= Solution
{c}
The <degree of a field extension> $[K:F]$ is the dimension of $K$ as a <vector space> over $F$. The <tower law> states that for finite extensions $F\subseteq E\subseteq K$,
$$
[K:F]=[K:E][E:F].
$$
If the left side is prime, one factor is one. Degree one forces equality of the corresponding fields, so \b[there is no proper intermediate field].
Let $K$ have 32 elements. Its <prime field> is $\mathbb F_2$, and its dimension over that field is five, since a dimension-$d$ <vector space> over $\mathbb F_2$ has $2^d$ elements. Every element outside $\mathbb F_2$ has <minimal polynomial> of degree five: its generated intermediate field has degree greater than one and dividing five.
Every element of $K$ satisfies $x^{32}=x$, by the <multiplicative group of a field> of order 31 for nonzero elements. Hence $X^{32}-X$ splits completely over $K$ and has no repeated roots, since its derivative is minus one. Conversely any <irreducible polynomial> $q$ over $\mathbb F_2$ gives a <quotient ring> $\mathbb F_2[X]/(q)$ with 32 elements. Its defining element satisfies $X^{32}-X$, so $q$ divides that <polynomial> and therefore has all five distinct roots in $K$.
The 30 elements of $K\setminus\mathbb F_2$ are thus partitioned into disjoint five-element root sets of the <irreducible polynomials>. Their number is
$$
\boxed{\frac{32-2}{5}=6.}
$$
No classification of the six individual <polynomials> is needed; the tower degree and separability already count them all.
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