5-40 Let $n$ be a positive integer, find the number of odd coefficients in the polynomial $P_{n}(x)=\left(x^{2}+x+1\right)^{n}$.

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[Solution] Let $P(x), Q(x)$ be polynomials with integer coefficients. If all the coefficients of $P(x)-Q(x)$ are even, we denote this as $P \sim Q$. If $P \sim Q$, then the number of odd coefficients of $P(x)$ and $Q(x)$ are the same. Below, we discuss several cases. Let $\beta(P(x))$ denote the number of odd coefficients of $P(x)$.
(1) $n=2^{q}, q \in \mathbb{Z}^{+}$, in this case, it is easy to prove by induction that
$$
P_{2^{q}}(x)=\left(x^{2}+x+1\right)^{2^{q}} \sim x^{2^{q+1}}+x^{2^{q}}+1,
$$

so $\quad \beta\left(P_{2^{q}}{ }^{q}(x)\right)=3$.
(2) $n=2^{m}-1, m \in \mathbb{N}$, first let $m=2 k+1, k \in \mathbb{N}$, then $n=2^{m}-1 \equiv 1(\bmod 3)$, consider the polynomial
$$
\begin{aligned}
R(x)= & (x+1)\left(x^{2 n-1}+x^{2 n-4}+\cdots+x^{n+6}+x^{n+3}\right)+x^{n+1} \\
& +x^{n}+x^{n-1}+(x+1) \times\left(x^{n-4}+x^{n-7}+\cdots+x^{3}+1\right) .
\end{aligned}
$$

Clearly, $\beta(R(x))=2 \times\left(\frac{n-4}{3}+1\right)+3+\left(\frac{n-4}{3}+1\right) \times 2=\frac{4 n+5}{3}$
$$
=\frac{1}{3}\left(2^{m+2}+1\right), \text { and } \\
R(x)\left(x^{2}+x+1\right) \sim & (x+1)\left(x^{2 n+1}+x^{2 n}+x^{2 n-1}+\cdots+x^{n+4}\right. \\
& \left.+x^{n+3}\right)+x^{n-1}\left(x^{4}+x^{2}+1\right)+(x+ \\
& 1)\left(x^{n-2}+x^{n-3}+\cdots+x+1\right) \\
\sim & \left(x^{2 n+2}+x^{n+3}\right)+\left(x^{n+3}+x^{n+1}+x^{n-1}\right)+ \\
& \left(x^{n-1}+1\right) \\
\sim & x^{2 n+2}+x^{n+1}+1,
\end{aligned}
$$

By the proof in (1), we also have
$$
P_{n}(x)\left(x^{2}+x+1\right) \sim x^{2 n+2}+x^{n+1}+1,
$$

Therefore, $\beta\left(P_{n}(x)\right)=\beta(R(x))=\frac{1}{3}\left(2^{m+2}+1\right)$.
If we set $m=2 k$, then $n \equiv 0(\bmod 3)$, consider the polynomial
$$
\begin{aligned}
Q(x)= & (x+1)\left(x^{2 n-1}+x^{2 n-4}+\cdots+x^{n+5}+x^{n+2}\right) \\
& +x^{n}+(x+1)\left(x^{n-3}+x^{n-6}+\cdots+x^{3}+1\right) .
\end{aligned}
$$

Similarly, we get
$$
\beta\left(P_{n}(x)\right)=\beta(Q(x))=\frac{1}{3}\left(2^{m+2}-1\right) .
$$

In summary, we have
$$
\beta\left(P_{2^{m}-1}(x)\right)=\frac{1}{3}\left(2^{m+2}-(-1)^{m}\right) \text {. }
$$
(3) Represent $n \in \mathbb{N}$ in binary as

where for each $i \geqslant 1, a_{i}>0$, for each $i \geqslant 2, b_{i}>0, b \geqslant 0$.
Let
$$
\begin{array}{l}
s_{1}=b_{1}, \\
s_{2}=b_{1}+a_{1}+b_{2}, \\
\cdots \cdots \\
s_{k}=b_{1}+a_{1}+b_{2}+a_{2}+\cdots+a_{k-1}+b_{k} .
\end{array}
$$

Then
$$
n=\sum_{i=1}^{k} 2^{s_{i}}\left(2^{a_{i}}-1\right)
$$

Thus
$$
\begin{aligned}
P_{n}(x) & =\prod_{i=1}^{k}\left(x^{2}+x+1\right)^{2_{i} s_{i}}\left(2^{a_{i}}-1\right) \\
& \sim \prod_{i=1}^{k}\left(x^{2_{i} s_{i}+1}+x^{2_{i}^{s_{i}}}+1\right)^{2^{a_{i}-1}} .
\end{aligned}
$$

Using the results already proven, we can obtain
$$
\beta\left(P_{n}(x)\right)=\prod_{i=1}^{k} \frac{1}{3}\left(2^{a_{i}+2}-(-1)^{a_{i}}\right) .
$$