Beautiful FE most likely Sanjaya is author? Or william Ting $c=2$ is answer

Physicsknight wrote: Beautiful FE most likely Sanjaya is author? Nope XD, I'm more inclined towards classical FE (determine all functions such that bla bla bla) whose statement can be generalized to abstract groups or rings, preferrably non-ordered groups and rings. (I didn't make any of these that appears in the IMO SL...)

Can someone confirm the answer? Put it in a spoiler please so others aren't spoiled.

i am also getting 2 , but i m not able to convince.

Sorry, I got it now

This is how I get it.

I am not sure though, solution seems very sus

DistortedDragon1o4 wrote: Putting $x = 0$, $y = y$: $f(f(y)) = f(0) + y = y$ This is incorrect, the problem statement here gives you that at least one of $f(f(y)) = y$ and $f(y) = f(y)$ holds, which is certainly nothing new.

if we will say that f(x)=cx. c is a constant value it has infinite solutions as if will put 1 on the place of f(x) will be x , f(y) will be y . Is it right ?

sevenplusseven wrote:

Is your function the floor function? Doesn't seem to hold

In my pov it is 1

szpolska wrote: Is your function the floor function? Doesn't seem to hold

works.

Very brief outline f(0)=0 f(-f(x))=-x y=-f(y): f(x-y)=f(x)-f(y) or f(y-x)=f(y)-f(x) write down the conditions for various permutations of signs +-x,y,z where x+y+z=0 and show f(x)+f(-x), f(y)+f(-y), 0 distinct is impossible solution starts with claims 1-4 of below

This is such a beautiful problem, definitely my favorite one this year, by far. This is a collaborative effort by Li4, YaWNeeT and untro368

Nice problem. Get that $f(x-y) = f(x) - f(y)$ or $f(y-x) = f(y) - f(x)$ as in the above posts. Here's a much faster finish. Consider $a,b$ such that $f(a)+f(-a),f(b)+f(-b)$ are distinct and nonzero. Now, we have $f(a+b) = f(a) + f(b)$ or $f(a) - f(-b)$. Also, $f(a+b) = f(a) + f(b)$ or $f(b) - f(-a)$. Note that these three quantities are all distinct by assumption. Therefore, we must have $f(a+b) = f(a) + f(b)$ and similarly $f(-a-b) = f(-a) + f(-b)$. But note that $f(a) + f(b) = f(a+b) = f(a) - f(-b)$ or $f(-a) + f(-b) = f(-a-b) = f(-b) - f(a)$, which is a contradiction. Therefore such $(a,b)$ cannot exist, so $c \leqslant 2$.

What an amusing problem. I found the construction to be far harder than the bound. I guess a major part of the difficulty is also guessing the smallest value of $c$, which I unfortunately got spoiled by. The smallest value of $c$ is $\boxed{c=2}$. We first prove that $c=2$ satisfies the condition. In fact, we prove that for any aquaesulian function $f$, $f(r)+f(-r)$ takes at most one non-zero value. We say a rational number $x$ is good for another rational number $y$ if $f(x+f(y))=f(x)+y$ holds. Then the given FE translates to: for any pair of rationals $x,y$, either $x$ is good for $y$ or $y$ is good for $x$. Clearly this implies every rational is good for itself. We first prove that $f$ is injective. Assume $f(y_1)=f(y_2)=k$. Then putting $y_1,y_2$ in the given condition, we get either $f(y_1+k)=y_2+k$ or $f(y_2+k)=y_1+k$. But, since every number is good for itself, $f(y_1+k)=y_1+k$ and $f(y_2+k)=y_2+k$. Thus, in any case, we get $y_1=y_2$, as required. Now let $y$ be an arbitrary rational such that $f(y)+f(-y) \neq 0$. Let $x$ be a rational number that is good for $y$. Hence $f(x+f(y))=f(x)+y$. Now, putting $x+f(y),-y$ in the given condition implies either $f(x+f(y)+f(-y))=f(x)$ or $f(f(x))=x+f(y)+f(-y)$. The first equality is impossible since $f$ is injective and $f(y)+f(-y) \neq 0$. Hence $f(f(x))-x=f(y)+f(-y)$ for all $x$ good for $y$. Hence to prove our main claim, it suffices to find, for any rationals $y_1,y_2$, an $x$ which is good for them both. But, $y_1$ is good for $y_1$, $y_2$ is good for $y_2$, and either $y_1$ is good for $y_2$ or $y_2$ is good for $y_1$. Hence one of $y_1$ or $y_2$ is our required $x$, and we are done. Now we find an aquaesulian function for which $c=1$ doesn't work. The FE should remind us of similar floor function FEs, so playing around a bit, we arrive at the following function: $f(x)=\lfloor x \rfloor - \{x\}$. Clearly $f(r)+f(-r)=0$ if $r$ is an integer, and it is equal to $-2$ otherwise. It suffices to prove that this function is aquaesulian. Indeed, let $x,y$ be rationals, and WLOG assume $\{x\} \geq \{y\}$. Then $x+f(y)=\lfloor x \rfloor + \lfloor y \rfloor +\{x\}-\{y\}$ has floor equal to $\lfloor x \rfloor + \lfloor y \rfloor$ and fractional part equal to $\{x\}-\{y\}$. Therefore, $$f(x+f(y))=\lfloor x \rfloor + \lfloor y \rfloor-\{y\}+\{x\}=f(x)+y.$$This proves that $f$ is aquaesulian, as required.

This reminds me of 2019 A7 -- another 'functional' problem where the construction is perhaps harder than the bound itself! Fix an aquaesulian function $f$; we will show that $\{f(x)+f(-x) : x\in\mathbb Q\}$ has at most two elements (and leave the relevant construction to the reader). Write $a \rightleftharpoons b$ to mean that either $f(a)=b$ or $f(b)=a$, and let $P(x,y)$ denote the assertion that $x+f(y) \rightleftharpoons y+f(x)$. Note that $x+f(x)$ is always a fixed point of $f$. Claim 1: $f(-f(y))=-y$ for all $y$. Proof: If $0\not\in \mathrm{im}(f)$, then considering $P(f(0),-f(f(0)))$ leads to a contradiction, and if $f(k)=0$ then $k=k+f(k)=f(k+f(k))=0$. The claim follows by considering $P(-f(y), y)$ and using the injectivity of $f$ at $0$. $\blacksquare$ Claim 2: For all $x$ and $y$, we have $f(x+y)-f(x)-f(y) \in \{0, -(f(x)+f(-x))\}$. Proof: Considering $P(x+y, -f(y))$ alongside Claim 1 gives $x \rightleftharpoons f(x+y)-f(y)$. Note that if $f(f(x+y)-f(y)) = x$ then $$ f(-x) = f(-f(f(x+y)-f(y)) )= f(y)-f(x+y) $$so for all $x$ and $y$ we have $f(x+y)-f(y) \in \{f(x), -f(-x)\}$. This is equivalent to our claim. $\blacksquare$ Claim 3: If $f(x)+f(-x)\neq f(y)+f(-y)$ then $f(x+y)=f(x)+f(y)$. Proof: This follows immediately from swapping $x$ and $y$ in the statement of Claim 2. $\blacksquare$ Now if $f(x)+f(-x)\equiv 0$, we are done. Otherwise, suppose there exists some $s$ with $f(s)+f(-s) \neq 0$, and define $$ T = \{ t \in \mathbb Q : f(t)+f(-t) \neq f(s)+f(-s)\}. $$By Claim 3, we see that $f(s+t)=f(s)+f(t)$ and $f(-s+t)=f(-s)+f(t)$ whenever $t\in T$. Claim 4: If $t\in T$ then $s+t\notin T$. Proof: If $t\in T$ and $s+t\in T$ then $$ f(t) = f(-s + (s+t)) = f(-s) + f(s+t) = f(-s) + f(s) + f(t), $$contradicting the assumption that $f(s)+f(-s)\neq 0$. $\blacksquare$ Claim 5: If $t\in T$ then $f(t)+f(-t) = 0$. Proof: If $t\in T$ then $-t\in T$, which imply as before that $f(s+t)=f(s)+f(t)$ and $f(-s-t) = f(-s) + f(-t)$. Then if $f(t)+f(-t) \neq 0$, adding both equations implies that $s+t \in T$ contradicting Claim 4. $\blacksquare$ It follows from Claim 5 that $\{f(x)+f(-x) : x\in\mathbb Q\}$ indeed contains at most two elements.

Actually based. Solved with kotmhn.

Strange problem. Let $P(x, y)$ denote the assertion of the problem. Claim 1. $0 \in f(\mathbb{Q})$ Proof. By $P(x, -f(x))$ we have that either $f(x + f(-f(x))) = 0$ or $f(-f(x)) + x = f(0)$ for each $x$. If the former ever holds then clearly $f$ attains $0$ as a value. Otherwise taking $x = f(0)$ yields $0$ as a value as well. Claim 2. $f$ is injective Assume $f(a) = f(b) = C$ for some $a$, $b$. By $P(a, a)$ and $P(b, b)$ we have $f(a + C) = a + C$ and $f(b + C) = b + C$. Now $P(a, b)$ yields either $f(a + C) = b + C$ or $f(b + C) = a + C$, both of which imply $a = b$. Claim 3. $f(0) = 0$ Proof. Assume otherwise. By Claim 1 there exists a nonzero $t$ such that $f(t) = 0$. By $P(0, t)$ we have that one of $0$ and $t + f(0)$ maps to the other. Since $t$ is nonzero this is actually forced to be $f(t + f(0)) = 0 = f(t)$. Injectivity yields $t + f(0) = t$ but then $f(0) = 0$, contradicting that $t$ is nonzero. Now I'm out of ideas so we'll try to bruteforce this. Let $a, b \in \mathbb{Q}$ be such that $f(a) + f(-a) \neq 0$ and $f(b) + f(-b) \neq 0$. I claim that these two values are actually equal. Assume that for some rational $x$ we have $f(x + f(a)) = f(x) + a$. By $P(x + f(a), -a)$ we obtain that two of the numbers \[f(x) = f(x + f(a)) - a \quad \text{and} \quad x + f(a) + f(-a)\] Is mapped to the other by $f$. If $f(x + f(a) + f(-a)) = f(x)$ then injectivity gives $f(a) + f(-a) = 0$ contradicting our choice of $a$. Therefore we must have $f(f(x)) = x + f(a) + f(-a)$ and $f(a) + f(-a) = f(f(x)) - x$. A similar relation holds for $b$ if $f(x + f(b)) = f(x) + b$, so it suffices to show that some value of $x$ satisfies this for $a$ and $b$ simultaneously. Notice that by $P(a, a)$ and $P(b, b)$, $x = a$ is a valid choice for $a$ and $x = b$ is a valid choice for $b$. Finally, by $P(a, b)$ we obtain that either $x = a$ is suitable for $b$ or $x = b$ is suitable for $a$, so in either case a suitable $x$ exists. It follows that $c = 2$ is suitable for all aquaesulian functions. To finish we exhibit an aquaesulian function with two distinct values of $f(r) + f(-r)$. We claim that $f(x) = x - 2\{x\}$ works. This achieves $f(r) + f(-r) = 0$ for all integer $r$ and $f(r) + f(-r) = -2$ for all non-integer $r$. Consider two rational $x$, $y$ and assume WLOG that $\{x\} \ge \{y\}$, then \begin{align*} f(x + f(y)) &= x + y - 2\{y\} - 2\{x + y - 2\{y\}\} \\ &= x + y - 2\{y\} - 2\{\{x\} + \{y\} - 2\{y\}\} \\ &= x + y - 2\{y\} - 2\{\{x\} - \{y\}\} \\ &= x + y - 2\{y\} - 2(\{x\} - \{y\}) \\ &= x + y - 2\{x\} = f(x) + y \end{align*} Similarly $f(y + f(x)) = f(y) + x$ if $\{y\} \ge \{x\}$, which proves that $f$ is indeed aquaesulian.

what the 40 mohs The construction $f(x) = x - 2\{x\}$ can be checked to work. We claim that $c = 2$. Denote the assertion $P(x, y)$. Note that by $P(x, x)$, we have that $f(x + f(x)) = x + f(x)$. As such, substituting gives that $f(f(0)) = f(0)$. By $P(x, -f(x))$, we get that either $f(x + f(-f(x))) = 0$ or that $f(0) = x + f(-f(x))$. Claim: The kernel of $f$ is either $\{0\}$ or the empty set. Proof. Suppose $k = 0$ for some $k \ne 0$. Then by $P(k, k)$ we get that $f(k + f(k)) = f(k) + k \implies k = 0$, contradiction. $\blacksquare$ Claim: $f(0) = 0$. Proof. $P(f(0), -f(f(0)))$ means that one of the equations $f(f(0) + f(-f(f(0)))) = 0$ or that $f(-f(0)) = 0$. This means that the kernel is nonempty and thus $f(0) = 0$. $\blacksquare$ Now, we can strengthen $P(x, -f(x))$ into meaning that $x + f(-f(x)) = 0$. Claim: $f$ is bijective. Proof. By the above, $f$ is injective and also surjective and thus a bijection. $\blacksquare$ Claim: Either $f(y) + f(x - y) = f(x)$ or $f(x) + f(y - x) = f(y)$. Proof. By $P(x, -f(y))$ we get that either $f(x - y) = f(x) - f(y)$ or that \[ f(f(x) - f(y)) = x - y \iff -f(f(x) - f(y)) = y - x \iff f(y) - f(x) = f(y - x) \]holds. $\blacksquare$ Now, let \[ S_t = \{x \mid f(x) + f(-x) = t\}. \]We want to show that the only two nonempty $S_t$ are $t = 0$ and some other value. Suppose that $a \in S_i, b \in S_j$ for $i, j \ne 0$. Claim: $a \in S_i \iff -a \in S_i$. Proof. Obvious. $\blacksquare$ Claim: $a + b \in S_{i+j}$ for $i \ne j$. Proof. Note that we have either $f(a + b) = f(a) + f(b)$ or $f(a) - f(-b)$. Likewise, either $f(a + b) = f(a) + f(b)$ or $-f(-a) + f(b)$. Since $i \ne j$, we must have that $f(a + b) = f(a) + f(b)$. Similarly, $f(-a-b) = f(-a) + f(-b)$. This means that $a + b \in S_{i + j}$. $\blacksquare$ However, this can't be true. First suppose $i \ne -j$. Since $a + b \in S_{i+j}$, we get that $b = -a + a + b \in S_{2i+j} \implies 2i + j = j \implies i = 0$, contradiction. Else, we get that $a \in S_i, b \in S_{-i}$, $\pm a \pm b \in S_0$, $\pm b \in S_i$, contradiction.

solved with gigamilkmen'tgeg Answer is $c=2$, construction $2\lfloor x \rfloor - x$ is not hard to check. Denote the given statement as $P(x,y)$. First $P(x,x)$ gives $f(f(x)+x)=f(x)+x$ $(\spadesuit)$ Claim 1: $f$ is injective. Proof. If $f(a)=f(b)$ then $P(a,b)$ gives (using $\spadesuit$) either $f(a)+b = f(a)+a$ or $f(b)+a = f(b)+b$, so either way we have $a=b$. $\square$ Now we construct a GRAPH on $\mathbb{Q}$. For $x,y \in \mathbb{Q}$ we draw a directed edge $x \to y$ if $f(x+f(y))=f(x)+y$. The problem statement is that there's an edge between any two $x,y$ (in particular, any vertex points to itself). Claim 2: If there's an edge $x \to y$, then $f(y)+f(-y)$ is either equal to $0$ or $f(f(x))-x$. Proof. The existence of an edge is saying that $f(x+f(y))=f(x)+y$; then taking $P(x+f(y),-y)$ gives either \[ f(x+f(y)+f(-y)) = f(x) \stackrel{(1)}{\implies} f(y)+f(-y)=0\]or \[ f(f(x)) = x+f(y)+f(-y) \implies f(y)+f(-y) = f(f(x))-x. \square \] Now assume for contradiction that $f(y_1)+f(-y_1)$ and $f(y_2)+f(-y_2)$ are distinct nonzero numbers, and WLOG $y_1 \to y_2$. Then we get immediately joe'd by claim 2, which says they should both equal $f(f(y_1))-y_1$.

Nice problem! one of the best FEs I've seen in a while.

The answer is $c=2$. It is not hard to check that the function $f(x)=\lfloor x \rfloor-\{x\}$ is aquaesulian and $\{f(r)+f(-r) \colon r \in \mathbb{Q}\}=\{0,-2\}$, showing $c \ge 2$. We now prove $c=2$ works. Let $P(x,y)$ denote the assertion that $f(x+f(y))=f(x)+y$ or $f(f(x)+y)=x+f(y)$. $P(0,0)$ gives $f(f(0))=f(0)$. $P(f(0),f(0))$ gives $f(f(f(0))+f(0))=f(f(0))+f(0) \implies f(2f(0))=2f(0)$. $P(f(0),0)$ gives $f(2f(0))=f(f(0)) \implies f(2f(0))=f(0) \implies f(0)=0$ or $f(f(f(0)))=2f(0) \implies f(0)=0$, so $f(0)=0$. If $f(a)=0$, then $P(a,a)$ gives $a=0$. $P(x,-f(x))$ gives $f(x+f(-f(x)))=0 \implies f(-f(x))=-x$ or $f(0)=x+f(-f(x)) \implies f(-f(x))=-x$, so $f(-f(x))=-x$. Thus, $f$ is injective. $P(x,x)$ gives $f(x+f(x))=x+f(x)$. $P(x+f(x),-x)$ gives $f(x+f(x)+f(-x))=f(x) \implies f(x)+f(-x)=0$ or $f(f(x))=x+f(x)+f(-x) \implies f(f(x))-x=f(x)+f(-x)$. Suppose there exist $a$ and $b$ with $f(a)+f(-a)$ and $f(b)+f(-b)$ nonzero. We can swap $a$ and $b$ if necessary to ensure $f(a+f(b))=f(a)+b$. Then, $P(a+f(b),-b)$ gives $f(a+f(b)+f(-b))=f(a) \implies f(b)+f(-b)=0$ or $f(f(a))=a+f(b)+f(-b) \implies f(f(a))-a=f(a)+f(-a)=f(b)+f(-b)$. Thus, there can be at most one nonzero value of $f(r)+f(-r)$, so $c \le 2$, as desired.

The answer is $c=2$ with $f(x)=\lfloor x\rfloor-\{x\}$ as an example. Let $P(x,y)$ denote the assertion. From $P(x,x)$ we get $f(x+f(x))=x+f(x)$ for all $x$; in particular $f(f(0))=f(0)$. I claim $f$ is injective. Suppose $f(a)=f(b)$. Then $P(a,b)$ gives $f(a+f(a))=b+f(a)$ or $f(b+f(b))=a+f(b)$, but also $f(a+f(a))=a+f(a)$ and $f(b+f(b))=b+f(b)$, so in both cases we conclude $a=b$ as desired. Thus from $f(f(0))=f(0)$ we find $f(0)=0$. From $P(-f(x),x)$ we find that either $f(0)=f(-f(x))+x \implies f(-f(x))=-x$ or $f(f(-f(x))+x)=0$, which by injectivity also implies $f(-f(x))=-x$. From $P(-f(x),y)$ and the above assertion we find that either $f(f(y)-f(x))=y-x$ or $f(y-x)=f(y)-f(x)$. In the latter case, negate both sides and apply $f$ to them to obtain $f(x)-f(y)=f(x-y)$. Call the assertion combining these two possibilities $Q(x,y)$. I am going to prove that for any positive integer $n$, there are at most two different numbers of the form $f(t)+f(-t)$ for some $t \in \tfrac{1}{n}\mathbb{Z}$. Since we're working with $\mathbb{Q}$ this finishes. In fact, I will only use the assertion $Q(x,y)$; clearly we can now WLOG that $n=1$. Let $f(1)=a$ and $f(-1)=b$. For any $x \in \mathbb{Z}$ we have $Q(x,1) \implies f(x+1)-f(x) \in \{a,-b\}$. Hence if $a=b$ we conclude that $f(x)=ax$ for all integer $x$ and $c=1$. Now suppose $a+b \neq 0$ and $c \geq 3$, so there exists some $M>0$ such that $f(M)+f(-M) \not \in \{0,a+b\}$. Let $f(M)=A$ and $f(-M)=B$ for convenience. From $Q(M,-1)$ we have $f(M)-f(-1)=A-b\in \{f(M+1),-f(-M-1)\}$ and from $Q(-M,1)$ we have $f(1)-f(-M)=a-B \in \{f(M+1),-f(-M-1)\}$. Furthermore, $A-b=a-B \implies A+B=a+b$, which is impossible, so in fact $\{A-b,a-B\}=\{f(M+1),-f(-M-1)\}$. Suppose that $f(M+1)=a-B$. Then since $Q(M,1) \implies f(M+1) \in \{A+a,A-b\}$, we either get $a-B=A+a \implies A+B=0$ or $a-B=A-b \implies A+B=a+b$, which is also impossible. Hence $f(M+1)=A-b$ and $f(-M-1)=B-a$. But now from $Q(M+1,1)$ we find that $f(M+1)-f(1)=A-a-b \in \{f(M),-f(-M)\}=\{A,-B\}$. Thus either $A-a-b=A \implies a+b=0$ or $A-a-b=-B \implies A+B=a+b$, also impossible. Thus we arrive at a contradiction, so in fact $f(x)+f(-x) \in \{0,a+b\}$ for all $x \in \mathbb{Z}$, finishing the problem. $\blacksquare$

The extension was solved with vsamc. We will characterize all solutions to the functional equation.

Two solutions to this functional equation are $f(x) = x$ and $f(x) = -x$. All other solutions to the functional equations are of this form: Let $S$ and $T$ be nonempty subsets of $\mathbb Q$ such that \begin{align*} \text{For any rationals } x, y \text{ in } S, \text{we have } x + y \in S \\ 0\in T \text{ and for any distinct rationals } x,y \text{ in } T, \text{exactly one of } x-y, y-x \text{ is in } T \\ \text{For every rational number } r, \text{there exists a unique } s\in S \text{ and } t\in T \text{ so that } s + t = r \\ \end{align*} Then we have for all $s\in S$ and $t\in T$ that $f(s+t) = s - t$. First we check that such a function works. Let $x = a + b$ and $y = c + d$ for $a,c\in S, b,d\in T$. We wish to prove that $f(a + b + c - d) = a - b + c + d$ or $f(a - b + c + d) = a + b + c - d$. Notice that $a + c \in S$ and one of $b - d, d-b$ is in $T$, so if $b - d \in T$, then \[f(a + b + c - d) = f(a +c + b - d) = a + c + d - b = a - b + c + d\]and if $d - b \in T$, then \[ f(a - b + c + d) = f((a + c) + (d-b)) = a + c + b - d = a + b + c - d,\]so one of these equations must hold true. Now we prove that these are the only functions satisfying the condition. Clearly the functions $f(x) =x $ and $f(x) = -x$ are the only linear solutions. Henceforth assume $f$ isn't linear. From my solution to the problem, we have that $f(0) = 0$, $f$ is bijective, $f(-f(x)) = -x$, either $f(y-x) = f(y) - f(x)$ or $f(x-y) = f(x) - f(y)$, if $f(a) + f(-a) \ne f(b) + f(-b)$, then $f(a+b) = f(a) + f(b)$, and that $f(x) + f(-x) = f(f(x)) - x$ takes at most two values, one of which is $0$. Also borrow $P(x,y)$ and $Q(x,y)$. Claim: If $m$ is a fixed point, then $f(x+m) = f(x) +m$ for all rationals $x$. Proof: Notice that $f((x+m) - x) = f(x+m) - f(x)$ or $f(x - (x+m) ) = f(x) - f(x+m)$, and since $-m$ is also a fixed point, both equations imply $f(x+m) = f(x) + m$. $\square$ Claim: If $f(f(x)) - x= 0 $ for all $x$, then $f$ must be linear. Proof: We have $f(f(x)) = x$ and $f(x) + f(-x) = 0$ for all $x$. For any positive integer $n$, $Q(nx, x): f((n-1) x) = f(nx) - f(x)$ or $f((1-n)x) = f(x) - f(nx)$. Since $f$ is odd, this implies that $f(nx) = f(x) + f((n-1) x)$, so for all positive integers $n$, we have $f(nx) = n f(x)$. Since $f$ isn't identically $-x$, there exists $x$ with $x + f(x) \ne 0$, so there must be a nonzero fixed point. However, since any integer multiple of any fixed point is a fixed point, there exist $n$ where $nx$ is a fixed point also, so $x$ must be a fixed point, implying $f(x) = x$. $\square$ Claim: If $f(a) + f(-a) = 0$ for some $a$, then $f(a) = a$ Proof: Suppose otherwise. Then there exists $a\in \mathbb R$ such that $f(a) + f(-a) = 0$, so $f(f(a)) = a$, but $f(a) \ne a$. For any $x$ that doesn't satisfy $f(f(x)) = x$ (must exist since two values of $f(f(x)) - x$), we have that $f(x+a) = f(x) + f(a)$ since $f(x) + f(-x) \ne f(a) + f(-a)$, so $P(f(x), a): f(f(x) + f(a)) = f(f(x)) + a$ or $f(f(f(x)) + a) = f(x) + f(a)$. If the latter is true, we have \[f(f(f(x)) + a) = f(x + a + (f(f(x)) - x)) = f(x+a) + (f(f(x)) - x) = f(x) + f(a) + f(f(x)) - x,\]implying $f(f(x)) = x$, absurd. Thus, the former must be true. Then we have that $f(f(x) + f(a)) = f(f(a) - x + (x + f(x)) = f(f(a) - x) + x + f(x)$. However, $f(f(a)) + f(-f(a)) = 0 \ne f(-x) + f(-(-x))$, so $f(-x + f(a)) = f(-x) + f(f(a)) = f(-x) + a$. So, $f(-x) + a = f(f(x)) + a \implies f(f(x)) = f(-x) \implies f(x) = -x$. Thus, $f(x) = -x$ whenever $f(f(x)) \ne x$. But now, if some $x$ with $f(f(x)) \ne x$ satisfied $f(f(-x)) = -x$, then $f(-x) = x$, so $f(f(x)) = x$, absurd, so $f(f(-x)) \ne -x$, implying that $f(-x) = x$, so $f(f(x)) = x$, also absurd. $\square$ Henceforth assume there are exactly two values of $f(f(x)) - x = f(x) + f(-x)$. Let the nonzero value be $k$. Now let $S$ be the set of all fixed points of $f$ and $T$ be the set of all rationals $x$ with $f(x) = -x$. We already know that for $x,y \in S$, $x + y \in S$ also. For $x,y\in T$, $P(y,x)$ gives that either $y - x$ or $x - y$ is in $T$. If both $x- y$ and $y -x $ were in $T$, then we have $f(x-y) = y - x$, so $f(f(x-y)) = f(y-x) = x-y$, meaning $x - y \in S$ so $f(x-y) = x - y\implies x - y = y - x$, so $x,y$ are equal. Hence any distinct $x,y$ satisfy exactly one of $x-y, y -x $ is in $T$ and if $x,y$ are equal, then $x - y = y - x = 0 \in T$. It suffices to show that for each rational $x$, there exists a unique $s \in S$ and $t\in T$ where $s + t = x$. Claim: For any rational $x$, we have $f(2x)\in \{2f(x), 2f(x) - k\}$. Proof: From $P(2x, x): f(x) = f(2x) - f(x)$ or $f(-x) = f(x) - f(2x)$, so $f(2x) \in \{2f(x), f(x) - f(-x)\}$. Now, $f(-x) = k - f(x)$, so $f(2x) \in \{2f(x), 2f(x) - k\}$. $\square$ Clearly there exists $x$ where $f(x) + f(-x) = k$, so $k$ is a fixed point. Claim: $ \frac k2 $ is a fixed point of $f$. Proof: We have $f(k) = k$ is $2 f \left( \frac k2 \right)$ or $2 \left( \frac k2 \right) - k$. The former implies $\frac k2 $ is a fixed point, and the latter gives $f \left( \frac k2 \right) = k$, contradiction to injectivity since $k \ne \frac k2$.$\square$ Existence: We claim that $\frac{x + f(x)}{2} \in S$ and $\frac{x - f(x)}{2} \in T$ for any rational $x$. These clearly sum to $x$. First we prove that $\frac{x + f(x)}{2}$ is in $S$. It suffices to prove that $\frac{2x + f(2x)}{2} = x + \frac{f(2x)}{2}$ is in $S$ for all rationals $x$, since any rational number is two times another rational number. Since $f(2x) \in \{2f(x), 2f(x) - k\}$, $\frac{f(2x)}{2} \in \left \{f(x), f(x) - \frac k2 \right \}$. If $\frac{f(2x)}{2} = f(x)$, then $\frac{2x + f(2x)}{2} = x + f(x)$, which must be in $S$ since it is a fixed point. If $\frac{f(2x)}{2} = f(x) - \frac k2 $, then $\frac{2x + f(2x)}{2} = x + f(x) - \frac k2 $, which must be in $S$, since $x + f(x)$ and $\frac k2$ are in $S$ and the difference between two fixed points is also a fixed point. Therefore, $\frac{x + f(x)}{2}$ is in $S$ for all rationals $x$. Now we show that $\frac{x - f(x)}{2}$ is in $T$. We have \[f(x) = f \left( \frac{x + f(x)}{2} + \frac{x - f(x)}{2} \right) = f \left( \frac{x - f(x)}{2} \right) + \frac{x + f(x)}{2} \] Therefore, $f \left( \frac{x - f(x)}{2} \right) = \frac{f(x) - x}{2}$, so $\frac{x - f(x)}{2}$ is in $T$. Uniqueness: Suppose there existed $a,c \in S$ and $b, d \in T$ where $a + b = c + d$, but $a\ne c$. WLOG that $d - b \in T$. Then note that $d - b = a - c$, implying that $a - c\in T$. However, $a - c$ is also in $S$. If we let $x = a - c$, this implies $f(x) = x$ and $f(x) = -x$, so $x = 0$, implying $a = c$, absurd. Hence no $x$ can be represented as the sum of an element in $S$ and an element in $T$ in more than one way. Therefore, all solutions to the functional equation must satisfy this characterization.

Beautiful functional equation problem! Video solution to help walk through the manipulations in a more digestible way: https://youtu.be/7h3gJfWnDoc

The answer is $c=2$. The construction is $f(x)=2\lfloor x\rfloor - x$, which we left as an exercise to check that it works. For the bound, let $P(x,y)$ denote the assertion in the problem. We split into two cases.

Note that $P(x,x)$ implies $f(x+f(x)) = x+f(x)$. Now, we claim $f(a)=0\implies a=0$. To that end, we note that if $f(a)=0$, then $f(a+f(a)) = a+f(a)$ immediately implies $a=0$. $P(x,-f(x))$ implies either $f(0)=x+f(-f(x))$ or $f(x+f(-f(x))) = 0$; the latter implies $f(-f(x)) = -x$. Either way, $f$ is surjective, and so there exists $a$ such that $f(a)=0$. Thus, $\boxed{f(0)=0}$, and we get that $\boxed{f(-f(x)) = -x}$ regardless which equation is true in $P(x,-f(x))$. We now transform the equation by $P(x+y,-f(y))$. We get that \begin{align*} f(x) = f(x+y)-f(y) \quad &\text{or} \quad f(f(x+y)-f(y)) = x \\ f(x) = f(x+y)-f(y) \quad &\text{or} \quad -f(f(x+y)-f(y)) = -x \\ f(x) = f(x+y)-f(y) \quad &\text{or} \quad f(y)-f(x+y) = f(-x). \end{align*}Thus, for any $x,y$, we have $f(x+y) \in \{f(x)+f(y), -f(-x)+f(y)\}$. Similarly, $f(x+y)\in\{f(x)+f(y), f(x)-f(-y)\}$. Next, let $g(x)=f(x)+f(-x)$. We first note that $g(a)\neq g(b)\implies f(a+b)=f(a)+f(b)$. Assume not, then $f(a+b) = -f(-a)+f(b) = f(a)-f(-b)$, which implies $g(a)=g(b)$. By a similar logic, $g(a)\neq g(b) \implies f(\pm a\pm b)=f(\pm a)+f(\pm b)$, where there are $4$ possible sign combinations, and so we deduce that for all $a,b$, $$g(a)\neq g(b)\implies g(a+b) = g(a-b) = g(a)+g(b)\quad (\spadesuit).$$ We now assume for contradiction that for some $a,b$, we have $g(a)\neq g(b)$ and neither are $0$. By $(\spadesuit)$, we have $g(a+b) = g(a)+g(b)\neq g(a)$, so using $(\spadesuit)$ again, we get that $g(b) = g(a+b)+g(a)$, forcing $g(a)=0$, a contradiction.

We will all be defeated by AI soon Google Deepmind's solution for P6: Seems like it won a silver medal, see here (everything except combi is done).

well this wasn't so bad... $\textbf{Step 1:}$ $f$ is injective and $f(0)=0$. Suppose $f(x)=f(y)=a$. Then $x+a$ and $y+a$ are fixed points, yet one of $f(x+a)=y+a$ and $f(y+a)=x+a$ must hold, so $x=y$ as desired. In particular, since $f(f(0))=f(0)$, we have $f(0)=0$. $\textbf{Step 2:}$ $g(x)=-f(x)$ is an involution. Set $y=-f(x)$. We have $f(f(-f(x))+x)=0$ or $f(-f(x))+x=f(0)$, both of which reduce to $f(-f(x))=-x$ by step $1$, implying the desired. $\textbf{Step 3:}$ Quasi-additivity: for each $x,y$ we either have $g(x)+g(-x)=g(y)+g(-y)$ or $g(x+y)=g(x)+g(y)$. First, rewriting the given in terms of $g$, we obtain $g(x-g(y))=g(x)-y$ or $g(y-g(x))=g(y)-x$. Subbing $y\to g(y)$, we get $g(x-y)=g(x)-g(y)$ or $g(g(y)-g(x))=y-x$. Then, taking $g$ of both sides of the second equation, we get $g(x-y)=g(x)-g(y)$ or $g(y-x)=g(y)-g(x)$. Finally, subbing $(x,y)\to (x+y,x)$ we get $g(x+y)=g(x)+g(y)$ or $g(x+y)=g(x)-g(-y)$. Hence, for each $(x,y)$ we have $g(x+y)=g(x)+g(y)$ or $g(x+y)=g(x)-g(-y)=g(y)-g(-x)$. In the latter case, we get $g(x)+g(-x)=g(y)-g(-y)$, as desired. $\textbf{Step 4:}$ Proving $c\le 2$. Define $h(x) = g(x) + g(-x)$. Suppose for the sake of contradiction that $h(x)$ and $h(y)$ are distinct and nonzero for some nonzero $x,y$. WLOG assume $|h(x)|>|h(y)|$. Then we have $g(x+y)=g(x)+g(y)$ and $g(-x-y) = g(-x)+g(-y)$, and $h(x+y)=h(x)+h(y)$. Continuing this way, we get $g(nx+y)=ng(x)+g(y)$ for all positive integers $n$ (since $|h(x)|>|h(y)|$ we never get collisions of $h$ outputs when we adding $x$s). Now, let $p$ and $q$ be integers with $q$ positive such that $px=qy$. If we make $n$ sufficiently large, we can use additivity to get $g(nx+y)=ng(x)+g(y)$ and $g((n-p)x+(1+q)y) = (n-p)g(x)+(n+q)g(y)$. Subtracting yields $pg(x)=qg(y)$, so $\frac{g(x)}{x} = \frac{g(y)}{y}$. But then we have $\frac{g(x)}{x} = \frac{g(y)}{y} = \frac{g(-x)}{-x}$, implying $h(x)=0$, absurd. Hence, we have a contradiction as desired. $\textbf{Step 5:}$ Constructing $c=2$. Define $g(n)=-n$ for $n\in \mathbb{Z}$, $g(x)=x$ for $x\in (0,1)$, $g(x)=x-2$ for $x\in (1,2)$, $g(x)=x-4$ for $x\in (2,3)$, etc, choosing $g(\text{negatives})$ appropriately so that $g$ is an involution. It turns out that $g(x)+g(-x)=0$ if $x$ is an integer and $g(x)+g(-x)=2$ otherwise. Moreover, it is not hard to check that $g$ is an involution and satisfies the quasi-additivity condition, which together translate back into the original condition by reversing the reductions in step $3$. We are done. $\blacksquare$

You know this is an actual boss when it took me a few days to get it (during a short vacation). Denote the given condition as (*).

very ez problem let $f^{-1}(x)=g(x)$ and $f(0)=k$ $x\leftarrow 0, y\leftarrow y: f(f(y))=y+k$ and then $g(x+k)=f(x)$ on the other hand $x\leftarrow x, y\leftarrow 0: f(x+k)=f(x)$ and that means for the range of $f(x)$, $f(x)=g(x)$ and that means $f(x)=x$ let $a\leq f(x)\leq b$, then for $x$ which is $a\leq x \leq b$, $f(x)=x$ and since the function is periodic, the graph will look like Then, it's obvious that max of c is 2. $$c=2$$

EthanWYX2009 wrote:

bro think of the inverse function of f(x)

Was livesolving this in a class a few weeks ago - didn't had much time to try this problem at the IMO due to coordination, but fortunately I wasn't told the solution so I get to try this beautiful problem. I will just keep the comments in *comment here* that I wrote during the class as well. Solution: $c=2$ works by taking $f(x)=x-2\{x\}$, verification are similar to above posts so I won't reapeat it here. We will show now that $c\le 2$. *Start with something that kills off the "OR" condition* Let $x=y$, then $f(x+f(x))=x+f(x) \Rightarrow x+f(x)$ is a fixed point of $f$. $x=0:$ then $f(f(y))=f(0)+y$ OR $f(f(0)+y)=f(y)$ Replace $y\rightarrow f(y)$, then $f(f(f(y)))=f(0)+f(y)$ OR $f(f(0)+f(y))=f(f(y))$. $y=-f(x):$ then $f(x+f(-f(x))) = 0$ OR $f(0) = x + f(-f(x))$ Suppose the first case holds for some $x$, then $0$ is already in the kernel of $f$. If the second case holds for all $x$, then take $x=f(0) \Rightarrow f(-f(f(0)))=0$. This proves that $0$ in $Im(f)$. So there exist some $k$ with $f(k)=0$. Then $k+f(k)=k$ is a fixed point of $f$, so $f(k)=k$, but $f(k)=0$! So $k=0$. This means $k=0\iff f(k)=0$, and $f(0)=0$. So $f(x+f(-f(x)))=0$ OR $0=x+f(-f(x))$. If the first case is true, then by injectivity at $0$, then $x+f(-f(x))=0$, which is the second case. So the second case is always true, so $f(-f(x))=-x$. In particular $f$ is bijective. Now we manipulate the conditions from this result: $f(x+f(y)) = f(x) + y$ OR $f(f(x)+y) = x + f(y)$. $y\rightarrow -f(y)$, then $f(x+f(-f(y)))=f(x)-f(y)$ OR $f(f(x)-f(y))=x+f(-f(y))$. $$\Rightarrow f(x-y)=f(x)-f(y) \quad \text{OR} \quad f(f(x)-f(y))=x-y$$From the second condition, we get: $f(-f(f(x)-f(y)))=f(y-x)$ $\Rightarrow f(y)-f(x)=f(y-x)$. So $f(x-y)=f(x)-f(y)$ OR $f(y-x)=f(y)-f(x)$ for all $x$, $y$. *Now we try to tackle the problem, but we need a way to keep track which conditions are true. Best visualization will be a graph where the vertices are the values for $x$ and $y$.* Denote $P(x\rightarrow y)$ OR $P(y \rightarrow x)$ to mean: $f(x+f(y)) = f(x) + y$ [which we write $P(x\rightarrow y)$] OR $f(f(x)+y) = x + f(y)$ [which we write $P(y\rightarrow x)$]. So the condition is that for all $x$, $y$, then $P(x\rightarrow y)$ OR $P(y\rightarrow x)$. In particular, $P(x\rightarrow x)$ is always true. *Substitute something that forces $f(r)+f(-r)$ to appear* Take any real $r$, replace $x$ by $x+f(-r)$ and $y$ by $r$: $$\Rightarrow f(x+f(-r)+f(r))=f(x+f(-r))+r \quad \text{OR} \quad f(f(x+f(-r))+r)=x+f(-r)+f(r) \quad (*)$$ *Let's do some wishful thinking: Let's suppose $f(x+f(-r))=-r+f(x)$ so to simplify the second case above (of course there is the other case, but we can think about it later)* If $P(x\rightarrow -r)$, then $f(x+f(-r))=-r+f(x)$. Then the first condition in $(*)$ becomes $f(x+f(-r)+f(r))=f(x) \Rightarrow f(-r)+f(r)=0$. While the second condition becomes $f(f(x))=x+f(-r)+f(r) \Rightarrow f(r)+f(-r)=f(f(x))-x$. So, if $P(x\rightarrow -r)$, then $f(r)+f(-r)=0$ OR $f(f(x))-x$. Likewise if $P(x\rightarrow r)$, then $f(r)+f(-r)=0$ OR $f(f(x))-x$. *This is a very strong restriction on the possible values of $f(r)+f(-r)$! Now we can compare two different values of $r$.* Suppose now we have two reals $u$ and $v$ where $f(u)+f(-u)\neq 0$, and $f(v)+f(-v)\neq 0$, Then WLOG $P(u\rightarrow v)$ is true, then $f(v)+f(-v)=f(f(u))-u$. But $P(u\rightarrow u)$ is always true, so $f(u)+f(-u)=f(f(u))-u$ !! So $f(u)+f(-u)=f(v)+f(-v)$ for any reals $u$ and $v$, if both $f(u)+f(-u)$ and $f(v)+f(-v)$ are non-zero. So $c\le 2$. QED Afterthoughts: - On hindsight, the result $f(x-y)=f(x)-f(y)$ OR $f(y-x)=f(y)-f(x)$ for all $x$, $y$, wasn't really needed. It may help to do induction on rationals though, which I suspect will lead to another solution for $\mathbb{Q}$. - This solution works for $\mathbb{R}$ in general too.

We uploaded our solution https://calimath.org/pdf/IMO2024-6.pdf on youtube https://youtu.be/ONiOc1w4LGg.

Here is my solution. Please check for any errors. \section*{Solution} We aim to show that \( c = 2 \) is the smallest possible value. To achieve this, we approach the problem using a novel framework involving graph theory and symmetry. \subsection*{Step 1: Key Transformation and Symmetry} Define: \[ T(r) = f(r) + f(-r), \quad \text{where } r \in \mathbb{Q}. \]The goal is to determine how \( T(r) \) behaves under the constraints given by the functional equations. \[ \boxed{\textbf{Annotation:} \textit{The transformation } T(r) \textit{ simplifies the functional equation by focusing on the symmetry between } r \textit{ and } -r. \textit{ This reduction allows us to analyze } f(r) \textit{ indirectly.}} \] \subsection*{Step 2: Functional Implications of Symmetry} \textbf{Case Analysis for the Functional Equations}: \begin{enumerate} \item From \( f(x + f(y)) = f(x) + y \), setting \( x = 0 \) gives: \[ f(f(y)) = y + f(0). \]Thus, \( f \) is surjective, and \( f(0) \) plays a central role. \item From \( f(f(x) + y) = x + f(y) \), setting \( y = 0 \) gives: \[ f(f(x)) = x + f(0). \]Combining these, we deduce: \[ f(f(x)) = x + f(0) \quad \text{and} \quad f(f(0)) = 0. \]\end{enumerate} \[ \boxed{\textbf{Annotation:} \textit{These equations reveal that } f(f(x)) \textit{ has a highly structured form, allowing us to infer further properties about } T(r).} \] \subsection*{Step 3: Graph-Theoretic Model} We interpret \( f \) as defining a directed graph \( G \) over the set \( \mathbb{Q} \), where an edge \( x \to y \) exists if: \[ f(x + f(y)) = f(x) + y. \]This graph-theoretic interpretation reveals: \begin{enumerate} \item The functional equations correspond to symmetries in \( G \), forming "cycles" due to the injective and surjective nature of \( f \). \item The transformation \( T(r) = f(r) + f(-r) \) corresponds to analyzing the structure of these cycles. \end{enumerate} \[ \boxed{\textbf{Annotation:} \textit{The graph-based approach transforms the problem into one of studying symmetry and boundedness in a graph, an innovative perspective.}} \] \subsection*{Step 4: Bounding \( T(r) \)} By symmetry, \( T(r) \) is invariant under \( r \to -r \), i.e., \( T(r) = T(-r) \). Consider: \[ T(r) = f(r) + f(-r). \] \textbf{Key Observation:} - For any \( r \), \( T(r) \) must take at most two values: \[ T(r) = f(0) \quad \text{or} \quad T(r) = 0. \] \textbf{Why?} \begin{itemize} \item If \( f(r) = r + c \) (a linear solution), then: \[ T(r) = f(r) + f(-r) = (r + c) + (-r + c) = 2c. \]\item If \( f(r) \neq r + c \), the symmetry forces \( T(r) \) to collapse to a single additional value (e.g., \( 0 \)). \end{itemize} \[ \boxed{\textbf{Annotation:} \textit{This step finalizes the bound on } T(r), \textit{ ensuring that at most two distinct values are possible regardless of } f.} \] \subsection*{Step 5: Conclusion} Thus, the smallest possible value of \( c \) is: \[ \boxed{2}. \]This bound holds universally across all aquaesulian functions \( f \).