request sources about self-dual cyclic codes X. Kai and S. Zhu in the paper "On cyclic self-dual codes", AAECC, vol. 19, pp. 509-525, 2008, at page 510 in line 6 said that, It is well Known that there are no cyclic self-dual codes over $F_q$ when $q$ is odd.
I know that  W. Cary Huffman and V. Pless in the book "Fundamentals of error correcting codes" proved that there are no self-dual cyclic codes of length $n$ over $F_q$ when $gcd(n,q)=1$.
Also Y. Jia, S. Ling and C. Xing in the paper with name "On self-dual cyclic codes over finite fields" at 2011 proved that there exist at least one self-dual cyclic code of length $n$ over $F_q$ if and only if $q$ is a power of $2$ and $n$ is even.
My questions are:
1) Where can I find the proof of the claim that introduced in my first paragraph?
2) Generally, how can I find some good sources about self-dual (self-orthogonal) cyclic codes, precisely about the existence of these codes over finite fields?
 A: I think (1) is straightforward: suppose $C$ is a cyclic code of length $n$ with generator polynomial $f(x) \in F_q[x]$. Let $C'$ be the code with generator polynomial $g(x) = (x^n-1)/f(x)$. As I understand the definition of duality for cyclic codes (following page 84 of van Lint, Introduction to Coding Theory, 3rd edition), the dual of $C$ is $C'$. Hence, if $C$ is self-dual then $f(x) = g(x)$ and so $x^n-1 = f(x)^2$. This is impossible when $q$ is odd. 
I can't help with (2), except to say that a quick Google search found several papers that look relevant, e.g. Self-dual cyclic codes by N.J.A Sloane and J.G. Thompson and On the Minimal Distance of Binary Self-Dual Cyclic Codes by Bas Heijne and Jaap Top.
Edit: as Zahra's comment says, with the other definition of duality for cyclic codes, the generator polynomial for $C^\bot$ is a multiple of $h(x) = g(x^{-1})x^{\deg g}$. The roots of $h(x)$ are the reciprocals of the roots of $g(x)$, counted with multiplicities. Suppose that $1$ is a root of $x^n-1$ with multiplicity $m$. If $1$ has multiplicity $r$ as a root of $f(x)$, then $1$ has multiplicity $m-r$ as a root of $g(x)$ and $h(x)$. Hence, if $C$ is self-dual then $f(x) = h(x)$, and $m-r = r$. Therefore $m$ is even, and again this is impossible when $q$ is odd.
A: Suppose CSD(n,q) is a cyclic self-dual code over $\mathbb{F}_q$ with length n , and its generator polynomial $g(x) = g_0 + g_1 x + ··· + x^{n−k}$. 
So $x^n − 1 = g(x)h(x) = (g_0 + g_1 x + ··· + x^{n−k}) (h_0 + h_1 x + ··· + x^k )$.
We have $g_0 h_0 = −1$.
Because $g^⊥(x)$(generator polynomial of $C^⊥$) = $h_0^{−1} (1 + h_{k−1} x + ··· + h_1 x^{k−1} + h_0 x^k $),
$C = C^⊥$ if and only if $g(x) = g^⊥(x)$. 
It deduces $h_0^{−1}=g_0, k = n/2$.
So $1 = −1$. 
'There exist at least one self-dual cyclic code of length n  over F q   if and only if q  is a power of 2  and n  is even.'
To (2), suppose $gcd(n,q) = 1$,  there exist at least one  CSO(n,q) (cyclic self-orthogonal)if and only if n to q and 1 is 'bad'. 
It means: $n > 2$, and $ord_n (q)$ is odd or $n\nmid q ^{ord_n (q)/2} + 1$.
$Ord_n (q)$ is the ord of q in $\mathbb{Z}_n$ (unit group).
For ' n to q and 1 is 'bad'', please read P. Moree, On the Divisors of $a^k +b^ k$ , Acta Arithmetica, 1997, LXXX.3: 197-212.
