Let $P(x,y)$ denote the given statement. Now, $P(0,0)\Rightarrow f(0)^2=2f(0)\Rightarrow f(0)=0,2$ If $f(0)=0$ then $P(x,0)\Rightarrow f(x)\equiv 0\forall x\in \mathbb{Z}$ which is a contradiction. So $f(0)=2$. Now $P(0,y)\Rightarrow f(y)=f(-y)$. We claim that $f(k)=2^{k}+\dfrac{1}{2^k} \forall k\geq 0$. We will prove it by induction. We can check that the claim is true for $k=0,1,2,3,4$. Let it be true for $k=m$ and $k=m-1$ when $m\geq 1$. Then consider $P(m,1)$ and it is a one-line calculation that gives us $f(m+1)=2^{(m+1)}+\dfrac{1}{2^{m+1}}$. Now note that we got $f(k)=2^{k}+\dfrac{1}{2^k} \forall k\geq 0$, then for some $k>0$ we have, $f(-k)=f(k)=2^{k}+\dfrac{1}{2^k}=\dfrac{1}{2^{-k}}+2^{-k} \forall k\geq 0$. So the answer is $f(x)=2^{x}+\dfrac{1}{2^{x}}\forall x\in \mathbb{Z}$. Thanks to Konigsberg for pointing to my mistake, that the answer doesn't contain $|x|$,rather it's just $2^{x}+\dfrac{1}{2^{x}}$

Isn't the answer just equal to $2^{x}+\frac{1}{2^{x}}$.

Mikasa wrote: Let $P(x,y)$ denote the given statement. Now, $P(0,0)\Rightarrow f(0)^2=2f(0)\Rightarrow f(0)=0,2$ If $f(0)=0$ then $P(x,0)\Rightarrow f(x)\equiv 0\forall x\in \mathbb{Z}$ which is a contradiction. So $f(0)=2$. Now $P(0,y)\Rightarrow f(y)=f(-y)$. We claim that $f(k)=2^{k}+\dfrac{1}{2^k} \forall k\geq 0$. We will prove it by induction. We can check that the claim is true for $k=0,1,2,3,4$. Let it be true for $k=m$ and $k=m-1$ when $m\geq 1$. Then consider $P(m,1)$ and it is a one-line calculation that gives us $f(m+1)=2^{(m+1)}+\dfrac{1}{2^{m+1}}$. Now note that we got $f(k)=2^{k}+\dfrac{1}{2^k} \forall k\geq 0$, then for some $k>0$ we have, $f(-k)=f(k)=2^{k}+\dfrac{1}{2^k}=\dfrac{1}{2^{-k}}+2^{-k} \forall k\geq 0$. So the answer is $f(x)=2^{x}+\dfrac{1}{2^{x}}\forall x\in \mathbb{Z}$. Thanks to Konigsberg for pointing to my mistake, that the answer doesn't contain $|x|$,rather it's just $2^{x}+\dfrac{1}{2^{x}}$ What is the idea to try $2^{x}+2^{-x}$ as function (before using induction) Does someone have a more direct approach to solve or can someone explain this motivation? Thanks

$f(0)=2,f(1)=\frac{5}{2}$ $f(x)f(1)=f(x+1)+f(x-1)$ or $f(x+1)=\frac{5}{2}f(x)-f(x-1)$ - reccurent sequence. $t^2-\frac{5}{2}t+1=0 \to t=2,\frac{1}{2}$ $f(x)=A*2^x+\frac{B}{2^x}$ Set $x=0,1$ we find $f(x)=2^x+\frac{1}{2^x}$