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Source: Netherlands Team Selection Test 2016 Day 1-Problem 4

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241 posts

#1 h Sep 22, 2016, 7:10 AM • 3 Y

Y by tiendung2006, Adventure10, Mango247

Find all funtions $f:\mathbb R\to\mathbb R$ such that: $f(xy-1)+f(x)f(y)=2xy-1$ for all $x,y\in \mathbb{R}$ .

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pco

#2 h Sep 22, 2016, 8:48 AM • 2 Y

Y by Bumblebee60, Adventure10

FabrizioFelen wrote:

Find all funtions $f:\mathbb R\to\mathbb R$ such that: $f(xy-1)+f(x)f(y)=2xy-1$ for all $x,y\in \mathbb{R}$ .

Let $P(x,y)$ be the assertion $f(xy-1)+f(x)f(y)=2xy-1$

If $f(0)\ne 0$ , $P(x,0)$ $\implies$ $f(x)=-\frac{f(-1)+1}{f(0)}$ constant, which is never a solution. So $f(0)=0$

$P(0,0)$ $\implies$ $f(-1)=-1$

Let $x\ne 0$ :
$P(x,\frac 1x)$ $\implies$ $f(\frac 1x)=\frac 1{f(x)}$

$P(x+1,\frac 1x)$ $\implies$ $(f(x+1)+1)f(\frac 1x)=\frac{x+2}x$ and so $f(x+1)=\frac{x+2}xf(x)-1$

$P(2,1)$ $\implies$ $f(1)(f(2)+1)=3$ and so $f(1)\ne 0$
$P(x+1,1)$ $\implies$ $f(x+1)=\frac{2x+1-f(x)}{f(1)}$

and so (two expressions for $f(x+1)$ ) : $\frac{x+2}xf(x)-1=\frac{2x+1-f(x)}{f(1)}$

And so assertion $Q(x)$ : $f(x)\frac{x(f(1)+1)+2f(1)}x=2x+1+f(1)$

If $f(1)=-1$ , this implies $f(x)=-x^2$ $\forall x\ne 0$ , still true when $x=0$
And so $\boxed{\text{S1 : }f(x)=-x^2\text{ }\forall x}$ which indeed is a solution

If $f(1)\ne -1$ , $Q(\frac{-2f(1)}{f(1)+1})$ $\implies$ $f(1)=1$

And $Q(x)$ becomes $f(x)=x$ $\forall x\notin\{0,-1\}$
We already know that $f(0)=0$ and $f(-1)=-1$
And so $\boxed{\text{S2 : }f(x)=x\text{ }\forall x}$ which indeed is a solution

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uraharakisuke_hsgs

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uraharakisuke_hsgs

#3 h Sep 22, 2016, 4:47 PM • 5 Y

Y by MABR, Pitagar, LoloVN, Mathlover_1, Adventure10

$f(0) = 0$
Plug in $x = 0$ so $f(-1) = -1$
Plug in $x-y-1$ so $f^2(1) = 1$ . Here we have $2$ cases
$\blacksquare$ If $f(1) = 1$ . Plug in $y=1$ so $f(x-1)+f(x) = 2x-1$ . Then $f(xy-1) = 2xy-1 - f(xy)$ $(1)$
Then $f(x)f(y) - f(xy) = 0$
So we have : $f(x)f(y) = f(xy)$ . But we also have : $f(x)+f(x+1) = 2x+1$
Plug in $x= y$ to $(1)$ then $f(x-1)f(x+1) +f^2(x) = 2x^2 - 1$
$\implies (2x-1-f(x))(2x+1-f(x))+f^2(x) = 2x^2-1$
$\implies (f(x)-x)^2 = 0$ so $f(x) = x$
$\blacksquare$ If $f(1) = -1$ . Plug in $y=1$ so $f(x-1)-f(x) = 2x-1$ . Do similary with case $1$
Then $f(x)f(y) = -f(xy)$
And : $-f(x-1)f(x+1) +f^2(x) = 2x^2-1$
Do similary with case $1$ we have $4x^2-1+2f(x) = 2x^2-1$ or $f(x) = -x^2$

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#4 h Apr 1, 2017, 11:15 AM • 1 Y

Y by Adventure10

Nice solution PCO.This is also Azerbaijan BMO 2016 TST.

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MABR

#6 h Jul 30, 2019, 10:43 AM • 2 Y

Y by Adventure10, Mango247

$(0,y)$ : $f(-1)+f(0)f(y)=-1$

$f(0)\ne0 \implies$ $f(x)=-\frac{f(-1)+1}{f(0)}$ which means function $f(x)$ is constant function. (contradiction)

So, $f(0)=0, f(-1)=-1$

$(x,1/x)$ $\implies$ $f(x)f(1/x)=1$

$(x,-1)$ $\implies$ $f(-x-1)-f(x)=-2x-1$ $....(*)$

and one more substitution here, ( $x$ $\implies$ $-xy$ )

We get $f(xy-1)-f(-xy)=2xy-1=f(xy-1)+f(x)f(y)$ (from original condition)

then, $-f(-xy)=f(x)f(y)$ $....(**)$

Put, $(1,1)$ : $f^2(1)=1$ yields two case

CASE1 $f(1)=1$
$y\implies -\frac{1}{x}$ at $(**)$ and multiply $f(1/x)$

Then, $-f(\frac{1}{x})=f(-\frac{1}{x})$ which means $-f(x)=f(-x)$

Apply this to $(*)$ : $f(x+1)+f(x)=2x+1$

from $f(0)=0, f(1)=1$ we easily can succeed induction $f(k)=k \in\mathbb N$

and odd function $f$ , So, $f(k)=k \in\mathbb Z$

Expansion to $\mathbb R$

By comparing two substitutions, $(x,y+\frac{t}{x})$ , $(x+\frac{t}{y},y)$

We get $f(x)f(y+\frac{t}{x})=f(y)f(x+\frac{t}{y})$ ....(A)

Then, there exist $x\notin\mathbb Z$ , $y\notin\mathbb Z$

but $t\in\mathbb R\implies x+\frac{t}{y}, y+\frac{t}{x}\in\mathbb Z$

Such $t$ must be exist by checking,

$t=xy\alpha,\alpha\in\mathbb R \quad x=\frac{x'}{1+\alpha},y=\frac{y'}{1+\alpha},\quad x', y'\in\mathbb Z$

Not to be too complicative check this : $x+\frac{t}{y}=x(1+\alpha),y+\frac{t}{x}=y(1+\alpha)$ , $\quad (t,x,y)$ is derived this idea.

So, $f(x+\frac{t}{y})=x(1+\alpha) \quad and \quad f(y+\frac{t}{x})=y(1+\alpha)$ $\quad where (x,y,t,\alpha) are on R$

Put these to (A), then, $f(x)y(1+\alpha)=f(y)x(1+\alpha)$ $\implies$ $f(x)y=f(y)x$ ( $x,y$ themselves are on $\mathbb R$ )

Now, independently take $y$ from $\mathbb Z$ and divide both side by $y$ .
Finally, we get $f(x)=x\in\mathbb R$

CASE2 $f(1)=-1$
Same way, induction, expansion $f(x)=-x^2$

Complement : $f(\alpha)=0$ , such $\alpha$ is only a value 0. If there is exception, contradiction comes from putting that value to $f(x)f(\frac{1}{x})=1$

So, we can freely divide both side except 0.

This post has been edited 5 times. Last edited by MABR, Jan 8, 2023, 6:28 AM
Reason: Typo

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#7 h Aug 28, 2019, 8:51 AM • 2 Y

Y by Pluto1708, Adventure10

Pluto1708 wrote:

Partial Solution

This was a partial solution given by Pluto1708 in another thread, could someone please continue on this idea?

I am having trouble solving the quadratic (and then finding $f(1)$ )

This post has been edited 1 time. Last edited by Delta0001, Aug 28, 2019, 8:51 AM

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#9 h May 25, 2021, 4:33 AM

Y by

Answer: $f(x)=x$ and $f(x)=-x^2$ for all $x\in \mathbb{R}.$
Proof: It's easy to see that these are indeed solutions. Let $P(x,y)$ denote the given assertion, we have $P(0,x): f(-1)+f(0)f(x)=-1.$ If $f(0)\ne 0$ , then $f\equiv \textrm{const.}$ which obviously isn't a solution to our FE. Therefore, $f(0)=0$ and so $f(-1)=-1$ and $P(1,1): f(1)^2=1 \implies f(1)=\pm 1.$ Consider two cases:
Case 1: $f(1)=-1$ .
We know $P\left(x, \frac{1}{x}\right)\implies f(x)f\left(\frac{1}{x}\right)=1$ for all $x\ne 0$ . So, when $x\ne 0$ , $P\left(x+1,\frac{1}{x}\right): f(x+1)=\frac{x+2}{x}f(x)-1.$ Then, as $f(x)=\frac{x+1}{x-1}f(x-1)-1\implies f(x-1)=\frac{x-1}{x+1}\left(f(x)+1\right)$ when $x\ne 1,$ $P(x,1): \frac{x-1}{x+1}\left(f(x)+1\right)-f(x)=2x-1 \implies f(x)=-x^2$ together with fact that $f(1)=-1,$ $\boxed{f(x)\equiv -x^2}.$
Case 2: $f(1)=1$ .
First, we know $P\left(x, \frac{1}{x}\right)\implies f(x)f\left(\frac{1}{x}\right)=1$ for all $x\ne 0$ . Then, $P(x+1,1): f(x)+f(x+1)=2x+1$ and $P(x+2,1): f(x+1)+f(x+2)=2x+3$ imply $f(x+2)=f(x)+2$ . From this, we find that $f(2)=f(0)+2=2$ and $f\left(\frac{1}{2}\right)=\frac{1}{2}.$ Also, comparing $P(x,1): f(x-1)+f(x)=2x-1$ and $P\left(2x,\frac{1}{2}\right): f(x-1)+\frac{1}{2}f(2x)=2x-1$ we get $f(2x)=2f(x)$ . Thus,
$\begin{align*} P\left(x+2,\frac{1}{2}\right)&: f\left(\frac{x}{2}\right)+\frac{1}{2}f(x+2)=x+1 \\ & \implies \frac{1}{2}f(x)+\frac{1}{2}(f(x)+2)=x+1 \\ & \implies \boxed{f(x)\equiv x}. \end{align*}$

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Miku3D

#10 h Jun 20, 2021, 3:13 PM

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Posting for storage purposes
My solution

This post has been edited 1 time. Last edited by Miku3D, Jun 20, 2021, 3:19 PM
Reason: Fixed LaTeX

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#11 h Nov 20, 2023, 3:28 PM • 2 Y

Y by monoditetra, Leparolelontane

f(xy-1)+f(x)f(y)=2xy-1

First look P(0,0);
x=y=0
f(-1)+f(0)×f(0)=-1
Assume f(0)=0, and prove it
And also if f(0)=0, f(-1)=-1
Let's look P(x,0)
f(-1)+f(x)*f(0)=-1
In that case, find f(x);
f(x)=-(1+f(-1))/f(0) In that time f(x) must be constant, but that's not correct.
So, f(0)=0, we proved.

Look P(x, 1/x);
f(0)+f(x)*f(1/x)=1
Use this claim; f(0)=0
We'll get
f(x)*f(1/x)=1
f²(1)=1 We have two cases:
i) f(1)=1
For this f(x)=x. For all real number x
ii) f(1)= -1
f(x)= -x² For all real number x

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#12 h Mar 25, 2024, 10:51 PM

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We claim that the two solutions are $f(x) \equiv x$ and $f(x) \equiv -x^2$ . Plugging these in, they work. We will show that no other solutions exist.

Plug in $(x, 0)$ to get $f(-1) + f(x)f(0) = -1.$ Suppose that $f(0) \ne 0$ . Then, $f$ must be a constant function, which is a contradiction. Thus, $f(0) = 0$ . Plug in $(0, 0)$ to get $f(-1) + f(0)^2 = -1 \implies f(-1) = -1.$ Plug in $(1, 1)$ to get $f(0) + f(1)^2 = 1 \implies f(1) = \pm 1.$ Plug in $\bigl(x, \frac{1}{x}\bigr)$ to get $f(x)f\bigl(\tfrac{1}{x}\bigr) = 1. \qquad (1)$ Notice that $(x + 1) \cdot \frac{1}{x} - 1 = \frac{1}{x}$ . Thus, plug in $\bigl(x + 1, \frac{1}{x}\bigr)$ to get $f\bigl(\tfrac{1}{x}\bigr) + f(x + 1)f\bigl(\tfrac{1}{x}\bigr) = 1 + \tfrac{2}{x}.$ Multiplying both sides by $xf(x)$ and substituting $(1)$ gets us $(x + 2)f(x) - xf(x + 1) = x.$ Plug in $(x + 1, 1)$ to get $f(x) + f(x + 1)f(1) = 2x + 1.$ Since $f(1) = \pm 1$ , we have two cases. We have two equations and two variables ( $f(x)$ and $f(x + 1)$ ). Solving, we get that $f(x) = x\text { or }f(x) = -x^2,$ as desired.

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lian_the_noob12

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lian_the_noob12

#13 h Jul 31, 2024, 8:19 PM

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It's easy to see that $f(0)=0,f(-1)=-1,f(1)=1,-1$

$P(x,\frac{1}{x}) \implies f(\frac{1}{x}) = \frac{1}{f(x)}$
$P(x+1,\frac{1}{x}) \implies 1+f(x+1)=f(x)[\frac{2}{x} +1]...(1)$
Then, $P(x+1,1) \implies f(x) + f(x+1)f(1)=2x+1$
If $f(1)=1$ then
$f(x+1)=2x+1-f(x)$
Plugging in $(1)$ gives $\boxed{f(x)=x}$

If $f(1)=-1$ then
$f(x+1)=f(x)-2x-1$
Plugging in $(1)$ gives $\boxed{f(x)=-x^2}$

This post has been edited 2 times. Last edited by lian_the_noob12, Jul 31, 2024, 8:20 PM

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#14 h Apr 14, 2025, 9:34 PM

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FabrizioFelen wrote:

Find all funtions $f:\mathbb R\to\mathbb R$ such that: $f(xy-1)+f(x)f(y)=2xy-1$ for all $x,y\in \mathbb{R}$ .

Let $P(x,y)$ be the assertion $f(xy-1)+f(x)f(y)=2xy-1$ .
$P(x,0)\Rightarrow f(x)f(0)=-1-f(-1)$
If $f(0)\ne0$ then $f$ is constant, but none of the constant solutions work, so $f(0)=0$ and $f(-1)=-1$ .
$P(1,1)\Rightarrow f(1)^2=1$

Case 1: $f(1)=1$
$P(x+1,1)\Rightarrow f(x+1)=-f(x)+2x+1\Rightarrow f(x+2)=f(x)+2$
Now we compare the following:
$P(x,y+2)\Rightarrow f(xy+2x-1)+f(x)f(y)+2f(x)=2xy+4x-1$
$P(x,y)\Rightarrow f(xy-1)+f(x)f(y)=2xy-1$
and get:
$f(xy+2x-1)+2f(x)=4x+f(xy-1).$ Set $y=\frac1x$ , then $f(2x)+2f(x)=4x$ for all $x\ne0$ (we can see that it's also true for $x=0$ ).
Set $y=0$ , then $f(2x-1)=f(2x)-1$ so $f(x+1)=f(x)+1$ . But since $f(x+1)=-f(x)+2x+1$ we must have $\boxed{f(x)=x}$ which fits.

Case 2: $f(1)=-1$
$P(x+1,1)\Rightarrow f(x+1)=f(x)-2x-1$
Now we compare the following:
$P(x,y+1)\Rightarrow f(xy+x-1)+f(x)f(y)-2yf(x)-f(x)=2xy+2x-1$
$P(x,y)\Rightarrow f(xy-1)+f(x)f(y)=2xy-1$
and get:
$f(xy+x-1)=f(xy-1)+2yf(x)+f(x)+2x.$ Set $y=\frac1x$ , then $f(x)=-x^2$ for $x\ne0$ , but we can see that it's also true for $x=0$ , so $\boxed{f(x)=-x^2}$ for all $x$ which fits.

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