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143 posts

#1 h Dec 26, 2014, 10:14 AM • 2 Y

Y by Adventure10, Mango247

Problem: Find all functions $f$ : $Z->Z$ such that:
$f(xf(y)+f(x))=2f(x)+xy$ for all x,y E $Z$

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39 posts

#2 h Dec 27, 2014, 4:46 PM • 1 Y

Y by Adventure10

$This\: \: \: \: function\: \: \: \:is\: \: \: \: bijective$

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39 posts

#3 h Dec 27, 2014, 5:19 PM • 3 Y

Y by BuiBaAnh, Adventure10, Mango247

$y\rightarrow \frac{-f(x)}{x}$
$f(xf(\frac{-f(x)}{x})+f(x))=f(x)-injective\Rightarrow xf(\frac{-f(x)}{x})+f(x)=x$
$x\rightarrow 0\Leftrightarrow f(0)=0$
$f(xf(y)+f(x))=2f(x)+xy$
$y\rightarrow 0\Rightarrow f(f(x))=f(x)$
$xf(\frac{-f(x)}{x})+f(x)=x$
$x\rightarrow -1\Rightarrow -f(f(-1))+f(-1)=-1\Rightarrow f(-1)=1$
$f(xf(y)+f(x))=2f(x)+xy$
$(x,y)\rightarrow (-1,-1)\Rightarrow 0=2f(-1)+1\Rightarrow f(-1)=\frac{-1}{2}$
$\frac{-1}{2}=f(-1)=1\: \: \: \: \: contradictions$

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#4 h Dec 27, 2014, 5:21 PM • 2 Y

Y by Adventure10, Mango247

$no\: \: \: \: \: \: solution$

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#5 h Dec 27, 2014, 6:21 PM • 2 Y

Y by Adventure10, Mango247

This isn't a true solution

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3258 posts

#6 h Dec 27, 2014, 7:33 PM • 3 Y

Y by alisherianvar, Adventure10, Mango247

@alish how do you know $\dfrac{-f(x)}{x}$ is an integer?

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#7 h Dec 28, 2014, 9:45 AM • 2 Y

Y by Adventure10, Mango247

f(x) = x+1, it also have short solution if f: R->R

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#8 h Dec 28, 2014, 12:56 PM • 2 Y

Y by Adventure10, Mango247

you can post solution if f: R->R, can't you?

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#9 h Dec 28, 2014, 2:49 PM • 2 Y

Y by Adventure10, Mango247

I was busy before

For f:Z->Z, note that f(y+6)=f(y)+6 and f(0)=1; f(1)=2;...;f(5)=6 deduce f(x)=x+1 for all x $\in Z$
For f:R->R.
*First, we have $f$ is bejective and $f(fx))=2f(x)-x$ (easy to prove)
*Next, with $f(-2)=-1$ , chose $y=-2$ we have $f(f(x)-x)=2(f(x)-x)$ .
Each $x \in R$ , exist $z$ such as $f(z)=f(x)-x \Rightarrow f(f(z))=2f(z) \Rightarrow z=0 \Rightarrow f(x)=x+1$

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#10 h Apr 29, 2025, 4:51 AM

Y by

mathslife wrote:

I was busy before

For f:Z->Z, note that f(y+6)=f(y)+6 and f(0)=1; f(1)=2;...;f(5)=6 deduce f(x)=x+1 for all x $\in Z$
For f:R->R.
*First, we have $f$ is bejective and $f(fx))=2f(x)-x$ (easy to prove)
*Next, with $f(-2)=-1$ , chose $y=-2$ we have $f(f(x)-x)=2(f(x)-x)$ .
Each $x \in R$ , exist $z$ such as $f(z)=f(x)-x \Rightarrow f(f(z))=2f(z) \Rightarrow z=0 \Rightarrow f(x)=x+1$

Why f(y+6)=f(y)+6 ? And why f(0)=1?

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11281 posts

#11 h Apr 29, 2025, 5:32 AM • 1 Y

Y by Blackbeam999

Blackbeam999 wrote:

mathslife wrote:

I was busy before

For f:Z->Z, note that f(y+6)=f(y)+6 and f(0)=1; f(1)=2;...;f(5)=6 deduce f(x)=x+1 for all x $\in Z$
For f:R->R.
*First, we have $f$ is bejective and $f(fx))=2f(x)-x$ (easy to prove)
*Next, with $f(-2)=-1$ , chose $y=-2$ we have $f(f(x)-x)=2(f(x)-x)$ .
Each $x \in R$ , exist $z$ such as $f(z)=f(x)-x \Rightarrow f(f(z))=2f(z) \Rightarrow z=0 \Rightarrow f(x)=x+1$

Why f(y+6)=f(y)+6 ? And why f(0)=1?

From $f(1)=2$ :
$P(1,x)\Rightarrow f(f(x)+2)=x+4$
so $f(x+6)=f(f(f(x)+2)+2)=f(x)+6$ .

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#12 h Apr 29, 2025, 5:51 AM

Y by

jasperE3 wrote:

Blackbeam999 wrote:

mathslife wrote:

I was busy before

For f:Z->Z, note that f(y+6)=f(y)+6 and f(0)=1; f(1)=2;...;f(5)=6 deduce f(x)=x+1 for all x $\in Z$
For f:R->R.
*First, we have $f$ is bejective and $f(fx))=2f(x)-x$ (easy to prove)
*Next, with $f(-2)=-1$ , chose $y=-2$ we have $f(f(x)-x)=2(f(x)-x)$ .
Each $x \in R$ , exist $z$ such as $f(z)=f(x)-x \Rightarrow f(f(z))=2f(z) \Rightarrow z=0 \Rightarrow f(x)=x+1$

Why f(y+6)=f(y)+6 ? And why f(0)=1?

From $f(1)=2$ :
$P(1,x)\Rightarrow f(f(x)+2)=x+4$
so $f(x+6)=f(f(f(x)+2)+2)=f(x)+6$ .

How can I find f(1) sorry for asking too much

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114 posts

#13 h Apr 29, 2025, 7:35 AM • 1 Y

Y by Blackbeam999

Blackbeam999 wrote:

How can I find f(1) sorry for asking too much

From $P(1,y):f(f(y)+f(1))=2f(1)+y$ , $f$ is surjective.
Therefore, $\exists c \in \mathbb{Z}$ s.t. $f(c)=0$ and $\exists d \in \mathbb{Z}$ s.t. $f(d)=1$ .
From $P(c,d): f(cf(d)+f(c))=2f(c)+cd$ , we get $0=cd$ .
From $P(d,c): f(df(c)+f(d))=2f(d)+cd$ , we get $f(1)=2+cd=2$ .

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