This is related to 2012A2 :p The main idea is to establish a linear recurrence $$ f(1)f(x+1) = f(2)f(x)-f(1)f(x-1) $$by a simple double counting. Then, fix the functional values of integral values and use the trick of A2 to derermine the functional values of others. The solutions are $ f(x) = k(a^{x}-a^{-x}) $ for some $ k,a $ or $ f(x) = bx $ for some $ b $ EDIT: sorry for typosssss

FEcreater wrote: This is related to 2012A2 :p The main idea is to establish a linear recurrence $$ f(1)f(x+1) = f(2)(f(x)+f(x-1)) $$by a simple double counting. Then, fix the functional values of integral values and use the trick of A2 to derermine the functional values of others. The solutions are $ f(x) = k(a^{x}-a^{-x}) $ for some $ k,a $ $f(x)=kx$ is also a solution.

Note that this has been posted before (hmm this remind me of P3 from this year IMO.) It is here Also,I believe there are another answer (see my post there) which now I'm not sure it is well-defined. If you think it's not, then please tell me, thank!

TLP.39 wrote: Note that this has been posted before (hmm this remind me of P3 from this year IMO.) It is here Also,I believe there are another answer (see my post there) which now I'm not sure it is well-defined. If you think it's not, then please tell me, thank! Nah, I think FEcreatoe is (almost) correct. The thread that you posted is really messy and I didn't see the conclusion, so let's wait for FEcreator for a solution

FEcreater wrote: This is related to 2012A2 :p The main idea is to establish a linear recurrence $$ f(1)f(x+1) = f(2)f(x)-f(1)f(x-1) $$by a simple double counting. Then, fix the functional values of integral values and use the trick of A2 to derermine the functional values of others. The solutions are $ f(x) = k(a^{x}-a^{-x}) $ for some $ k,a $ or $ f(x) = bx $ for some $ b $ EDIT: sorry for typosssss Maybe you mean 2012 A5? It seems that 2012 A2 is not an FE.

I suspect there are even more solutions. The three classes of solutions depends on the discriminant of the characteristic equation: if $\Delta >0$, we have the exponential solution, and $\Delta=0$ gives the linear solution. The final case gives some kind of periodic solution. However, these solutions only hold for integer multiples (i.e. we are only certain that for fixed $x$, $f(nx)$ is of the abovementioned form). There is obviously no "mixing" of the three different types of solution (since the growth rates are all different, and for rational $p,q$ we may consider $r$ where $p,q$ are both multiples of $r$, so both sequence are subsequences of a single such sequence). It is not too difficult to establish also that for the periodic case, the amplitudes must be the same. Now we get to the difficult part. Write $$f(x)=A\sin (2\pi g(x))$$Then we have that $g$ satisfies the Cauchy Equation mod 1. I have not solved this completely, but it is known that not all solutions are linear. Let me know if I've made a mistake somewhere.

Rickyminer wrote: FEcreater wrote: This is related to 2012A2 :p The main idea is to establish a linear recurrence $$ f(1)f(x+1) = f(2)f(x)-f(1)f(x-1) $$by a simple double counting. Then, fix the functional values of integral values and use the trick of A2 to derermine the functional values of others. The solutions are $ f(x) = k(a^{x}-a^{-x}) $ for some $ k,a $ or $ f(x) = bx $ for some $ b $ EDIT: sorry for typosssss Maybe you mean 2012 A5? It seems that 2012 A2 is not an FE. It is 2012 A2, where the key idea is to first solve the problem restricted on $\mathbb{Z}$ and then solve the general problem by considering the problem restricted on $\frac{1}{n}\mathbb{Z}$ for all $n\in\mathbb{N}$.

DVDthe1st wrote: I suspect there are even more solutions. The three classes of solutions depends on the discriminant of the characteristic equation: if $\Delta >0$, we have the exponential solution, and $\Delta=0$ gives the linear solution. The final case gives some kind of periodic solution. However, these solutions only hold for integer multiples (i.e. we are only certain that for fixed $x$, $f(nx)$ is of the abovementioned form). There is obviously no "mixing" of the three different types of solution (since the growth rates are all different, and for rational $p,q$ we may consider $r$ where $p,q$ are both multiples of $r$, so both sequence are subsequences of a single such sequence). It is not too difficult to establish also that for the periodic case, the amplitude must be the same. Now we get to the difficult part. I'll finish this in a while. I agree. The periodic part is more complicated than just $k(a^x-a^{-x})$. But basically it can be solved by the idea that FEcreator gives, although the final solution should be in a somewhat ugly form...... EDIT: Oh and there comes the solution at #7 :p

@#5 I guess you think I mean my first post, as you said that it is 'messy' (oops) and that it has no conclusion. If so, then sorry for not clarifying before. I mean my second post. This is one of the answer that I came up with: TLP.39 wrote: So,the general way to construct the function in this case is: 0)Let $f(0)=0$. 1)Arbitrarily choose nonzero real $p$ and non-real $v$ such that $v+\frac{1}{v}\in\mathbb{R}$ and let $f(1)=p,\frac{f(2)}{f(1)}=v+\frac{1}{v}$. 2)For any prime number $h$,choose a complex number $w_h$ such that $w_h^h=v$.Then let $f(\frac{1}{h})=\frac{p(w_h-\frac{1}{w_h})}{v-\frac{1}{v}}$ and $\frac{f(\frac{2}{h})}{f(\frac{1}{h})}=w_h+\frac{1}{w_h}$. 3)For any prime $h$,choose an infinite sequence $w_{h^2},w_{h^3},w_{h^4},...$ such that $w_{h^{n+1}}^h=w_{h^n}$ for all natural number $n$. Then let $f(\frac{1}{h^n})=\frac{p(w_{h^n}-\frac{1}{w_{h^n}})}{v-\frac{1}{v}}$ and $\frac{f(\frac{2}{h^n})}{f(\frac{1}{h^n})}=w_{h^n}+\frac{1}{w_{h^n}}$. 4)For any natural number $h=p_1^{a_1}p_2^{a_2}...p_k^{a_k}$,define $w_h$ to be a complex number such that $w_h^{\left(\frac{h}{p_j^{a_j}}\right)}=w_{p_j^{a_j}}$ for all $j=1,2,...,k$ (its existant can be proved by CRT).Then let $f(\frac{1}{h})=\frac{p(w_h-\frac{1}{w_h})}{v-\frac{1}{v}}$ and $\frac{f(\frac{2}{h})}{f(\frac{1}{h})}=w_h+\frac{1}{w_h}$. 5)For any nonzero integer $m,n$ where $n>0$,let $f(\frac{m}{n})=\frac{p(w_n^m-\frac{1}{w_n^m})}{v-\frac{1}{v}}$. It can be proved that the definition didn't contradict itself and that the function is real-valued and satisfied the recurrence,and thus the assertion. To make it short, I randomly 'define' the rational powers of a complex number $v$ with amplitude $1$, and make the answer $f(x)\equiv \frac{k(v^x-v^{-x})}{v-v^{-1}}$.

TLP.39 wrote: @#5 I guess you think I mean my first post, as you said that it is 'messy' (oops) and that it has no conclusion. If so, then sorry for not clarifying before. I mean my second post. This is one of the answer that I came up with: TLP.39 wrote: So,the general way to construct the function in this case is: 0)Let $f(0)=0$. 1)Arbitrarily choose nonzero real $p$ and non-real $v$ such that $v+\frac{1}{v}\in\mathbb{R}$ and let $f(1)=p,\frac{f(2)}{f(1)}=v+\frac{1}{v}$. 2)For any prime number $h$,choose a complex number $w_h$ such that $w_h^h=v$.Then let $f(\frac{1}{h})=\frac{p(w_h-\frac{1}{w_h})}{v-\frac{1}{v}}$ and $\frac{f(\frac{2}{h})}{f(\frac{1}{h})}=w_h+\frac{1}{w_h}$. 3)For any prime $h$,choose an infinite sequence $w_{h^2},w_{h^3},w_{h^4},...$ such that $w_{h^{n+1}}^h=w_{h^n}$ for all natural number $n$. Then let $f(\frac{1}{h^n})=\frac{p(w_{h^n}-\frac{1}{w_{h^n}})}{v-\frac{1}{v}}$ and $\frac{f(\frac{2}{h^n})}{f(\frac{1}{h^n})}=w_{h^n}+\frac{1}{w_{h^n}}$. 4)For any natural number $h=p_1^{a_1}p_2^{a_2}...p_k^{a_k}$,define $w_h$ to be a complex number such that $w_h^{\left(\frac{h}{p_j^{a_j}}\right)}=w_{p_j^{a_j}}$ for all $j=1,2,...,k$ (its existant can be proved by CRT).Then let $f(\frac{1}{h})=\frac{p(w_h-\frac{1}{w_h})}{v-\frac{1}{v}}$ and $\frac{f(\frac{2}{h})}{f(\frac{1}{h})}=w_h+\frac{1}{w_h}$. 5)For any nonzero integer $m,n$ where $n>0$,let $f(\frac{m}{n})=\frac{p(w_n^m-\frac{1}{w_n^m})}{v-\frac{1}{v}}$. It can be proved that the definition didn't contradict itself and that the function is real-valued and satisfied the recurrence,and thus the assertion. To make it short, I randomly 'define' the rational powers of a complex number $v$ with amplitude $1$, and make the answer $f(x)\equiv \frac{k(v^x-v^{-x})}{v-v^{-1}}$. Oh oops, I think I just read the post too fast (with my phone) and missed something. Sorry about that :p

The above solution could be well-defined when the set $\mathbb{F}_0$ of all zeros of $f$ satisfy $\mathbb{F}_0=\{dn, n\in \mathbb{Z}\}$ for some rational number $d$. However, I think there are a lot of other hard-to-define solutions. Take $S$ as any set consisting of primes (can be empty, can consist of infinitely many prime numbers), and take any function $h: S\rightarrow \mathbb{Z}$ such that $h(p)>0$ only for finitely many $p\in S$. I think there should be a function $f$ satisfying the original equality, also with: $$\mathbb{F}_0 = \{ x\in \mathbb{Q} | v_p(x)\geq h(p) \quad \forall p\in S\}$$ Note that $f\equiv 0$ correspond to $S=\{ \}$. Also, $\mathbb{F}_0=\{dn, n\in \mathbb{Z}\}$ happens only when we take $S$ as the set of all primes and $h(p)=0$ for all but finitely many prime number $p$. The idea of taking $\mathbb{F}_0$ with that form is that $\mathbb{F}_0$ is a closed group under addition, and (hopefully) with Bezout lemma, $\mathbb{F}_0$ must take the described form. And, below is the proof. Claim: $\mathbb{F}_0$ is a closed group under addition. Proof. Assume that $a, b\in \mathbb{F}_0$. By substitution $x=a+b, y=b$ into the original equation, we get $f(a+b)^2=0$, so $a+b\in \mathbb{F}_0$. Now, by interchanging $y$ with $-y$, we obtain that $f(y)^2=f(-y)^2$ for every $y\in \mathbb{Q}$, so if $y\in \mathbb{F}_0$, then $-y\in \mathbb{F}_0$. Both this property concludes the claim. We are done. Extra: Whoops, seems that I forgot to abuse the simple fact that squares of real numbers are always non-negative. $f(x)=k(a^x-a^{-x})$ for some $k, a\in \mathbb{R}$ with $k>0$ are the only possible functions if $f$ is not bounded. The above idea can be possibly used only on the bounded case (and yes, there's non-periodic bounded solutions; $f(x)=\sin{x}$ as an example).

Bumping this. Any complete solution?

Another bump on this.