<s>[INST] Number of answers : $f(x)=f^{-1}(x)$ 
let $f(x)= 1+\sqrt{x+k+1}-\sqrt{x+k} \ \ k \in \mathbb{R}$
Number of answers :
$$f(x)=f^{-1}(x) \ \   \ \ :f^{-1}(f(x))=x$$

MY Try :
$$y=1+\sqrt{x+k+1}-\sqrt{x+k} \\( y-1)^2=x+k+1-x-k-2\sqrt{(x+k+1)(x+k)}\\(y-1)^2+k-1=-2\sqrt{(x+k+1)(x+k)}\\ ((y-1)^2+k-1)^2=4(x^2+x(2k+1)+k^2+k)$$
now what do i do ?
 [/INST] Let $y=f(x)=f^{-1}(x)$ and then
\begin{eqnarray}
y=1+\sqrt{x+k+1}-\sqrt{x+k},\tag{1}\\
x=1+\sqrt{y+k+1}-\sqrt{y+k}.\tag{2}
\end{eqnarray}
Subtracting (1) from (2) gives
\begin{eqnarray}
x-y&=&(\sqrt{y+k+1}-\sqrt{x+k+1})-(\sqrt{y+k}-\sqrt{x+k})\\
&=&\frac{y-x}{\sqrt{y+k+1}+\sqrt{x+k+1}}-\frac{y-x}{\sqrt{y+k}+\sqrt{x+k}}
\end{eqnarray}
and hence
$$ (x-y)\left[\frac{1}{\sqrt{y+k+1}+\sqrt{x+k+1}}-\frac{1}{\sqrt{y+k}+\sqrt{x+k}}\right]=0.$$
Thus one has either
$$ x=y $$
or
$$ \frac{1}{\sqrt{y+k+1}+\sqrt{x+k+1}}=\frac{1}{\sqrt{y+k}+\sqrt{x+k}}\tag{3}. $$
Since (3) does holds, one must have $x=y$ and hence
$$ x=1+\sqrt{x+k+1}-\sqrt{x+k}. $$
It is easy to see that $x$ is the real roots of the following equation
$$x(x^3-8x^2+12x-4)-4k(x+1)^2=0.\tag{4}$$
The number of answers depends on the real solutions of (4), either 1,2,3 or 4, depending on $k$.
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