find all rational functions $f(x)$ such that $(f(x))^2-f(x^2)=constant$, Find all  $f(x)$ such that $$(f(x))^2-f(x^2)=constant$$
 A: Assume that $a$ is a pole of $f$ or order $n$ then $f^2$ has $a$ as a pole of order $2n$ and therefore so does $f(x^2)$. Therefore $a^2$ is a pole of $f$ of order $2n$. Therefore $f$ has no poles different from zero.  
Let $g(x):=f(1/x)$. Notice that $g(x)$ also satisfy the equation. Subtract the two equations to get $$(f(x)-g(x))(f(x)+g(x))=f(x^2)-g(x^2).$$
Therefore, for the function $h(x):=f(x)-g(x)$ we have $h(x)(f(x)+g(x))=h(x^2)$. If $a$ is a zero of $h(x)$ of order $n$, then $a^2$ is a zero of $h(x)$ too. Therefore, either $a=0$ or $h(x)=constant$. 
Now we have $f(x)=P(x)/x^k$ for some polynomial $P$. Then $P(x)/x^k-P(1/x)/x^{-k}=C$.
If $n$ is the degree of $P$, comparing degrees we get that $n=2k$. Comparing coefficients we get that $P$ is a symmetric polynomial (the coefficients can be put in reverse order), i.e. $P(x)=a_{2k}x^{2k}+a_{2k-1}x^{2k-1}+\ldots+a_{2k-1}x+a_{2k}$.
So, so far we have that $f$ is a linear combination of $x^k+1/x^k$. Plugging a linear combination into $f(x)^2-f(x^2)=constant$, 
... to be continued, but from there you should get the solutions ...
A: The earlier answers have forgotten $f(x)=x+1/x$
