Let $f(x)$ satisfy the equation $$f^2(x)=1+xf(x+1)$$ and the inequality $$x+1<2f(x)<4(x+1)$$ for all $x>1$. Find $f(x)$. Here, $f^2(x)$ is $\big(f(x)\big)^2$.
I tried with assuming polynomial, but of no help. Is assuming polynomial correct?
If $f(x)=x+1$, then $$f^2(x)=(x+1)^2=x^2+2x+1=1+x(x+2)=1+xf(x+1).$$ When $x>1$, we have $x+1>0$, so $$x+1<2(x+1)<4(x+1)$$ is true. Therefore, $f(x)=x+1$ is a solution. Are there other solutions?
Yes, there is a unique polynomial solution to the functional equation
$$f^2(x)=1+xf(x+1)\tag{1}$$
It is :
(moreover compatible with the given inequations).
Existence : Let us look for a (at most) first degree polynomial solutions $f(x)=ax+b$ ($a$ can be zero).
$$(ax+b)^2=1+x(a(x+1)+b)$$
Expanding, one obtains :
$$a^2x+2abx+b^2=ax^2+(a+b)x+1 \ \ \iff \ \ \begin{cases}a^2&=&a\\2ab&=&a+b\\b^2&=&1\end{cases}$$
The first equation has solutions $a=0$ or $a=1$. The case $a=0$ has to be excluded because it gives a contradiction with the other equations (we would have $b=0$ and $b=1$ !).
Therefore, plugging $a=1$ into the other equations gives $b=1$, and we have obtained solution (2).
Unicity: No other polynomial solution $f(x)=ax^n+...$ with a higher degree (i.e., $n>1$) can exist. Two reasons for that:
The functional equation would have a LHS beginning by $a^2 x^{2n}$ whereas the RHS would begin by $ax^{n+1}$ : impossible for $n>1$.
The inequations couldn't hold, knowing the growth of a polynomial with a degree $n$ higher than $1$ (imagine for example graphically a parabola "confined" between a pair of non vertical lines).
Edit : it remains to know if there are other (non polynomial) solutions.
We can look for them under the form :
$$f(x)=x+1+u(x)\tag{3}$$
with $u$ is non identically $0$ (we exploit here the particular solution we have found).
Plugging (3) into (1), we get :
$$(x+1+u(x))^2=1+x(x+2+u(x+1))$$
and finally :
$$\dfrac{u(x+1)}{u(x)}=\frac{u(x)}{x}+\frac{2}{x}+2\tag{4}$$
with the accompanying condition
$$-\frac12(x+1)<u(x)<x+1\tag{5}$$
We can obtain a contradiction with $u \neq 0$ by considering a non-zero value $u(x_0)$. Due to condition (5), (4) gives
$$\left|\dfrac{u(x_0+n+1)}{u(x_0+n)}\right|>\dfrac{3}{2}+\dfrac{2}{x_0+n}$$
providing an exponential growth for the sequence $u(x_0+n)$ in contradiction with (5).
Warning : we have assumed implicitly that all values $u(x_0+n)$ are non zero. A rigorous argument is needed for proving that $x_0$ can be chosen in this way.