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Statement

Notation as on the Theorem 1 page (p. 2); irreducibility is in $\mathbb Z[x]$.

Corollary (p. 3, unnumbered). Let $f(x),g(x)\in\mathbb Z[x]$ with $f(0)\neq0$, $g(0)\neq0$ and $\gcd_{\mathbb Z}(f(x),g(x))=1$, and let $r_1$ and $r_2$ be the numbers of non-zero terms of $f$ and $g$. Put

$$N=2\lVert f\rVert^2+2\lVert g\rVert^2+2r_1+2r_2-7.$$

If

$$n\ \ge\ \max\Bigl\{2\times5^{2N-1},\ 2\max\{\deg f,\deg g\}\Bigl(5^{N-1}+\frac14\Bigr)\Bigr\},$$

then the non-reciprocal part of $f(x)x^n+g(x)$ is irreducible or identically $1$ or $-1$, unless one of the following holds:

(i) $-f(x)g(x)$ is a $p$th power for some prime $p$ dividing $n$;

(ii) for $\varepsilon=1$ or for $\varepsilon=-1$, one of $\varepsilon f(x)$ and $\varepsilon g(x)$ is a 4th power, the other is 4 times a 4th power, and $4\mid n$.

The paper adds (p. 3) that when (i) or (ii) holds the non-reciprocal part of $f(x)x^n+g(x)$ is not irreducible, so the exceptions are genuine. It credits the case $f=1$ (equivalently $g=1$), without an explicit bound on $n$, to Schinzel (Acta Arith. 11 (1965), Theorem 5; Acta Arith. 13 (1967), Lemma 4).

The case $g=1$ (read off here; the paper does not write it out). Then the conditions $g(0)\ne0$ and $\gcd_{\mathbb Z}(f,1)=1$ hold automatically, $N=2\lVert f\rVert^2+2r_1-3$, and exception (ii) reduces to $f$ being 4 times a 4th power with $4\mid n$, since $\pm1$ is not 4 times a 4th power and $-1$ is not a 4th power in $\mathbb Z[x]$.

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 Corollary and the remarks after it on p. 3, its proof on p. 10. The authors' 1999 lecture states an abbreviated form, compared on the talk page.

Read depth. Claims checked: the statement and the remarks after it were read clause by clause on the page images. The proof was read but not checked step by step, and the case $g=1$ above is an observation of this page, not the paper's. Nothing here is independently reviewed.

Proof pointer

P. 10. Apply Theorem 2 to $F=f(x)x^n+g(x)$ with $k_0=2\max\{\deg f,\deg g\}$; the norm and term count of $F$ turn Theorem 2's $N$ into the one above. Since $k\ge k_0$, the terms coming from $g$ get $y$-exponent $0$, and condition (i) of Theorem 2 forces the terms coming from $f$ to share one $y$-exponent $\ell$, positive by the remark after Theorem 2. So $x^{-m}G(x,y)=f(x)x^dy^\ell+g(x)x^{d'}$, and Capelli's theorem on binomials over $\mathbb Q(x)$ (cited from Schinzel's Selected Topics on Polynomials) shows it irreducible unless (i) or (ii) holds.

Dependencies

Theorem 2 of the same paper; Capelli's theorem.

Bears on

• Problem 7: the paper recalls (p. 1) Schinzel's result that a polynomial $f\in\mathbb Z[x]$ with $f(1)\ne-1$ and $f(x)x^n+1$ reducible for every positive integer $n$ would force an odd covering of the integers, and says its approach gives factorization information on $f(x)x^n+1$ sufficient to carry out that connection (p. 2). The case $g=1$ of the Corollary supplies that information with an explicit range of $n$. Neither the paper nor this page constructs a covering or decides whether an odd covering exists.