IZHO 2017 Functional equations

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user01

59 posts

user01

#1 h Jan 14, 2017, 8:38 AM • 4 Y

Y by Davi-8191, muradmurad, Adventure10, Mango247

Find all functions $f:R \rightarrow R$ such that $(x+y^2)f(yf(x))=xyf(y^2+f(x))$ , where $x,y \in \mathbb{R}$

This post has been edited 2 times. Last edited by user01, Jan 14, 2017, 8:39 AM

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kk108

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kk108

#2 h Jan 14, 2017, 9:19 AM • 2 Y

Y by Iiiaaannn16789, Adventure10

Awesome problem .....!

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MMike

105 posts

MMike

#3 h Jan 14, 2017, 9:21 AM • 3 Y

Y by Brilli, Nothing000, Adventure10

My solution.
Let $P(x,y)$ be the assertion of $(x+y^2)f(yf(x))=xyf(y^2+f(x))$ (1)
$P(x,0): f(0)=0$
$P(x,1): (x+1)f(f(x))=x(f(f(x)+1)$ . From there, we can see that $f$ is injection.
$P(-y^2,y): -y^3f(y^2+f(-y^2))=0=f(0)$ . Then, $f(-y^2)=-y^2 or f(x)=x$ $\forall x \leq0$ (2)
If $f(x)=0$ $\forall x \in \mathbb{R}$ . We can see that $\boxed{f(x)=0}$ is a solution.
If exist $x_0 \in \mathbb{R}$ such that $f(x_0)\neq 0$ , then $P(x_0,-y)$ and regard $P(x_0,y)$ , we have $f(x)=-f(-x)$ $\forall x \in \mathbb{R}$ (3)
(2)(3), we have $\boxed{f(x)=x}$ $\forall x \in \mathbb{R}$

This post has been edited 1 time. Last edited by MMike, Jan 14, 2017, 9:23 AM
Reason: a

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tarzanjunior

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tarzanjunior

#4 h Jan 14, 2017, 9:23 AM • 2 Y

Y by Adventure10, Mango247

MMike wrote:

$P(x,1): (x+1)f(f(x))=x(f(f(x)+1)$ . From there, we can see that $f$ is injection.

How? Can you explain?

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MMike

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MMike

#5 h Jan 14, 2017, 9:28 AM • 2 Y

Y by Adventure10, Mango247

If $f(x_1)=f(x_2)$ , then $\frac{f(f(x_1)+1)}{f(f(x_1))}=\frac{f(f(x_2)+1)}{f(f(x_2))}$ , then $1+ \frac{1}{x_1}=1+ \frac{1}{x_2} \rightarrow x_1=x_2$

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Garfield

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Garfield

#6 h Jan 14, 2017, 9:49 AM • 3 Y

Y by Adventure10, Mango247, DEKT

If $f(x)=const$ we have $f(x)=0$ .Now we handle the case $f(x)\neq const$ .
We have $f$ is injective,because $f(x_{1})=f(x_{2})$ we have:
$\frac{x_{1}+y^{2}}{x_{1}y}=\frac{x_{2}+y^{2}}{x_{2}y}$ so $y^{2}(x_{1}-x_{2})=0$ so $x_{1}=x_{2}$ .
now $P(\frac{y^{2}}{y-1},y)$ so we have $f(yf(x))=f(y^{2}+f(x))$ so $yf(x)=y^{2}+f(x)$ so
$f(\frac{y^{2}}{y-1})=\frac{y^{2}}{y-1}$ now set set $\frac{y^{2}}{y-1}=z$ so setting $y_{1}=\frac{-z+\sqrt{z^{2}+4z}}{2}$ we have $f(z)=z$ for all $z$ except $-4<z<0$ .Now we will prove $f(x)=x$ $\forall x\in R$ .
$P(x,y)$ where $-4<x<0$ , we could pick large enough $|y|$ such that both $yf(x)$ and $y^{2}+f(x)$ are not in interval $(-4,0)$ so $f(yf(x))=yf(x)$ and $f(y^{2}+f(x))=y^{2}+f(x)$ so we have:
$y^{3}f(x)=y^{3}x$ so $f(x)=x$ $\forall x\in R$ .

This post has been edited 1 time. Last edited by Garfield, Jan 14, 2017, 9:50 AM

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tenplusten

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tenplusten

#7 h Jan 14, 2017, 9:50 AM • 3 Y

Y by Omeredip, Adventure10, Mango247

$\text{My Solution}$

Let $P(x,y)$ be the assertion of this equation
$P(0,0)$ $\implies$ $f(0)=0$
So there exists $a\in R$ s.t $f(a)=0$ Two cases exist
$i)$ $f(y^2)=0$ for all $y$
So $P(4,2)$ $\implies$ $f(4)=4$
$P(4,\frac{x}{4})$ and $P(4,\frac{-x}{4})$ $\implies$ $f$ is odd.
So $f(-y^2)=-f(y^2)=0$
So $f(x)=0$ for all $x\in R$
$ii)$ There exist $x_0\neq 0$ such that $f((x_0)^2)\neq 0$
$P(x_0,a)$ $\implies$ $a\cdot x_0\cdot f((x_0)^2)=0$ so $a=0$
We got $f(a)=0$ $\iff$ $a=0$
So $P(-x^2,x)$ $\implies$ $f(-x^2)=-x^2$ .So for all nonnegative $x$
$f(x)=x$ .
Now let's select $y$ -positive and $x$ -negative $P(x,y)$ $\implies$
$f(y^2+x)=y^2+x$
For all $x$ we can find $y$ such that $y^2+x$ is positive and gets all values in $R^{+}$ $f(x)=x$ for all positive $x$
So the all solutions are $f(x)=x$
and $f(x)=0$ .

This post has been edited 6 times. Last edited by tenplusten, Jan 14, 2017, 2:15 PM

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mgrulz

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mgrulz

#8 h Jan 14, 2017, 10:00 AM • 1 Y

Y by Adventure10

Can somebody please post other problems from day 1?

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baysa

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baysa

#9 h Jan 14, 2017, 10:48 AM • 3 Y

Y by mgrulz, Adventure10, Mango247

see http://matol.kz/olympiads/499

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mgrulz

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mgrulz

#10 h Jan 14, 2017, 10:59 AM • 2 Y

Y by Adventure10, Mango247

Thanks!!

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TJK1

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TJK1

#11 h Jan 14, 2017, 11:20 AM • 1 Y

Y by Adventure10

I think some cases are mising.Be carefull.

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Zoom

77 posts

Zoom

#12 h Jan 14, 2017, 11:24 AM • 2 Y

Y by Garfield, Adventure10

I think that all the solutions are wrong. Edit for the correct one is coming soon.
The correct is
$f(x)=0$
$f(x)=x$
$f(-x_0)=\sqrt x_0$ for a fixed $x_0>0$
And all other $f(x)=0$
And $f(-x_0)=-\sqrt x_0$ for a fixed $x_0\ge1$ And all other $f(x)=0$

This post has been edited 2 times. Last edited by Zoom, Jan 14, 2017, 11:30 AM
Reason: X0 can be 1

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aleksam

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aleksam

#13 h Jan 14, 2017, 11:55 AM • 3 Y

Y by Zoom, Adventure10, Mango247

MMike wrote:

If $f(x_1)=f(x_2)$ , then $\frac{f(f(x_1)+1)}{f(f(x_1))}=\frac{f(f(x_2)+1)}{f(f(x_2))}$ , then $1+ \frac{1}{x_1}=1+ \frac{1}{x_2} \rightarrow x_1=x_2$

$f(f(x_1))$ could be zero if I am not wrong?

Garfield wrote:

We have $f$ is injective,because $f(x_{1})=f(x_{2})$ we have:
$\frac{x_{1}+y^{2}}{x_{1}y}=\frac{x_{2}+y^{2}}{x_{2}y}$ so $y^{2}(x_{1}-x_{2})=0$ so $x_{1}=x_{2}$ .

I think this has the same problem, division by zero.

Murad.Aghazade wrote:

Clearly true that $f$ is injective.

Would you mind explaining?

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tenplusten

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tenplusten

#15 h Jan 14, 2017, 12:35 PM • 1 Y

Y by Adventure10

So to get $f$ is injective is impossible?

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Anar24

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Anar24

#16 h Jan 14, 2017, 1:05 PM • 2 Y

Y by Adventure10, Mango247

MMike wrote:

If $f(x_1)=f(x_2)$ , then $\frac{f(f(x_1)+1)}{f(f(x_1))}=\frac{f(f(x_2)+1)}{f(f(x_2))}$ , then $1+ \frac{1}{x_1}=1+ \frac{1}{x_2} \rightarrow x_1=x_2$

but there is one thing:f(f(x)) can be zero.You need to observe this case(because denominator can be zero)

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yayups

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yayups

#43 h Aug 21, 2019, 3:26 AM • 7 Y

Y by aops29, ultralako, bero2005, starchan, Adventure10, Mango247, khina

Let $P(x,y)$ be the given FE. Note that
$P(1,0)\implies f(0)=0.$ Now, suppose there was some $a$ such that $f(a)=0$ and $a\ne 0$ . Then,
$P(a,y)\implies 0=ayf(y^2),$ so $f(x)=0$ for all $x\ge 0$ . Based on this, we'll split into two cases.

Case 1: $f(x)=0\iff x=0$ .

We have
$P(-y^2,y)\implies 0=-y^3f(y^2+f(-y^2))\implies f(-y^2)=-y^2,$ so $f(x)=x$ for all $x\le 0$ . Pick $x,y<0$ such that $y^2+x<0$ . Then, the FE gives
$(x+y^2)f(yx)=xy(y^2+x)\implies f(xy)=xy.$ For any given $b>0$ , we can choose $y=-\epsilon$ and $x=-b/\epsilon$ for small enough $\epsilon$ such that $y^2+x<0$ , so $f(b)=b$ for all $b>0$ as well. So in this case, we see that $f\equiv\mathrm{id}$ .

Case 2: $f(x)=0$ for all $x\ge 0$ .

Suppose there was an $a<0$ such that $f(a)>0$ . Pick $y\ne -\sqrt{-a}$ , $y<0$ . Then,
$P(a,y)\implies (a+y^2)f(yf(a))=ayf(y^2+f(a))=0,$ so $f(yf(a))=0$ . In particular, this means $f$ is $0$ everywhere, except the point $-\sqrt{-a}f(a)$ , so we must have
$a=-\sqrt{-a}f(a)\implies f(a)=\sqrt{-a}.$ Thus, we have the solution $f(x)=0$ for all $x\ne a$ and $f(a)=\sqrt{-a}$ for any fixed negative value of $a$ .

Now, WLOG, suppose that $f(x)\le 0$ for all $x<0$ . Pick any $x,y<0$ . The FE then gives
$xyf(y^2+f(x))=0\implies f(y^2+f(x))=0.$ This implies $f(n)=0$ for all $n>f(x)$ . In particular, we also have $f(y^2+f(a))=0$ for $a<0$ and $y>0$ . Thus, the FE gives
$(a+y^2)f(yf(a))=0.$ So if $f(a)<0$ , then we get $f(x)=0$ for all $x$ except $x=\sqrt{-a}f(a)$ , so again $a=\sqrt{-a}f(a)$ , so $f(a)=-\sqrt{-a}$ and $f(x)=0$ for all other $x$ .

Thus, we see that either $f(x)\equiv x,0$ , or $f$ is $0$ at all but one negative point $a$ where the value is $\pm\sqrt{-a}$ . We check that the solutions $x$ and $0$ work.

We write the other solution as $f(x)=0$ for all $x\ne -a^2$ and $f(-a^2)=a$ for some nonzero value of $a$ . Firstly, if we plug in any $x\ne -a^2$ , then both sides are $0$ , so we only need to verify for the case when $x=-a^2$ . In this case, we see that
$(-a^2+y^2)f(ay)=-a^2yf(y^2+a).$ If $y\ne -a$ , then the left side is $0$ so we have
$yf(y^2+a)=0.$ If $-a^2-a>0$ , then there is a nonzero value of $y$ which makes this fail, so the FE is satisfied if and only if $a^2+a\le 0$ , or $a\in(-\infty,-1]\cup[0,\infty)$ . So our solutions are $f(x)\equiv x,0$ , or $0$ everywhere except $f(-a^2)=a$ for some fixed $a\in(-\infty,-1]\cup[0,\infty)$ . We checked that these work, so we're done.

This post has been edited 1 time. Last edited by yayups, Aug 21, 2019, 6:23 AM

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Mida

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Mida

#44 h Dec 30, 2019, 4:23 PM • 1 Y

Y by Adventure10

I didn't have time to read all the other solutions , but i think this is a new approach :
Let p(x,y) indicate the main equation.
If f(x) = f(y) ( and not equal to 0 ) :
P(x , $\frac{1}{f(x)}$ ) and p(y , $\frac{1}{f(y)}$ ) says that our function is injective.
Now for every pair that satisfies $y^2 - yx +x =0$ by injectivity we have $y^2 - y f(x) + f(x) = 0$ and vice versa. Therefore the roots of these equations are equal, therefore is their sum and product , therefore f(x) = x for every real x ≥4 .
Now p(x ≥4 , y ) : $f(xy) = xy$ therefore $f(z) = z$ for every real z and we're done .
But in the case that f(X) = f(Y ) =0 at least one of them is not 0 , as example X , now p(X,y) : f( $y^2$ ) =0
So f(z) = 0 for every positivie real z . And that
's where i can't handle it anymore. I'd be glad if anyone could finish it .

This post has been edited 3 times. Last edited by Mida, Apr 1, 2020, 2:02 PM
Reason: Latex edit

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Math-wiz

6107 posts

Math-wiz

#45 h Dec 30, 2019, 4:38 PM • 1 Y

Y by Adventure10

user01 wrote:

Find all functions $f:R \rightarrow R$ such that $(x+y^2)f(yf(x))=xyf(y^2+f(x))$ , where $x,y \in \mathbb{R}$

Here is the short summary for the key steps to prove this.
1. $f(0)=0$
2. $f(-x)=-f(x)$ (odd function)
3. $f(a)=f(b)\implies a=b$ (injectivity) if $f(x)\neq 0\forall x$
4. $f(y^2-f(y^2))=0\implies f(x)=x\forall x\geq 0$
5. Using 2, $f(x)=x\forall x$ or $f(x)=0$

This post has been edited 1 time. Last edited by Math-wiz, Dec 30, 2019, 4:38 PM

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itslumi

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itslumi

#46 h May 13, 2020, 5:05 AM • 1 Y

Y by Mango247

How did you show injectivity ?

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Idio-logy

206 posts

Idio-logy

#47 h Aug 17, 2020, 2:48 AM • 2 Y

Y by Nathanisme, MiraclesINmaths

Solution

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EulersTurban

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EulersTurban

#48 h Aug 17, 2020, 3:37 PM

Y by

We insert $x=-y^2$ and we have that:
$f(y^2+f(-y^2))=0$ Now we insert $x=t^2+f(-t^2)$ and we have that:
$(t^2+y^2+f(-t^2))f(0)=(t^2+f(-t^2))yf(y^2)$ we now insert $y=0$ from which we have that:
$(t^2+f(-t^2))f(0)=0$ in either case we have that $f(0)=0$
Now when we return to the original equation, we set $y=1$ , from which we get that $f$ is injective.
Which means that:
$f(-t^2)=-t^2$ so it follows that for all $x \leq 0$ we have that $f(x)=x$ .
Thus we set $y > 0$ and we set $x = \epsilon$ , where $\epsilon$ is the greatest negative real number.
Now we have that:
$f(y^2-\epsilon)=y^2-\epsilon$ so obviously we have that $y^2-\epsilon \geq 0$ , and since it is possible to express every $\mathbb{R}^{+}$ in this manner we have that $f(x)=x$ for all $x > 0$ .

Thus the only answer is $f(x)=x$

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Olympikus

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Olympikus

#49 h Aug 17, 2020, 4:04 PM

Y by

@above there is no greatest negative real number, because the Supremum of the negative reals is 0, but you can correct this by fixing some $t$ and plug in $y=t+\delta$ and $x=\epsilon<0$ such that $(t+\delta)^2+\epsilon=t$ .

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jj_ca888

2726 posts

jj_ca888

#50 h Dec 28, 2020, 11:27 PM • 1 Y

Y by kawazmlekiem

Let $P(x, y)$ denote the assertion. $P(1, 0)$ yields $f(0) = 0$ . Also, $P(-x, \sqrt{x})$ for positive reals $x$ yields $f(x + f(-x)) = 0$ .

If $f$ is injective at $0$ , then we get $f(-x) = -x$ hence $f(x) = x$ for all nonpositive $x$ . Next, we will consider $P(-1, \sqrt{x + 1})$ for $x > 0$ . This gives $xf(-\sqrt{x + 1}) = -\sqrt{x+1}f(x) \implies f(x) = x$ since $f(-\sqrt{x+1}) = -\sqrt{x+1}$ . Hence, $\boxed{f(x) = x \text{ for all } x \in \mathbb{R}}$ is a solution.
If is not injective at $0$ , then suppose where $c \neq 0$ . Then yields hence for all $y > 0$ , we have $f(y^2) = 0$ . Thus, it follows that for all (clearly $f(0) = 0$ ). Next we compare with $P(x, -y)$ :and note that one of and is nonnegative, so is $0$ . So for nonzero $x, y$ , it follows that $f(y^2 + f(x)) = 0$ . Check that if $x = 0$ , then clearly $f(y^2 + f(x)) = 0$ . Hence,for all $x, y \in \mathbb{R}$ .
- If there does not exist $d$ for which $f(d) \neq 0$ , then we arrive at the solution $\boxed{f(x) = 0}$ for all real $x$ .
- If there does exist $d$ , which must be negative, for which $f(d) \neq 0$ , then for any $y \neq \pm \sqrt{-d}$ it follows that $f(yf(d)) = 0.$ Since $f(d)$ is constant, it thus follows that $f(x) = 0$ for all reals $x$ except when $x \in \{-f(d)\sqrt{-d}, f(d)\sqrt{-d}\}$ . Since $f(d) \neq 0$ , it follows that $d \in \{-f(d)\sqrt{-d}, f(d)\sqrt{-d}\} \implies f(d) \in \{\sqrt{-d}, -\sqrt{-d}\}.$ This sceneario yield all solutions of the form $\boxed{f(x)=\begin{cases} 0 &\text{ if } x \neq d \\ \sqrt{-d} &\text{ if } x = d \end{cases}} \text{ and } \boxed{f(x)=\begin{cases} 0 &\text{ if } x \neq d \\ -\sqrt{-d} &\text{ if } x = d \end{cases}}$ for some negative constant $d$ . Testing such solutions by plugging them back into the original assertion $P(x, y)$ actually yields necessarily that $\sqrt{-d} \leq {-d}$ , so $\boxed{d \leq -1}$ is additionally required.

The aforementioned solutions all work upon being plugged back into the original functional equation, so we are done. $\blacksquare$

This post has been edited 3 times. Last edited by jj_ca888, Dec 28, 2020, 11:48 PM

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EpicNumberTheory

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EpicNumberTheory

#51 h Dec 29, 2020, 10:18 AM

Y by

Let the assertion in the given problem be denoted by $P(x,y)$ .
$P(x,0) \implies xf(0) = 0 \implies f(0)=0$ .
Let $t$ be such that $f(t) = 0$ . Let $s$ be any positive integer.
$P(t,\sqrt{s}) \implies \sqrt{s} \cdot t \cdot f(s) = 0 \implies t = 0 \text{ or } f(s) = 0$ .

Case 1: $t=0$
This implies: $f(x) = 0 \iff x = 0$ .
$P(x,\sqrt{s}) \implies (x+s)f(\sqrt{s}f(x)) = \sqrt{s}xf(f(x)+s)$
Let $x = -s$ $\implies \sqrt{s}xf(f(x)+s) = 0 \implies f(x) = -s$ . So $f(r) = r$ $\forall r < 0 \in \mathbb{R}$ .
Let $x,y < 0$ and $y^2 + x < 0$ . $P(x,y) \implies \underbrace{(x+y^2)}_{<0}\underbrace{f(xy)}_{xy > 0}=\underbrace{xy}_{>0}\underbrace{(y^2+x)}_{<0}$ .
$\implies f(xy) = xy$ . Now since $xy$ can be made to be any non-negative real even under the condition $y^2 + x < 0$ we have our first Solution which does indeed work.
$\boxed{\text{S1:} f(x) = x \text{ } \forall x \in \mathbb{R}}$

Case 2: $f(s) =0$
Hence $\forall$ non-negative numbers $s_1$ we have that $f(s_1) = 0$ .

Exact Same proof for this case as @above.
$\boxed{\text{S2: }f(x) = 0 \text{ } \forall x \in \mathbb{R}}$
$\boxed{\text{S3: }f(x) = \sqrt{-r} \text{ for some negative constant r, and } f(x) = 0 \text{ otherwise.}}$
$\boxed{\text{S4: }f(x) = -\sqrt{-r} \text{ for some negative constant r, and } f(x) = 0 \text{ otherwise.}}$

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DottedCaculator

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DottedCaculator

#53 h Sep 30, 2021, 7:10 PM • 2 Y

Y by teomihai, centslordm

Solution

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lazizbek42

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lazizbek42

#54 h Dec 13, 2021, 8:15 AM

Y by

very nice problem.

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naonaoaz

329 posts

naonaoaz

#55 h Jun 2, 2024, 3:09 AM

Y by

Let $P(x,y)$ be the assertion. $P(x,0)$ implies $f(0) = 0.$ Now letting $x = -y^2$ gives
$xyf(y^2+f(x)) = 0 \implies f(-x+f(x)) = 0 \text{ for all $x \le 0$}$ For each $x$ , define $a$ to be $f(x) - x$ . We take $2$ cases:

If all $a = 0$ , then $f(x) = x$ for all $x \le 0$ . Choose any $(x,y)$ such that $x+y^2 < 0$ , so
$f(yf(x)) = xy$ Taking $x \to -\infty$ , then $y$ can be anything and still $x+y^2<0$ . Thus
$f(xy) = xy \implies f(x) = x \text{ for all $x$}$ by taking any negative $y$ .
If there's an $a \neq 0$ , then $P(a,y)$ implies
$0 = ay \cdot f(y^2+f(a)) \implies f(y) = 0 \text{ for all $y \ge 0$}$ since $f(a) = 0$ by the first line. Now, first assume there's some $k$ such that $f(k) > 0$ , so $k<0$ . $P(k,y)$ (for $y<0$ ) implies
$(k+y^2)f(yf(k)) = 0 \implies f(y) = 0 \text{ for all $y \neq -\sqrt{-k} \cdot f(k)$}$ Since $f(k) > 0$ , we must have that the one input of $f$ that's gives a non zero $f(x)$ be $k$ . So
$f(k) = \frac{k}{-\sqrt{-k}} = \sqrt{-k} \implies f(x) = \begin{cases} 0 & x \neq k \\ \sqrt{-k} & x = k \\ \end{cases}$ for some constant $k<0$ . Now assume there's some $k$ such that $f(k)<0$ , so we must have $k<0$ . Take any $y<0$ to get $P(k,y)$ implies
$0 = ky \cdot f(y^2+f(k)) \implies f(y) =0 \text{ for all $y^2>f(k)$}$ This implies, for any nonzero $y$ ,
$(k+y^2)f(yf(k)) = ky \cdot f(y^2+f(k)) = 0 \implies f(y) = 0 \text{ for all $y \neq \sqrt{-k} \cdot f(k)$}$ where we take the positive root since that ensures the one ``bump" in $f$ is at a negative $y$ value (since $f(y) = 0$ for any $y \ge 0$ ). We must have
$f(k) = \frac{k}{\sqrt{-k}} = -\sqrt{-k} \implies f(x) = \begin{cases} 0 & x \neq k \\ -\sqrt{-k} & x = k \\ \end{cases}$ Lastly, if $0> k > -1$ , $P(k,\sqrt{k+\sqrt{-k}})$ gives a contradiction, so we must have $k \le -1$ .
Finally, we can deduce the answer of
$\boxed{f(x) = x, \text{ } f(x) = \begin{cases} 0 & x \neq k \\ \sqrt{-k} & x = k \\ \end{cases}, \text{ and } f(x) = \begin{cases} 0 & x \neq k \\ -\sqrt{-k} & x = k \\ \end{cases}}$ where the middle function is for any real constant $k<0$ and the second one is for any real constant $k \le -1$ . It's easy (but painful) to check that these work.

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Ilikeminecraft

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Ilikeminecraft

#56 h Mar 8, 2025, 11:50 PM

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Pick $y = 0$ and we get $xf(0) = 0.$ Hence, $f(0) = 0.$
We do casework on if $f$ is injective at 0 or not.

If $f$ isn't injective at 0, assume $f(k) = 0.$ Take $a = 0,$ and we get $f(y^2) = 0,$ so $f\equiv 0$ for all nonnegative reals. Clearly, $f\equiv 0$ is a solution. Thus, assume there exists $k$ such that $f(k)\neq 0.$

If $f(k) > 0,$ take $x = k$ and we get that $f\equiv 0$ unless $y = \sqrt{-x},$ and thus, $f(k) = \sqrt{-k}.$

If $f(k) < 0,$ take $x =k, y<0$ and we get $f(y^2 + f(k)) = 0.$ Hence, $f(x) = 0$ for $x > f(k).$ However, take $y > 0$ and we get $f(yf(k)) = 0$ unless $y = -\sqrt{-k}.$ Thus, $f(k) = -\sqrt{-k}$ and is 0 for all other $x.$

If $f$ is injective at 0, let $A, B$ be the set of reals so that $f(A) \geq0, f(B)\leq 0.$ Clearly, $A\cup B = \mathbb R, A\cap B = \{0\}.$ It can be seen that $f$ is odd by comparing $(x, -y), (x, y)$ and then fixing the single edge case by noting that $f$ is injective at 0.

Hence, $A = -B.$ For $x\in B,$ pick $y^2 = -f(x).$ We get $(x - f(x))f(\sqrt{-f(x)}f(x)) = 0.$ Hence, $f(x) = x$ or $f(x) = 0.$ Thus, $f(x) = x$ for $x \in B.$ However, since $A = -B$ and $A\cup B = \mathbb R, A\cap B = \{0\},$ we conclude $f\equiv x.$

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HamstPan38825

8853 posts

HamstPan38825

#57 h Mar 16, 2025, 9:47 PM

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The answers are $f \equiv x$ , $f \equiv 0$ , and for any positive real number $c \geq 1$ , the function $f(x) = 0$ for all $x \neq -c$ and $f(-c) \in \left\{\sqrt c, -\sqrt c\right\}$ .

Verification: The first two functions clearly work. For the last function, observe that when $y = -c$ , both the left and right sides equal $0$ , and when $x = -c$ , $f(yf(x)) \neq 0$ if and only if $y^2 = c$ , in which case the left side is $0$ anyways. Similarly, $y^2+f(x) \geq -\sqrt c \geq -c$ with equality only when $y = 0$ , so the right side will be $0$ too.

Restriction: By setting $y = 0$ we get $f(0) = 0$ . We first have the following claim:

Claim: $f$ is injective, else $f(x) = 0$ for all $x \geq 0$ .

Proof: Suppose first that $f(x) \neq 0$ for all $x \neq 0$ . Then, let $x_1 \neq x_2$ and assume for the sake of contradiction that $f(x_1) = f(x_2)$ . For a sufficiently large value of $y$ , $y^2+f(x_1) > 0$ , so it follows that $\frac 1y + \frac y{x_1} = \frac{f\left(y^2+f(x_1)\right)}{f(yf(x_1))} = \frac 1y + \frac y{x_2}$ or $x_1 = x_2$ . Thus in this case, $f$ is injective.

Suppose that $f(x_1) = 0$ for some $x_1 \neq 0$ . Then setting $x = x_1$ , we must have $f\left(y^2+f(x_1)\right) = f\left(y^2\right) = 0$ for all nonzero $y$ . So $f(y) = 0$ for all $y > 0$ and we have the result. $\blacksquare$

Now we split into cases.

First Case: Assume that $f$ is injective. Then setting $x = -y^2$ yields $f\left(y^2+f\left(-y^2\right)\right) = 0$ , so $f(y) = y$ for all $y < 0$ . Now, for all $x < 0$ , we have $\left(x+y^2\right)f(xy) = xyf\left(x+y^2\right).$ In particular, by letting $y > 0$ we obtain $f\left(x+y^2\right) = x+y^2$ as $xy < 0$ . This deterines $f \equiv x$ on all real numbers.

Second Case: Assume that $f(x) = 0$ for all $x \geq 0$ , $f$ is not the zero function, and that there exists a value $x_0 < 0$ such that $f(x_0) = c > 0$ . Then $f\left(y^2+f(x_0)\right) = 0$ for all real numbers $y$ , hence $f(yc) = 0$ for all $y \neq \sqrt{-x_0}$ , i.e. $f$ is zero on all values but $f\left(c\sqrt{-x_0}\right)$ .

It follows that $c\sqrt{-x_0} = x_0$ holds, hence $x_0 = -c^2$ and $f\left(-c^2\right) = c$ . This yields the first solution, which works only for $c \geq 1$ .

Third Case: Assume that $f(x) = 0$ for all $x \geq 0$ , $f$ is not the zero function, and that there exists a value $x_0 < 0$ such that $f(x_0) = -c < 0$ . Setting $x = x_0$ , we get $\left(x_0+y^2\right) f(-cy) = x_0 y f\left(y^2-c\right).$ Then, for all $y \leq 0$ , $f(-yc) = 0$ , hence by the reverse logic as the above, it follows that $f\left(y^2-c\right) = 0$ for all $y < 0$ , i.e. $f(x) = 0$ for all $x > -c$ .

On the other hand, fix some $y \geq \sqrt c$ in this equation. Unless $y = \sqrt{-x_0}$ , $f\left(y^2-c\right) = 0$ , so $f(-cy) = 0$ . Thus, $f(x) = 0$ for all $x \leq -c^{3/2}$ except for $f\left(-c\sqrt{-x_0}\right)$ .

Finally, take $0 < y < \sqrt c$ in this equation. Since $y^2 - c > -c$ , it follows that either $y = \sqrt{-x_0}$ or $f(-cy) = 0$ . In any case, $f(x) = 0$ for all $-c^{3/2} < x < 0$ from this equation. So $f \equiv 0$ except for at $-c\sqrt{-x_0}$ .

By the same logic as above, it follows that $-c\sqrt{-x_0} = x_0$ , which implies $x_0 = -c^2$ and $f\left(-c^2\right) = -c$ . This yields the second solution, which once again only works for $c \geq 1$ .

We have exhausted all cases, so the proof is complete.

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lksb

163 posts

lksb

#58 h Mar 16, 2025, 10:05 PM

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$y=0$ : $xf(0)=0\iff \boxed{f(0)=0}$ $x=-y^2$ : $0=-y^3f(y^2+f(-y^2))$ If $f(-y^2)\neq -y^2$ , then $f$ has an infinite amount of roots, therefore, $\boxed{f(x)=0\quad \forall x\in\mathbb{R}}$ or $\boxed{f(x)=x\quad \forall x\in\mathbb{R}_{\leq 0}}$
By choosing $x<0$ and $y>0$ with $x+y^2>0$ : $(x+y^2)xy=xyf(x+y^2)\iff f(x+y^2)=x+y^2\iff \boxed{f(x)=x\quad \forall x\in \mathbb{R}_{\geq0}}$ With so, $\boxed{f(x)=0\quad\forall x\in\mathbb{R}}$ and $\boxed{f(x)=x\quad\forall x\in\mathbb{R}}$

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