The case of n = p^a q.

Factors of D_n are F_p, F_q, F_pq, F_p²q, ..., F_{p^b q}.... I state some formulas:

F_{p^b q}(X) = F_pq (X^{p^b-1}) = 1 - X^{p^b-1}+... . face=symbol>F_p^k (X) =

p-1
ĺ
i = 0

X^{i p^k-1}

The first one comes from a more general identity: F_{p^k q^l} (X ) = F_pq (X^{p^k-1q^l-1}), which follows from the wonder formula

F_n(X) =

'
d\mid n

(X^d - 1)^m(n/d) (W)

it being the Möbius inversion of the factorization of D_n.

Now in a factorization of D_n = A×B, F_pq is a factor of (say) P, and so may be F_{p or q}.

ˇ If it is not so, then F_pq |P and F_p.F_q | Q, the other factors being the F_{p^b q} (all are polynomials in X^p by the preceding Lemma). But then Q must be wrong:

Q(X) = F_p(X)F_q(X)×R(X^p) = (1+2X+...)(1+k >X^p + ...) = 1+2X + ...

and this is not a 0-1 polynomial.

ˇ Say P contains the factors F_pq and F_q together: then P admits the factor F_pq.F_q = 1+X^p+...+X^p(q-1) (see (W)), and >we get two subcases:

* F_p also divides P: then Q is only a product of the face=symbol>F_{p^b q} and hence a polynomial in X^p,

* F_p does not divide P: then P is a polynomial in X^p.

ˇ Last case: [(F_pq.F_p = F_p(X^q) \buildrel\sevenrm (see (W) ))/( = >1+X^q+...+X^q(p-1) \mid P)], hence

P(X) = F_p(X^q).S(X^p) while Q(X) = F_q(X)×R(X^p) = (1+X+X²+... >X^q-1) R(X^p).

But the completing polynomials R and S (in X^p) are made of

F_{pb. q}(X) = F_pq(X^{p^b}) = (1 - X^{p^b}+ ... ) face=symbol>b ł 1

Hence S(X^p) = 1-X^{p^b}+... for some b ł 1, and P is not a 0-1 polynomial as P(X) = (1+X^q +... )×(1-X^{p^b}+...) and q š p^b.

The case of n = pa q.

The case of n = p^a q.