<s>[INST] Find the polynomial of integral coefficient with minimum degree and root $z+z^3+z^9$. 
Let $z$ be a 13th root of unity $(z\neq 1)$. Find the  polynomial of integral coefficient with minimum degree and root $z+z^3+z^9$.

My idea:
since $z$ such
$$0=z^{12}+z^{11}+\cdots+z+1=\prod_{k=1}^{12}(z-z_{k})$$ maybe use 
$$\sum_{k=1}^{n}z_{k}=-1,\sum_{i<j}z_{i}z_{j}=1,\sum_{i<j<k}z_{i}z_{j}z_{k}=-1,\cdots$$
then we must find
$$\sum_{k=1}^{12}(z_{k}+z^3_{k}+z^9_{k})=?$$
$$\sum_{k<i}(z_{k}+z^3_{k}+z^9_{k})(z_{i}+z^3_{i}+z^9_{i})$$
$$\cdots$$
I think it's too complicated, and it seems like an interesting question, how to think correctly
 [/INST] The hint.
Let $z+z^3+z^9=a$, $z^2+z^5+z^6=b$, $z^4+z^{10}+z^{12}=c$ and $z^7+z^8+z^{11}=d$.
Now, show that:
$$a+b+c+d=-1,$$
$$ac=3+b+d,$$
$$bd=3+a+c,$$
$$bc=a+b+c$$ and get the polynomial:
The second equation gives $$ac=2-a-c$$ or $$c=\frac{2-a}{a+1}.$$
From the fourth equation we obtain:
$$b(c-1)=a+c,$$ which gives
$$b=\frac{a^2+2}{1-2a}.$$ 
Also, $$d=-1-a-b-c=\frac{a^3+3a-5}{(a+1)(1-2a)}.$$
Thus, from the third equation we obtain:
$$\frac{a^2+2}{2a-1}\cdot\frac{a^3+3a-5}{(a+1)(2a-1)}=3+a+\frac{2-a}{a+1}$$ or
$$(a-5)(a^4+a^3+2a^2-4a+3)=0$$ or
$$a^4+a^3+2a^2-4a+3=0.$$
For example, $$ac=3+b+d$$ because
$$ac=(z+z^3+z^9)(z^4+z^{10}+z^{12})=$$
$$=z^5+z^{11}+1+z^7+1+z^2+1+z^6+z^8=$$
$$=3+z^2+z^5+z^6+z^7+z^8+z^{11}=3+b+d.$$
</s>