Two exercises on characters on Marcus (Part 2) I am trying to solve exercises 17 and 18 on Marcus book (page 210, chapter 7). Let's look at the second. This exercise is essential in order to deduce the class number formula in Marcus approach.
18)Let $\chi$ be a primitive character mod $m\ge 3$ such that $\chi(-1)=-1$. Set:
$u=\sum_{1\le k<m}\chi(k)k, v=\sum_{1\le k<m/2}\chi(k)k, w=\sum_{1\le k<m/2}\chi(k)$
a) Show that $u=2v-mv$

The claim to prove is $\sum_{1\le k<m}\chi(k)k=\sum_{1\le k<m/2}\chi(2k)-\sum_{1\le k<m/2}\chi(k)m$. Now one side gives me $\sum_{1\le k<m/2}\chi(k)k+\sum_{m/2\le k<m}\chi(k)k$. The other side gives me $\sum_{1\le k<m/2}\chi(k)(2k-m)=-\sum_{1\le k<m/2}\chi(k)(m-k)+\sum_{1\le k<m/2}\chi(k)k$. So it remains to prove that $\sum_{m/2\le k<m}\chi(k)k=-\sum_{1\le k<m/2}\chi(k)(m-k)$. But this is true because setting $k=m-t$ you obtain $\sum_{m/2\le k<m}\chi(k)k=\sum_{1\le t\le m/2}\chi(m-t)(m-t)=-\sum_{1\le t\le m/2}\chi(t)(m-t)$.

b) Suppose m is odd. Show that $u=4\chi(2)v-m\chi(2)w$ Hint: Replace odd values of $k$ by $m-k$, $k$ even.

I tried but I fail at some point. Thanks to part a), I get that the statement is equal to prove that $u=\chi(2)(u+2v)$. I obtain from one side $\sum_{1\le 1<m, k even}\chi(k)k-\sum_{1\le k<m,k even}\chi(k)(m-k)$. From the other side $\sum_{1\le k<m}\chi(2k)k+\sum_{1\le k<m/2}\chi(2k)2k$. So it remains to prove that $\sum_{1\le k<m}\chi(2k)k=\sum_{1\le k<m, k even}\chi(k)(k-m)$ which seems false to me. Any help??

c)Show that if $m$ is odd, then $u=mw/(\overline{\chi(2)}-2)$
d)Now suppose $m$ is even. Use exercise 17d at the link Two exercises on characters on Marcus (part 1) to prove that $u=-mw/2$.
 A: Your calculation for a) is essentially correct, but it misses an argument why
$$\sum_{1 \leqslant t \leqslant m/2}\chi(t)(m-t) = \sum_{1 \leqslant t < m/2} \chi(t)(m-t).\tag{$\ast$}$$
The argument is simple of course: If $m$ is odd, then for an integer $t$ we have $t \leqslant m/2 \iff t < m/2$, and if $m$ is even, then $\gcd(m,m/2) > 1$, so $\chi(m/2) = 0$. Nevertheless, the argument must be made.
Also, you have mistyped what you want to prove. At the statement of a), it must be $u = 2v - mw$, not $u = 2v - mv$, and at the start of your calculation you wrote $\sum \chi(2k)$ instead of $\sum \chi(k)2k$.
Let's repeat the calculation laid out differently:
\begin{align}
\sum_{1 \leqslant k < m} \chi(k)k
&= \sum_{1 \leqslant k < m/2} \chi(k)k + \sum_{m/2 \leqslant k < m} \chi(k)k \\
&= \sum_{1 \leqslant k < m/2} \chi(k)k + \sum_{m/2 < k < m} \chi(k)k \tag{$\ast$} \\
&= \sum_{1 \leqslant k < m/2} \chi(k)k + \sum_{1 \leqslant t < m/2} \chi(m-t)(m-t) \tag{$t = m-k$} \\
&= \sum_{1 \leqslant k < m/2} \chi(k)k + \sum_{1 \leqslant t < m/2} -\chi(t)(m-t) \tag{$\chi(m-t) = -\chi(t)$} \\
&= \sum_{1\leqslant k < m/2} \chi(k)k + \sum_{1 \leqslant k < m/2} \chi(k)(k-m) \\
&= 2 \sum_{1 \leqslant k < m/2} \chi(k)k - m \sum_{1 \leqslant k < m/2} \chi(k).
\end{align}
The calculation for part b) is similar. Following the hint (also strongly suggested by the appearance of $\chi(2)$ as a common factor), we start by splitting into sums for even $k$ and for odd $k$:
\begin{align}
\sum_{1 \leqslant k < m} \chi(k)k
&= \sum_{\substack{1\leqslant k < m \\ k \equiv 0 \pmod{2}}} \chi(k)k + \sum_{\substack{1\leqslant k < m \\ k \equiv 1 \pmod{2}}} \chi(k)k \\
&= \sum_{1 \leqslant r < m/2} \chi(2r)2r + \sum_{1 \leqslant s < m/2} \chi(2s-1)(2s-1) \\
&= \sum_{1 \leqslant r < m/2} \chi(2r)2r + \sum_{1 \leqslant r < m/2} \chi(m - 2r)(m-2r) \tag{$r = \frac{m+1}{2}-s$}\\
&= \sum_{1 \leqslant r < m/2} \chi(2r)2r + \sum_{1 \leqslant r < m/2} \chi(2r)(2r-m) \tag{$\chi(m-t) = -\chi(t)$} \\
&= 4 \sum_{1 \leqslant r < m/2} \chi(2r)r - m\sum_{1\leqslant r < m/2}\chi(2r) \\
&= 4\chi(2)\sum_{1 \leqslant r < m/2} \chi(r)r - m\chi(2)\sum_{1\leqslant r < m/2} \chi(r). \tag{$\chi(2r) = \chi(2)\chi(r)$}
\end{align}
For part c), we use the results of a) and b) together with $\overline{\chi(2)}\chi(2) = 1$ for odd $m$:
\begin{align}
\bigl(\overline{\chi(2)} - 2\bigr)u &= \overline{\chi(2)}u - 2u \\
&= \overline{\chi(2)}\bigl(\underbrace{4\chi(2)v-m\chi(2)w}_{b)}\bigr) - 2\bigl(\underbrace{2v-mw}_{a)}\bigr) \\
&= (4v-mw) - (4v-2mw) \\
&= mw,
\end{align}
and dividing by $\overline{\chi(2)} - 2$ (which is nonzero) yields the result.
For part d), we again split the sum at $m/2$, and use $\chi(m/2+k) = -\chi(k)$ [17.d) gives that for odd $k$, and for even $k$ that follows from $\chi(k) = -\chi(m/2+k) = 0$ because $m/2$ is even by 17.a)]:
\begin{align}
\sum_{1 \leqslant k < m} \chi(k)k
&= \sum_{1 \leqslant k < m/2} \chi(k)k + \sum_{m/2 \leqslant r < m} \chi(r)r \\
&= \sum_{1 \leqslant k < m/2} \chi(k)k + \sum_{m/2 < r < m} \chi(r)r \tag{$\ast$} \\
&= \sum_{1 \leqslant k < m/2} \chi(k)k + \sum_{1 \leqslant k < m/2} \chi(m/2 + k)\biggl(\frac{m}{2} + k\biggr) \\
&= \sum_{1 \leqslant k < m/2} \chi(k)k - \sum_{1 \leqslant k < m/2} \chi(k)\biggl(\frac{m}{2} + k\biggr) \tag{17.d)} \\
&= -\frac{m}{2}\sum_{1 \leqslant k < m/2} \chi(k).
\end{align}
Note that since $\chi(2) = 0$ for even $m$, the results of c) and d) can both be written in the same way:
$$u = \frac{mw}{\overline{\chi(2)} - 2}.$$
The derivation however depends on the parity of $m$.
