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#1 h Jul 7, 2020, 3:40 PM • 1 Y

Y by alexey_phenichniy

f:R+--->R+:
f(x^3+f(y))=y+(f(x))^3

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#2 h Jul 7, 2020, 4:06 PM

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johnlp1234 wrote:

$f:R+--->R+$ :
$f(x^3+f(y))=y+(f(x))^3$

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#3 h Jul 7, 2020, 5:03 PM • 3 Y

Y by abhradeep12, CatsMeow12, Wisphard

johnlp1234 wrote:

f:R+--->R+:
f(x^3+f(y))=y+(f(x))^3

Can you please say what is the question,please dont post incomplete problems
please read this post's point number c Rules

This post has been edited 3 times. Last edited by Aritra12, Jul 7, 2020, 5:06 PM

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#4 h Jul 7, 2020, 6:04 PM • 1 Y

Y by abhradeep12

johnlp1234 wrote:

Find all function $f:\mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$ such that :
$f(x^3+f(y))=y+(f(x))^3$

FTFY

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#5 h May 17, 2025, 3:06 AM

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johnlp1234 wrote:

f:R+--->R+:
f(x^3+f(y))=y+(f(x))^3

Let $P(x,y)$ be the assertion $f\left(x^3+f(y)\right)=y+f(x)^3$ .
$P(f(x),y^3+f(z))\Rightarrow f\left(f(x)^3+y+f(z)^3\right)=f(f(x))^3+y^3+f(z)$
Swapping $x,z$ , we get that $f(f(x))^3=f(x)+c$ for all $x>0$ where $c\in\mathbb R$ is some constant, that is, $f(x)=\sqrt[3]{x+c}$ for all $x\in f(\mathbb R^+)$ .
$P(1,x)\Rightarrow f(1+f(x))=x+f(1)^3$ so $\left(f(1)^3,\infty\right)\subseteq f(\mathbb R^+)$ and $f(x)=\sqrt[3]{x+c}$ for all $x>f(1)^3$ .

Let $u>\max\left\{f(1)^3,8647\right\}$ be sufficiently large, then $u^3+f(x)>f(1)^3$ for all $x>0$ and $u>f(1)^3$ , so:
$P(u,x)\Rightarrow\sqrt[3]{u^3+f(x)+c}=x+u+c$
So $f$ is a cubic polynomials, however since $f(x)=\sqrt[3]{x+c}$ for all $x>f(1)^3$ it cannot be a cubic polynomial. Hence no solutions.

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#6 h May 17, 2025, 9:11 AM

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But $f(x)=x$ clearly satisfies...

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#7 h May 17, 2025, 9:34 AM

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Let $P(x,y)$ denote the assertion
Fixing $x$ and varying $y$ we see that $f$ is surjective
Let $f(a)=f(b)$ for some $a$ , $b$ then
$P(x,a)-P(x,b) \implies f(x^3+f(a))-f(x^3-f(b))=a-b \implies a=b$ thus $f$ is bijective
To be continued
Ι found $f(f(x))=x \: \forall \: x \in \mathbb{R_+}$ using the substitution jasper chose above. Too lazy to writeup.

This post has been edited 1 time. Last edited by GreekIdiot, May 17, 2025, 9:49 AM

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#8 h May 17, 2025, 4:41 PM

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surjective in R+ or for values greater than (f(c))^3 ?

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#9 h May 17, 2025, 4:57 PM

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I didnt use surjectivity anyways so it shouldnt matter...
$f$ is non constant thus picking $x$ to minimize $f(x)$ we get that $f$ is surjective on interval $(min^3\{f\}, \infty)$

This post has been edited 1 time. Last edited by GreekIdiot, May 17, 2025, 4:57 PM

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#10 h May 17, 2025, 4:58 PM

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Actually $f$ is surjective because it's an involution.

Should I post about this FE here instead of the identical https://artofproblemsolving.com/community/c6h2179422?

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#11 h May 17, 2025, 5:08 PM

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Hi. $P(f(x), y^3+ f(z)) : f(f(x)^3 + z + f(y)^3) = f(f(x))^3 + y^3 + f(z).$
And therefore by switching $x, y$ and comparing, $f(f(x))^3 = x^3 + c.$ $P(f(x), y) : f(f(x)^3 + f(y)) = y + x^3 + c$
$P(x, f(y)) : f(x^3 + f(f(y))) = f(y) + f(x)^3$
which implies $f(x^3 + \sqrt[3]{y^3 + c}) = f(y) + f(x)^3$
Take function of both sides, and substitute into $P(f(x), y)$ , get $f(f(x^3 + \sqrt[3]{y^3 + c})) = y + x^3 + c$
Cube both sides and apply some stuff, now we have $(x^3 + \sqrt[3]{y^3 + c})^3 + c = (x^3 + y + c)^3$ for all $x, y$ . This is equivalent to $(x^3 + y + c)^3 - (x^3 + \sqrt[3]{y^3 + c})^3 = c$ . Fix a $y$ and let $a = y + c$ and $b = \sqrt[3]{y^3 + c}$ . Both $a$ and $b$ are positive. Then we have $(x^3 + a)^3 - (x^3 + b)^3 = c$ . When we factor, we get $(a - b)((x^3 + a)^2 + (x^3 + a)(x^3 + b) + (x^3 + b)^2) = c.$ Notice that, when increasing $x$ , the LHS gets arbitrarily large (and therefore not constant) if $a - b > 0$ , and gets arbitrarily small (and therefore not constant) if $a - b < 0$ . The only possibility is $a - b = 0$ and therefore $c = 0$ . Therefore, $f(f(x))^3 = x^3 \implies f(f(x)) = x.$ This implies $f$ is bijective. If we take $t=x^3 + f(y)$ , then $f(t) > y = f(f(y))$ . Now $t$ can be anything greater than $f(y)$ . Because $f$ is surjective, $f(y)$ can be anything and $f$ is strictly increasing.

If, for some $x$ we have $f(x) < x$ , then $x = f(f(x)) > f(x)$ contradicting the fact that $f$ is strictly increasing. If, for some $x$ we have $f(x) > x$ , then $f(x) > f(f(x)) = x$ which also contradicts $f$ strictly increasing. Therefore $f(x) = x$ for all $x$ , and it works.

This post has been edited 7 times. Last edited by maromex, May 17, 2025, 8:16 PM
Reason: Done!

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#12 h May 17, 2025, 5:14 PM

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maromex wrote:

Actually $f$ is surjective because it's an involution.

Yeah you are right I forgot I proved that $f(f(x))=x$ :blush:

:blush:

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