This is a qualifying problem. I cannot understand how the inclusion exclusion principle work here in detail.
However, I have an argument which leads to a different answer. I am not sure my answer is correct. Perhaps, I made some stupid mistake. My answer is as follows:
*Statement:*It is known that for $p$ prime the polunomial $x^{p^n}-x$ is precisely the product of all the distinct irreducible polynomial in $\mathbb F_p[x]$ of degree $d$ where $d$ runs through all divisors of $n$. I am not sure whether this statement is also true for $q=p^n$.
Now, suppose it's indeed true, we would have:
$$x^{q^{12}}-x=\prod_{\deg f=d|12,f \ \text{irreducible}} f$$ $$x^{q^6}-x=\prod_{\deg f=d|6,f \ \text{irreducible}} f$$ So, we would have $$\prod_{\deg f=12,f \ \text{irreducible}}f=\frac{x^{q^{12}}-x}{x^{q^6}-x}$$ where the right side is a polynomial with degree $q^{12}-q^{6}$. So, my answer would be $\frac{q^{12}-q^{6}}{12}$.
Please help me to check both answer to the problem, and tell me whether the statement I used here is also true for $q=p^n$. Thank you very much!
You need to also divide out the irreducibles of degree dividing $4$. Since we divided out twice by the irreducibles of degree dividing $2$, we need to multiply by them once.
So the product of all degree $12$ irreducibles should be: $$\frac{(x^{q^{12}}-x)(x^{q^2} - x)}{(x^{q^{4}}-x)(x^{q^{6}}-x)}$$
The degree is $q^{12} - q^6 - q^4 + q^2$, as desired.