The seventy-sixth question: The function $G(n): N \rightarrow N$ satisfies $G(0)=0, G(n)=n-G(G(n-1))$, find all $G(n)$ that satisfy the condition.
The 76th question,
Solution: According to the conditions, $G(1)=1-G(G(0))=1$. Clearly, $G(n) \leq n$. If $G(0), G(1), \ldots, G(n)$ are all chosen, then $G(n+1)=n-G(G(n))$ is also fixed. This indicates that the function $G(n)$ is uniquely determined.

Next, we prove that $G(n)=\left[\frac{\sqrt{5}-1}{2}(n+1)\right]$ satisfies the conditions. That is, we need to prove that for any non-negative integer $n$, we always have $\left[\frac{\sqrt{5}-1}{2}(n+1)\right]=n-\left[\frac{\sqrt{5}-1}{2}\left(\left[\frac{\sqrt{5}-1}{2} n\right]+1\right)\right]$.

For convenience, let $\alpha=\frac{\sqrt{5}-1}{2}$. We need to prove that $[\alpha(n+1)]=n-[\alpha([\alpha n]+1)]$. At this point, $\alpha^{2}+\alpha-1=0$, and $\frac{1}{\alpha}=\alpha+1$, so
\[
\begin{array}{l}
{[\alpha(n+1)]-(n-[\alpha([\alpha n]+1)])} \\
=[\alpha(n+1)]+[\alpha([\alpha n]+1)]-n \\
=[\alpha n]+[\{\alpha n\}+\alpha]+\left[\alpha^{2} n-\alpha\{\alpha n\}+\alpha\right]-n \\
=[\alpha n]+[\{\alpha n\}+\alpha]+[(1-\alpha) n-\alpha\{\alpha n\}+\alpha]-n \\
=[\alpha n]+[\{\alpha n\}+\alpha]+[-\alpha n-\alpha\{\alpha n\}+\alpha] \\
=[\{\alpha n\}+\alpha]+[[\alpha n]-\alpha n-\alpha\{\alpha n\}+\alpha] \\
=[\{\alpha n\}+\alpha]+[-\{\alpha n\}-\alpha\{\alpha n\}+\alpha] \\
=[\{\alpha n\}+\alpha]+\left[\frac{1}{\alpha}(1-\alpha-\{\alpha n\})\right]
\end{array}
\]

Notice that $\{\alpha n\}+\alpha<1+\alpha$. If $0 \leq\{\alpha n\}+\alpha<1$, then $[\{\alpha n\}+\alpha]+\left[\frac{1}{\alpha}(1-\alpha-\{\alpha n\})\right]=0+0=0$; if $1 \leq\{\alpha n\}+\alpha<1+\alpha$, then $[\{\alpha n\}+\alpha]+\left[\frac{1}{\alpha}(1-\alpha-\{\alpha n\})\right]=1-1=0$. Therefore, we always have $[\alpha(n+1)]=n-[\alpha([\alpha n]+1)]$.
In summary, the function $G(n)=\left[\frac{\sqrt{5}-1}{2}(n+1)\right]$ satisfies the conditions.