Since it's been unanswered for so long, here's a quick proof, albeit non-elementary: Let $f(z)=z^n+az+p,\ g(z)=z^n+p$. We have $|f-g|=|az|$. For $z$ on the unit circle, this is $=|a|<-1+p=-|z^n|+p\le|z^n+p|=|g|$. By Rouche's Theorem, $f$ and $g$ must have the same number of roots inside the unit disk, and since $g$ has none, all the roots of $f$ are outside the closed unit disk. This proves the irreducibility of $f$, since if it were reducible, one of its factors would have to have free term equal to $\pm 1$, so it would have to have a root with absolute value $\le 1$.

We can also use triangular inequality to prove the fact that all roots are outside the unit disk. This makes it more elementary.

Can we use Rouche theorem in olympiad?

nmd27082001 wrote: Can we use Rouche theorem in olympiad? Should be proved I assume