答案：B
A 不一定正確。留意 $f(k) = k^2 -k +1$。
$\begin{array}{rcl}
f(-k) & = & (-k)^2 -(-k) +1\\
f(-k) & = & k^2 +k +1
\end{array}$
所以，$f(k) \not \equiv f(-k)$。
B 必為正確。
$\begin{array}{rcl}
f(1-k) & = & (1-k)^2 -(1-k) +1 \\
f(1-k) & = & 1 -2k +k^2 -1+k+1 \\
f(1-k) & = & k^2 -k +1
\end{array}$
所以，$f(k) \equiv f(1-k)$。
C 不一定正確。留意 $f(1) = 1^2 -1 +1 =1$。
$\begin{array}{rcl}
f(k+1) & = & (k+1)^2 -(k+1) +1 \\
f(k+1) & = & k^2+2k+1-k-1+1 \\
f(k+1) & = & k^2 +k+1 \\
\end{array}$
所以，$f(k+1) \not \equiv f(k) +f(1)$。
D 不一定正確。
$\begin{array}{rcl}
f(k-1) & = & (k-1)^2 -(k-1) +1 \\
f(k-1) & = & k^2 -2k +1 -k+1 +1\\
f(k-1) & = & k^2 -k +3
\end{array}$
所以，$f(k-1) =f(k) -f(1)$。
