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Statement

Setting (p. 143). $f(x)$ is a polynomial with rational integer coefficients, $n$ runs through $0,1,2,3,\ldots$, and $P_n$ is the largest prime factor of $f(n)$. Equation (3) of the paper is $\lim_{n\to\infty}P_n=\infty$.

Satz II (p. 144, quoted). "Ist $f(x)$ ein irreduzibles Polynom vom zweiten Grade, so gilt (3)."

In English: if $f(x)$ is an irreducible polynomial of degree two, then $P_n\to\infty$; equivalently, for every finite set of primes only finitely many $n$ make $f(n)$ a product of primes from that set.

The paper notes (p. 144) that Satz II contains Størmer's result for $x^2+1$, the last of the three polynomials of its equation (2). Pólya offers Satz II, not Thue's Satz I, as his contribution to the question raised on p. 144: for which polynomials (3) holds, and for which only the classical statement (1), that the largest prime factor of $f(n)$ is unbounded.

Remark (p. 147). The paper adds that Thue's results on Diophantine equations also give (3) for $x^n-1$, and more generally for polynomials in which a certain number of the coefficients following the leading one are zero (citing Thue's third paper, p. 304), and that the full answer to the general question of § 1 probably needs another source.

Source. Georg Pólya, Zur arithmetischen Untersuchung der Polynome, Mathematische Zeitschrift 1 (1918), 143-148, doi:10.1007/BF01203608: the setting on p. 143, Satz II on p. 144, the proof in § 3 on pp. 145-147. The edition read is identified on the source card.

Read depth. Claims checked: the statement, its setting and the remark were read clause by clause on the printed pages. The proof was followed for structure and not verified. Nothing here is independently reviewed.

Proof pointer

§ 3, pp. 145-147. If Satz II failed for $ax^2+bx+c$ it would fail for $x^2+bx+ac$, by the identity $af(x)=f^*(ax)$, so $f$ may be taken monic, $f(x)=(x-\alpha)(x-\alpha')$ with conjugate algebraic integers of degree two in the field generated by $\sqrt D$. The paper keeps in view the case of real $\alpha,\alpha'$, which it calls the more involved one because of the infinitely many units. Let $1,\omega$ be an integral basis, $h$ the class number, $\varepsilon$ the fundamental unit, and $\mathfrak p_1,\ldots,\mathfrak p_l$ the prime ideals above the finitely many primes allowed. If infinitely many $n$ had $f(n)$ built from those primes, the ideal $(n-\alpha)$ would factor over the $\mathfrak p_i$; reducing the exponents modulo $3h$ leaves a factor that is the cube of a principal ideal $(x+\omega y)$, which gives (10), $n-\alpha=\beta(x+\omega y)^3$, where the ideal $(\beta)$ takes at most $3^lh^l$ values and, for a given ideal, $\beta$ is one of six numbers $\pm\beta,\pm\varepsilon\beta,\pm\varepsilon^2\beta$. Subtracting the conjugate equation gives (11): $(\alpha'-\alpha)/(\omega'-\omega)=F(x,y)$, a binary cubic form with rational integer coefficients that is not the cube of a linear form, because the three roots of $F(z,1)=0$ are distinct. Thue's theorem then allows only finitely many solutions.

Dependencies

Thue's theorem as the paper states it in § 2 (pp. 144-145), in its second form: if $c\ne0$ and $F(x,y)=c$ has infinitely many integer solutions, then the binary form $F$ is, up to a constant factor, a power of a linear form or of an indefinite quadratic form. The paper cites it from Thue's papers and does not prove it. The proof also uses the finiteness of the class number and the structure of the unit group of a quadratic field.

Bears on

No Erdős problem page in the corpus consumes Satz II. The paper's link to Problem 891 runs through equation (14), the reformulation of Satz I.