Let q=p^e, where p is a prime and e is a positive integer. Let a be in GF(q). Show that f(x)=x^p-x+a is irreducible over GF(q) if and only if f(x) has no root in GF(q)
The Attempt at a Solution
One of the directions seems obvious. Namely, if f is irreducible, the f has no roots: if f had a root, say b in GF(q), then f(x)=(x-b)*g(x), where g is a poly with coefficients from GF(q).
The other direction: if f has no root, then f is irreducible. Or equivalently, f is reducible, then f has a root in GF(q). Frankly, I don't have any ideas. The only "observation" I made is that if the power of f were q rather than p and it had a root, say b, then b+1, b+2,...,b+p-1 would be roots, too. I think the key point here is that I need to "guess" what the linear factorization of f in its splitting field looks like.