Let $f:\mathbb{R}\to\mathbb{R}$ be a continuous function such that $f(x)f(x+2)+f(x+1)=0$ for all $x\in\mathbb{R}$. Prove that there are infinitely many real values of $x$ such that $f(x)=0$.
Here is my approach (by inspection):
For $x=0$ we have $f(0)f(2)+f(1)=0$ (Eq1).
Now, if $f(0)=0$, then $f(1)=0$. From $f(1)f(3)+f(2)=0$ for $x=1$ we can also conclude that $f(2)=0$. In this case it's not hard to prove that $f(n)=0\ \ \forall n\in\mathbb{N}$.
On the other hand, if $f(0)>0$, then Eq1 let us conclude that $f(1)$ and $f(2)$ have opposite signs, say $f(1)<0<f(2)$. In this case, the IVT guarantes that there exists a $c\in (1,2)$ such that $f(c)=0$. Again, it's not hard to prove that $f(n+c)=0\ \ \forall n\in\mathbb{N}$.
I am stuck with the case where $f(0)<0$. In this case $f(1)$ and $f(2)$ have the same sign and I couldn't figure out what to do next. Any hint to this case or a different approach would be great. Thanks in advance.
$f(x)f(x+2)+f(x+1)=0 $
Suppose there are only finitely many $x$ such that $f(x)=0$. Then there is a largest such $x$. Call it $x_0$.
Then $f(x_0)=0$ and, for all $x > x_0$ $f(x) \ne 0$. Since $f$ is continuous, $f(x)$ is of constant sign for all $x > x_0$.
Assume $f(x)>0$ for $x > x_0$. This works just as well for $f(x) < 0$.
We have $0 =f(x_0)f(x_0+2)+f(x_0+1) $. Since $f(x_0)=0$, $f(x_0+1)=0$ which contradicts the definition of $x_0$.
Therefore there is no largest $x$ such that $f(x)=0$.
(added later)
In particular, this shows that if $f(x)=0$ then $f(x+1)=0$ so $f(x+n)=0$ for all positive integers $n$.