Find the smallest odd value of $n$ for which there is a proper linear cyclic code of length $n$ and dimension $k = 5$.
For a proper code, $k < n$. So $n \geq 7$. My notes say we need a proper factor of degree $d = n-k$ of $1+x^n$ over $\mathbb{F}_2$.
I've been given the irreducible polynomials:
$1+x^7 = (1+x)(1+x+x^3)(1+x^2+x^3)$
$1+x^9 = (1+x)(1+x+x^2)(1+x^3+x^6)$
$1+x^{11} = (1+x)(1+x+x^2+\dots+x^{10})$
$1+x^{13} = (1+x)(1+x+x^2+\dots+x^{12})$
$1+x^{15} = (1+x)(1+x+x^2)(1+x+x^4)(1+x^3+x^4)(1+x+x^2+x^3x^4)$
I don't understand why $1+x^{15}$ "has a proper factor" of degree $10$ but the others don't have factors of their respective degree.
For $n$ odd, any $x^n+1$ can be factored as $\prod_{i=0}^{n-1}(x-\alpha^i)$ where $\alpha$ is an element of multiplicative order $n$ in an extension field of the binary field $\mathbb F_2$. Unfortunately, choosing $k$ factors from these $n$ factors arbitrarily does not result in a polynomial with binary coefficients and so one cannot use such a polynomial as the (parity-check) polynomial of a binary cyclic code. We need to find a binary polynomial divisor of $x^n+1$, and such binary polynomials $h(x)$ have the property that if $h(\alpha^i)=0$, then $h(\alpha^{2i})$ also equals $0$. $h^{2i}$ is called a conjugate of $\alpha^i$. This is similar to the more familiar fact that if $f(x)$, a polynomial with real coefficients has a complex root $z$, then $z^*$, the conjugate of $z$ is also a root of $f(x)$. In finite fields, an element can have many conjugates, not just one as in the complex field.
For your specific question, you are given the factorization of $x^n+1$ into binary polynomials for $n=7,9,11,13,15$. To construct a binary cyclic code of dimension $5$, pick some subset of these factors whose degrees add up to $5$ and use this as the parity-check polynomial of your binary cyclic code of dimension $5$. A quick look shows that you can't find polynomials whose degrees add up to $5$ for $n=7,9,11,13$, but you can find such a set when $n=15$ (e.g. $x^4+x+1$ and $x+1$, can you find another?), and you are done.