Recently, while scribbling around, I came up with the functional equation $f(x+1)=xf(x)f\left(\frac1x\right)$. It is somewhat similar to the functional equation for which the Gamma function is a solution, since $\Gamma(x+1)=x\Gamma(x)$, but it has the extra factor of $f\left(\frac1x\right)$.

If we introduce the restriction that $x\geq 1$, then I noticed that there is *at least* one solution (with this restriction put in place). That solution turns out to be, amazingly, $f(x)=\{x\}$, or the fractional part of $x$.

**Edit:** In fact, if $x\geq 1$, then $f(x)=\{x+a\}$ is a solution for all $a\in\mathbb{R}$ (I'm using the convention that $\{x\}=x-\lfloor x\rfloor$, where $\lfloor x\rfloor$ is defined as the greatest integer that is lower than $x$, therefore $a$ can be negative as well).

I have no idea how to approach this problem at all. The only thing that I was able to deduce is that for $n\in\mathbb{N}$, the following holds:

$$f(n)=f(1)\cdot (n-1)!\prod_{k=1}^{n-1}f\left(\frac1k\right)$$

I don't know if this helps in any way, but it might be a good start.

Knowing that when $x$ is restricted to be greater or equal to $1$, $\{x\}$ is a solution, I wonder if there can even exist a solution that is continuous. However, I don't know how to prove or disprove this, nor do I know how to find *any* solutions for the equation.

Is there a way to at least gain insight on how a possible solution for the equation must behave?