Functions are awesome!!!

Let $P(x,y)$ be the statement $f(x+y)=f(x)f(y)f(xy)$ .

$\displaystyle P(x,0)\implies f(x)=f(x)f(0)^2\implies \begin{cases}f(x)=0,\:\forall x\in\mathbb R\\\text{or}\\f(0)=\pm 1\end{cases}$

So there's a solution $\boxed{f(x)=0}$ $\forall x\in\mathbb R$ (we can check it and it works). Let the function be different from this one. Then $f(0)=\pm 1$ .

$\displaystyle P(x,-x)\implies f(0)=\pm 1 =f(x)f(-x)f(-x^2),\:\forall x\in\mathbb R$ $\displaystyle\implies \not\exists a\in\mathbb R : f(a)=0\:\:(1)$

$\displaystyle P(x,-1)\implies f(x-1)=f(x)f(-x)f(-1),\:\forall x\in\mathbb R\:\:(2)$

$\displaystyle P(x-1,1)\implies f(x)=f(x-1)^2 f(1)$ $\displaystyle\stackrel{(2)}=f(x)^2f(-x)^2f(-1)^2f(1),\:\forall x\in\mathbb R\:\:(3)$

Replace $x$ with $-x$ in $(3)$ . Then we have: $\displaystyle f(-x)=f(x)^2f(-x)^2f(-1)^2f(1),\:\forall x\in\mathbb R\:\:(4)$

$\displaystyle (3)(4)\implies f(x)=f(-x),\:\forall x\in\mathbb R\:\:(5)$

\displaystyle P(x,1)\stackrel{(5)}\implies f(x+1)=f(x)^2f(1),\:\forall x\in\mathbb R\:\:(6)

\displaystyle (2)\stackrel{(5)}\implies f(x-1)=f(x)^2f(1)\:\:(7)

If we put $x$ in place of $x-1$ in $(7)$ , then $\displaystyle f(x)=f(x+1)^2f(1),\:\forall x\in\mathbb R\:\:(8)$

$\displaystyle (6)(8)\implies f(x+1)=f(x)^2f(1)=f(x+1)^4f(1)^3$ \displaystyle \stackrel{(1)}\implies f(x+1)^3=\frac{1}{f(1)^3}

$\displaystyle \implies f(x+1)=\frac{1}{f(1)}=c\implies f(x)=c,\:\forall x\in\mathbb R,c\in\mathbb R, c\neq 0$

This means $f(x)$ is a constant function, i.e. $\displaystyle f(x)=c,\:\forall x\in\mathbb R, c\in\mathbb R, c\neq 0$

\displaystyle f(x+y)=f(x)f(y)f(x+y),\:\forall x,y\in\mathbb R\implies c=c^3\stackrel{c\neq 0}\implies c^2=1 $\displaystyle \implies \begin{cases}c=1\\\text{or}\\c=-1\end{cases}\implies \begin{cases}\boxed{f(x)=1}\:\forall x\in\mathbb R\\\text{or}\\ \boxed{f(x)=-1}\:\forall x\in\mathbb R\end{cases}$

After checking these $2$ possible functions, we see they work.

Therefore, there are only $3$ possible functions (all the ones I've boxed). $\square$

If $f$ is constant, then let it be $c$ , then $c^3=c \Rightarrow c=-1,0,1$ . We can easily check that these are only solutions. So we assume the rest of the solutions are non constant.

First, $f(x)=f(x-1)^2f(1)=(f(x)f(-x)f(-1))^2f(1)=f(-x-1)^2f(1)=f(-x)$ . So we have $f(x)=f(x+1)f(-(x+1)f(-1)=f(x+1)^2f(-1)=(f(x)f(-x)f(1))^2f(-1)=(f(x)^2f(1))^2f(-1)=f(x)^4f(1)^3 \Rightarrow f(x)^3=f(1)^3 \Rightarrow f(x)=f(1)$ , which imply $f$ a constant function, contradiction.

So $f(x)=0$ , $f(x)=-1$ , $f(x)=1$ are only solutions.

Having: $f(1)=f(0)^2f(1)$ $f(2)=f(1)^3$ $f(3)=f(1)f(2)^2=f(1)^7$ $f(n)=f(1)^{2^n-1}$ $\Rightarrow f(1)^{2^{x+y}}=f(1)^{2^x+2^y+2^{xy}-2}$ $\Rightarrow f(1)=1|-1|0, \Rightarrow answer=\boxed{3}$

Let the degree of the LHS be d, we see that the degree of the RHS has to be 3d. Thus, we conclude that the function must be constant. Try $y=0$ , the equation becomes $f(x) = f(x)f(0)^2$ . Thus we find $f(0)=1$ .

Letting $x=y=2$ , we see that $f(4) = f(2)^2f(4)$ . Leading us to solutions $f(2) = -1, f(2) = 1$ . Another easy solution is to notice that $f(x) = 0$ works by a simple substitution. Thus we have $\boxed{3}$ solutions

Substituting $y\rightarrow 0$ we get

$f(x) = f(x)f(0)^{2}$

$f(0) = \pm 1 \text{ or } \boxed{f(x) = 0}$

Substituting $y \rightarrow x$ we get

$f(2x) = f(x)^{2}f(x^{2})$ __(1)

Let $deg(f(x)) = n$

(1); $n = n^{4}$

$n = 0, 1$

Let $f(x) = ax+b$

(1); $2ax+b = a^{3}x^4+2a^{2}bx^{3}+a^{2} b x^{2}+a b^{2} x^{2}+2 a b^{2} x+b^{3}$

Compare the coefficients $x^{4}$ we get $a^{3} = 0 \rightarrow a = 0$ .

Therefore, $f(x) = c$ .

By observing, $\boxed{f(x) = 1, -1}$ .

There are only $\boxed{3}$ solution available. ~~~