I am trying to solve exercises 17 and 18 on Marcus book (page 210, chapter 7). Let's look at the first. This should be solved and correct now.
17) Let $m$ be even, $m\ge3$ and suppose $\chi$ is a primitive character mod $m$. Set $n=m/2$.
a) Show that $n$ must be even.
I found the answer here: Primitive characters mod k
b) Show that $(n+1)^2\equiv 1\pmod m$, and for odd $k$, $(n+1)k\equiv (n+k)\pmod m$
Point b) is clear assuming a)
c)Prove that $\chi(n+1)=-1$. Hint: use (b) and the fact that $\chi$ is primitive
Well, by b) we have $1=\chi((n+1)^2)=\chi(n+1)\chi(n+1)=\chi^2(n+1)$ and so $\chi(n+1)=+-1$. Now suppose by absurd that $\chi(n+1)=1$, and consider every element $a\in \mathbb Z$ such that $(a,m)=1$ and $a\equiv 1\pmod n$. We have that every element $a$ has the form $1+nk$ for some $k$. Now if $k$ is even then $a=1+mt$ for some $t$ and so $\chi(a)=1$. If $k$ is odd then $a=1+mt+n$ which is congruent to $(n+1)\pmod d$. Hence again $\chi(a)=1$.
d) Prove that $\chi(n+k)=-\chi(k)$.
Trivial for even $k$, easy for odd $k$ assuming c)
Hints are welcome.