This is A5.

From $P(x,f(x)+y), P(x,2f(x)+y),\dots,P(x,-f(x)+y),P(x,-2f(x)+y),\dots$, one can easily know, by induction, that $f(x+y)+f(x-y)=f(x+nf(x)+y)+f(x-nf(x)-y)\forall x,y,n \in \mathbb{Z}$. Let's call this equation $Q(x,y,n)$. Let $k=lcm(f(0),f(1))$, and suppose that $k=af(0)=bf(1)$. $Q(0, -x, a):f(-x)+f(x)=f(-x+k)+f(x-k)$ $Q(1, -x-1, b):f(-x)+f(x+2)=f(-x+k)+f(x+2-k)$ $\Rightarrow f(x+2)-f(x)=f(x+2-k)-f(x-k)......(1)$ For convenience, let $g(x,y)=f(x+y)-f(x)$, then $g(x,2)=g(x-k,2)\forall x\in \mathbb{Z}.......(2)$. By $(2)$, for all integer $x$, $f(x+2k)-f(x)=g(x,2k)=\sum_{i=0}^{k-1} g(x+2i,2)=\sum_{i=0}^{k-1} g(x+2i \mod k, 2)=\sum_{i=0}^{k-1} g(i, 2)=g(0,2k)=const......(3)$ Let $k_{\min}$ be the smallest positive integer satisfying $(3)$ and $c=g(0,2k_{\min})\in 2\mathbb{Z}$, then from $P(x+2k_{\min},y)$ and $P(x,y+c)$, $f(x+y)+f(x-y)+2c=f(x+2k_{\min}+y)+f(x+2k_{\min}-y)$ $=f(x+2k_{\min}+f(x+2k_{\min})+y)+f(x+2k_{\min}-f(x+2k_{\min})-y)$ $=f(x+f(x)+y+c)+f(x-f(x)-y-c)+2c=f(x+y+c)+f(x-y-c)+2c$ That is, $g(x+y,c)=g(x-y-c,c)$, so there exist two constants $c_1,c_2\in 2\mathbb{Z}$ such that $g(2s+1,c)=c_1,g(2s,c)=c_2\quad \forall s\in\mathbb{Z}$. Hence, $2k_{\min} c_2 = \sum_{i=0}^{2k_{\min} -1} g(ci,c) = g(0, 2k_{\min} c) = \sum_{i=0}^{c-1} g(2k_{\min} i, 2k_{\min} ) = c^2$. Similarly, $2k_{\min} c_1 = c^2$, so $c_1=c_2$ (note that $k_{\min} > 0$). Accordingly, $2k_{\min} =c$ also satisfies $(3)$. Hence, by the definition of $k_{\min}$, $2k_{\min} | c$. Assume that $c=2k_{\min}t$. Also, since $f(x)\in 2\mathbb{Z}+1$, there exists $A\in \mathbb{Z}$ such that $k_{\min}f(x)=A\times 2k_{\min}+k_{\min}$. Hence, $Q(x,y,k_{\min}):f(x+y)+f(x-y)=f(x+k_{\min}f(x)+y)+f(x-k_{\min}f(x)-y)$ $=f(x+k_{\min}+y)+2Ak_{\min}t+f(x-k_{\min}-y)-2Ak_{\min}t=f(x+k_{\min}+y)+f(x-k_{\min}-y)$. That is, $g(x+y,k_{\min})=g(x-y-k_{\min},k_{\min})$. By the definition of $k_{\min}$, $k_{\min}|k\in 2\mathbb{Z}+1$, so one can easily prove that $g(x,k_{\min})$ is a constant, which is $k_{\min}t$ Assume that $k^{'}$ is the smallest positive integer such that $g(x,k^{'})$ is a constant, then $k^{'}|k_{\min}\in 2\mathbb{Z}+1$ and $g(x,k^{'})=k^{'}t.....(4)$. Now, our target is to prove that $k^{'}|f(x)\forall x$. Otherwise, there must exist a prime $p$ and a integer $x_0$ satisfying $p|\frac{k^{'}}{f(x_0)}$. Since $k^{'}$ is odd, $p$ is an odd prime. Let $X=\{x|p\mid \frac{k^{'}}{f(x)}\}$. From $(4)$, $x\in X\Leftrightarrow x\mod k^{'} \in X $. If $x\in X$, then there exist two integers $A,B$ s.t. $Af(x)-Bk^{'}=\frac{k^{'}}{p}$. $Q(x,y,An):f(x+y)+f(x-y)=f(x+Anf(x)+y)+f(x-Anf(x)-y)$ $=f(x+\frac{nk^{'}}{p}+y)+B k^{'}t+f(x-\frac{n k^{'}}{p}-y)-B k^{'}t=f(x+\frac{nk^{'}}{p}+y)+f(x-\frac{n k^{'}}{p}-y)......R(x,y,n)$ Lemma 1 At least $p-1$ integers in $x_0,x_0+\frac{k^{'}}{p},\dots,x_0+\frac{(p-1)k^{'}}{p}$ are elements of $X$. proof: Otherwise, there must be two integers $0\leq a<b\leq p-1$ s.t. $x_0+ \frac{ak^{'}}{p},x_0+\frac{bk^{'}}{p}\not\in X$. For convenience, let $Y=\{i\mid x_0+ \frac{ik^{'}}{p}\in X\}$. For any $0\leq i,j\leq p-1$, if $i\in Y$, then from $R(x_0+ \frac{ik^{'}}{p}, 0,j)$, one can get that at least one of $i-j,i+j\in Y$. Since $x\in X\Leftrightarrow x\mod k^{'} \in X $, one of $(i-j)\mod p, (i+j)\mod p \in Y$. Hence, if $i,j\not\in Y$, then $\frac{i+j}{2}\mod p\not\in Y$. Since $p$ is an odd prime, one can obtain that $a+\frac{s(b-a)}{2^l}\not\in Y$ where $0\leq s \leq 2^l$. If $2^l > p$, then there must exist $s$ s.t. $a+\frac{s(b-a)}{2^l}=0$, so $0\not\in Y$, a contradiction. So the lemma is proved. Back to the main problem, if $f(x)$ is a solution, then $h(x)=f(x+s)$ is also a solution. So WLOG $x_0 = 0$ and from lemma, $0,\dots,p-2 \in Y$. From (4), it's easy to get that $g(x, \frac{k^{'}}{p})=g(x\mod k^{'}, \frac{k^{'}}{p})\forall x$......(5).For all $y\in\mathbb{Z},0\leq i<p-1$, from $R(\frac{ik^{'}}{p},y+\frac{(i-1)k^{'}}{p},1)$, we can know that $g(y+\frac{(2i-1)k^{'}}{p},\frac{k^{'}}{p})=g(-y,\frac{k^{'}}{p})$ Put $y=y+\frac{2k^{'}}{p},i=p-2$, $\Rightarrow g(y+\frac{(2p-3)k^{'}}{p},\frac{k^{'}}{p})=g(-y+\frac{2k^{'}}{p},\frac{k^{'}}{p})$. Put $y=-y+\frac{k^{'}}{p},i=1$, $\Rightarrow g(y+\frac{(2p-3)k^{'}}{p},\frac{k^{'}}{p})=g(-y+2\frac{k^{'}}{p},\frac{k^{'}}{p})=g(y-\frac{k^{'}}{p},\frac{k^{'}}{p})=g(-y,\frac{k^{'}}{p})$ So from $(5)$, $g(y+\frac{ik^{'}}{p},\frac{k^{'}}{p})=g(-y,\frac{k^{'}}{p})\quad \forall 0\leq i\leq p-1$, and from $(4)$, $g(y+\frac{ik^{'}}{p},\frac{k^{'}}{p})=\frac{k^{'}t}{p}\quad \forall 0\leq i\leq p-1$. In particular, when $i=0$, $g(y,\frac{k^{'}}{p})=\frac{k^{'}t}{p}$, a contraction with the definition of $k^{'}$. In conclusion, $f(x+k^{'})-f(x)=k^{'}t$ and $k^{'}|f(x)\forall x$. One can easily obtain that such $f$ is a solution to the equation, so we're done.

va2010 wrote: Let $2\mathbb{Z} + 1$ denote the set of odd integers. Find all functions $f:\mathbb{Z} \mapsto 2\mathbb{Z} + 1$ satisfying \[ f(x + f(x) + y) + f(x - f(x) - y) = f(x+y) + f(x-y) \]for every $x, y \in \mathbb{Z}$. Let $T(x,y)$ be a function defined by $T(x,y)=f(x+y)+f(x-y)$. Then,the assertion is equivalent to $$\boxed{P(x,y)\, ; \,T(x,y)=T(x,f(x)+y)}$$We claim that the function $f(x)$ satisfy the assertion if and only if there exists two integers $S,P$ such that $f(x+S)=f(x)+P$ $S|f(x)\,\forall x\in\mathbb{Z}$ It is easy to see that any function that satisfy the above conditions works. Now, we will prove that these are indeed all solutions. Let $a_x$ denote the smallest positive integer that can replace $f(x)$ in the assertion. That is, the smallest positive integer such that $T(x,y)=T(x,a_x+y)$ for all integers $y.$ Note that $a_x|f(x)$ for all integers $x,$ meaning that $a_x$ are odd for all $x.$ Claim 1 : $a_{x-y}|lcm[a_x,a_{x+y}]$ for all integers $x,y.$ Proof : Note that $T(x,y+z)+T(x,y-z)=T(x+y,z)+T(x-y,z)$ for all $x,y,z.$ Thus $$T(x,y+z)+T(x,y-z)=T(x+y,z)+T(x-y,z)$$$$T(x,y+z+lcm[a_x,a_{x+y}])+T(x,y-z-lcm[a_x,a_{x+y}])=T(x+y,z)+T(x-y,z)$$$$T(x+y,z+lcm[a_x,a_{x+y}])+T(x-y,z+lcm[a_x,a_{x+y}])=T(x+y,z)+T(x-y,z)$$$$T(x+y,z)+T(x-y,z+lcm[a_x,a_{x+y}])=T(x+y,z)+T(x-y,z)$$$$T(x-y,z+lcm[a_x,a_{x+y}])=T(x-y,z)$$Hence,it is easy to see that the claim is true and we are done. Now,from the special case of Claim 1 where $y=\pm 1$,we can easily get the following corollary : Corollary 2 : $Z=lcm[…,a_{-2},a_{-1},a_0,a_1,a_2,…]$ exists, and is equal to $lcm[a_i,a_{i+1}]$ for all integers $i.$ This corollary implies that $Z$ is the smallest positive integer such that $T(c,d)=T(c,Z+d)$ for all integers $c,d.$ Claim 3 : The first condition can holds for $S=Z$, and for any pair of $S,P$ such that the first condition holds,we must have $S\ge Z.$ Proof : Since $T(c,d)=T(c,Z+d)$,we have $$T(c,0)=T(c,Z)\implies 2f(c)=f(c+Z)+f(c-Z)\,\forall c\in\mathbb{Z}$$This means that all $Z$ sequences of the form $\{f(c+nZ)\}_{n\in\mathbb{Z}}$ are in arithmetic progression. Also, since $Z$ is odd, we have $$T(c+\frac{Z+1}{2},\frac{1-Z}{2})=T(c+\frac{Z+1}{2},\frac{1+Z}{2})\implies f(c+Z)-f(c)=f(c+1+Z)-f(c+1)$$And hence all such arithmetic sequences have the same difference $D$ and the first part of this claim is proved. For the second part, just note that if $S$ satisfies the first condition, then $T(c,d)=T(c,S+d)$ and we are done. Now,it is clear that if the second condition holds for $S=Z$ then we are done. Let $b_x=gcd(f(x),Z)$,we wish to prove that $b_x=Z$ for all $x.$ Fix the integer $x$,we will show that the following claims hold. Claim 4 : $a_{x+nb_x}|b_x$ for all integers $n.$ Proof : First, since $a_x|f(x)$ and $a_x|Z$, we can see that $a_x|b_x.$ Assume that for some $n$,we have $a_{x+nb_x}\nmid b_x,$ then there must exists odd prime power $p^k$ such that $p^k|a_{x+nb_x}$ but $p^k\nmid b_x\implies p^k\nmid a_x.$ Now, from Claim 1, we have $a_{x+nb_x}|lcd[a_x,a_{x-nb_x}]\implies p^k|a_{x-nb_x}.$ And thus we have $p^k|f(x-nb_k),f(x+nb_x),Z.$ But then $$T(x,y)=T(x,y+a_x)\implies T(x,y)=T(x,y+nb_x)$$$$\implies 2f(x)=f(x+nb_x)+f(x-nb_x)$$$$\implies p^k|2f(x)\implies p^k|f(x)$$Hence $p^k|gcd(f(x),Z)=b_x$ which is a contradiction and we are done. Claim 5 : The first condition can hold for $S=b_x.$ Proof : From Claim 4, we can see that $$T(x+nb_x,y)=T(x+nb_x,y+a_{x+nb_x})\implies T(x+nb_x,y)=T(x+nb_x,y+mb_x)$$for all integers $n,m.$ Thus,by expanding,we can see that $$\left(\frac{Z}{b_x}\right)(T(x+b_x,y)-T(x,y+b_x))=\sum_{i=0}^{\frac{Z}{b_x}-1}(T(x+b_x,y+ib_x)-T(x,y+ib_x))$$$$\implies \frac{Z}{b_x}(f(x+b_x-y)-f(x-b_x-y))=(f(x+y+Z)-f(x+y))+(f(x-y+b_x)-f(x-y+b_x-Z))$$$$\implies f(x+b_x-y)-f(x-b_x-y)=\frac{b_x}{Z}(2D)=\frac{2Db_x}{Z}$$$$\implies f(z+2b_x)-f(z)=\frac{2Db_x}{Z}\,\forall z\in\mathbb{Z}$$$$\implies f(z+(Z+1)b_x)-f(z)=\left(\frac{Z+1}{2}\right)\left(\frac{2Db_x}{Z}\right)=Db_x+\frac{Db_x}{Z}$$$$\implies f(z+b_x)-f(z)=(Db_x+\frac{Db_x}{Z})-b_xD=\frac{Db_x}{Z}\,\forall z\in\mathbb{Z}$$And the result follow. Hence, from Claim 3 and Claim 5, we have $b_x\ge Z \implies b_x=Z$ and we are done. $\square$

Thinking of regular Z-gon (Z is same as TLP39’S Z) will help this problem.

Pathological wrote: Set $r = \frac{P}{p},$ notice that $q | r.$ Sorry about the (probably dumb) question. How do we get $q|r$?

stroller wrote: Pathological wrote: Set $r = \frac{P}{p},$ notice that $q | r.$ Sorry about the (probably dumb) question. How do we get $q|r$? Dear $\mathbf{stroller}$, Sorry, I was actually wrong about that, as far as I can tell there is no way to get $q|r$. Fortunately, this is easily patched by defining $q = \gcd(f(P) - f(0), P)$ instead. I think the rest of the proof of the lemma now works fine. Thanks! --Pathological

Here is a solution which solves the FE without any substitution . Call a function $g$ quasi-periodic if it can be written as the sum of a periodic function and a linear function. The period of that periodic function is called quasi-period. The answers are any quasi-periodic function with quasi-period dividing any element in $f(\mathbb{Z})$. It's easy to show that these work. So the main part is to prove that these are all solutions. Define a block as a closed interval whose endpoints are integers. A value of block $[x,y]$ is $v[x,y] = f(y)-f(x)$. Note that $v[a,b] = v[a,c]+v[c,b]$. Under this terminology, the functional equation now reads ISL 2015 A5, rephrased wrote: The value of any block of length (multiple of) $f(x)$ is equal to the value of its reflection across point $x$. The proof now proceeds in three steps. Step 1: Show that $f$ is quasi-periodic with odd quasi-period. Consider a block of length $L=\mathrm{lcm}(f(0),f(1))$. Reflecting across $0$ and across $1$ preserves its value. But the composition of these two transformations is translation of distance $2$. Thus we have $$v[x,x+L] = v[x+2,x+L+2]$$for any $x$. Thus the function $g(x)=v[x,x+L]$ is periodic with period $2$. Moreover, from the functional equation, $v[0,L]=v[-L,0]$ which means $g(0)=g(-L)$. But $L$ is odd hence $g$ must be constant. Thus $f$ is quasi-periodic with odd quasi-period dividing $L$. Step 2: Analyze one reflection Let $T$ be the minimal quasi-period of $f$. We analyze a reflection further and claim that. Claim: The value of any block of length $\gcd(T,f(a))$ is equal to the value of the block's reflection across $a$. Proof: Define function $g(x) = f(x)+f(2a-x)$. By the functional equation, $g$ is periodic with period $f(a)$. However, from Step 1, $g$ is periodic with period $T$. Thus $g$ is periodic with period $\gcd(T,f(a))$ hence the claim $\blacksquare$. Step 3: Show that the quasi-period must divide $f(a)$ for any $a$. We now mix the reflections to derive the final result. Assume that there exists a prime power $p^e\| T$ that does not divide $f(a)$. We prove the following claim. Claim: $p^e$ must divide $f\left(a\pm f(a)\right)$. Proof: We will prove only the plus sign. The minus follows analogously. Let $b=a+f(a)$. Assume for the contradiction that $p^e\nmid f(b)$. By Number Theory, we get $\gcd(T,f(a))\mid\tfrac Tp$ and $\gcd(T,f(b))\mid\tfrac Tp$. From Step 2, reflecting block of length $\tfrac Tp$ through $a$ and $b$ preserve its value, and their composition is translating by length $2(b-a)$, we get $$v\left[x,x+\tfrac Tp\right] = v\left[x+2(b-a),x+2(b-a)+\tfrac Tp\right].$$Thus function $g(x) = f(x+\tfrac Tp)-f(x)$ is periodic with period $2(b-a)=2f(a)$. However, it has period $T$ from step 1. Hence the minimal period must divide $\gcd(2f(a),T)\mid\tfrac Tp$. However, note that $$f(x+T)-f(x) = g(x)+g\left(x+\tfrac Tp\right)+g\left(x+\tfrac{2T}p\right)+\hdots + g\left(x+\tfrac{(p-1)T}{p}\right) = pg(x)$$The LHS is constant thus the RHS must be constant too. This gives $f(x+\tfrac Tp)-f(x)$ is constant, or $f$ has quasi-period $\tfrac Tp$, contradiction to minimality. $\blacksquare$ To finish, we note that $f(a+f(a)) + f(a-f(a)) = 2f(a)$. The left hand side is divisible by $p^e$, contradiction.

Very hard problem. The answer is all functions such that there exist $T$ such that $f(x)-f(x-T)=Tk$ for some constant $k,$ and $T\mid f(x)$ for all $x.$ They clearly work, and I will show they are the only functions. Let $P(x,y)$ denote the assertion $f(x + f(x) + y) + f(x - f(x) - y) = f(x+y) + f(x-y).$ By $P(x,f(x)+y),$ et cetera, we can easily see $f(x + kf(x) + y) + f(x - kf(x) - y) = f(x+y) + f(x-y),$ for which we will denote $Q(x,y,k)$. Let $[a,b]$ denote the lcm of $a,b$ Claim: $f(x+2)-f(x)$ has period dividing $T=\gcd_{x\in Z} [f(x),f(x+1)]$ Proof. Let $A=[f(x),f(x+1)]$. Consider $Q(x,y,\frac{A}{f(x)})$ and $Q(x+1,y-1,\frac{A}{f(x+1)})$. We can see $$f(x+A+y)+f(x-A-y)=f(x+y)+f(x-y)$$ and $$f(x+A+y)+f(x-A-y+2)=f(x+y)+f(x-y+2)$$ Therefore, substituting $z=x-y$, for all $z\in \mathbb{Z}, f(z+2)-f(z)=f(z+A+2)-f(z+A)$ as desired. Claim: $f(x+1)-f(x)=f(x+2T+1)-f(x+2T)$. Proof. Let $g(x)=f(x+1)-f(x)$, then $g(x)-g(x+T)=g(x+T+1)-g(x+1)$. This implies $g(x)-g(x-T)=(-1)^xM$ for some integer $M$. This implies $g(x)=g(x-T)+(-1)^xM=g(x-2T)+M((-1)^x+(-1)^{x-T})=g(x-2T)$ as $T$ is odd. Let $U$ be the minimal period of $f(x)-f(x-1).$ We know $U\mid 2T$ exists. I first show $U|f(x)-f(x-U)=C$ Comparing $P(x,0)$ and $P(x-U,0)$ we get $f(x+C)-f(x)=f(x)-f(x-C)=\frac 1U (f(x)-f(x-UC))=\frac{C^2}{U}$ is a constant, so it follows that $U\mid C$. Claim: $\frac{U}{2^{\nu_2(U)}}\mid f(x)$ for all $x$. Suppose $p^e\mid U, p^e\nmid f(x).$ This implies $\gcd(f(x),U)\mid \frac Up$. Observe $$f(x+kf(x)+y)+f(x-kf(x)-y)=f(x+(kf(x) \bmod\; U) + y)+f(x-(kf(x) \bmod\; U)-y$$ Let $g(y)=f(x+\frac{yU}{p})$. Let $S=\{ x: U\nmid g(x): x\in \mathbb{Z}_p\}$. We can see if $x\notin S,$ then for all $y$, $y$ or $2x-y\notin S$. Since $S$ is nonempty, we assume $S$ is not full. Suppose $a\notin S, a+m\in S.$ This would imply $a+2^km\in S,$ and we can use divide and conquer to easily see that for all $y\ne a, x\in S.$ Therefore, either all elements of $\mathbb{Z}_p$ are contained in $S$, or exactly one of them is not in $S$. If all of them are contained in $S$, then we can see that $g$ is actually linear, which is a contradiction. Therefore, one of them is not in $S$. We can still see $g$ is linear. Therefore, $f(x+k\frac Up)=f(x)+k\frac Cp$ for all $k\in \mathbb{Z}$. $P(x',a): f(x+\frac Up+a)+f(x-\frac Up-a)=f(x+a)+f(x-a)$ $P(x'-\frac Up,a): f(x-\frac Up+a)+f(x-\frac Up-a)=f(x+a)+f(x-\frac{2U}{p}-a)$ Subtraction yields $f(x'+\frac Up+a)-f(x'+a)=f(x'-\frac{U}{p}-a)-f(x'-\frac{2U}{p}-a)$ for all $\frac Up\mid x'-x, a\in \mathbb{Z}$ We can also see $f(x'+3\frac Up+a)-f(x'+2\frac Up+a)=f((x'+\frac Up)+\frac Up + (a+\frac Up))-f((x'+\frac Up)+ (a+\frac Up))=f(x'-\frac{U}{p}-a)-f(x'-\frac{2U}{p}-a)=f(x'+\frac Up+a)-f(x'+a)$ Let $M=f(x'+(2k+1)\frac Up+a)-f(x'+2k\frac Up+a), N=f(x'+2k\frac Up+a)-f(x'+(2k-1)\frac Up+a)$. Since $f(x)-f(x-U)=C,$ we can see that $M=f(x'+(2k+1)\frac Up+a)-f(x'+2k\frac Up+a)=f(x'+(2k+p+1)\frac Up+a)-f(x'+(2k+p)\frac Up+a)=f(x'+2(k+\frac{p+1}{2})\frac Up+a)-f(x'+(2(k+\frac{p+1}{2})-1)\frac Up+a)=N.$ Since $f(x)-f(x-U)=pM, f(x)-f(x-\frac Up)=\frac Cp$ for all $x\in \mathbb{Z}$. If $U$ is odd, I'm done. It remains to show $U$ cannot be even. Let $U=2U'.$ Let $f(x)=(2m+1)U'$, then $P(x,0)$ gives $2f(x)=f(x+(2m+1)U')+f(x-(2m+1)U')=f(x+U')+mC+f(x-U')-mC=f(x+U')+f(x-U')=2f(x)$. Therefore, $f(x+U')-f(x)=f(x)-f(x-U')=\frac{C}{2}$ for all $x$, as desired.

If we define $A(x, y) = f(x + y) + f(x - y)$ then the condition becomes $A(x, y) = A(x, y + f(x))$. In other words, $y \to A(x, y)$ for fixed $x$ is periodic with period $f(x)$. Iterating the equation gives that \[ A(x, y + kf(x)) = A(x, y) \]Let $T_{a,b} = \text{lcm}(f(a), f(b))$. Let $m_{2n}$ be the minimal period such \[ f(x + 2n) - f(x) = f(x + 2n + m_{2n}) - f(x + m_{2n}) \] Claim: Let $T_{a,b} = \text{lcm}(f(a), f(b))$. Then $f(x + 2b - a) - f(x + a)$ is periodic with period dividing $T$. Proof. It follows that \[ A(a, x) = A(a, x + T_{a,b}) \]and \[ A(b, x + b - a) = A(b, x + b - a + T_{a,b}). \]Subtracting the two equations gives \[ f(x + 2b - a + T_{a,b}) - f(x + a + T_{a,b}) \]$\blacksquare$ Notably, when $b = 1$ and $a = 0$ it follows that that $f(x + 2) - f(x)$ is periodic as $x$ varies. Then, summing means that \[ \sum_{k=0}^{T_{0,1}-1} f(x + 2k + 2) - f(x + 2k) = \sum_{k=0}^{T_{0,1}-1} f(x + 2k + T_{0,1} + 2) - f(x + 2k + T_{0,1}) \]or that \[ f(x + 2T_{0,1}) - f(x) = f(x + 3T_{0,1}) - f(x + T_{0,1}) \]which means that \[ 2f(x) = f(x + T_{0,1}) + f(x - T_{0,1}) \]or that $A(x, 0) = A(x, T_{0,1})$ for all $x$. As such, the sequence \[ [x] = \{ f(x - 2T_{0,1}), f(x - T_{0,1}), f(x), f(x + T_{0,1}), \dots \} \]is an arithmetic progression. Claim: All such arithmetic progressions have a common difference. In other words, $f(x + T_{0,1}) - f(x)$ is constant. Proof. Since $T_{0,1}$ is odd, we can take the sequences $\pmod{T_{0,1}}$. Since $f(x + 2) - f(x)$ is periodic, it follows that $[a]$ and $[a + 2]$ have the same common difference. $\blacksquare$ Let $T \mid T_{0,1}$ be minimal such $2f(x) = f(x + T) + f(x - T)$ holds. As such, it follows that \[ f(x) = k \cdot \left\lfloor \frac{x}{T} \right\rfloor + g(x \pmod{T}) \]for some function $g$ and $k$. Claim: If $A(x, y) = A(x, y + p)$ for all $x$ and $y$, it follows that $T \mid p$. Proof. This implies that for any $m$ and $n$, \[ A(x, 0) = A(x, mT) = A(x, mT - nk). \]By Bezout's and minimality it must follow that $T \mid \gcd(T, k) \mid k$. $\blacksquare$ Since $A(x, y) = A(x, y + f(x))$ it follows that $T \mid f(x)$ for all $x$. Claim: $T \mid k$ Proof. It follows from simplifying $A(x + T, y + T)$ that $A(x, y + f(x)) = A(x, y + f(x) + k)$. $\blacksquare$ This in turn implies that $T \mid g(x)$ for all $x$. As such, due to the domain, the solution set is thus \[ f(x) = T \cdot \left(k \cdot \left\lfloor \frac{x}{T} \right\rfloor + g(x \pmod{T})\right) \]for some $T > 0$, $g \colon {\mathbb Z}/t{\mathbb Z} \to {\mathbb Z}$ and even $k$. This can easily be checked to work.

Who proposed this ?

$$f(x+2k)-f(x)$$$$=g(x,2k)$$$$=\sum_{i=0}^{k-1} g(x+2i,2)$$$$=\sum_{i=0}^{k-1} g(x+2i \mod k, 2)$$$$=\sum_{i=0}^{k-1} g(i, 2)$$$$=g(0,2k)$$$$=const......(3)$$

technically, $2\mathbb{Z}$ is just a coset of $\mathbb{Z}$

Known as the hardest FE in IMOSL history. I think I spent about 4-5 hours in total(not include typing), solution is long but each step is motivated. Use $[x,y]$ to represent $\text{lcm} [x,y]$ for all integer $x,y.$ For integer $x,y,$ define $F(x,y):=f(x+y)+f(x-y).$ The condition is equivilant to $F(x,y+f(x))=F(x,y),$ so $F(y)$ is $|f(x)|$-periodic, which gives $F(x,y+nf(x))=F(x,y)\cdots (*)$ for all integer $n.$ Now for $a,b,y_1,y_2\in\mathbb Z,$ by $(*)$

substract $(1)$ and $(2)$ $$f(y+a-b)-f(y+b-a)=f(y+a-b-[f(a),f(b)])-f(y+b-a-[f(a),f(b)]).$$Therefore $f(y+a-b)-f(y+b-a)$ have period $[f(a),f(b)].$ In perticular $f(y+1)-f(y-1)$ is periodic, take minimum period $L.$ Since $f(x+L+2)-f(x+L)=f(x+2)-f(x),$ $f(x+L+2)-f(x+2)=f(x+L)-f(x),$ so $f(x+L)-f(x)$ is 2-periodic. By $f(x+2L)-f(x)=(f(x+2L)-f(x+L))+(f(x+L)-f(x))$ is also 2-periodic, however it have period $[f(0),f(L)]$ which is odd, this gives $f(x+2L)-f(x)\equiv C$ where $C$ is constant. Take the smallest positive integer $L_0$ such that $f(x+L_0)-f(x)$ is constant(we proved $L_0$ exist). Then for integer $k,0\le r<L_0,$ $f(L_0k+r)=C_0k+f(r).$ Consider $x=k_1L_0+r_1,$ $y=k_2L_0+r_2$ and plug in the original function: \begin{align*}0&=f(x + f(x) + y) + f(x - f(x) - y) -f(x+y) - f(x-y)\\ &=f(k_1L_0+r_1+k_2L_0+r_2+f(k_1L_0+r_1))+f(k_1L_0+r_1-k_2L_0-r_2-f(k_1L_0+r_1))\\&-f(k_1L_0+r_1+k_2L_0+r_2)-f(k_1L_0+r_1-k_2L_0-r_2)\\&=C(k_1+k_2)+f(r_1+r_2+Ck_1+f(r_1))+C(k_1-k_2)+f(r_1-r_2-Ck_1-f(r_1))\\&-C(k_1+k_2)-f(r_1+r_2)-C(k_1-k_2)-f(r_1-r_2)\\&=F(r_1,r_2+Ck_1+f(r_1))-F(r_1,r_2)\\&=F(r_1,r_2+Ck_1)-F(r_1,r_2).\end{align*}Take $k_1=1$ we get $F(r_2)$ is $C$-periodic. Note that it is also $L_0,f(r_1)$-periodic, so $L_0\mid C,f(r_1)$ ($L_0$ is minimum period), therefore all possible solutions to the function are $$f(L_0k+r)=Ck+f(r)=L_0tk+L_0g(r),\quad\forall k,r\in\mathbb Z,0\le r<L_0$$where $L_0,g(r)$ are odd, and $t$ is even. To prove all these functions hold, by above it is equivilant to show that $$F(r_1,r_2+Ck_1+f(r_1))=F(r_1,r_2)$$which is obvious.$\Box$