Math Problems Vault

Problem 8.116

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

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Problem 8.117

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

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Problem 8.118

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

\frac{\left(n-\left[\frac n2\right]\right)!}{2^{n-\left[n/2\right]}} ,

then $P(x)$ is irreducible.

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Remark.

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Cf. Problem 70 of Part VI.

Problem 8.119

The bound given in 118 may be replaced by

\left(\frac d2\right)^{n-\left[n/2\right]}\left(n-\left[\frac n2\right]\right)! ,

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

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Problem 8.120

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

x(x-(n!+1))(x-2(n!+1))\cdots (x-(n-1)(n!+1))+n! = x^n+\cdots

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

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Problem 8.121

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

(x-a_1)(x-a_2)\cdots (x-a_n)-1

is always irreducible.

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Problem 8.122

(continued). Investigate the cases in which

(x-a_1)(x-a_2)\cdots (x-a_n)+1

is reducible.

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Problem 8.123

(continued). The function

(x-a_1)^2(x-a_2)^2\cdots (x-a_n)^2+1

is irreducible.

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Problem 8.124

(continued). The function

(x-a_1)^4(x-a_2)^4\cdots (x-a_n)^4+1

is also irreducible.

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Problem 8.125

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

F((x-a_1)(x-a_2)\cdots (x-a_n))

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.

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Problem 8.126

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

A(x-a_1)^4(x-a_2)^4\cdots (x-a_n)^4+1

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

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Problem 8.127

(1)(2)(3)

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:

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(1)

The zeros of $P(x)$ lie in the half-plane $\Re x<n-\tfrac 12$ .

(2)

$P(n-1)\neq 0$ .

(3)

$P(n)$ is a prime number.

Then $P(x)$ is irreducible.

Problem 8.128

Let

p = a_0a_1\ldots a_m = a_010^m+a_110^{m-1}+\cdots +a_m

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

a_0x^m+a_1x^{m-1}+\cdots +a_{m-1}x+a_m

is irreducible.

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Remark.

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Cf. Problem 127, Problem 24 of Part III.

Problem 8.129

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

r\equiv 1 \pmod8 \qquad \text{and} \qquad \left(\frac rs\right) = \left(\frac sr\right) = 1 .

(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

P(x) = (x-\sqrt r-\sqrt s)(x-\sqrt r+\sqrt s)(x+\sqrt r-\sqrt s)(x+\sqrt r+\sqrt s)

= x^4-2(r+s)x^2+(r-s)^2

\begin{align} (1) \qquad & = (x^2+r-s)^2-4rx^2 \\ (2) \qquad & = (x^2-r+s)^2-4sx^2 \\ (3) \qquad & = (x^2-r-s)^2-4rs \end{align}

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).

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