Thread: Using eisenstein

I know how to prove this for a general case using Eisenstein criterion but not for all of p and q

If p and q are prime numbers and p does not = q, then x^5-p^5*q contained in Z(x) is irreducible in Q(x).

Originally Posted by chadlyter

I know how to prove this for a general case using Eisenstein criterion but not for all of p and q

If p and q are prime numbers and p does not = q, then x^5-p^5*q contained in Z(x) is irreducible in Q(x).

It is necessary and sufficient to show that x^5-p^5*q is irreducible in Z[x] for that will imply that it is irreducible in Q[x].

Use Eisenstein Criterion with q, then q|p^5*q obviously.
But q^2 does not divide p^5*q since p!=q.
Q.E.D.