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#1 h Mar 31, 2018, 9:17 AM • 6 Y

Y by Hamel, Amir Hossein, tiendung2006, Adventure10, Mango247, Rounak_iitr

Find all functions $f : \mathbb{R} \rightarrow \mathbb{R}$ , where $\mathbb{R}$ is the set of real numbers, such that
$f(x)f(yf(x) - 1) = x^2 f(y) - f(x) \quad\forall x,y \in \mathbb{R}$

This post has been edited 2 times. Last edited by dominicleejun, Aug 16, 2019, 9:16 PM

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Hamel

#2 h Mar 31, 2018, 9:53 AM • 2 Y

Y by Adventure10, Mango247

Some steps I used

This post has been edited 4 times. Last edited by Hamel, Mar 31, 2018, 9:55 AM

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TLP.39

#3 h Mar 31, 2018, 9:59 AM • 3 Y

Y by Tawan, Adventure10, Mango247

Let $P(x,y)$ be the assertion.
First,it is easy to see that $\boxed{f(x)\equiv 0}$ is a solution.
Now,we will assume that $f(x)\not\equiv 0$ .
$P(1,1)\implies f(1)f(f(1)-1)=0\implies \exists\, a\in\mathbb{R},f(a)=0$ .
$P(a,0)\implies a^2f(0)=0\implies f(0)=0$ .
$P(x,0)\implies f(x)f(-1)=-f(x)\implies f(-1)=-1$ .
$P(-1,-y)\implies f(y-1)=-f(-y)-1^{(*)}$ .
Hence the assertion become
$P(x,y);f(x)[-f(-yf(x))-1]=x^2f(y)-f(x)\implies \boxed{f(x)f(-yf(x))=-x^2f(y)}$ .
If $f(d)=0$ ,then $P(d,y)\implies d=0$ .Hence $d\neq 0\implies f(d)\neq 0$ .
Now,for any $x\neq 0$ ,we have $P(x,x)\implies f(-xf(x))=-x^2$ .Hence all nonpositive real numbers are in the range of $f(x)$ .
From the new assertion,it is easy to see that $f(a)=f(b)\implies a=\pm b$ .
But if $f(a)=f(-a)\,\exists a\neq 0$ ,then
$(*)\implies f(a-1)=f(-a-1)\implies a-1=\pm (-a-1)$ which is a contradiction.Hence $f(x)$ is injective.
Thus $P(x,x)\implies f(-xf(x))=-x^2\,\forall x\neq 0\implies f(-xf(x))=f(xf(-x))=-x^2\implies -f(x)=f(-x)\,\forall x\neq 0$ .
Hence $f(x)$ is surjective and $f(y-1)=f(y)-1$ .
Thus the assertion now become
$P(x,y);\boxed{f(x)f(yf(x))=x^2f(y)}$ .
$P(x,1)\implies f(x)f(f(x))=x^2f(1)=-x^2f(-1)=x^2$ .
Hence $f(x)f(yf(x))=f(x)f(f(x))f(y)\implies f(yf(x))=f(f(x))f(y)\,\forall x\neq 0\implies f(yf(x))=f(f(x))f(y)\,\forall x,y\in\mathbb{R}$ .
But since $f(x)$ is surjective,we get that $f(x)$ is multiplicative.
$P(x,y+1)-P(x,y)\implies f(x)[f(yf(x)+f(x))-f(yf(x))]=x^2=f(x)f(f(x))\implies f(z+a)-f(z)=f(a)\,\forall a\neq 0$ .
Now it is easy to see that $f(x)$ is addictive,too.Hence $\boxed{f(x)\equiv x}$ .(Since we have exclude $f(x)\equiv 0$ .)
Hence all solution are $f(x)\equiv 0$ and $f(x)\equiv x$ which clearly satisfied the assertion.

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#4 h Mar 31, 2018, 10:01 AM • 2 Y

Y by Adventure10, Mango247

Incorrect

This post has been edited 5 times. Last edited by AnArtist, Mar 31, 2018, 10:33 AM

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#5 h Mar 31, 2018, 10:25 AM • 2 Y

Y by paragdey01, Adventure10

AnArtist wrote:

Claim 2) $f(\frac{1}{x})= \frac{f(x)}{x^2}$ $\forall x \neq 0$ .
Proof
$P( x, \frac{1}{f(x)})$ .

$P(x,\frac {1}{f(x)})$ gives $f(\frac {1}{f(x)})=\frac {f(x)}{x^2}$

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gghx

#7 h Nov 2, 2020, 5:19 AM

Y by

Here is a bit of improvement on the solutions above.

If $f$ is non-constant, $P(0,y): f(0)=0$ .
If $f(c)=0, P(c,y): c=0$
$P(1,1): f(1)f(f(1)-1)=0$
Thus, $f(1)=1$ .
$P(1,y): f(y+1)=f(y)+1$
Hence the original equation becomes $f(x)f(yf(x))=x^2f(y)$

We get $f(x)f(f(x))=x^2$ and $f(x+1)=f(x)+1$ , we can also do:
$x^2+2x+1=(x+1)^2=f(x+1)f(f(x+1))=(f(x)+1)(f(f(x))+1)=f(x)f(f(x))+f(x)+f(f(x))+1=x^2+f(x)+f(f(x))+1$
And so $f(x)+f(f(x))=2x$ .

Lwt $f(x)=a, f(f(x))=b$ , $ab=x^2, a+b=2x$ so $b=2x-a \implies a(2x-a)=x^2 \implies (x-a)^2=0$
Thus, $f(x)=a=x$

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Mr.C

#8 h Nov 2, 2020, 5:43 AM • 1 Y

Y by Mango247

dominicleejun wrote:

Find all functions $f : \mathbb{R} \rightarrow \mathbb{R}$ , where $\mathbb{R}$ is the set of real numbers, such that
$f(x)f(yf(x) - 1) = x^2 f(y) - f(x) \quad\forall x,y \in \mathbb{R}$

let $p(x,y)=f(x)f(yf(x)-1)=x^2f(y)-f(x)$ clearly $p(0,y)$ either gives $f(x)=0$ or $f(0)=0$ so assume $f(0)=0$
then $p(x,0)=f(x)f(-1)=-f(x)$ so since $f(x)$ is not the zero constent $f(-1)=-1$
now $p(-1,y)=f(-y-1)=-f(y)-1$
so $f(y-1)=-f(-y)-1$ so
$f(yf(x)-1)=-f(-yf(x))-1$ so rewrite as
$p(x,y)=-f(x)f(-yf(x))=x^2f(y)$ .
now $p(x,-1)=f(x)f(f(x))=x^2$ . now if $f(x)=0$ then
$p(x,y)=x^2f(y)=0$ so $x=0$ .
now rewrite as
$p(x,y)=f(-yf(x))=-f(y)f(f(x))$
now if $f(a)=f(b)$ then $a^2=b^2$ also as we had $f(-y-1)=-f(y)-1$ we have
$f(-a-1)=f(-b-1)$ so
$(b+1)^2=(a+1)^2$ which gives us $a=b$ .
now $p(x,x)=f(-xf(x))=-x^2=p(-x,-x)$ so
$-xf(x)=xf(-x)$ which gives $-f(x)=f(-x)$
so rewrite equations as
$f(y+1)=f(y)+1$
$p(x,y)=f(yf(x))=f(y)f(f(x))$ as $x \in D_f$ for $x<0$ and clearly if $a \in D_f$ then $-a \in D_f$ we have $f$ is surgective.
so rewrite as
$p(x,y)=f(xy)=f(x)f(y)$
so $f((x+1)y)=f(x+1)f(y)=f(x)f(y)+f(y)=f(xy)+f(y)$
so $f(a+b)=f(a)+f(b)$ .
these give $f(x)=x$ for all real $x$ .

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#9 h Aug 11, 2021, 12:02 AM • 1 Y

Y by sk2005

Let $P(x,y)$ be the assertion $f(x)f(yf(x)-1)=x^2f(y)-f(x)$ .
$P(1,1)\Rightarrow f(1)f(f(1)-1)=0$
If $f(1)=0$ , then $P(1,x)\Rightarrow\boxed{f(x)=0}$ .
Otherwise, there is a $j$ with $f(j)\ne0$ and $f(f(1)-1)=0$ .
$P(f(1)-1,j)\Rightarrow f(1)=1$
$P(1,x+1)\Rightarrow f(x+1)=f(x)+1$
$P(x,1)\Rightarrow f(x)f(f(x))=x^2$
Taking $x\mapsto x+1$ in this, we see that $f(x)+f(f(x))=2x$ . Then $f(x)(2x-f(x))=x^2$ , so $\boxed{f(x)=x}$ .

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ZETA_in_olympiad

#10 h Jun 19, 2022, 6:02 PM

Y by

As per usual, we use $P(x,y).$ Assume not $f\equiv 0$ (which works.) We get $f(1)(f(1)-1)=0$ from $P(1,1).$ Take $f(z)=0,$ then $P(z,y)$ implies $z=0.$ In particular $f(1)=1.$ $\begin{align*} P(1,x)\implies &f(x-1)=f(x)-1 \\ P(a,b)\implies &f(a)f(bf(a))=a^2f(b) \\ P(a,1)\implies & f(a)f(f(a))=a^2f(1) \\ P(a+1,1)\implies & f(f(a))=2a-f(a) \\ P(a,1)\implies & f(a)\equiv a \quad \text{which fits.} \end{align*}$

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#11 h Jun 20, 2022, 11:53 AM • 3 Y

Y by PRMOisTheHardestExam, Mango247, Rounak_iitr

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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#13 h Sep 2, 2024, 4:28 PM

Y by

The answers are $f \equiv 0$ and $f \equiv x$ , which clearly work.

Let $P(x, y)$ denote the assertion. We start by making the following claim:

Claim: For all $x \in \mathbb{R}$ , we have
$f(x)f(f(x)) = x^2.$
Proof We first note that $P(x, 0)$ gives us
$f(x)f(-1) = -f(x) \implies f(x)(f(-1) + 1) = 0.$ So either $f(x) = 0$ for all $x \in \mathbb{R}$ or $f(-1) = -1$ . Note that $f \equiv 0$ is indeed
a solution, so assume $f(-1) = -1$ . Now we observe that:

$\bullet$ $P(-1, y) \implies f(-y - 1) = -(f(y) + 1)$ .
$\bullet$ $P(x, -1) \implies f(x)f(-f(x) - 1) = -x^2 - f(x)$ .

Comparing the first and second bullets, we get that:
$f(x)[f(f(x)) + 1] = x^2 + f(x) \implies f(x)f(f(x)) = x^2,$ as claimed.

Now, we show that if there exists a $z \neq 0$ such that $f(z) = 0$ , then $f$ is zero
everywhere. Indeed, by $P(z, x)$ ,
$z^2f(x) = 0 \implies f(x) = 0,$ as needed. Now $P(0, x)$ gives us that
$f(0)f(xf(0) - 1) = -f(0) \implies f(0)[f(f(0)x - 1) + 1] = 0.$ If $f(0) \neq 0$ , then $f(x) = -1$ for all $x$ , which is not a solution. So indeed $f(0) = 0$ .
Thus we can write
$f(f(x)) = \frac{x^2}{f(x)},$ for all $x \neq 0$ . We assume $x \neq 0$ from now on.

We also realize that $P(f(x), f(x))$ gives us
$f(f(x))f(f(f(x))f(x) - 1) = f(x)^2f(f(x)) - f(f(x)),$ $\frac{x^2}{f(x)}f(x^2 - 1) = x^2f(x) - \frac{x^2}{f(x)},$ $f(x^2 - 1) = f(x)^2 - 1.$
This gives us $f(1) = \pm 1$ . Note that $f(1) = -1$ does not work as $P(-1, 1)$ gives
$f(-2) = 0$ . So $f(1) = 1$ .

Now, $P(1, x)$ gives us that
$f(x - 1) = f(x) - 1$ for all natural $n$ . Therefore, the equation becomes
$f(x)f(yf(x)) = x^2f(y),$ which we now denote by $P(x, y)$ . Now, $P(x, x)$ gives
$f(xf(x)) = x^2.$ Finally, $P(xf(x), y)$ implies
$f(x^2y) = f(x)^2f(y).$ Note that $y = 1$ gives
$f(x^2) = f(x)^2,$ so the equation becomes
$f(x^2y) = f(x^2)f(y).$ Holding $y$ non-negative, we get that $f$ is multiplicative in $\mathbb{R}^+$ , but as $f$ is odd,
we have $f$ is multiplicative everywhere. To finish, as we have
$f(x + 1) = f(x) + 1,$ let $x \rightarrow x/y$ for $y \neq 0$ , you get
$f(x + y)f\left(\frac{1}{y}\right) = f(x)f\left(\frac{1}{y}\right) + 1,$ $f(x + y) = f(x) + f(y),$ as $f(1/y) = 1/f(y)$ . Therefore, $f$ is additive as well. But as $f(x^2) = f(x)^2 \geq 0$ ,
by Cauchy, we have $f(x) = cx$ for all $x \in \mathbb{R}$ . Plugging it back in, we get $c \in \{0, 1\}$ ,
giving us the solutions.

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#14 h Oct 31, 2024, 6:38 PM

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$f(x)(f(yf(x) - 1)+1)=x^2f(y)$ Answers are $f\equiv 0$ and $f(x)=x$ . Assume that $f$ is non-constant. Plugging $x=0$ yields $f(0)(f(yf(0)-1)+1)=0$ . $f(0)\neq 0$ contradicts with our assumption hence $f(0)=0$ . $y=0$ implies $f(x)f(-1)+f(x)=0$ thus, $f(-1)=-1$ . Note that if $f(a)=0$ , then $x=a$ gives $a^2f(y)=0$ which is impossible for $a\neq 0$ . $x=y$ implies $f(xf(x)-1)=x^2-1$ which gives that $f$ is surjective over positive reals. Also for $x=1$ , $f(f(1)-1)=0$ or $f(1)=1$ . Plugging $x,f(y)$ , by symmetry we get $\frac{x^2f(f(y))}{f(x)}=\frac{y^2f(f(x))}{f(y)}$ thus, $\frac{f(f(x))f(x)}{x^2}$ is constant. For $x=1$ , it's equal to $1$ hence $x^2=f(x)f(f(x))$ . Plug $y=1$ to see that
$f(x)(f(f(x)-1)+1)=x^2\iff f(f(x)-1)=f(f(x))-1\overset{\text{surjectivity for positive reals}}{\implies} f(x^2-1)=f(x^2)-1$ $P(-1,-y^2)$ gives $-f(y^2)=-f(y^2-1)-1=f(-y^2)$ hence $f$ is odd which proves the surjectivity for all reals. Now we can choose $f(x)$ as any real on equation $f(f(x)-1)=f(f(x))-1$ so $f(x-1)=f(x)-1\iff f(x+n)=f(x)+n$ .
$f(x)f(yf(x))=x^2f(y)$ Compare $P(x,y)$ with $P(x,y+1)$ to observe $f(x)f(yf(x)+f(x))=x^2f(y)+x^2=f(x)f(yf(x))+x^2$ . After plugging $x,\frac{y}{f(x)}$ we get $f(x)(f(y+f(x))-f(y))=x^2$ or $f(y+f(x))-f(y)=f(f(x))$ Since $f$ is surjective, $f(x+y)=f(x)+f(y)$ hence $f$ is additive.
$(f(x)+f(z))(f(yf(x))+f(yf(z)))=(x+z)^2f(y)\iff 2xzf(y)=f(x)f(yf(z))+f(z)f(yf(x))$ Plug $y=z=1$ to verify $2x=f(x)+f(f(x))$ . Also $f(x)f(f(x))=x^2$ hence $x^2=f(x)(2x-f(x))\iff (x-f(x))^2=0$ which implies $f(x)=x$ for all reals as desired. $\blacksquare$

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