Example 2.5.6 For each natural number $n$, $f(n)$ represents the number of non-negative integer solutions $(x, y, z)$ to the indeterminate equation $x+2 y+5 z=n$. Try to find all $n$ such that $5 f(n)$ is a perfect square, and find the square expression at this time.
The generating function is $f(x)=\left(1+x+x^{2}+\cdots\right)\left(1+x^{2}+x^{4}+\cdots\right)\left(1+x^{5}+\right.$
$$
\begin{array}{l}
\left.x^{10}+\cdots\right)=\frac{1}{1-x} \frac{1}{1-x^{2}} \frac{1}{1-x^{5}}, \\
\sum_{n=0}^{\infty} f_{n} x^{n}\left(1-x^{5}\right)=f(x) \cdot\left(1-x^{5}\right) \\
=\frac{1}{1-x} \frac{1}{1-x^{2}} \\
=\frac{\frac{1}{2}}{(1-x)^{2}}+\frac{\frac{1}{4}}{1-x}+\frac{\frac{1}{4}}{1+x} \\
=\sum_{n=0}^{\infty} \frac{n+1}{2} x^{n}+\frac{1}{4} \sum_{n=0}^{\infty} x^{n}+\frac{1}{4} \sum_{n=0}^{\infty}(-1)^{n} x^{n} \\
=\sum_{n=0}^{\infty} \frac{2 n+3+(-1)^{n}}{4} x^{n}, \\
\end{array}
$$

We have $f_{n}-f_{n-5}=\frac{2 n+3+(-1)^{n}}{4}$.
Let $n=5 k+r(0 \leqslant r<5)$,
$$
\begin{array}{l} 
f_{n}-f_{r}=f_{5 k+r}-f_{r} \\
=\sum_{i=1}^{k} \frac{2(5 i+r)+3+(-1)^{5 i+r}}{4} \\
=\frac{5}{2} \frac{k(k+1)}{2}+\frac{r}{2} k+\frac{3}{4} k+\frac{(-1)^{r}}{4} \frac{(-1)^{k}-1}{2}, \\
\begin{aligned}
f_{n}= & \frac{5 k^{2}+8 k+2 r k}{4}+\frac{(-1)^{r}}{4} \frac{(-1)^{k}-1}{2}+f_{r}, \\
4 \cdot 5 f_{n} & =25 k^{2}+40 k+10 r k+\frac{(-1)^{r} \cdot 5}{2}\left[(-1)^{k}-1\right]+20 f_{r} \\
& =(5 k+4+r)^{2}-(4+r)^{2}+20 f_{r}+\frac{5}{2}(-1)^{r}\left[(-1)^{k}-1\right] .
\end{aligned}
\end{array}
$$
$4-5 f_{n}$ is a perfect square, and the right side is also a perfect square.
By $f_{0}=1, f_{1}=1, f_{2}=2, f_{3}=2, f_{4}=3$ and ten cases of $k$ being odd or even, it is known that the requirement is met if and only if $k$ is odd and $r=1$, at which time
$$
5 f_{10 k^{\prime}+6}=\frac{\left[5\left(2 k^{\prime}+1\right)+4+1\right]^{2}}{4}=\left(5 k^{\prime}+5\right)^{2},
$$

Therefore, if and only if $n=10 k^{\prime}+6$, $5 f(n)=25\left(k^{\prime}+1\right)^{2}=\left(\frac{n+4}{2}\right)^{2}$.