Let . Prove that is irreducible over for all .

[There is probably a more direct way to do this.]

We begin with a combinatorial lemma.

Lemma: Let be an even integer. Then there does not exist a partition of such that and . Proof: Note that . If , then since , one of and is odd and the other even. Mod 2 we have , a contradiction. Thus . In particular, for each odd prime , if then all multiples of are in ; likewise for . Also, one of or contains only odd numbers except for one, which is maximally divisible by .

Now we will consider how 2, 3, 4, and 5 might be distributed between and . Suppose and are in different classes. Then and are in the same class, and so and are in the same class. But then 4 divides and , a contradiction. Thus 3 and 4 are in the same class. Suppose now that 5 is in the same class as 3 and 4. Now the class containing 3, 4, and 5 must contain all multiples of 3 and 5, as well as all multiples of 2 except for one. Since is even and greater than 5, this class contains elements, a contradiction since and must have the same cardinality. Thus it must be the case that 3 and 4 are in the same class and 5 is in the other class. Then 2 must be in the same class as 5. But now 10 cannot be in either class, since it is both even and a multiple of 5. So .

Suppose . We are forced to have 3 and 4 in one class and 5 and 2 in the other; thus 6 is forced to be in the class with 3 and 4 and 1 with 5 and 2. But then the products and are 72 and 10; neither difference of these is 2.

Suppose . Again, the divisibility criteria force the partition , but now and are 70 and 576, a contradiction.

Thus no such partition exists.

We are now prepared for the main problem.

Let and . Suppose is reducible in , with . Note that for each , we have . Since , we have for all such ; thus , and moreover . Let and . Certainly then for all . By a lemma to this previous exercise, we have and . Considering degrees and the monicness of and , we have . Now , and thus . If , then the left hand side of this equation has positive degree while the right is constant, a contradiction. Thus , and in particular is even. Now evaluating at , we have , and thus or , depending on whether is even or odd. In either case, no such partition exists for , so that .

Note that for , . This polynomial is Eisenstein at 3 and thus irreducible. For , we have . Taking the partition , we let and ; indeed, is reducible.

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A more direct way for is to assume for reductio that is reducible, and to reduce as you did to the case that . Then must be even, so . Evaluating the equation at we have that the RHS is , but the LHS is , a contradiction. So is irreducible for all .