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Source: Serbian Mathematical Olympiad 2021, P5

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#1 h May 14, 2021, 11:22 AM • 1 Y

Y by HWenslawski

Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that for every $x,y\in\mathbb{R}$ the following equality holds: $f(xf(y)+x^2+y)=f(x)f(y)+xf(x)+f(y).$

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gghx

#2 h May 14, 2021, 11:53 AM • 1 Y

Y by HWenslawski

$P(x,y):f(xf(y)+x^2+y)=f(x)f(y)+xf(x)+f(y)$
$P(0,0): f(0)=0$
$P(x,0): f(x^2)=xf(x)$ . This means $f$ is odd.
$P(-x,-y): f(xf(y)+x^2-y)=f(x)f(y)+xf(x)-f(y)$
Thus, $f(xf(y)+x^2+y)=f(xf(y)+x^2-y)+2f(y)$
As such, $f(x+y)=f(x-y)+2f(y)$ for any $x\ge 0$ (Since the function $x^2+cx$ always covers all non-negative integers [to prove note $x=-c$ gets $0$ and the function is continuous])
Since $f$ is odd, $f(-x^2+y)=-f(x^2-y)=-f(x^2+y)-2f(-y)=f(-x^2-y)+2f(y)$ anyways.
Thus $f(x+y)+f(y-x)=2f(y)$ for any reals $x,y$ . This easily transforms to Jensen with $f(0)=0$
As such $f$ additive and $xf(x)=f(x^2)$ implies $f(x)=xf(1)$ , which always works.

@below oh yah what was I thinking lol

This post has been edited 12 times. Last edited by gghx, May 17, 2021, 11:11 AM

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#3 h May 14, 2021, 12:47 PM

Y by

gghx wrote:

$P(x,-f(x)): f(x)f(-f(x))+xf(x)=0$

I don't think you can get this by plugging $y=-f(x)$ .

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#4 h May 15, 2021, 7:14 AM

Y by

gghx wrote:

@below oh yah what was I thinking lol

Hope this is correct:

$P(x,y):f(xf(y)+x^2+y)=f(x)f(y)+xf(x)+f(y)$
$P(0,0): f(0)=0$
$P(x,0): f(x^2)=xf(x)$ . This means $f$ is odd.
$P(-x,-y): f(xf(y)+x^2-y)=f(x)f(y)+xf(x)-f(y)$
Thus, $f(xf(y)+x^2+y)=f(xf(y)+x^2-y)+2f(y)$
As such, $f(x+y)=f(x-y)+2f(y)$ for any $x\ge 0$ (Since the function $x^2+cx$ always covers all non-negative integers [to prove note $x=-c$ gets $0$ and the function is continuous])
Since $f$ is odd, $f(-x^2+y)=-f(x^2-y)=-f(x^2+y)-2f(-y)=f(-x^2-y)+2f(y)$ anyways.
Thus $f(x+y)+f(y-x)=2f(y)$ for any reals $x,y$ . This easily transforms to Jensen with $f(0)=0$
As such $f$ additive and $xf(x)=f(x^2)$ implies $f(x)=xf(1)$ , which always works.

Can you explain how to transform Jensen and how it implies that function is additive? This is the first time I have seen Jensen in functional equation.

This post has been edited 1 time. Last edited by SerdarBozdag, May 15, 2021, 7:15 AM

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#5 h May 15, 2021, 11:57 AM • 2 Y

Y by SerdarBozdag, Mango247

@above

$f(x+y)+f(x-y)=2f(x)$
Set $x=\frac{a+b}{2},y=\frac{a-b}{2}$ , $f(a)+f(b)=2f(a+b)$
Let $P(x,y): f(x)+f(y)=2f(\frac{x+y}{2})$
$P(x+y,0): f(x+y)=2f(\frac{x+y}{2})$ . Using this in $P(x,y)$ , $f(x+y)=f(x)+f(y)$ so $f$ is additive.

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#6 h May 20, 2021, 1:27 AM • 1 Y

Y by Timmy456

gghx wrote:

@above

$f(x+y)+f(x-y)=2f(x)$
Set $x=\frac{a+b}{2},y=\frac{a-b}{2}$ , $f(a)+f(b)=2f(a+b)$
Let $P(x,y): f(x)+f(y)=2f(\frac{x+y}{2})$
$P(x+y,0): f(x+y)=2f(\frac{x+y}{2})$ . Using this in $P(x,y)$ , $f(x+y)=f(x)+f(y)$ so $f$ is additive.

Just a quick addendum: An alternate way of doing it is by plugging in $x=y$ into
$f(x+y)+f(x-y)=2f(x)$ from which you quickly obtain $f(2x)=2f(x)$ . From there just swap the RHS and set $x+y=a$ and $x-y=b$ .
Also, to clarify, $f(x)=xf(1)$ is directly obtained from $f(x^2)=xf(x)$ by plugging in $x+1$ for $x$ and using additivity.

There is also a second solution to the problem, which doesn't use additivity.

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#7 h May 20, 2021, 1:38 AM

Y by

NZP_IMOCOMP4 wrote:

There is also a second solution to the problem, which doesn't use additivity.

I'm interested.

What is it?

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gvole

#8 h May 28, 2021, 4:20 PM • 1 Y

Y by BorivojeGuzic123

I was so perplexed by the formulation of P4 that I ended up wasting almost 4 hours until I was sure I didn't fake solve, and I ended up not having enough time to figure this one out. Solved it at home with the idea I had on contest in about 20 minutes :blush:

:blush:

Plugging in $y = 0$ yields $f(x^2) = xf(x)$ so $f$ is odd.
Now plugging in $x = 0$ we obtain $f(0) = 0$ .
For $y$ such that $f(y)\neq 0$ , plug in $x = \frac{-y}{f(y)}$ and we figure out $f(\frac{-y}{f(y)}) = 1$ .
We will now prove that this implies $f(y) = cy$ .

Assume $f(a) = f(b) = 1$ .
By plugging in $(x,\ y) = (a,\ b^2)$ and $(x,\ y) = (b,\ a^2)$ we obtain $a = b$ , so $f(y) = cy$ whenever $f(y) \neq 0$

We will now prove $f \equiv 0$ or $f(y) = 0 \implies y = 0$ .
If $f(y) = 0$ we get $f(x^2 + y) = f(x^2)$ $\forall x \in \mathbb{R}$ , and clearly, if $y \neq 0$ this implies $f$ is periodic, but since $f(x) = cx,\ c\neq 0$ whenever $f(x) \neq 0$ this is impossible, proving the claim above.
Now the only possible solutions are $f(x) \equiv cx,\ c \in \mathbb{R}$ , which are easily checked to work.

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#9 h Mar 1, 2022, 8:52 AM

Y by

Any other solution? I think 2 above solutions aren’t correct

This post has been edited 1 time. Last edited by ILOVEMYFAMILY, Mar 1, 2022, 8:53 AM
Reason: Mistake

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#10 h Mar 2, 2022, 1:51 AM

Y by

ILOVEMYFAMILY wrote:

Any other solution? I think 2 above solutions aren’t correct

My solution has been fixed.

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#11 h May 22, 2022, 9:26 AM

Y by

Let $P(x,y)$ be the given assertion, and consider $P=\{p\in \mathbb{R}\;|\;f(x+p)=f(x)\;\forall x\in \mathbb{R}\}$ be the set of all periods of $f.$
Clearly, $P$ is closed under addition and negation, and $0\in P.$ Let $Q=\{u\;|\;f(u)=0\}$ be the set of zeros of $f.$
$P(0,0)$ gives $f(0)=0,$ so $P(x,0)$ gives that $^{(*)}\;\;f(x^{2})=xf(x)\;\forall \;x\in\mathbb{R},$ hence, $f$ is odd.
Now $0\in Q,$ so $P\subseteq Q.$ Moreover, $(*)$ implies $Q$ is closed under squaring and negation, so if $u\in Q$ and $x\ge 0,$ we see that $P(\sqrt{x},u)$ implies $f(x+u)=\sqrt{x}f(\sqrt{x})=f(x)\;\forall \;x\ge 0,$ and if $x<0$ then $P(\sqrt{-x},-u)$ implies (remember $-u\in Q$ and $f$ is odd)
$f(-x-u)=\sqrt{-x}f(\sqrt{-x})=f(-x)\;\Rightarrow f(x+u)=f(x)\;\forall \;x<0$ that is, $u\in P,$ hence $P\equiv Q$ (the periods of $f$ are the zeros, and the set is closed under addition, subtraction, and squaring). Now let's consider two cases:
$Case1.\;\;P\ne \{0\},$ then pick a nonzero $u\in P,$ we get from $(*):$ (again, $2u,u^{2}\in P$ )
$f(1)=f(1+2u+u^{2})=f((1+u)^{2})=(1+u)f(1+u)=f(1)+uf(1)$ so $1\in P$ , now plug $P(-1-f(x),x),$ we get
$f(f(x)+x)=f(1+f(x)+x)=-f(-1-f(x))+f(x)=f(f(x))+f(x)$ but $P(1,x)$ implies $f(f(x)+x)=f(f(x)+1+x)=f(x)$ , so $f(x)\in P\;\forall \;x\in \mathbb{R}.$ Now $P(x-f(x),x)$ gives
$f(-xf(x)+x^{2}+x)=f(x-f(x))x+f(x)=xf(x)+f(x)$ while $P(-x,x)$ implies $f(-xf(x)+x^{2}+x)=-f(x)^{2}+xf(x)+f(x).$ \newline Therefore, $f(x)=0\;\forall \;x\in \mathbb{R},$ which is indeed a solution.
$Case2.\;\;P\equiv \{0\},$ let's prove $f$ is injective. Indeed, let $f(a)=f(b)$ for some $a\ge b,$ then $a-b\ge 0\ge -\frac{f(b)^{2}}{4},$ in particaular, the equation $x^{2}+xf(b)=a-b$ has a solution $x_{0}\in \mathbb{R},$ now plug $P(x_{0},b),$ we get
$f(a)=f(x_{0}f(b)+x_{0}^{2}+b)=f(x_{0})(f(b)+x_{0})+f(b)$ then $f(x_{0})(x_{0}+f(b))=0,$ and so $x_{0}=0$ or $-f(b),$ in both cases, $a-b=x_{0}(x_{0}+f(b))=0,$ hence $f$ is injective.
Now let $x\neq 0$ and plug $P(-\frac{x}{f(x)},x),$ the LHS well cancel with the middle term of the RHS (because of $(*)$ ) and so
$0=f(x)(f(-\frac{x}{f(x)})+1)\;\Rightarrow f(\frac{x}{f(x)})=1\;\forall \;x\neq 0$ and so by injectivity we conclude that $f(x)=cx\;\forall \;x\in \mathbb{R},$ where $c$ is a nonzero constant, we can check that this is indeed a solution. So we are done.
(we can merge the two solutions into one family: $f(x)=kx$ , for some $k\in \mathbb{R}$ .)

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#12 h Jan 11, 2023, 10:38 PM • 1 Y

Y by NZP_IMOCOMP4

1) first, putting $x=y=0$ into the equation we get $f(0)^2=0$ $\Rightarrow$ $f(0)=0$ .

2) now, choosing $y=0$ leads us to $f(x^2)=xf(x)$ for all $x\in \mathbb{R}$ . And, this forces $f$ to be an odd function. Therefore, it would be enough if we could find $f(x)$ for $x\ge 0$ .

3) Let's put $(-x,-y)$ to the original equation: $f(xf(y)+x^2-y) = f(x)f(y)+xf(x)-f(y).$ So $f(x^2+xf(y)-y)+2f(y) = f(x^2+xf(y)+y)$ holds for all $x,y \in \mathbb{R}$ . For any $y>0$ we have $z:=x^2+xf(y)-y = \left(x+\frac{f(y)}{2}\right)^2 - \frac{f(y)^2}{4}-y$ can be any non-negative real number. When $z=0$ (we choose such $x$ ) we obtain $f(2y)=2f(y)$ for all $y\ge 0$ . Thus, for any non-negative real numbers $z$ and $w$ , we get $f(z)+f(w)=f(z+w)$ (just letting $w=2y$ and choosing appropriate $x$ for $z=x^2+xf(y)-y$ ).

Now from $f(x^2)=xf(x)$ , for any $x\ge 0$ we get that $f((x+1)^2) =(x+1)f(x+1) \quad \Rightarrow \quad f(x^2)+2f(x)+f(1)=xf(x)+xf(1)+f(x)+f(1)$ and $f(x)=xf(1)$ .

Since $f$ is odd, we see that $f(x)=xf(1)$ for all $x\in \mathbb{R}$ .

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#13 h Jul 31, 2024, 9:20 PM

Y by

The only solution is $f(x) = cx$ for some real constant $c$ . These can be checked to work. Now we prove they are the only solutions. Let $P(x,y)$ denote the given assertion.

$P(0,0): f(0) = 0$ .

$P(x,0): f(x^2) = xf(x)$ .

Now comparing $x$ and $-x$ for $x\ne 0$ here gives that $f(x) = -f(-x)$ . Combined, with $f(0) = 0$ , we have $f$ is odd.

$P(x,y)$ compared with $P(-x, -y)$ gives that $f(xf(y) + x^2 + y) - f(y) = f(xf(y) + x^2 - y) + f(y)$ . Now, $xf(y) + x^2$ can take any nonnegative real as $x$ varies for a fixed $y$ , so for each real number $y$ , we have $f(x + y) - f(y) = f(x - y) + f(y)$ for all reals $x \ge 0$ , so $f(x+y) - f(x-y) = 2f(y)$ .

Now since $f(x+y) - f(x-y) = 2f(y)$ implies $f((-x) + (-y)) - f((-x) - (-y)) = 2f((-y))$ , so it is true for $-x, -y$ , implying $f(x+y) - f(x-y) = 2f(y)$ for all reals $x,y$ .

Now setting $x = y$ in this equation gives $f(2x) = 2f(x)$ , so $f(x+y) - f(x-y) = f(2y)$ , meaning $f(x-y) + f(2y) = f(x + y)$ , so $f$ is clearly additive.

Now combined with $f(x^2) = xf(x)$ , we have that $f(x) = cx$ for some constant $c$ , as desired (the proof of it is using the fact that $f((x+1)^2) = (x+1) f(x+1)$ and i'm too lazy to write it up again )

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#14 h Mar 24, 2025, 7:01 AM

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Let $P(x,y)$ denote the assertion $f(x^2+f(x)+y)=xf(x)+f(x+y)$ .

Claim 1: $f(0)=0$
$P(0,x)\Rightarrow f(x+f(0))=f(x)\Rightarrow f(f(0))=f(0)=f(-f(0))$
$P(f(0),x)-P(-f(0),x)\Rightarrow f(0)=0$

Claim 2: If $f(x+a)=f(x)$ for all $x$ , where $a$ is some nonzero constant, then $f(x)=0$ .
$P(x+a,y-x^2-f(x))-P(x,y-x^2-f(x))\Rightarrow f(2xa+y)=f(y)+af(x)\Rightarrow f(2xa)=af(x)$
Setting $x=\frac12$ and using the fact that $f(a)=0$ , we have $f\left(\frac12\right)=0$ . Then $f(2xa+y)=f(y)+f(2xa)$ , so $f$ is additive.
$P(x+y,0)-P(x,0)-P(y,0)\Rightarrow f(2xy)=xf(y)+yf(x)$ , setting $y=\frac12$ we obtain $\boxed{f(x)=0}$ , which fits.

Assume that $f$ is not identically zero, so $f(x+a)=f(x)$ implies $a=0$ .
Assume $f(k)=0$ for some $k$ .
$P(k,x-k)\Rightarrow f(x+k^2-k)=f(x)\Rightarrow k\in\{0,1\}$
$P(-1,0)\Rightarrow f(f(-1)+1)=0\Rightarrow f(-1)\in\{-1,0\}$

If $f(-1)=0$ , then $P(-1,x+1)\Rightarrow f(x+2)=f(x)$ , contradiction.

Now assume $f(-1)=-1$ :
$P(-1,x+1)\Rightarrow f(x+1)=f(x)+1$
$P(x+1,y-x^2-f(x))-P(x,y-x^2-f(x))\Rightarrow f(2x+y)=f(x)+x+f(y)$ , setting $y=-x$ gives the solution $\boxed{f(x)=x}$ , which is a solution.

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