# $f\left ( x+m \right )\left ( f\left ( x \right )+\sqrt{m+1} \right )=-\left ( m+2 \right ),\forall x\in \mathbb{R},m\in \mathbb{Z^+}$

Find all the function $f:\mathbb{R}\rightarrow \mathbb{R}$ sastisfied that $f$ continuous on $\mathbb{R}$ and $$f\left ( x+m \right )\left ( f\left ( x \right )+\sqrt{m+1} \right )=-\left ( m+2 \right ),\forall x\in \mathbb{R},m\in \mathbb{Z^+}$$ I tried to delete $f(x+m)$ or $f(x)$ but I can't

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Pick a sufficient large $m \in \mathbb{Z}^+$ so that $f(0)+\sqrt{m+1}>0$. Then $f(m)=\frac{-(m+2)}{f(0)+\sqrt{m+1}}<0$.
If $f(a)=0$ for some $a \in \mathbb{R}$, then $0=f(a)(f(a-1)+\sqrt{2})=-3$, a contradiction. Thus $f(x) \not =0 \, \forall x \in \mathbb{R}$.
If $f(x)>0$ for some $x \in \mathbb{R}$, then by intermediate value theorem, since $f(x)$ is continuous, $f(a)=0$ for some $a \in \mathbb{R}$, a contradiction. Thus $f(x)<0 \, \forall x \in \mathbb{R}$.
$(f(x)+\sqrt{2})=\frac{-3}{f(x+1)}>0$ so $0>f(x)>-\sqrt{2}$, so $|f(x)|<\sqrt{2}$ and $|f(x)+\sqrt{2}|<\sqrt{2}$ for all $x \in \mathbb{R}$.
$3=|f(x+1)||f(x)+\sqrt{2}|<\sqrt{2}\sqrt{2}$, a contradiction. Thus no such function exist.