Integer-valued polynomials

The polynomial

is integral-valued, but it does not have integral coefficients if $m\geq 2$.

Every polynomial $P(x)$ of degree $m$ may be expressed in the form

$P(x)$ is integral-valued if and only if the numbers $b_0,b_1,\ldots ,b_{m-1},b_m$ are integers.

If the polynomial $P(x)$ of degree $m$ is integral-valued, then $m!P(x)$ has integral coefficients.

If a polynomial of degree $m$ assumes integral values for $m+1$ consecutive integral values of the variable, then it assumes integral values for all integral values of the variable.

Every odd polynomial $P(x)$ of degree $2m-1$ may be written in the form

$P(x)$ is integral-valued if and only if the numbers $c_1,c_2,\ldots ,c_{m-1},c_m$ are integers.

Every even polynomial $P(x)$ of degree $2m$ may be written in the form

$P(x)$ is integral-valued if and only if the numbers $d_0,d_1,\ldots ,d_m$ are integers.

There exist polynomials with integral coefficients of degree $m$, whose absolute value is equal to 1 at $m+1$ integral points if $m\leq 3$, but there exist no such polynomials if $m\geq 4$.

If a rational integral function (a polynomial) of degree $m$ assumes rational values at $m+1$ integral points, then its coefficients are rational numbers.

If a rational function that is not a polynomial assumes rational values at all positive integral points, then it is the quotient of two relatively prime polynomials with integral coefficients.

If a rational function has integral values for infinitely many integral values of the variable, then it is a rational integral function (a polynomial).

Let $P(x)$ be an integral-valued polynomial. Is it possible that all the terms of the sequence of numbers

are prime numbers? (With the particular choice of $P(x)=x^2-x+41$ the first 40 terms are prime numbers, as Euler noted.)

The prime divisors of the polynomial $x^2+1$ are $2,5,13,17,\ldots$, i.e. 2 and the odd prime numbers of the form $4n+1$; the prime numbers of the form $4n+3$ are not prime divisors of $x^2+1$.

Determine the prime divisors of $x^2+15$.

There are infinitely many prime numbers that are not prime divisors of a given irreducible polynomial of second degree with integral coefficients.

If an integral-valued polynomial that does not vanish identically has a rational zero, then every prime number is a prime divisor of the polynomial with the possible exception of finitely many primes.

Every prime number is a prime divisor of the polynomial

which has no rational zeros.

An odd prime divisor of $K_m(x)$, the $m$th cyclotomic polynomial [36], is either a divisor of $m$ or $\equiv 1 \pmod m$.

Let $a,b,c$ be integers, $a\geq 0$, $b\geq 2$, $c\geq 0$. The integral-valued function $ab^x+c$ has infinitely many prime divisors.

Let $P(x)$ be an integral-valued polynomial that is not a constant. Then $P(x)$ has infinitely many prime divisors.

The fact stated in 108 may be expressed as follows: If $P(x)$ is an integral-valued polynomial that is not a constant, then not all the terms of the sequence of numbers

can be constructed from finitely many prime numbers. Is it possible that infinitely many terms of the sequence can be constructed from finitely many prime numbers?

Let the two integral-valued non-constant polynomials $P(x)$ and $Q(x)$ be relatively prime. Then there exist arbitrarily large integers $n$ such that the least common multiple of the numbers $P(n)$ and $Q(n)$ contains prime factors that are not contained in their greatest common divisor.

Let $P(x)$ be an irreducible integral-valued polynomial that is not a constant. (See § 3, the introduction to Problem 116.) Then there exist arbitrarily large integers $n$, such that in $P(n)$ there occurs at least one simple prime factor $p$; i.e. $P(n)\equiv 0 \pmod p$, $P(n)\not\equiv 0 \pmod{p^2}$.

Let $P(x)$ be an integral-valued polynomial. Let the zero of $P(x)$ of smallest multiplicity have multiplicity $m$. (Then all the prime factors of $P(n)$, with the possible exception of finitely many, must when raised to the power $m$ also be factors of $P(n)$, $n=0,1,2,\ldots$.) There exist arbitrarily large integers $n$ such that in $P(n)$ at least one prime factor occurs not more than $m$ times.

If the polynomial $P(x)$ assumes for every positive integer $x$ a value that is the square of an integer, then $P(x)$ is the square of a polynomial. (Analogous results are true for higher powers.)

Let $b_1,b_2,\ldots ,b_k$ be distinct positive integers, $0<b_1<b_2<\cdots <b_k$ and $P_1(x),P_2(x),\ldots ,P_k(x)$ polynomials with integral coefficients. Then the function

has infinitely many prime divisors.

Let the power series in descending powers of $z$

converge outside some circle. Let the numbers $b_1,b_2,\ldots ,b_m$ be rational. If the series assumes integral values for infinitely many integral values of $z$, then $b_0$ is rational and

(Generalization of 93.) [Hurwitz--Courant, p. 30.]

The series

represents an integer for infinitely many integral values of $z$.

Deduce 114 from 188.

Let the power series

that is not everywhere divergent represent integers for all sufficiently large integral values of $z$. Then $F(z)$ is an integral-valued polynomial.

If two polynomials are such that they assume integral values at the same points of the complex plane, then either their sum or their difference is a constant. (This constant is naturally an integer.)

If $m$ is the degree of $y=f(x)$ and $n$ the degree of $z=g(x)$, then we have for sufficiently large $\left|y\right|$

where all $m$ determinations are admissible for $y^{\frac1m}$.

Let the series in descending powers of $z$

be convergent outside some circle, and let its values for infinitely many rational integral values of $z$ be rational integral. If the coefficients $a_m,a_{m-1},\ldots ,a_1,a_0,a_{-1},a_{-2},\ldots$ are all rational numbers, then the series represents a rational entire function (it terminates, 188); if the coefficients all belong to a finite field, then the power series represents an algebraic function. (Example: 189.)

Prove that a polynomial of degree $<n$, which takes integer values at $n$ consecutive integer values of the variable, takes integer values at all integer values of the variable.

Is it true that such a polynomial has integer coefficients?