answer is $f(x)=x$ and $f \equiv 0$ only, which both work. Hence suppose $f \not \equiv 0$ . Let $P(x,y)$ denote the assertion.

From $P(x,0)$ we have $f(x)(f(-1)+1)=x^2f(0)$ . If $f(-1)+1 \neq 0$ , then $f(x)=cx^2$ for some nonzero constant $c \in \mathbb{R}$ . But this clearly doesn't work, so $f(-1)=-1$ and thus $f(0)=0$ . Then from $P(-1,y)$ we obtain $f(-y-1)=-f(y)-1$ .

If $f(x)=0$ for some $x$ , by picking $y$ such that $f(y) \neq 0$ we find that $x=0$ . Then for $x \neq 0$ , $P(x,x)$ gives $f(xf(x)-1)=x^2-1$ , hence the range of $f$ contains $(-1,\infty)$ . Since $f(-y-1)=-f(y)-1$ , the range of $f$ also contains $(-\infty,1)-1$ , so $f$ is surjective. Additionally, $P(-1,-1)$ implies $f(-2)=-2$ , hence $-f(1)-1=f(-2) \implies f(1)=1$ .

Using $f(-y-1)=-f(y)-1$ , from $P(x,-1)$ we obtain $f(x)(f(-f(x)-1)+1)=-x^2 \implies f(x)f(f(x))=x^2$ . Thus we can rewrite the assertion as $f(yf(x)-1)=f(f(x))f(y)$ . From surjectivity, this becomes $f(xy-1)=f(x)f(y)-1$ . Comparing $(x,y)$ with $(xy,1)$ and using $f(1)=1$ implies that $f$ is multiplicative. On the other hand, by plugging in $y=1$ we also get $f(x-1)=f(x)-1$ .

Since $f$ is multiplicative, we have $f(x^2)=f(x)^2>0$ for all $x>0$ , hence $f$ sends positive reals to positive reals. Thus let $g(x)=\log f(e^x)$ sending real numbers to real numbers, so $g$ is additive (by only considering the multiplicativity of $f$ over $\mathbb{R}^+$ ). On the other hand, $f(x-1)=f(x)-1$ implies that $f(x)>1$ for $x>1$ , hence $g(x)>0$ for $x>0$ . Therefore $g$ is linear, so $f(x)=x^k$ for some $k \in \mathbb{R}$ . Then we have $2^k-1=1$ , hence $k=1$ and $f(x)=x$ as desired. $\blacksquare$

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