Irreducible Polynomials Which Divide Trinomials Over GF(2). - The ...

Note that when r = 2 and p = 7, neither of the f i (x)’s is self-reciprocal, whereΦ 7 (x) = x 6 + x 5 + x 4 + x 3 + x 2 + x + 1 = (x 3 + x 2 + 1)(x 3 + x + 1)has only 7 terms, and the two factors of Φ 7 (x) are already trinomials.Theorem 10. Let p be a prime and Φ p (x) = (xp −1)(x−1)= f 1(x)f 2 (x)f 3 (x)f 4 (x) be a productof four irreducible polynomials (i.e. r = 4). Then the f i (x)’s divide no trinomials.Proof. If any one of the f i (x)’s is self-reciprocal, then the f i (x)’s divide no trinomialsby Theorem 8. Otherwise, the f i (x)’s come in pairs. Let α be a root of f 1 (x) andα −1 be a root of f 2 (x). Similarly, let β = α u be a root of f 3 (x) and β −1 be a root off 4 (x). Suppose f 1 (x) divides some trinomial t 1 (x) (including the case that f 1 (x) itself isa trinomial). Then we can writet 1 (x) = x m + x a + 1 = f 1 (x)g 1 (x),with 1 ≤ a < m < p. Replacing α by α −1 for all roots of t 1 (x), we getf 2 (x)g ∗ 1(x) = t ∗ 1(x) = x m + x m−a + 1.22