t. 5 points, find all functions $f: Z \rightarrow Z$, such that for all $\therefore$ we have $f[f(n)]+f(n)=2 n+3$, and $f(0)=1$.

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When $n=0$, $f[f(0)]+f(0)=f(1)+f(0)=$ $f(1)+1=2 \times 0+3$, so $f(1)=2$.
Let $n=1$, then
$$
f[f(1)]+f(1)=2 \times 1+3=5,
$$

which means $f(2)+2=5$, so $f(2)=3$.
From this, we conjecture: $f(n)=n+1$.
We need to prove this by induction for all $n$ in the set of integers $\mathbb{Z}$.
For $n \geqslant 0$, if $n \leqslant k$, then $f(n)=n+1$, so
$$
\begin{array}{l}
f[f(k)]+f(k)=2 k+3, \\
f(k+1)+k+1=2 k+3, \\
f(k+1)=k+2.
\end{array}
$$

Thus, for $n \geq 0$, and $n \in \mathbb{Z}$, $f(n)=n+1$.
Assume for some $k \leqslant 0$, when $n \geqslant k$, we have
$$
f(n)=n+1.
$$

We need to prove that when $n=k-1$, $f(k-1)=k$.
$$
\begin{array}{l}
\text{Let } f(k-1)=a \in \mathbb{Z}, \text{ then} \\
\begin{aligned}
f(a) & =f[f(k-1)] \\
& =2(k-1)+3-f(k-1) \\
& =2 k+1-a.
\end{aligned}
\end{array}
$$

Thus, $f(2 k+1-a)=f[f(a)]$
$$
\begin{array}{l}
=2 a+3-(2 k+1-a) \\
=3 a-2 k+2,
\end{array}
$$
$$
\begin{array}{l}
f(3 a-2 k+2) \\
=f[f(2 k+1)-a] \\
=2(2 k+1-a)+3-f(2 k+1-a) \\
=6 k-5 a+3.
\end{array}
$$

If $2 k+1-a \geqslant k$, by the induction hypothesis,
$$
f(2 k+1-a)=2 k+2-a,
$$

then
$$
\begin{array}{l}
3 a-2 k+2=2 k+2-a, \\
a=k.
\end{array}
$$

If $3 a+2-2 k \geqslant k$, by the induction hypothesis,
$$
f(3 a+2-2 k)=3 a+3-2 k=6 k-5 a+3,
$$
$a=k$.
If $\left\{\begin{array}{l}2 k+1-ak+1, \\ a<k-\frac{2}{3},\end{array}\right.$ a contradiction arises.
Thus, $f(k-1)=k$.
In conclusion, for all $n \in \mathbb{Z}$, $f(n)=n+1$.