TITLE: Find monotonically increasing function $f$ on $[ 1,+\infty )$ such that $ x ( f ( x^{2} ) + 1 ) = f ( x ) ( x^{2}+1 ) $?
QUESTION [5 upvotes]: Find all monotonically increasing functions $f$ on $\left[1,+\infty\right)$ such that
$$x\left( f \left( x^{2} \right) + 1 \right) = f \left( x \right) \left( x^{2}+1 \right) $$
Does there only exist the unique solution $f(x)=x$?
At first time, I think that $f$ is monotonically increasing is necessary and meaningful here.
Thanks for the comments, there are some strange solutions beyond my thought. 
And now it seems the property of monotonically increasing is not important, maybe it is because there exists a closed and beautiful form of the solutions.

REPLY [6 votes]: Clearly, $f(1)=1$. Now, the functional equation is equivalent to
$$f(x^2)=\left(x+\frac1x\right)f(x)-1$$
This the same as
$$f(x^2)-x^2=\left(x+\frac1x\right)(f(x)-x)$$
Or, if we define $g(x)=\dfrac{f(x)-x}{x-1/x}$ for $x>1$, we get
$$g(x^2)=g(x)\tag1$$
Now the functions $g$ that satisfy $(1)$ are functions of form $g(x)=h(\ln(\ln x))$ where 
$ h:\mathbb{R}\to\mathbb{R}$ is an arbitrary periodic function having $\ln(2)$ as period.
So, the general solution of the proposed functional equation is
$$f(x)=x+\left(x-\frac1x\right)h(\ln(\ln x))$$
Where $ h:\mathbb{R}\to\mathbb{R}$ is an arbitrary periodic function having $\ln(2)$ as period.
Now, the monitonicity condition adds some conditions on our choices for $h$.