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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.

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    a) A non-primitive character $\bmod m$ is of the form $\chi(k) = \psi(k) 1_{gcd(k,m) = 1}$ where $\psi$ is a character $\bmod d | m$ (note $1_{gcd(k,m) = 1}$ is the trivial character $\bmod m$). If $m = 2(2a+1)$ then ... – 2017-01-19
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    But is the form you are suggesting trivial (I mean you have a short proof of that fact)? Because Marcus should be self contained and If I'm not wrong, he never mentions this property. Or are there other ways to do this exercise? Anyway thank you very much – 2017-01-19
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    What is the definition of a non-primitive character ? – 2017-01-19
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    Actually I cannot see your point. Could you give me more details please? I know that idea is to find a character $\chi'$ that induces $\chi$. I tried $\chi'$(k mod (2a+1)=$\chi$(k mod m) but this is not well defined. – 2017-01-19
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    take an example https://www.wolframalpha.com/input/?i=Table%5BDirichletCharacter%5B10,+2,+n%5D,+%7Bn,+10%7D%5D – 2017-01-19
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    You don't see a $\psi \bmod 5$ in this $\chi \bmod 10$ ? – 2017-01-19
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    Yes, I see it for values 1,3,7,9 but I'm not getting how to prove the situation in general – 2017-01-19
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    No, you see it for $5,1,7,3,9$ – 2017-01-19
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    For the general situation, I understood that I need to show that for every $a$mod $m$, $(a,m)=1$ such that $a$mod $(2a+1)$=$1$ mod $(2a+1)$, we have $\chi(a$ mod $m)=1$. Any hints/proofs? – 2017-01-20

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