2.3. Determine all functions $f: \mathbb{Z} \longrightarrow \mathbb{Z}$, where $\mathbb{Z}$ is the set of integers, such that
$$
f(m+f(f(n)))=-f(f(m+1))-n
$$
for all integers $m$ and $n$.
2.3 Replacing $m$ by $f(f(m))$, we have
$$
f(f(f(m))+f(f(n)))=-f(f(f(f(m))+1))-n .
$$

Interchanging $m$ and $n$ in (1), we get
$$
f(f(f(m))+f(f(n)))=-f(f(f(f(n))+1))-m .
$$

Put $m=1$ in the original functional equation and denote for simplicity $f(f(2))$ by $k$. We obtain $f(f(f(n))+1)=-k-n$. Using this and equating (1) and (2), we get $f(-m-k)-$ $f(-n-k)=m-n$. Letting $m=-p+k$ and $n=-p+k+1$, we have $f(p)-f(p-1)=-1$. Inductively, we obtain $f(p)=f(0)-p$. Thus, $f(f(p))=f(f(0)-p)=f(0)-(f(0)-p)=p$. Substituting this into the original functional equation, we have $f(m+n)=-m-1-n$. In other words, $f(n)=-n-1$ for all $n$. Indeed, $f(n)=-n-1$ satisfies the given functional equation.
(2nd solution by Teo Wei Hao) Setting $m=0$, we obtain $f^{3}(n)=-f^{2}(1)-n$. This functional relation immediately shows that $f$ is bijective, because $f\left(f^{2}(n)\right)=-f^{2}(1)-n$ and $f(p)=f(q) \Longrightarrow f^{3}(p)=f^{3}(q) \Longrightarrow p=q$. Now we may let $f^{2}\left(n_{0}\right)=1$ for some $n_{0}$. The original functional equation becomes $f(m+1)=-f^{2}(m+1)-n_{0}$. Using the fact that $f$ is bijective, we may let $f(m+1)=x$, so that $f(x)=-x-n_{0}$. Applying this on $f^{2}\left(n_{0}\right)=1$ gives $n_{0}=1$. Therefore, $f(x)=-x-1$.
(3rd solution by Colin Tan) Let $f^{2}(1)=c$. Put $n=1, m=0$ and replace $n$ by $f(n)$, we get, respectively,
$$
\begin{aligned}
f(m+c) & =-f^{2}(m+1)-1 \\
f^{3}(n) & =-c-n \\
f\left(m+f^{3}(n)\right) & =-f^{2}(m+1)-f(n)
\end{aligned}
$$

From (1) and (3), we get
$$
f(m+c)+1=f(m-c-n)+f(n)
$$

Put $n=-2 c+1$ and replace $m$ by $m+1-c$ in (4), we get
$$
f(m+1)+1=f(m)+f(-2 c+1) .
$$

Thus $f$ is linear. Write $f(x)=a x+b$ and from the original equation, we get $a=b=-1$. Thus $f(x)=-x-1$. From the above computation, we see that this function satisfies the original functional equation.