This is so nice and easy!! Take $z$ a real root of $p$ and observe that it must also be a root for $q,r$. But since $p$ is irreducible, it follows that $p$ divides $q,r$ and the conclusion follows.

harazi wrote: Take $z$ a real root of $p$ and observe that it must also be a root for $q,r$. But since $p$ is irreducible how did you reach that conclusion? it is impossible for me to understand

You see $p$ has odd degree. So it has a real root like $z$. $p$ is irreducible so it's the minimal polynomial $z$ so for each $f\in \mathbb R[x]$ if $f(z)=0$ then $p(x)|f(x)$. Now $q(z)^2+r(z)^2+q(z)r(z)=0$. Therefore $q(z)=r(z)=0$ and $p|q,r$ and everything is done. Also there is a longer solution with using the statemaent $q^2+r^2+qr=(q+\omega r)(q-\omega r)$

Omid Hatami wrote: $p$ is irreducible so it's the minimal polynomial $z$ so for each $f\in \mathbb R[x]$ if $f(z)=0$ then $p(x)|f(x)$ How do we know that $p$ is the minimal polynomial of $z$? We know that it is irreducible in $\mathbb Q$, but does that suffice? (why can't we decompose it?) The rest I've understood. Thanks.

Minimal polynomial and irreducibility are terms in respect to the field we work with. Here both is meant in relation to $\mathbb{Q}$, and in general the following holds: The minimal polynomial of an element is irreducible and an irreducible polynomial is the minimal polynomial for all it's roots. Prove: Let be $p'$ the (rational) minimal polynomial of $z$. Take division of polynomials to get $p(x)=p'(x)q(x)+r(x)$ with $deg(r)<deg(p')$. Since $p(z)=p'(z)=0$, we get that $r(z)=0$, but as $p'$ is minimal, we get that $r(x) \equiv 0$. This gives that $p(x)=q(x)p'(x)$, and when $q$ wouldn't be constant we get a contradiction to the irreducibility of $p$, so $p$ is minimal.

Lemma: Suppose $z\in \mathbb C$ is a root of $p(x)$. Then $p(x)$ is irreducible if and only if it minimal polynomial of $z$. Proof: If $z$ is the minimal polynomial of $p$ then obviously $p$ is irreducible.(if $p(x)=s(x).t(x)$ then $s(z)=0$ or $t(z)=0$). And if $p(z)$ is irreducible. Suppose $T(x)$ is the minimal polynomial of $z$. Then $z$ is a root of $U(x)=gcd(p(x),T(x))$. Now $U(x)=1$ or $U=p$ (because $p$ is irreducible). $U=1$ is not possible beacause $z$ is the root of $U(x)$. So $U=p(x)$. and $deg\ p\leq deg\ T$. So $p(x)=T(x)$ and everything is done.

Oh ZetaX I didn't see your proof.