Is this FE solvable?

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Is this FE solvable?

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#1 h Apr 1, 2025, 6:58 PM

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Find all $f:\mathbb{R} \rightarrow \mathbb{R}$ such that $f(2x+y) + f(x+f(2y)) = f(x)f(y) - xy$ for all reals $x$ and $y$ .

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#2 h Apr 2, 2025, 12:56 PM • 3 Y

Y by HuongToiVMO, Mathdreams, Sedro

Mathdreams wrote:

Find all $f:\mathbb{R} \rightarrow \mathbb{R}$ such that $f(2x+y) + f(x+f(2y)) = f(x)f(y) - xy$ for all reals $x$ and $y$ .

Let $P(x,y)$ be the assertion $f(2x+y)+f(x+f(2y))=f(x)f(y)-xy$
Let $a=f(0)$

1) $a\ne 1$
Proof

2) $f(x)$ is injective
Proof

3) $a=3$ and $f(-3)=0$
Proof

4) $\boxed{f(x)=x+3\quad\forall x}$
Proof

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#3 h Apr 2, 2025, 2:55 PM

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That is the answer I designed this problem around, but I didn't know if it was solvable.

Thanks!

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#4 h Apr 3, 2025, 1:20 AM

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Mathdreams wrote:

Find all $f:\mathbb{R} \rightarrow \mathbb{R}$ such that $f(2x+y) + f(x+f(2y)) = f(x)f(y) - xy$ for all reals $x$ and $y$ .

Let $P(x,y)$ be the assertion $f(2x+y)+f(x+f(2y))=f(x)f(y)-xy$ , and let $c=f(0)$ .

Claim 1: $f(-c)=0$
$\begin{align*} P(0,0)&\Rightarrow f(c)=c^2-c\\ P(0,-c)&\Rightarrow f(f(-2c))=(c-1)f(-c)\\ P(c,-c)&\Rightarrow f(c+f(-2c))=(c^2-c)f(-c)+c\\ P(f(-2c),0)&\Rightarrow f(2f(-2c))+f(c+f(-2c))=(c^2-c)f(-c)\Rightarrow f(2f(-2c))=-c\\ P(0,f(-2c))&\Rightarrow c(c-2)f(-c)=0 \end{align*}$ If $cf(-c)=0$ we get $f(-c)=0$ , so from now on for this claim suppose $c=2$ . We have:
$\begin{align*} P(0,0)&\Rightarrow f(2)=2\\ P(0,1)&\Rightarrow f(1)=2\\ P(-1,1)&\Rightarrow f(-1)=1\\ P(-1,0)&\Rightarrow f(-2)=0 \end{align*}$ so in this case $f(-c)=0$ as well.

Claim 2: $c=3$
Suppose $c=0$ , in this case we have:
$\begin{align*} P(x,0)&\Rightarrow f(2x)=-f(x)\Rightarrow f(4x)=-f(2x)=f(x)\\ P(0,x)&\Rightarrow f(x)+f(f(2x))=0 \end{align*}$ Comparing these we get $f(f(2x))=f(2x)$ , or $f(f(x))=f(x)$ . Using these properties, we have:
$\begin{align*} P(x,2f(y))&\Rightarrow f(2x+2f(y))+f(x+f(4f(y)))=f(x)f(2f(y))-2xf(y)\\ &\Rightarrow-f(x+f(y))+f(x+f(f(y)))=-f(x)f(f(y))-2xf(y)\\ &\Rightarrow-f(x+f(y))+f(x+f(y))=-f(x)f(y)-2xf(y)\\ &\Rightarrow(f(x)+2x)f(y)=0 \end{align*}$ So either $f(x)=-2x\forall x$ or $f(x)=0\forall x$ , neither of which are solutions, which forces $c\ne0$ .

$P\left(0,-\frac c2\right)\Rightarrow (c-1)f\left(-\frac c2\right)=c$
If $c=1$ then $c=0$ , impossible, so $f\left(-\frac c2\right)=\frac c{c-1}$ .
$P\left(-\frac c2,0\right)\Rightarrow f\left(\frac c2\right)=\frac{c^2}{c-1}$
$P\left(\frac c2,-\frac c2\right)\Rightarrow\frac{2c^2}{c-1}=\frac{c^3}{(c-1)^2}+\frac{c^2}4$
So after simplifying, we get $c=3$ .

Finish:
Note $f(-3)=f(-c)=0$ . Then we calculate:
$\begin{align*} P(0,0)&\Rightarrow f(3)=6\\ P(3,0)&\Rightarrow f(6)=9\\ P(-3,0)&\Rightarrow f(-6)=-3 \end{align*}$ So comparing:
$\begin{align*} P(x+3,-3)&\Rightarrow f(2x+3)+f(x)=3x+6\\ P(x,3)&\Rightarrow f(2x+3)+f(x+12)=6f(x)-3x \end{align*}$ we get $f(x+12)=7f(x)-6x-6$ .

Using this, we have:
$f(2x+27)=7f(2x+15)-12x-96=49f(2x+3)-96x-264$ and so:
$\begin{align*} P(x+3,-3)&\Rightarrow f(2x+3)+f(x)=3x+9\\ P(x+15,-3)&\Rightarrow f(2x+27)+f(x+12)=3x+45\\ &\Rightarrow49f(2x+3)-96x-264+7f(x)-6x-6=3x+45\\ &\Rightarrow7f(2x+3)+f(x)=15x+45 \end{align*}$ and solving this system in $f(2x+3)$ and $f(x)$ , we get $\boxed{f(x)=x+3}$ which satisfies $P(x,y)$ .

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#5 h Apr 3, 2025, 3:09 AM • 2 Y

Y by alexanderhamilton124, megarnie

I should have saved this FE for a competition huh

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