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Statement

Notation (p. 2). For a real $u$, $[u]$ is the greatest integer at most $u$; $a\bmod k$ is the integer $b\in[0,k)$ with $a\equiv b\pmod k$. For $F(x)=\sum_{j=0}^r a_jx^{d_j}$, $\lVert F\rVert=(\sum_{j=0}^r a_j^2)^{1/2}$ and $\widetilde F(x)=x^{\deg F}F(1/x)$; $F$ is reciprocal if $F=\pm\widetilde F$. The non-reciprocal part of $F$ is $F$ divided by the product of its irreducible reciprocal factors in $\mathbb Z[x]$ with positive leading coefficient, each taken to the multiplicity with which it divides $F$. Reducibility and irreducibility are in $\mathbb Z[x]$ unless stated otherwise, and $1$ and $-1$ are neither (p. 1).

Theorem 1 (p. 2). Let $F(x)=\sum_{j=0}^r a_jx^{d_j}\in\mathbb Z[x]$ with $0=d_0<d_1<\cdots<d_r$ and $a_0a_1\cdots a_r\neq0$, and let $k_0\ge2$ be real. Put $N=2\lVert F\rVert^2+2r-5$ and suppose

$$\deg F\ \ge\ \max\Bigl\{2^{N+9\times2^{N-1}}+2^{9\times2^{N-2}},\ k_0\times2^{9\times2^{N-2}}\Bigr\}.$$

If the non-reciprocal part of $F$ is reducible in $\mathbb Z[x]$, then there is a positive integer $k\in[k_0,\deg F]$ such that

$$G(x,y)=\sum_{j=0}^r a_jx^{\bar d_j}y^{\ell_j},\qquad \bar d_j=d_j\bmod k,\quad d_j=k\ell_j+\bar d_j,$$

is reducible in $\mathbb Z[x,y]$.

The paper remarks (p. 2) that the converse nearly holds: a factorization of $G(x,y)$ gives one of $F(x)=G(x,x^k)$, but a nontrivial factorization of $F$ need not make its non-reciprocal part reducible.

Source. M. Filaseta, K. Ford and S. Konyagin, On an irreducibility theorem of A. Schinzel associated with coverings of the integers, Illinois J. Math. 44 (2000), no. 3, 633--643, doi:10.1215/ijm/1256060421, read in the author manuscript identified on the source card, whose pages are numbered 1 to 10 and carry no journal pagination: the notation and Theorem 1 on p. 2, Lemmas 1 and 2 on pp. 4--5, the proof in Section 3 on pp. 8--9.

Read depth. Claims checked: the notation and the statement were read clause by clause on the page images. The proof was read but not checked step by step. Nothing here is independently reviewed.

Proof pointer

Pp. 8--9, with Lemma 2 (p. 5). If the non-reciprocal part is reducible, $F$ factors as $uv$ with $u$ and $v$ both non-reciprocal; then $W=u\widetilde v$ satisfies $F\widetilde F=W\widetilde W$, the four polynomials being distinct of degree $d_r$. Comparing the coefficient of $x^{d_r}$ gives $\lVert W\rVert=\lVert F\rVert$, so $W$ has at most $\lVert F\rVert^2$ terms. The exponents of $F$ and $W$ and their distances from $d_r$ form a set of at most $N$ positive integers, and Lemma 2, an elementary statement on residues, gives an integer $k\in[k_0,d_r]$ for which each of them has residue below $k/2$. Splitting exponents modulo $k$ then lifts $F$, $\widetilde F$, $W$ and $\widetilde W$ to two-variable polynomials whose $x$-degrees stay below $k/2$, so the identity lifts without carries, and unique factorization in $\mathbb Z[x,y]$ forces the lift $G$ of $F$ to be reducible.

Dependencies

Lemmas 1 and 2 of the same paper (pp. 4--5), summarized on the source card.

Bears on

• Problem 7: the paper presents its approach as an alternative way to obtain the factorization information on $f(x)x^n+1$ that Schinzel's link between such polynomials and odd coverings uses (pp. 1--2). The theorem is a statement about polynomials; it constructs no covering and proves nothing about whether an odd covering exists.