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#1 h Jun 25, 2020, 6:15 PM • 42 Y

Y by Aryan-23, XbenX, Com10atorics, Functional_equation, plagueis, Math-wiz, Imayormaynotknowcalculus, Desargues123, math_pi_rate, bobjoe123, AmirKhusrau, bjh0411, Atpar, Granville-Austin, Bunrong123, Bumblebee60, mijail, amar_04, Valmir, Ard_11103, Kar-98k, srijonrick, Mr.C, Wizard0001, Pluto1708, OlyMan, Abhaysingh2003, Eliot, Bassiskicking, hsiangshen, TheBarioBario, Supercali, Nathanisme, OlympusHero, magicarrow, Abidabi, Euler1728, jasperE3, tigerzhang, Sedro, EpicBird08, straight

Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that for all irrational numbers $x, y$ and $z$ ,
$f(x+y)f(z)=f(xz)+f(yz)$
Some stories about this problem. This problem it is proposed by me (Dorlir Ahmeti) and Valmir Krasniqi. We did proposed this problem for IMO twice, on 2018 and on 2019 from Kosovo. None of these years it wasn't accepted and I was very surprised that it wasn't selected at least for shortlist since I think it has a very good potential. Anyway I hope you will like the problem and you are welcomed to give your thoughts about the problem if it did worth to put on shortlist or not.

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Functional_equation

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Functional_equation

#2 h Jun 25, 2020, 6:58 PM

Y by

The question seems very difficult (it looks beautiful)

. I wonder why they did not accept this problem?

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#3 h Jun 25, 2020, 7:03 PM • 2 Y

Y by Eliot, OlympusHero

Functional_equation wrote:

The question seems very difficult (it looks beautiful)

. I wonder why they did not accept this problem?

Come on, as long as great and hard problems like 2019\1 full of insights and new ideas are out there we don't need any other problem.

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#4 h Jun 25, 2020, 7:08 PM • 2 Y

Y by matinyousefi, Nathanisme

matinyousefi wrote:

Functional_equation wrote:

The question seems very difficult (it looks beautiful)

. I wonder why they did not accept this problem?

Come on, as long as great and hard problems like 2019\1 full of insights and new ideas are out there we don't need any other problem.

I believe this comment is said ironically.

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#5 h Jun 25, 2020, 7:18 PM

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how hard is it supposed to be? (imo sl difficulty)

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#6 h Jun 25, 2020, 7:25 PM • 1 Y

Y by Nathanisme

anyone__42 wrote:

how hard is it supposed to be? (imo sl difficulty)

I believe it should be like A6 or A7. Hard to say but I think it is for hard problem this is just my opinion.

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Functional_equation

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Functional_equation

#7 h Jun 25, 2020, 7:39 PM

Y by

Sorry

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Imayormaynotknowcalculus

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#8 h Jun 25, 2020, 7:40 PM • 2 Y

Y by Mango247, Mango247

@above, $x,y,z\not\in\mathbb{Q}$ .

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#10 h Jun 25, 2020, 9:57 PM • 6 Y

Y by mijail, AmirKhusrau, Imayormaynotknowcalculus, amar_04, Nathanisme, OlympusHero

We claim that the only answers are $f \equiv 2$ , $f=\text{id}$ and $f(x)=0$ for $x \neq 0$ , with $f(0)$ any random value. One can easily check that these work, so now we show that these are the only ones. Let $\mathbb{I}$ represent the set of irrational numbers, and suppose $P(x,y,z)$ is the given assertion. Throughout the solution, $x,y,z$ are irrationals, while $q,r$ represent a non-zero rational number. Also we'll repeatedly use the fact that if $x$ is irrational, then $\frac{1}{x}$ and $rx$ is also irrational.

First assume $f(0) \neq 0$ . Note that $P \left(\frac{1}{x},\frac{-1}{x},x \right) \Rightarrow f(0)f(x)=f(1)+f(-1) \Rightarrow f(x)=\frac{f(1)+f(-1)}{f(0)}$ This is true for all $x \in \mathbb{I}$ . Then $P(x,-x,x) \Rightarrow f(1)+f(-1)=\frac{2}{f(0)}(f(1)+f(-1)) \Rightarrow f(1)+f(-1)=0 \text{ or } f(0)=2$ Also, $P \left(x,x,\frac{r}{x} \right) \Rightarrow 2f(r)=\left(\frac{f(1)+f(-1)}{f(0)} \right)^2$ Taking $r=\pm 1$ above gives $f(1)=f(-1)$ . If $f(1)+f(-1)=0$ (i.e. $f(1)=f(-1)=0$ ), then the above two relations directly give one of the solution sets. Else if $f(0)=2$ (and $f(1)=f(-1) \neq 0$ ), then taking $r=1$ in the above expression yields $f(1)=2$ . Then combining the above two relations, we get the solution $f \equiv 2$ .

Now let $f(0)=0$ . If $f(1)=0$ , then $P \left(\frac{1}{x},\frac{1}{x},x \right) \Rightarrow f(x)f \left(\frac{2}{x} \right)=2f(1)=0 \Rightarrow f(x)=0 \text{ or } f \left(\frac{2}{x} \right)=0$ But, if $f \left(\frac{2}{x} \right)=0$ with $x \neq \pm \sqrt{|r|}$ , then $P \left(\frac{1}{x},\frac{1}{x},x^2 \right) \Rightarrow f \left(\frac{2}{x} \right)f(x^2)=2f(x) \Rightarrow f(x)=0$ In particular, we must have $f(x)=0$ for all $x \neq \pm \sqrt{|r|}$ . Then one can easily adapt the proof in the above para to get the solution $f(a)=0$ for $a \neq 0$ . So we can assume $f(1) \neq 0$ . This also implies that $f(x) \neq 0$ for any $x \in \mathbb{I}$ (See the first expression of this para). Also, the above fact along with $P \left(x,x,\frac{r}{x} \right)$ gives that $f(r) \neq 0$ for non-zero rationals. Thus, $f(k)=0 \Leftrightarrow k=0$ . Finally, note that $P(x,-x,x)$ and $P \left(x,-x,\frac{r}{x} \right)$ directly gives that $f$ is an odd function.

Now, $P(x,x,2x) \Rightarrow f(2x)^2=2f(2x^2) \Rightarrow \text{ Replacing } x \text{ by } \frac{1}{2x}, \text{ } f \left(\frac{1}{x} \right)^2=2f \left(\frac{1}{2x^2} \right)$ But, $P \left(x,x, \frac{1}{x} \right) \Rightarrow f(2x)f \left(\frac{1}{x} \right)=2f(1) \Rightarrow 4f(1)^2=f(2x)^2f \left(\frac{1}{x} \right)^2=4f(2x^2)f \left(\frac{1}{2x^2} \right)$ Also, for $z \neq \pm \sqrt{|r|}$ , we can write $P \left(z^2,z^2, \frac{1}{2z^2} \right) \Rightarrow f(2z^2)f \left(\frac{1}{2z^2} \right)=2f \left(\frac{1}{2} \right) \Rightarrow 2f \left(\frac{1}{2} \right)=f(1)^2$ Then $P \left(\frac{y}{2},\frac{y}{2},\frac{1}{y} \right) \Rightarrow f(y)f \left(\frac{1}{y} \right)=2f \left(\frac{1}{2} \right) \Rightarrow f(y)f \left(\frac{1}{y} \right)=f(1)^2$

Now, $P \left(x,y,\frac{1}{x} \right) \Rightarrow f(x+y)f \left(\frac{1}{x} \right)=f(1)+f \left(\frac{y}{x} \right)=f(1)+\frac{f(1)^2}{f \left(\frac{x}{y} \right)}=\frac{f(1)}{f \left(\frac{x}{y} \right)} \left(f(1)+f \left(\frac{x}{y} \right) \right)$ Comparing with $P \left(x,y, \frac{1}{y} \right)$ we get $f(x+y)f \left(\frac{1}{x} \right)f \left(\frac{x}{y} \right)=f(1)f(x+y)f \left(\frac{1}{y} \right)$ Replacing $y$ by $\frac{1}{y}$ , and using $f(x)f \left(\frac{1}{x} \right)=f(1)^2$ , we have $f(x)f(y)=f(1)f(xy) \text{ or } x+\frac{1}{y}=0$ Here, if the second condition holds, then $f(x)f(y)=f(x)f \left(\frac{-1}{x} \right)=-f(x)f \left(\frac{1}{x} \right)=-f(1)^2=f(1)f(-1)=f(1)f(xy)$ So we can say that $f(x)f(y)=f(1)f(xy)$ holds true for all $x,y \in \mathbb{I}$ . Plugging this back in the RHS of $P(x,y,z)$ , we get that $f(1)f(x+y)=f(x)+f(y) \quad \forall x,y \in \mathbb{I}$ Doing $x \mapsto -x$ and putting $y=2x$ in the above relation gives $f(1)(f(1)f(x)+f(x))=f(1)f(2x)=2f(x) \Rightarrow f(1)^2+f(1)=2$ So $f(1)=1$ or $f(1)=-2$ . If $f(1)=-2$ , then $y=z-x$ (for $z \neq x$ ) in the above relation gives $0=4f(z)+2f(x)+2f(z-x)=4f(z)+2f(x)-(f(z)+f(-x)) \Rightarrow f(z)+f(x)=0 \Rightarrow f(z+x)=0$ which is a contradiction for $z \neq -x$ . So $f(1)=1$ , and thus $f$ is additive as well as multiplicative over irrational input.

Now, Multiplicativity over irrationals gives $f(rx) f \left(\frac{1}{x} \right)=f(r) \Rightarrow f(rx)=f(r)f(x)$ Then, we get $f(qr)f(x)=f(qrx)=f(q)f(rx)=f(q)f(r)f(x) \Rightarrow f(qr)=f(q)f(r)$ Since $f(0)=0$ was already assumed, so the above expressions give that $f(ab)=f(a)f(b)$ is true for all $a,b \in \mathbb{R}$ . Similarly, additivity can be used to give $f(-x)+f(r+x)=f(r) \Rightarrow f(r+x)=f(r)+f(x)$ Then we get $f(q+r)+f(x)=f(q+r+x)=f(q)+f(r+x)=f(q)+f(r)+f(x) \Rightarrow f(q+r)=f(q)+f(r)$ Again, combined with $f(0)=0$ this implies that $f(a+b)=f(a)+f(b)$ holds for all $a,b \in \mathbb{R}$ . Thus, $f$ is additive as well as multiplicative. It is well known that $f$ must then either be the identity function or the zero function. Both of these solutions are already in our solution set. $\blacksquare$

REMARK

REMARK 2

This post has been edited 6 times. Last edited by math_pi_rate, Jun 26, 2020, 9:20 AM
Reason: Separated one para into two

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Imayormaynotknowcalculus

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Imayormaynotknowcalculus

#11 h Jun 26, 2020, 2:59 AM • 2 Y

Y by math_pi_rate, Nathanisme

Solution
Comments

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#12 h Jun 26, 2020, 5:39 AM • 3 Y

Y by Imayormaynotknowcalculus, Nathanisme, Mango247

Using substitution $f(x) = 2g\left(\frac{x}{2}\right)$ , and changing $z$ to $2z$ , the equation becomes
$g\left(\frac{x + y}{2}\right) g(z) = \frac{g(xz) + g(yz)}{2}$ Denote this equation by $P(x, y, z)$ . Now,
$P(x, x, z) : g(x)g(z) = g(xz) \; \forall x, z \in \mathbb{R} \setminus \mathbb{Q} \tag{1}$ Then, $P(x, y, z)$ becomes
$g\left(\frac{x + y}{2}\right) g(z) = \frac{g(x) + g(y)}{2} g(z)$ Thus, either $g(z) = 0$ for all irrational numbers $z$ , or
$g\left(\frac{x + y}{2}\right) = \frac{g(x) + g(y)}{2} \; \forall x, z \in \mathbb{R} \setminus \mathbb{Q} \tag{2}$
We solve the first case, $g(z) = 0$ for all $z \in \mathbb{Q}$ . Then, (1) yields $g(xz) = 0$ for all $x, z \in \mathbb{R} \setminus \mathbb{Q}$ . In particular, $g(y) = 0$ for all $y \neq 0$ ; if $y$ is irrational then this is immediate, while if $y$ is rational, then we can write $g(y) = g(y\pi)g(\pi^{-1}) = 0$ since $y \pi$ must be irrational. This gives the solution $g(x) = 0, x \neq 0$ (while $g(0)$ can take any value) for $g$ , and
$\boxed{f(x) = \begin{cases} 0, & x \neq 0 \\ c, & x = 0 \end{cases}}$ where $c$ is a fixed real constant.

For the second case, both (1) and (2) are true. Now, we attempt to generalize both to $\mathbb{R} \setminus \{0\}$ ; the latter would also include $0$ .
Claim 1, for equation (2)
Proof
Claim 2, for equation (1)
Proof

Hence, $g$ is both Jensen and almost multiplicative; $g(xy) = g(x) g(y)$ holds whenever $x$ and $y$ are both non-zero. This is sufficient to conclude that $g(x) \geq 0$ for all $x > 0$ , and thus $g$ is linear. Going back to multiplicativity, we obtain either $g \equiv 0$ , $g \equiv 1$ , or $g \equiv Id_{\mathbb{R}}$ . They lead to $\boxed{f \equiv 0}$ , $\boxed{f \equiv 2}$ , $\boxed{f \equiv Id_{\mathbb{R}}}$ .

Comment:
This is actually a pretty good functional equation that directs to extending equations. However, I feel that the main reason this problem does not get a place in the IMO shortlist is that they end up being additivity (or Jensen/something similar) + multiplicativity. An example where this is a possible solution is ISL 2016 A7, but as far as I remember, the additivity + multiplicativity solution is the second solution listed there. Also, another reason is that extending to $\mathbb{R}$ would require quite some detail, as mentioned by @above.

Are there any solutions that do not direct to this additive (or Jensen in this case) + multiplicative?

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#13 h Jun 26, 2020, 5:50 AM • 3 Y

Y by Imayormaynotknowcalculus, amar_04, Nathanisme

BlazingMuddy wrote:

Comment:
This is actually a pretty good functional equation that directs to extending equations. However, I feel that the main reason this problem does not get a place in the IMO shortlist is that they end up being additivity (or Jensen/something similar) + multiplicativity. An example where this is a possible solution is ISL 2016 A7, but as far as I remember, the additivity + multiplicativity solution is the second solution listed there. Also, another reason is that extending to $\mathbb{R}$ would require quite some detail, as mentioned by @above.

Are there any solutions that do not direct to this additive (or Jensen in this case) + multiplicative?

I agree that the end part (additivity+multiplicativity) is the same in all solutions (which we know till now), however I still feel that the problem still has a lot of material for being an ISL problem. The point to note is that all the 3 solutions above use different methods to deal with $f(0) \neq 0$ , and have completely different pathways to reach the end step (with mine being the longest :oops:

:oops:

). The extension from irrationals to all reals is well known (and easy to find on your own, as well), so I think that wouldn't have been a problem. All in all, this could have been a treat on the ISL of last year (which I didn't like much personally, except for a few questions).

P.S. It would be really nice if @dangerousliri could tell us how he came up with this problem, as well as the solution that was submitted for Shortlist consideration.

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#14 h Jun 26, 2020, 7:20 AM • 8 Y

Y by math_pi_rate, BlazingMuddy, Imayormaynotknowcalculus, amar_04, Aryan-23, Gaussian_cyber, Atpar, Nathanisme

This is the solution which was presented by the commite when they send their feedback. Probably this is the cleanest solution for this problem. Soon I will post also my solution that I did send.

Answer: The equation is satisfied by three types of functions:
$f(x)=0$ for all $x\in\mathbb{R}\setminus \{0\}$ and $f(0)=c$ where $c$ is any real constant or $f(x)=2$ for all $x\in\mathbb{R}$ or $f(x)=x$ for all $x\in\mathbb{R}$ .
Solution.
Lets write with $(*)$ initial condition. The restrition to irrational numbers is inconvenient to work with, so we first show that $(*)$ is true for a wider class of arguments.
Suppose that $t_1+t_2=t_3+t_4$ , with all $t_i\neq 0$ . Then we can find irrational $z$ such that all $t_i/z$ are also irrational. Substituting $(t_1/z,t_2/z,z)$ and $(t_3/z,t_4/z,z)$ in $(*)$ , we conlude that $f(t_1)+f(t_2)=f(t_3)+f(t_4)$ , so $f(a)+f(b)$ it depends only on the value of $a+b$ provided both $a$ and $b$ nonzero. Since for any nonzero $x$ and $y$ we
can find irrational $x$ and $y$ with the same sum, it follows that $(*)$ holds for all nonzero $x$ and $y$ and irrational $z$ .
Putting $x=-y$ , we see that $f(0)f(z)$ does not depend on the value of $z$ for $z$ irrational. We now divide into two cases, according to whether $f$ takes the same value for all irrational arguments.
Case 1. $f$ is constant on the irrationals.
Say $f(z)=c$ for all irrationals $z$ . For any irrational $z$ and rational $q\neq 0$ , we have $2c=f(z/2)+f(z/2)=f(z-q)+f(q)=c+f(q)$ , so $f(q)=c$ and so $f(x)=c$ for all nonzero $x$ . Putting $x=y=z$ irrational gives $c^2=2c$ , so either $c=2$ or $c=0$ , while putting $x=-y$ nonzero gives $cf(0)=2c$ . So if $c=2$ we have $f(x)=2$ for all $x$ , which satisfy $(*)$ , while if $c=0$ any value of $f(0)$ results in a function satisfying $(*)$ .
Case 2. $f$ is not constant on the irrationals.
We thus have $f(0)=0$ , and $x=-y$ gives that $f$ is an odd function. We have some irrational $z$ with $f(z)\neq 0$ . We now substitute various nonzero values of $x$ and $y$ , with that particular $z$ .
$\bullet$ $x=y=1$ gives $f(2)f(z)=2f(z)$ , so $f(2)=2$ .
$\bullet$ $x=2$ and $y=-1$ gives $f(1)f(z)=f(2z)+f(-z)=f(2z)-f(z)$ (since $f$ is odd), so $f(2z)=(1+f(1))f(z)$ .
$\bullet$ $x=2$ and $y=1$ gives $f(3)f(z)=f(2z)+f(z)=(2+f(1))f(z)$ , so $f(3)=2+f(1)$ .
Now $2+2=1+3$ , so $4=f(2)+f(2)=f(1)+f(3)=2+2f(1)$ , so $f(1)=1$ and $f(3)=3$ . Repeatedly using $f(n)+f(1)=f(2)+f(n-1)$ yields $f(n)=n$ for all positive integers (and thus all integers) n.
Putting $x=n$ and $y=-1$ then gives $f(nz)=nf(z)$ for all $z$ irrational and $n$ a nonzero integer (and thus for any integer $n$ ). In turn, this gives $f(qz)=qf(z)$ for all z irrational and $q$ rational, and now putting $x=y=q/2$ gives $f(q)=q$ for all rational numbers $q$ .
If $x=y=z$ is irrational, we have $f(2x)f(x)=2f(x^2)$ , and $f(2x)=2f(x)$ , so $f(x^2)\geq 0$ . All positive irrational numbers are the squares of irrational numbers, so $f(x)\geq 0$ for all positive irrational $x$ .
Suppose $f(x)\neq x$ for some $x$ , which must be irrational. Negating if necessary, we have some $x$ with $f(x)<x$ ; say $f(x)=x-t$ for some positive $t$ . Note that because $f(a)+f(b)$ only depends on $a+b$ for $a$ and $b$ nonzero, we have that $f(x)=2f(x/2)=f(x+q)+f(-q)=f(x+q)-q$ , so $f(x+q)=f(x)-q$ for all rationals $q$ . So picking some rational $q$ with $-x<q<t-x$ , we have positive $x+q$ with negative $f(x+q)$ , a contradiction. So in this case we must have $f(x)=x$ for all positive irrational $x$ , and since function $f$ is odd, hold for all irrationals $x$ , which means also for all real number $x$ , which satisfies $(*)$ .

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#15 h Jun 26, 2020, 7:50 AM • 5 Y

Y by math_pi_rate, Imayormaynotknowcalculus, Kanep, Nathanisme, Mango247

math_pi_rate wrote:

P.S. It would be really nice if @dangerousliri could tell us how he came up with this problem, as well as the solution that was submitted for Shortlist consideration.

I was one day with my friend Valmir thinking about a problem to send at IMO. And I said lets do something that it never had been in IMO a functional and he said to do something with irrationals. So I was looking to do something like from irrational to irrational function but it is hard to do like that since you can't get something like $x+y$ since it could be rational. So I was thinking then to do something from R to R but condition to be with irrationals and after some try I write this problem and it did seems doable and with some help from my friend we did finish. So this way we did create the problem.

Now I will present my solution that I did send to IMO.

Answer: The equation is satisfied by three types of functions:
$f(x)=0$ for all $x\in\mathbb{R}\setminus \{0\}$ and $f(0)=c$ where $c$ is any real constant or $f(x)=2$ for all $x\in\mathbb{R}$ or $f(x)=x$ for all $x\in\mathbb{R}$ .

Solution.
Let $\mathbb{I}$ and $\mathbb{Q}$ denote the set of irrational numbers and rational numbers, respectively. Let $P(x,y,z)$ denote the given equation. We distinguish the following cases.

Case 1. $f\left(\sqrt{2}\right)=0$ .

Taking $P\left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2},\sqrt{2}\right)$ we have $f\left(\sqrt{2}\right)^2=2f(1)\Rightarrow f(1)=0$ .

Taking $P\left(x,\frac{\sqrt{2}}{2},\sqrt{2}\right)$ we have
$f\left(x+\frac{\sqrt{2}}{2}\right)f\left(\sqrt{2}\right)=f\left(x\sqrt{2}\right)+f(1)\Rightarrow f\left(x\sqrt{2}\right)=0$ for all $x\in\mathbb{I}$ , hence we have $f(x)=0$ for all $x\in\mathbb{R}\setminus\mathbb{A}$ where $\mathbb{A}=\left\{ q\sqrt{2}|q\in\mathbb{Q}\right\}$ .

Taking $P\left(\frac{q\sqrt{2}}{2}-1,\frac{q\sqrt{2}}{2}+1,q\sqrt{2}\right)$ where $q\in\mathbb{Q}\setminus \{0\}$ we obtain
$f\left(q\sqrt{2}\right)^2=f\left(q^2-q\sqrt{2}\right)+f\left(q^2+q\sqrt{2}\right)=0\Rightarrow f\left(q\sqrt{2}\right)=0.$ Since $q^2-q\sqrt{2},q^2+q\sqrt{2}\in\mathbb{R}\setminus\mathbb{A}$ it implies that $f(x)=0$ for all $x\in\mathbb{A}\setminus \{0\}$ . Therefore $f(x)=0$ for all $x\in\mathbb{R}\setminus \{0\}$ .

Substituting it to the first equation (as it holds for all $x,y,z\in\mathbb{I}$ ) we obtain $x,xz,yz\in\mathbb{R}\setminus \{0\}$ which implies $f(x)=f(xz)=f(yz)=0$ thus
$f(x+y)f(z)=f(xz)+f(yz)$ therefore it does not depend on $f(0)$ .

Case 2.1. $f\left(\sqrt{2}\right)\neq 0$ and $f(0)\neq 0$ .

Taking $P\left(\frac{1}{x},-\frac{1}{x},x\right)$ we have
$f(0)f(x)=f(1)+f(-1)\Rightarrow f(x)=\frac{f(1)+f(-1)}{f(0)}\Rightarrow f(x)=a$ for all $x\in\mathbb{I}$ and since $f\left(\sqrt{2}\right)\neq 0$ we also have $a\neq 0$ .

Taking $P\left(q-\sqrt[4]{2},\sqrt[4]{2},\frac{1}{q\sqrt[4]{2}-\sqrt{2}}\right)$ where $q\in\mathbb{Q}$ we have
$f(q)f\left(\frac{1}{q\sqrt[4]{2}-\sqrt{2}}\right)=f\left(\frac{1}{\sqrt[4]{2}}\right)+f\left(\frac{1}{q-\sqrt[4]{2}}\right)\Rightarrow af(q)=2a\Rightarrow f(q)=2$ (since $\frac{1}{q\sqrt[4]{2}-\sqrt{2}},\frac{1}{\sqrt[4]{2}},\frac{1}{q-\sqrt[4]{2}}\in\mathbb{I}$ ) hence $f(x)=2$ for all $x\in\mathbb{Q}$ and since $f(x)=\frac{f(1)+f(-1)}{f(0)}$ for all $x\in\mathbb{I}$ we have $f(x)=2$ for all $x\in\mathbb{I}$ .
Therefore $f(x)=2$ for all $x\in\mathbb{R}$ .

Case 2.2. $f\left(\sqrt{2}\right)\neq 0$ and $f(0)= 0$ .
Taking $P\left(x,x,\sqrt{2}\right)$ we have
$f(2x)f\left(\sqrt{2}\right)=2f\left(x\sqrt{2}\right)\Rightarrow f\left(x\sqrt{2}\right)=\frac{f(2x)f\left(\sqrt{2}\right)}{2}$ for all $x\in\mathbb{I}$ .

Taking $P\left(x,y,\sqrt{2}\right)$ we have

$f(x+y)f\left(\sqrt{2}\right)=f\left(x\sqrt{2}\right)+f\left(y\sqrt{2}\right)=\frac{f(2x)f\left(\sqrt{2}\right)}{2}+\frac{f(2y)f\left(\sqrt{2}\right)}{2}$
$\Rightarrow f(x+y)=\frac{f(2x)+f(2y)}{2}\Rightarrow f(2x)+f(2y)=2f(x+y)$
for all $x,y\in\mathbb{I}$ .

Since for all real numbers $x$ exist numbers $r,s\in\mathbb{I}$ we have
$f(x)=f(r+s)=\frac{f(2r)+f(2s)}{2}.$ now for all $x,y\in\mathbb{R}$ exists numbers $r,s,t,u\in\mathbb{I}$ such that $r+s=x$ , $t+u=y$ and $r+t,s+u\in\mathbb{I}$ hence we have

$\begin{align*} f(2x)+f(2y)&=f(2r+2s)+f(2t+2u) &=\frac{f(4r)+f(4s)}{2}+\frac{f(4t)+f(4u)}{2} &=\frac{f(4r)+f(4t)}{2}+\frac{f(4s)+f(4u)}{2} &=f(2r+2t)+f(2s+2u) &=f(2(r+t))+f(2(s+u)) &=2f(r+t+s+u) &=2f(r+s+t+u) &=2f(x+y) \end{align*}$
Hence $f(2x)+f(2y)=2f(x+y)$ for all $x,y\in\mathbb{R}$ .

Taking $y=0$ in the last relation we have
$f(2x)+f(0)=2f(x)\Rightarrow f(2x)=2f(x)$ hence last equation transforms to $f(x)+f(y)=f(x+y)$ .

Taking $y=-x$ in last equation we have
$f(x)+f(-x)=f(0)=0\Rightarrow f(-x)=-f(x)$ hence the function is odd.

Taking $P\left(\sqrt{\frac{x}{2}},\sqrt{\frac{x}{2}},\sqrt{2x}\right)$ where $x\in\mathbb{I}^+$ we have
$f\left(\sqrt{2x}\right)^2=2f(x)\Rightarrow f(x)\geq 0$ for all $x\in\mathbb{I}^+$ .

Taking $P\left(\sqrt{2},\sqrt{2},\frac{x}{\sqrt{2}}\right)$ where $x\in\mathbb{Q}^+$ we have
$f\left(2\sqrt{2}\right)f\left(\frac{x}{\sqrt{2}}\right)=2f(x)\Rightarrow f(x)\geq 0$ for all $x\in\mathbb{Q}^+$ since $2\sqrt{2},\frac{x}{\sqrt{2}}\in\mathbb{I}^+$ .

Hence we have $f(x)\geq 0$ for all $x\in\mathbb{R}^+$ .

Since the function is bounded in $\mathbb{R}^+$ , is odd and satisfies the Cauchy equation we have that $f(x)=bx$ for all $x\in\mathbb{R}$ where $b\neq 0$ is real constant (since $f\left(\sqrt{2}\right)\neq 0$ ).

By substituting to the first equation we obtain
$b(x+y)bz=bxz+byz\Rightarrow b^2(x+y)z=b(x+y)z\Rightarrow (b^2-b)(x+y)z=0$ for all $x,y,z\in\mathbb{I}$ hence $b^2=b\Rightarrow b=1$ hence $f(x)=x$ for all $x\in\mathbb{R}$ .

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Mr.C

#16 h Sep 29, 2020, 7:06 PM • 2 Y

Y by Abhaysingh2003, jam10307

i just wanted to say , a really good problem , you clearly are one of the best FE proposers , and also i like your problems verry much as they are challenging and hard , i wished this was on shortlist

any ways a bump to this one as i think many people should see this beautifull FE

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#17 h Sep 29, 2020, 7:16 PM • 5 Y

Y by niyu, Nathanisme, OlympusHero, Math_olympics, v4913

dangerousliri wrote:

We did proposed this problem for IMO twice, on 2018 and on 2019 from Kosovo.

I believe the IMO regulations say "They [problem proposals] should be new and may not have been suggested for or used in any other mathematics competition". So I don't think it is allowed to submit a problem to two different years of IMO. Am I mistaken?

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#18 h Sep 29, 2020, 7:25 PM • 2 Y

Y by Valmir, Nathanisme

v_Enhance wrote:

dangerousliri wrote:

We did proposed this problem for IMO twice, on 2018 and on 2019 from Kosovo.

I believe the IMO regulations say "They [problem proposals] should be new and may not have been suggested for or used in any other mathematics competition". So I don't think it is allowed to submit a problem to two different years of IMO. Am I mistaken?

I didn't proposed at different competetions. I just did it proposed twice on same competetion on different years plus I didn't publish anywhere.

Also can you tell me where it says it shouldn't had been proposed before on IMO.

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#19 h Sep 29, 2020, 7:29 PM • 4 Y

Y by niyu, Nathanisme, Math_olympics, v4913

My understanding of the rule had been that you can't submit to IMO twice either (as in, I believe "proposed for other mathematics competitions" includes previous IMO instances). At least in the USA once someone submits a problem to IMO of year $n$ I do not take it again for year $n+1$ .

Maybe I am wrong about this in which case it would be nice to know.

EDIT: It sounds like I am probably wrong about this! Hearing a bit of different things from others privately.

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#21 h Sep 29, 2020, 9:07 PM • 8 Y

Y by dangerousliri, v_Enhance, Mr.C, Valmir, DVDthe1st, CantonMathGuy, pavel kozlov, OlympusHero

v_Enhance wrote:

My understanding of the rule had been that you can't submit to IMO twice either (as in, I believe "proposed for other mathematics competitions" includes previous IMO instances).

This is my experience. Last year I proposed a problem and we submit it to IMO from Iran. The PSC didn't select it and they told us that the problem is excellent but they had a better problem with the same difficulty (I think they were talking about G3 and I like G3 more too) and they suggested us to submit it again next year. So it is definitely possible.

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Mr.C

#23 h Sep 29, 2020, 11:20 PM

Y by

my solution to this beautiful FE .
.
let $p(x,y,z)=f(x+y)f(z)=f(xz)+f(yz)$ let $x+y=0,xz=1$ so if $f(0)$ is non_zero then
$f(z)=\frac{f(1)+f(-1)}{f(0)}=C$
so all irrational numbers are constant so rewrite $p$ as
$Cf(x+y)=2C$
to do this we need both $xz,yz$ to be irrational which can be done easily .
so either $C=0$ or $f(x)=2$ for any $x$ so one can get $f(x)=2,f(x)=0$ as constant answers .
now if $f(0)=0$ we have
$f(xz)+f(-xz)=0$ since any real number can be writen as the product of two iirrational numbers then $f(x)+f(-x)=0$ now we have $y$ is irrational iff $-y$ is irrational so , summing
$p(x,y,z)+p(x,-y,z)$ we have
$(f(x+y)+f(x-y))f(z)=2f(xz)$ now if $f(z)=0$ we have $f(xz)=0$ and this emply that atmost one set of liners have a nonzero value on $f$ and that is the liners of $z$ but we can have the same result for the liners of $xz$ so we have $f$ is constant .
so we can devide through $f(z)$ without any worry .
from $p(x,x,z)$ we have $f(2x)f(z)=2f(xz)$
so
$f(x+y)+f(x-y)=f(2x)$ set $y=3x$ so
$f(4x)=2f(2x)$
so
$f(2x)=2f(x)$ for irrational $x$
we have again $f(x)f(z)=f(xz)$ we prove this also holds for any real $a,b$ we have $a=ts,b=uv$ do this in such a way that $tu,sv$ are both irrational
$f(ab)=f(tsuv)=f(tu)f(sv)=f(t)f(u)f(s)f(v)=f(ts)f(uv)=f(a)f(b)$ so $f$ is multyplactive , now rewrite $p$ as
$f(x+y)=f(x)+f(y)$
we also prove this for all reals $a,b$ we have if
$a=i+j,b=k+t$ for irrational $i,j,t,k,k+i,j+t$
$f(a+b)=f(i+j+k+t)=f(i+k)+f(j+t)=f(i)+f(j)+f(k)+f(t)=f(i+j)+f(k+t)=f(a)+f(b)$
so we have $f(x)=x$ for any real $x$ and done!
.
i can't agree for this to be $A_6,A_7$ tbh , maybe a $A_4?$

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Mr.C

#24 h Sep 30, 2020, 4:01 AM

Y by

smt that can be found in my proof , its not proven yet , but i guess its true.
(i was awake the whole night so i can't think straight here so i might be totally wrong here) if we have a set $S \in R$ such that
$R=\{x+y| x,y \in S\}=\{xy|x,y \in S \}$
then the problem can be solved with $x,y,z \in S$ in the original problem
.
@below , that was hands down the best thing i heard about $2020,P_1$

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#26 h Oct 4, 2020, 11:33 AM • 1 Y

Y by Mango247

dangerousliri wrote:

Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that for all irrational numbers $x, y$ and $z$ ,
$f(x+y)f(z)=f(xz)+f(yz)$
Some stories about this problem. This problem it is proposed by me (Dorlir Ahmeti) and Valmir Krasniqi. We did proposed this problem for IMO twice, on 2018 and on 2019 from Kosovo. None of these years it wasn't accepted and I was very surprised that it wasn't selected at least for shortlist since I think it has a very good potential. Anyway I hope you will like the problem and you are welcomed to give your thoughts about the problem if it did worth to put on shortlist or not.

This problem is artificial and contrived! Why should the equation only hold for irrational numbers?? It formulates some kind of distributive rule, and then restricts it to irrational values. Why? In real life mathematics the equation would hold for all reals and then the problem is not challenging. Is this artificial constraint just added to make the problem look more interesting?

We all must be thankful to the PSC 2018 and to the PSC 2019 that they have protected IMO against such an unnatural and artificial creature!

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gghx

#27 h Oct 4, 2020, 12:15 PM • 12 Y

Y by Valmir, Mr.C, dangerousliri, RevolveWithMe101, GorgonMathDota, IndoMathXdZ, amar_04, OlympusHero, Functional_equation, mathleticguyyy, Mango247, megarnie

TelMarin wrote:

Why should the equation only hold for irrational numbers?? It formulates some kind of distributive rule, and then restricts it to irrational values. Why? In real life mathematics the equation would hold for all reals and then the problem is not challenging.

I believe there is nothing wrong with restricting it to irrationals, even when you think it is artificial/contrived. After all, many other problems also look the same, especially FEs, where you sometimes try to fit the equation to the solution. But that does not compromise the difficulty or beauty of the problem. In fact, you say it makes it more interesting, which i think is the entire point of math olympiad. What is the point of solving boring questions over and over?

As for the point about real life application, what does that have to do with math olympiad? Which imo prblem has real life application? The aim is to challenge your thinking abilities, which an unorthodox question like this does well. Math olympiad should be enjoyed.

And nice problem @dangerousliri

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#28 h Oct 4, 2020, 1:40 PM • 5 Y

Y by DVDthe1st, Aryan-23, test20, Mango247, Mango247

Well, I think the interesting part of the problem is it holds only for the irrationals. So, it deserves attention only because of that (congrats for the authors!). But all that remains after removing this obstacle is just one routine FE. And that's the issue. I think, it was perhaps not included in the SL because many people are fed up with standard FE, inequalities, etc. I'm happy the PSC in the recent years, trend to include interesting problems which bear ideas based on different subjects. For example problems like IMO 2020 p3, p6; IMO 2019 SL, A3, A4, C9, N3, N6, G8, etc. Even that inequality (IMO 2020 p2) is somehow not a standard one. The more ideas based on different subjects a problem includes, the more interesting it becomes.

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#29 h Oct 16, 2020, 11:05 AM

Y by

gghx wrote:

As for the point about real life application,

I was talking about real life mathematics and not about real life application.

Real life mathematics is mathematics for grown-ups, in contrast to olympiad mathematics for highschool children. If you show this functional
equation to a grown-up mathematician, the first reaction will be: "Why this weird restriction to irrational numbers?". And the only honest answer will be "Because there is a number of bad olympiad problems that are extremely artificial and contrived."

I once again want to thank the PSC 2018 and the PSC 2019 for protecting IMO against such problems.

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#30 h Oct 16, 2020, 11:25 AM • 1 Y

Y by Nathanisme

TelMarin wrote:

gghx wrote:

As for the point about real life application,

I was talking about real life mathematics and not about real life application.

Real life mathematics is mathematics for grown-ups, in contrast to olympiad mathematics for highschool children. If you show this functional
equation to a grown-up mathematician, the first reaction will be: "Why this weird restriction to irrational numbers?". And the only honest answer will be "Because there is a number of bad olympiad problems that are extremely artificial and contrived."

I once again want to thank the PSC 2018 and the PSC 2019 for protecting IMO against such problems.

Apperantely you have no idea what you are talking about.

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#31 h Oct 16, 2020, 11:31 AM

Y by

TelMarin wrote:

gghx wrote:

As for the point about real life application,

I was talking about real life mathematics and not about real life application.

Real life mathematics is mathematics for grown-ups, in contrast to olympiad mathematics for highschool children. If you show this functional
equation to a grown-up mathematician, the first reaction will be: "Why this weird restriction to irrational numbers?". And the only honest answer will be "Because there is a number of bad olympiad problems that are extremely artificial and contrived."

I once again want to thank the PSC 2018 and the PSC 2019 for protecting IMO against such problems.

I would hate to believe that the things you claim are true. First of all, not many people have solved every problem from this highschool children problems competition you are claiming irrespective of whether they have tried or not. I want to say in short that not many people, just a small fraction of a small fraction of people can do this problem.

Besides we have math researchers that do these problems and struggle a lot. Real life mathematics does have a huge correspondence to Olympiad level Mathematics though being great at latter doesn't imply you'll be great in the previous directly. You don't experience beauty that often. Moreover, it holds for irrational numbers. Have you ever seen such a beautiful condition? I haven't seen yet a beautiful FE from you, I'm not attacking I'm just telling I haven't seen a nice problem.

You know, people keep functions from $\mathbb{Z}_p \rightarrow \mathbb{Z}_p$ , or from irrational to irrational, how does that matter? As long the problem is beautiful. I again do not want to attack few SL Problem Proposers because I am neither excellent in math nor experienced but this would've been a gem in between few boring functions that have been proposed in the past

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#32 h Oct 16, 2020, 11:39 AM

Y by

DapperPeppermint wrote:

You know, people keep functions from $\mathbb{Z}_p \rightarrow \mathbb{Z}_p$ , or from irrational to irrational, how does that matter? As long the problem is beautiful. I again do not want to attack few SL Problem Proposers because I am neither excellent in math nor experienced but this would've been a gem in between few boring functions that have been proposed in the past

So did you see problem A7 on the 2019 shortlist? That functional equation is a real beauty, and also grown-up mathematicians from the real world are going to appreciate it.

The PSC 2019 did a fantastic job when it selected the most beautiful functional equations and rejected the artificial and contrived ones.

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#33 h Oct 16, 2020, 12:16 PM

Y by

dangerousliri wrote:

Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that for all irrational numbers $x, y$ and $z$ ,
$f(x+y)f(z)=f(xz)+f(yz)$
Some stories about this problem. This problem it is proposed by me (Dorlir Ahmeti) and Valmir Krasniqi. We did proposed this problem for IMO twice, on 2018 and on 2019 from Kosovo. None of these years it wasn't accepted and I was very surprised that it wasn't selected at least for shortlist since I think it has a very good potential. Anyway I hope you will like the problem and you are welcomed to give your thoughts about the problem if it did worth to put on shortlist or not.

My dolution :

Let $P(x, y)$ be the assertion. First thing to see js constant functions, and there exists two constant functions : $\boxed{f(x)=0}$ and $\boxed{f(x)=2}$ for all reals $x$ . Now let us assume that $f$ is not constant.

Observe that $P\left ( \dfrac{1}{x}, \dfrac{-1}{x}, x\right ) \implies f(x)f(0)=f(1)+f(-1)$ and in order to evade constant value for $f(x)$ , $f(0)=0$ and $f(1)=-f(-1)$ must hold. Now, it can be seem generally that $f$ is odd using $P(x, -x, k)$ .

We see that $f$ is injective at $0$ .

$P(x, x, 2x) \implies f(2x)^2=2f(2x^2)$ . $P\left (x, x, \dfrac{1}{x}\right ) \implies 4f(1)^2 = f(2x)^2f\left ( \dfrac{1}{x}\right )$ . Let $abs (z)$ not be root of a rational. $P\left (z^2, z^2, \dfrac{1}{2z^2}\right )$ and $P\left ( \dfrac{z}{2}, \dfrac{z}{2}, \dfrac{1}{z}\right )$ imply that $2f\left (\dfrac{1}{2}\right ) = f(1)^2 = f(z) f\left ( \dfrac{1}{z}\right )$ . It can be seen that $f(x) = \dfrac{1}{f\left ( \dfrac{1}{z}\right )}$ generally holds. Now comparing $P\left ( x, y, \dfrac{1}{x}\right )$ and $P\left ( x, xy, \dfrac{1}{y}\right )$ give that $f(x+y)f\left ( \dfrac{1}{x}\right ) f\left ( \dfrac{x}{y}\right ) = f(1)f(x+y) f\left ( \dfrac{1}{y}\right )$ . Now changing $y$ by reciprocal of $y$ and using above fact we get that $f(x)f(y) = f(1)f(xy)$ or $y = \dfrac{-1}{x}$ . Second condition also lead to first one so $f(x)f(y)=cf(xy)$ , $f(1)=c$ . It can now be seen that $f$ is additive due to this condition ( $f(1)f(x) = f(-x)+f(2x)$ and add this to original condition for a different $y$ and eventually you'll get that $f(t)=0$ for $t \neq 0$ , I'm skipping details because class time) and due to this additive condition, $f(1)=1$ . Neat. So $f$ is properly multiplicative and additive over irrationals. I'm yet to deal with rationals, I forgot to tell.

Observe that $f$ is additive over irrationals mean that it is also additive over rationals. Why? Just see that if two rationals $p, q$ exist, $f(p)+f(q) = f(p-\pi ) + f(\pi ) + f(q-\pi ) + f(\pi ) = f(p+q - 2\pi ) + f(2 \pi ) = f(p+q)$ . Neat. Now if $p, q$ are two rationals then $f(p)f(q) = f\left ( \dfrac{p}{e}\right ) f(e)^2 f\left ( \dfrac{q}{e}\right ) = f\left ( \dfrac{pq}{e^2}\right ) f(e^2) = f(pq)$ . Neater.

So $f$ is multiplicative and additive all over $\mathbb{R}$ . We know now that $f(x)=x$ due to famous lemma : If $f$ is both multiplicative and additive over reals then it is the identity function or $0$ .
Proof : $f(x)=0$ is trivial. It can be seen that $f(x)=x$ is true for all rationals. Moreover $f$ is odd so we'll show for positive reals. Without loss of generality let $f(x) = x + \epsilon$ for $x > 1$ . It can be seen that $f$ is positive for all positive reals. This will be extremely crucial. Let $b \in \mathbb{R}^+$ such that $b \{ \epsilon \} > 1$ and $b$ is positive integer, $\{ k \}$ denotes non integer part of $k$ . Then $f(xb) =f(x)f(b) > xb+1 > x+1$ . Now $xb$ is irrational so there exist an integer between $xb, xb+1$ . Let us say that positive integer is $n$ . $f(xb) + f(n-xb) = f(n)$ so $f(n-xb)$ is negative, which contradicts the fact that for all positive reals $x$ , $f(x)$ is positive.

Hence $\boxed{f(x)=x}$ is the last function

TelMarin wrote:

DapperPeppermint wrote:

You know, people keep functions from $\mathbb{Z}_p \rightarrow \mathbb{Z}_p$ , or from irrational to irrational, how does that matter? As long the problem is beautiful. I again do not want to attack few SL Problem Proposers because I am neither excellent in math nor experienced but this would've been a gem in between few boring functions that have been proposed in the past

So did you see problem A7 on the 2019 shortlist? That functional equation is a real beauty, and also grown-up mathematicians from the real world are going to appreciate it.

The PSC 2019 did a fantastic job when it selected the most beautiful functional equations and rejected the artificial and contrived ones.

That problem is wonderful but I can think why people may hate this problem, but it is beautiful in the sense of how to properly use irrational numbers

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#36 h Apr 1, 2025, 1:32 AM

Y by

dangerousliri wrote:

Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that for all irrational numbers $x, y$ and $z$ ,
$f(x+y)f(z)=f(xz)+f(yz)$

Let $\mathbb I=\mathbb R\setminus\mathbb Q$ . If $f(x)=0$ for all $x\in\mathbb I$ then setting $x=y$ in the equation gives $f(xz)=0$ for $x,z\in\mathbb I$ . Every nonzero real can be expressed as the product of two irrationals, so $\boxed{f(x)=\begin{cases}0&\text{if }x\ne0\\c&\text{if }x=0\end{cases}}$ which satisfies the equation for any $c\in\mathbb R$ .
Otherwise, suppose there is some $j\in\mathbb I$ with $f(j)\ne0$ .

Part 1: $f(x+y)=f(x)+f(y)$ and $f(xy)=f(x)f(y)$ for $x,y\in\mathbb I$
Again set $x=y$ in the original equation, we get $f(xz)=\frac12f(2x)f(z)$ .
Swapping $x,z$ and setting $z=j$ , we get that $f(2x)=2af(x)$ for some $a\in\mathbb R$ (we got $a=\frac{f(2j)}{2f(j)}$ ). So we have $f(xy)=af(x)f(y)$ , and substituting we get:
$f(x+y)f(z)=af(x)f(z)+af(y)f(z),$ or, setting $z=j$ and cancelling,
$f(x+y)=af(x)+af(y).$ If $a=0$ , then setting $x=y=\frac j2$ immediately gives $f(j)=0$ , impossible, so $a\ne0$ .

Suppose $a\ne1$ as well. Choose any $x,y\in\mathbb I$ , then we can find a $c\in\mathbb I$ such that $c+x,c+y\in\mathbb I$ . So:
$a^2f(c)+af(x)+a^2f(y)=af(c+y)+af(x)=f(c+x+y)=af(c+x)+af(y)=a^2f(c)+a^2f(x)+af(y).$ Comparing these, we get $a^2f(x)-af(x)=a^2f(y)-af(y)$ , so $f(x)=f(y)$ since $a\notin\{0,1\}$ , that is, $f$ is constant over $\mathbb I$ . Because we can express any real as the sum of two irrationals, we must have
$f(r_1)=af(i_1)+af(i_2)=af(i_3)+af(i_4)=f(r_2)$ for any $r_1,r_2\in\mathbb R$ where $i_1+i_2=r_1$ and $i_3+i_4=r_2$ for some $i_1,i_2,i_3,i_4\in\mathbb I$ .
Testing in the original equation (remembering that $f(j)\ne0$ ) we get the solution $\boxed{f(x)=2}$ . To search for functions other than these, we can proceed under the assumption that $a=1$ which gives $f(x+y)=f(x)+f(y)$ and $f(xy)=f(x)f(y)$ for $x,y\in\mathbb I$ .

Part 2: $f(x+y)=f(x)+f(y)$ and $f(xy)=f(x)f(y)$ for $x,y\in\mathbb R$
Fix any $x\in\mathbb I$ , we can find some $y\in\mathbb I$ with $x+y\in\mathbb I$ . Then for any $q\in\mathbb Q$ , we have:
$f(x+q)=f(x+y)+f(q-y)=f(x)+f(y)+f(q-y)=f(x)+f(q).$ So for $q,r\in\mathbb Q$ we have:
$f(j)=f(j+q+r)+f(-q-r)=f(j+q)+f(r)+f(-q-r)=f(j)+f(q)+f(r)+f(-q-r).$ Set $r=0$ , then $f$ is odd over rationals, so this equation simplifies to $f(q+r)=f(q)+f(r)$ .

So for any $(x,y)\in\mathbb R^2$ , whether $(x,y)\in\mathbb I^2$ , $\mathbb I\times\mathbb Q$ , $\mathbb Q\times\mathbb I$ , $\mathbb Q^2$ we must have $f(x+y)=f(x)+f(y)$ .

Similarly, let $x\in\mathbb I$ and choose $y\in\mathbb I$ such that $xy\in\mathbb I$ , we have
$f(xq)=f(xy)f\left(\frac qy\right)=f(x)f(y)f\left(\frac qy\right)=f(x)f(q)$ for $q\ne0$ . Since $f(0)=0$ from $f(x+y)=f(x)+f(y)$ it holds for $q=0$ as well. In addition:
$f(j)=f\left(\frac j{qr}\right)f(qr)=f\left(\frac jq\right)f\left(\frac1r\right)f(qr)=f(j)f\left(\frac1q\right)f\left(\frac1r\right)f(qr)$ Setting $r=1$ , we get $f\left(\frac1q\right)=\frac1{f(q)}$ , so $f(qr)=f(q)f(r)$ if $qr\ne0$ . Since $f(0)=0$ it must also hold when $qr=0$ .

So for any $(x,y)\in\mathbb R^2$ , whether $(x,y)\in\mathbb I^2$ , $\mathbb I\times\mathbb Q$ , $\mathbb Q\times\mathbb I$ , $\mathbb Q^2$ we must have $f(xy)=f(x)f(y)$ .

Now it's well known that the only solution to $f(x+y)=f(x)+f(y)$ and $f(xy)=f(x)f(y)$ over $\mathbb R$ will be $\boxed{f(x)=x}$ , which satisfies the equation, and we're done.

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#37 h Apr 1, 2025, 6:16 PM

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dangerousliri wrote:

Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that for all irrational numbers $x, y$ and $z$ ,
$f(x+y)f(z)=f(xz)+f(yz)$
Some stories about this problem. This problem it is proposed by me (Dorlir Ahmeti) and Valmir Krasniqi. We did proposed this problem for IMO twice, on 2018 and on 2019 from Kosovo. None of these years it wasn't accepted and I was very surprised that it wasn't selected at least for shortlist since I think it has a very good potential. Anyway I hope you will like the problem and you are welcomed to give your thoughts about the problem if it did worth to put on shortlist or not.

Very beautiful, thanks for the amazing problem

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