Polynomials with Integral Coefficients and Integral-Valued Functions

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).

In the sequence

the terms are pairwise relatively prime. (We can deduce from this that there exist infinitely many primes.)

In the sequence of numbers

there exist several, in fact infinitely many, terms with the same prime factors.

In the sequence

every term has a prime factor that is $\equiv -1 \pmod6$.

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 $\equiv 1 \pmod m$.

If $p$ is a prime number, $p>2$, $p$ not a divisor of $m$ and $K_m(a)\equiv 0 \pmod p$, then $a$ is relatively prime to $p$ and $m$ is the order of $a \pmod p$, i.e. $m$ is the smallest integer such that $a^m\equiv 1 \pmod p$.

There are infinitely many prime numbers of the form $6n-1$.

There are infinitely many prime numbers of the form $4n-1$.

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 $m$ be a positive integer. In the arithmetical progression

there are infinitely 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.

If a polynomial with integral coefficients is reducible, then it is the product of polynomials with integral coefficients of lower degree.

A polynomial with integral coefficients $f(x)$ cannot vanish at an integral value of $x$, if $f(0)$ and $f(1)$ are both odd.

If the polynomial $P(x)$ of degree $n$ with integral coefficients assumes at $n$ distinct integral values of $x$ values that are all different from 0 and in absolute value less than

then $P(x)$ is irreducible.

The bound given in 118 may be replaced by

where $d$ is the minimum distance between any two of the $n$ integral values of $x$ where $P(x)$ assumes the integral values considered.

In order that the value of the polynomial $P(x)$ with integral coefficients should be equal to a prime number for infinitely many integral values of $x$, $P(x)$ must be irreducible and the greatest common divisor of the coefficients of $P(x)$ must be 1. The converse of this obvious statement is not true. The polynomial

is irreducible but it does not represent a prime number for any integral value of $x$, if $n\geq 3$.

Let $a_1,a_2,\ldots ,a_n$ be distinct integers. Prove that the function

is always irreducible.

(continued). Investigate the cases in which

is reducible.

(continued). The function

is irreducible.

(continued). The function

is also irreducible.

Investigate the reducibility of $F((x-a_1)(x-a_2)\cdots (x-a_n))$, where $F(z)=z^4+Az^3+Bz^2+Az+1$ is a positive definite irreducible polynomial with integral coefficients (continuing the study of 121--124). If $F(z)=z^4+Az^3+Bz^2+Az+1$ is a positive definite irreducible polynomial with integral coefficients, then

is reducible only if $F(z)=z^4-z^2+1$ (the 12th cyclotomic polynomial) and $(x-a_1)(x-a_2)\cdots (x-a_n) = (x-\alpha )(x-\alpha -1)(x-\alpha -2)$ where $\alpha$ is an integer.

(continued). If $A$ is positive and integral, then

is reducible only if $A/4$ is the fourth power of an integer.

Let $P(x)$ be a polynomial with integral coefficients and assume that there exists an integer $n$ such that the following three conditions are satisfied:

Let

be a prime number expressed in the decimal system, $0\leq a_\nu \leq 9$, $\nu =0,1,\ldots ,m$; $a_0\geq 1$. Then the polynomial

is irreducible.

Let $r$ and $s$ be odd prime numbers, such that

(For definition of the Legendre symbol $\left(\dfrac np\right)$ see Problem 45 of Part V.) Such prime numbers are for example $r=41$ and $s=5$. The polynomial

is irreducible but it is ``reducible'' (it can be expressed as the product of two factors of the 2nd degree) modulo $m$, where $m$ is an arbitrary integer, as can be shown in an elementary manner by formulae (1), (2) and (3).