6.44 For a prime $p \geqslant 3$, define
$$
F(p)=\sum_{k=1}^{\frac{p-1}{2}} k^{120}, \quad f(p)=\frac{1}{2}-\left\{\frac{F(p)}{p}\right\},
$$

where $\{x\}=x-[x]$ denotes the fractional part of $x$. Find the value of $f(p)$.
(China National Team Selection Test, 1993)
[Solution] Perform the division of 120 by $p-1$ with remainder:
$$
120=q(p-1)+r, \quad 0 \leqslant r \leqslant p-2 .
$$

Since the prime $p \geqslant 3$, $p$ is odd, and since 120 and $p-1$ are even, $r$ is even.

Define
$$
G(p)=\sum_{k=1}^{\frac{p-1}{2}} k^{r} .
$$

According to Fermat's Little Theorem, for $k=1,2, \cdots, \frac{p-1}{2}$, we have

Thus,
$$
\begin{aligned}
k^{p-1} & \equiv 1 \quad(\bmod p), \\
F(p) & \equiv G(p) \quad(\bmod p) . \\
f(p) & =\frac{1}{2}-\left\{\frac{F(p)}{p}\right\} \\
& =\frac{1}{2}-\left\{\frac{G(p)}{p}\right\} .
\end{aligned}
$$

We will discuss the following two cases.
Case I: $r=0$.
In this case,
$$
\begin{aligned}
G(p) & =\sum_{k=1}^{\frac{p-1}{2}} k^{0}=\frac{p-1}{2} . \\
f(p) & =\frac{1}{2}-\left\{\frac{G(p)}{p}\right\} \\
& =\frac{1}{2}-\frac{p-1}{2 p} \\
& =\frac{1}{2 p} .
\end{aligned}
$$

Case II: $r \neq 0$.
Since $r$ is even, for $\bmod p$, we have
$$
\begin{aligned}
G(p) & =1^{r}+2^{r}+\cdots+\left(\frac{p-1}{2}\right)^{r} \\
& \equiv(p-1)^{r}+(p-2)^{r}+\cdots+\left(p-\frac{p-1}{2}\right)^{r} \\
& =(p-1)^{r}+(p-2)^{r}+\cdots+\left(\frac{p+1}{2}\right)^{r} \quad(\bmod p) . \\
2 G(p) & =G(p)+G(p) \\
& \equiv \sum_{i=1}^{p-1} i^{r}(\bmod p) .
\end{aligned}
$$

Also, the congruence equation
$$
x^{r} \equiv 1(\bmod p) \text {. }
$$

has at most $r$ distinct solutions, $0 \leqslant r \leqslant p-2$, so there exists at least one
$$
a \in\{1,2, \cdots, p-1\}
$$

such that
$$
a^{r} \not \equiv 1(\bmod p) .
$$

We have
$$
\begin{aligned}
2 a^{r} G(p) & =(1 \cdot a)^{r}+(2 \cdot a)^{r}+\cdots+((p-1) a)^{r} \\
& \equiv 2 G(p) \quad(\bmod p)
\end{aligned}
$$
(because $1 \cdot a, 2 \cdot a, \cdots,(p-1) a$ are pairwise incongruent modulo $p$, forming a complete residue system modulo $p$), thus
$$
2\left(a^{r}-1\right) G(p) \equiv 0 \quad(\bmod p) .
$$

Since $2\left(a^{r}-1\right) \not \equiv 0 \quad(\bmod p)$,
we have $G(p) \equiv 0 \quad(\bmod p)$.
In this case, $f(p)=\frac{1}{2}-\left\{\frac{G(p)}{p}\right\}=\frac{1}{2}$.
Now, we seek the primes $p \geqslant 3$ that belong to Case I, i.e., those satisfying $p-1 \mid 120$. These primes are
$$
3,5,7,11,13,31,41,61 \text {. }
$$

All other primes belong to Case II.
In summary, the answer to this problem is
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|}
\hline$p$ & 3 & 5 & 7 & 11 & 13 & 31 & 41 & 61 & other odd primes \\
\hline$f(p)$ & $\frac{1}{6}$ & $\frac{1}{10}$ & $\frac{1}{14}$ & $\frac{1}{22}$ & $\frac{1}{26}$ & $\frac{1}{62}$ & $\frac{1}{82}$ & $\frac{1}{122}$ & $\frac{1}{2}$ \\
\hline
\end{tabular}