Cubic function from Olymon

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Cubic function from Olymon

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Source: Olymon Volume 11 2010 663

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#1 h Mar 21, 2025, 5:25 AM

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Find all $f:\mathbb{R}\rightarrow\mathbb{R}$ such that
$x^2y^2(f(x+y)-f(x)-f(y))=3(x+y)f(x)f(y)$ $\forall$ $x,y \in \mathbb{R}$

This post has been edited 2 times. Last edited by Adywastaken, Mar 21, 2025, 5:35 AM

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#2 h Mar 21, 2025, 6:40 AM

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Let $S(x,y)$ be the set of all substitution made on $(E): x^2 y^2(f(x+y)-f(x)-f(y))=3(x+y)f(x)f(y)$

Consider $S(x\ne 0,0): \implies 3xf(x)f(0)=0$ hence we have $f(0)=0$ or $f(x)=0$ for all real number $x\ne 0$ .

Case 1: If $f(0)\ne 0$ then $f(x) =0$ for all non zero real number then by setting $S(x,-x): x^4\cdot f(0)=0\implies f(0)=0$ which is a contradiction.

Case 2: if $f(0)=0$ we get by setting $S(x,-x): f(-x)=-f(x)$ meaning that $f$ is odd.
By plugging $S(x+y,-y)$ such that $x,y\ne 0$ we get :
$(x+y)^2y2\left(f(x)+f(y)-f(x+y)\right)=-3xf(x+y)f(x)$ $\iff (x+y)^2y^2\left(-\frac{3(x+y)f(x)f(y)}{x^2y^2}\right)=-3xf(x+y)f(y)$ $\iff \frac{(x+y)^3}{x^2}\cdot f(x)f(y)=xf(x+y)f(y)$ $\iff (x+y)^3f(x)f(y)=x^3f(x+y)f(y)$ Swapping $x$ and $y$ we get that :
$x^3f(x+y)f(y)=y^3f(y+x)f(x)\iff f(x+y)\left(y^3f(x)-x^3f(y)\right)=0$ For $y=1$ we get : $f(x+1)(f(x)-x^3f(1))=0$ hence for all real number $x\ne 0$ we have $f(x+1)=0$ or $f(x)=x^3f(1)$ i.e we have $f(x)=0$ or $f(x)=x^3f(1)$ .
If $\exists a,b\ne0$ such that $f(a)=a^3f(1)$ and $f(b)=0$ we get by setting $S(a,b)$ that : $a^2b^2(f(a+b)-a^3f(1))=0\implies f(a+b)=a^3f(1)$ Having that $f(a+b)$ is either $0$ or $(a+b)^3f(1)$ , to not contradict that $b\ne 0$ we have to have $f(1)=0$ hence $f(a)=0$ .

Conclusion: we have to function to verify $f\equiv 0$ and $f\equiv c\cdot x^3$ for some constant $c$ .

Verification:
1)- Suppose that $f\equiv 0$ : this is a valid solution which is ez to verify

2)- Suppose that $f\equiv c\cdot x^3$ for some constant $c$ . Hence :
$LHS=c\cdot x^2y^2((x+y)^3-x^3-y^3)=c\cdot (x+y) (x^2+y^2+2xy-x^2-y^2+xy)=3c\cdot (x+y)xy$
While $RHS=3(x+y)c\cdot x^3 \cdot c\cdot y^3=3c^2(x+y)xy$
The equility $LHS=RHS$ holds if and only if $c\in\{0,1\}$

Hence $f\equiv 0$ and $f\equiv x^3$ is the only solution to $(E)$

This post has been edited 1 time. Last edited by Ahmed_mallek, Mar 21, 2025, 7:18 AM
Reason: Made a mistake while swapping x and y that s.o made me aware of thank you

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#3 h Mar 21, 2025, 6:58 AM

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Ahmed_mallek wrote:

$\iff (x+y)^2y^2\left(-\frac{3(x+y)f(x)f(y)}{x^2y^2}\right)=-3xf(x+y)\color{red}{f(x)}$

You actually have $-3xf(x+y)\color{red}f(y)\color{black}\xRightarrow{\hspace{25pt}}x^3f(y)-y^3f(x)=0\xRightarrow{\hspace{25pt}}\boxed{f\equiv 0,x^3}$ , which are the solutions.

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#4 h Mar 21, 2025, 7:08 AM

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Yeahh sorry didnt pay much attention i knew sth is wrong but i could find it
Anyways i'll edit it

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#5 h Mar 21, 2025, 11:46 AM

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Let $P(x,y)$ denote the assertion.
$P(x,0) \implies 3xf(x)f(0)=0 \implies f(0)=0$ or $f(x)=0 \forall x \in \mathbb{R}/{0}$
If $f(x)=0 \forall x \in \mathbb{R}/{0}$ , then $P(x,-x)\implies f(0)=0$ too
So, $P(x,-x) \implies f$ is odd
$P(x,x) \implies x^4(f(2x)-2f(x))=6xf(x)^2$
Let $f(x)=u, f(2x)=v$
Rearranging, $v=\frac{6u^2}{x^3}+2u$
$P(2x,-x) \implies 4x^4(f(x)-f(2x)-f(-x))=3xf(2x)f(-x)$
Rearranging, $v=\frac{8ux^3}{4x^3-3u}$
Thus, $\frac{8ux^3}{4x^3-3u}=\frac{6u^2}{x^3}+2u$
$8x^6=24ux^3-18u^2+8x^6-6ux^3$
$u^2-ux^3=0$
So, $f(x)(f(x)-x^3)=0$
Hence, $f\equiv0$ or $f\equiv x^3$

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