Let $P(x,y)$ denote the assertion. $P(1,1): f(1)=f(1)^2f(2)\implies f(1)=0\text { or }f(1)f(2)=1$. $P(2,2): f(4)=f(2)^2f(4)\implies f(4)=0\text { or }f(2)=\pm 1$. Suppose $f(k)=0$ for some positive real $k$. $P(k,x): f(kx)=0\implies \boxed{f\equiv0}$, which works. Henceforth assume no $k$ satisfies $f(k)=0$. Then $f(1)f(2)=1$ and $f(2)=\pm 1$. Case 1: $f(2)=1$. Then $f(1)=1$. $P(1,x): f(x)=f(x)f(x+1)\implies f(x+1)=1$. So the only solution here is $\boxed{f\equiv1}$, which works. Case 2: $f(2)=-1$. Then $f(1)=-1$. $P(1,x): f(x)=-f(x)f(x+1)\implies f(x+1)=-1$. So the only solution here is $\boxed{f\equiv-1}$, which works. Edit: I made a little mistake, it’s only true for x>1, but $P(x,\frac{1}{x})$ gives f is constant Edit: Thanks Jupiter123

Case 1: $f(k)=0$ for some $k$ $P\left(\frac xk,k\right)\Rightarrow\boxed{f(x)=0}$ which is a solution. Case 2: $f(x)\ne0\forall x$ Let $c=\frac1{f(1)}$ which is defined. $P(x,1)\Rightarrow f(x+1)=c\Rightarrow f(x)=c\forall x>1$ Let $x\le1$ now. Then $x+\frac1x+1\ge\frac1x+1>1$, so: $P\left(x,\frac1x+1\right)\Rightarrow f(x)=c$ Testing solutions of the form $f(x)=c$, we have the two new solutions $\boxed{f(x)=-1}$ and $\boxed{f(x)=1}$.

megarnie wrote: Case 1: $f(2)=1$. Then $f(1)=1$. $P(1,x): f(x)=f(x)f(x+1)\implies f(x+1)=1$. So the only solution here is $\boxed{f\equiv1}$, which works. Case 2: $f(2)=-1$. Then $f(1)=-1$. $P(1,x): f(x)=-f(x)f(x+1)\implies f(x+1)=-1$. So the only solution here is $\boxed{f\equiv-1}$, which works. True for all $x > 1$. Shouldn't you also prove it for $0 < x \leq 1$?

Oh shoot I’ll edit when I get home P(x,1/x) should work

MarkBcc168 wrote: Determine all functions $f : \mathbb R^+ \to \mathbb R$ that satisfy the equation $$f(xy) = f(x)f(y)f(x+y)$$for all positive real numbers $x$ and $y$. Let $P (x, y)$ be the statement $f (xy) = f (x) f (y) f (x + y)$ we do $P (1, y)$: $f (x) = f (1) f (x) f (x + 1)$ Now we have $2$ cases: Case 1: $f (x) = 0$ then $P (x, y)$ holds for all x and y, therefore $f (x) = 0$ is a solution. Case 2: $f (x) \not 0$ then $1 = f (1) f (x + 1)$, and we do $P (x + 1, 1)$: $f (x + 1) = f (1) f (x + 1) f (x + 2)$ $f (x + 1) = f (x + 2)$, we replace and we have that $1 = f (1) f (x + 2)$, then by induction we obtain that $1 = f (1) f (n) \forall n$ we make $n = 1$, and then we have $f (1) ^ 2 = 1$, so $f (1) = 1$ or $f (1) = - 1$, this implies that $f (x) = - 1$ or $f (x) = 1$, so in this case the solutions are $f (x) = 1$ or $f (x)= -1$

The answers are $f\equiv0$, $f\equiv1$ and $f\equiv-1$ which is not hard to see that they work. Let $P(x,y)$ be the functional equation. If $f(x_0)=0$ for some $x_0>0$, then $P(\frac{x}{x_0},x_0)$ gives $f(x)=0$ for each $x\in\mathbb{R^+}$. So, the rest is to prove that if $f(x)\neq0$ for each $x$, then $f\equiv1$ or $f\equiv-1$. $P(2,2)$, gives $f(2)^2=1$. Plug $y=1$ yields $f(x)=\frac{1}{f(1)}$ for all $x>1$. So $f(x)$ is constant for each $x\geq 1$. assume $x_1\geq 1$, $P(x_1,\frac{1}{x_1})$ gives $f(1)=f(\frac{1}{x_1})$. So $f$ is a constant function implies $f(x)^2=1\implies f\equiv 1$ or $f\equiv-1$ as desired.

By setting y = 1 we get f(x) = f(x)f(1)f(x+1). If there exists such x that f(x) = 0 then for any possitive real y f(xy) = 0, thus f(x) = 0 for any positive real x is our first solution. If there does not exist such x we can cancel out f(x) on both sides and divide our equation by f(1). We get f(x+1) = 1/f(1). Thus f(x) = c, where c satisfies c = c^3. Polynomial c^3 - c has exactly 3 roots which are -1,0,1, hence f(x) = -1 or 0 or 1 for any positive real x. Sorry for not using latex, I'm new, so I can't

If $f$ is zero at any point then it is zero everywhere. If not then setting $y=1$ implies $f(x)=1/f(1)=k$ for all $x>1.$ But $y=1/x$ implies $f$ is constant everywhere whence testing we get $k\in \{1,-1\}.$ All of them work.

Solved with the AoPS user @Taco12. We claim the only functions that satisfy the given equation are $f(x) = 0$, $f(x) = 1$, and $f(x) = -1$. Let $P(x,y)$ denote the assertion. First, notice that $f(x) = 0$ is a solution. $P(x, 1)$: $f(x) = f(x)f(1)f(x+1) \Rightarrow f(1)f(x+1) = 1$. (We can do this since we already acknowledged $f(x) = 0$ is a solution.) $P(1, 1)$: $f(1) = f(1)^2 \cdot f(2)$. Notice that we can divide $f(1)$ from both sides as we know it is not equal to $0$ because then $f(1)f(x+1) = 1$ would not hold true. So, $f(1)f(2) = 1$. Substituting, we have $f(x+1) = f(2) = c$. Hence, $f(x) = c$, where $c$ is a fixed real number. So, we see that $c = c^3 \Rightarrow c^3 - c = 0 \Rightarrow c(c^2-1) = 0 \Rightarrow c = 0, \pm 1$, and we are done. $\blacksquare$

@above I believe you miss solutions in the interval $(0,1]$