Number of Solutions of a Functional Equation

If we consider the original functional equation together with the equation that results by interchanging $x$ and $y$ into that equation, we get $f(x+y)+2f(x-y)=2f(x)f(y)$ $f(y+x)+2f(y-x)=2f(y)f(x)$ Subtracting the second from the first, we get that $f(x-y)-f(y-x)=0,$ and making $y=0,$ the equation $f(-x)=f(x).$ So any solution of the given original functional equation is an even function.

Making $x=\frac{t}{2}$ and $y=\frac{t}{2}$ into the given equation again, we get $f(t)+2f(0)=2f(\frac{t}{2})f(\frac{t}{2})\quad\quad\quad(1)$ Making $x=\frac{t}{2}$ and $y=-\frac{t}{2},$ we get $f(0)+2f(t)=2f(\frac{t}{2})f(-\frac{t}{2})=2f(\frac{t}{2})f(\frac{t}{2})\quad\quad\quad(2)$

Therefore, $f(t)+2f(0)=f(0)+2f(t),$ that implies that $f(t)=f(0).$ So, the only possible solutions of the given functional equation are constant function of the form $f(x)=f(0)$ . Making the substitution $x=y=0$ into the given functional equation, it results that $3f(0)=2f(0)^2.$ Therefore, the only possible values of $f(0)$ are 0 or $\frac{3}{2}.$ Then the only possible continuous solutions of the given functional equation are $f(x)=0$ for all real values of $x$ and the function $f(x)=\frac{3}{2}.$ for all real values of $x.$ Then the number of possible solutions is $\boxed{2}.$

We do not need continuity here. Putting $y=0$ gives us that $3f(x) = 2f(x)f(0)$ for all $x$ , so either $f(x) = 0$ for all real $x$ or else $f(0) = \tfrac32$ .

Suppose that $f(0) = \tfrac32$ . Putting $x=0$ gives us that $f(y) + 2f(-y) = 3f(y)$ for all real $y$ , so that $f(-y) = f(y)$ for all real $y$ . Putting $x=y$ gives $f(2x) + 3 \; = \; 2f(x)^2$ for all real $x$ . Putting $y=-x$ gives $\tfrac32 + 2f(2x) \; = \; 2f(x)f(-x) \; = \; 2f(x)^2$ for all real $x$ . Hence $f(2x) + 3 \; = \; \tfrac32 + 2f(2x)$ for all real $x$ , so that $f(2x) = \tfrac32$ for all real $x$ .

Thus, without assuming that $f$ is continuous, we deduce that either $f \equiv 0$ or $f \equiv \tfrac32$ , and so there are exactly $\boxed{2}$ solutions.

Let's examine this functional equation for two cases:

CASE 1 ( $f(x)$ is constant): This case yields $C + 2C = 2C^2 \Rightarrow 0 = C(2C-3) \Rightarrow C = 0, \frac{3}{2}$

So $\boxed{f(x) = 0, \frac{3}{2}}$ are acceptable solutions.

CASE 2 ( $f(x)$ is non-constant): Assuming the continuous function $f: \mathbb{R} \rightarrow \mathbb{R}$ is diffentiable over all real numbers, let us first differentiate the above functional equation with respect to $x$ and $y$ :

$f'(x+y) + 2f'(x-y) = 2f'(x)f(y)$ (i) AND $f'(x+y) - 2f'(x-y) = 2f(x)f'(y)$ (ii)

and adding (ii) to (i) gives: $2f'(x+y) = 2[f'(x)f(y) + f(x)f'(y)]$ (iii). Now, we differentiate (iii) same as before to obtain:

$2f''(x+y) = 2[f''(x)f(y) + f'(x)f'(y)]$ (iv) AND $2f''(x+y) = 2[f'(x)f'(y) + f(x)f''(y)]$ (v)

which ultimately yields the differential equation: $f''(x)f(y) = f(x)f''(y) \Rightarrow f''(x) = K \cdot f(x) \Rightarrow \boxed{f(x) = ae^{\sqrt{K}x} + be^{-\sqrt{K}x}}$ (vi) (where $a,b,K \in \mathbb{R}$ ).

Substitution of (vi) back into our original functional equation produces:

$ae^{\sqrt{K}(x+y)} + be^{-\sqrt{K}(x+y)} + 2[ae^{\sqrt{K}(x-y)} + be^{-\sqrt{K}(x-y)}] = 2[ae^{\sqrt{K}x} + be^{-\sqrt{K}x}][ae^{\sqrt{K}y} + be^{-\sqrt{K}y}]$ ;

or $ae^{\sqrt{K}(x+y)} + be^{-\sqrt{K}(x+y)} + 2[ae^{\sqrt{K}(x-y)} + be^{-\sqrt{K}(x-y)}] = 2[a^{2}e^{\sqrt{K}(x+y)} + abe^{-\sqrt{K}(x-y)} + abe^{\sqrt{K}(x-y)} + b^{2}e^{-\sqrt{K}(x+y)}]$ ; (vii)

By matching the coefficients in each exponential term, we obtain:

$a = 2a^2 \Rightarrow a = 0, \frac{1}{2};$ ,

$b = 2b^2 \Rightarrow b = 0, \frac{1}{2}$ ,

$2a = 2ab \Rightarrow b = 1$ ,

$2b = 2ab \Rightarrow a = 1$

The only solution that satisfies all four of these coefficient conditions is $a = b = 0$ , thus no non-constant function $f(x)$ satisfies the original functional equation above.

Conclusion: $\boxed{f(x) = 0, \frac{3}{2}}$ are the only solutions.