Some Problems on Algebraic Integers

If $\alpha$ is an integer, then $\sqrt\alpha$ is also an integer.

If $r$ and $s$ are rational numbers and $r+s\sqrt{-1}$ is an integer, then $r$ and $s$ must be rational integers.

If $r$ and $s$ are rational numbers and $r+s\sqrt{-5}$ is an integer, then $r$ and $s$ must be rational integers.

If $r$ and $s$ are rational numbers and $r+s\sqrt{-3}$ is an integer, then $2r$ and $2s$ must be rational integers and $2r\equiv 2s \pmod2$; $r$ and $s$ need not necessarily be rational integers.

If both $\vartheta /\pi$ and $\cos \vartheta$ are rational, then $\cos \vartheta$ must have one of the following five values:

If both $\vartheta /\pi$ and $\cos \vartheta$ are rational, then $\cos \vartheta$ must have one of the following five values:

Consider the interior angle included between two contiguous faces of a regular solid and call it

according as the solid is a tetra-, hexa-, octa-, dodeca-, or icosahedron (a hexahedron is a cube). Show that the ratios

are irrational.

Consider the interior angle included between two contiguous faces of a regular solid and call it

according as the solid is a tetra-, hexa-, octa-, dodeca-, or icosahedron (a hexahedron is a cube). Show that the ratios

are irrational.

The numbers $\alpha_1,\alpha_2,\ldots ,\alpha_m$ are called linearly dependent if there exist rational integers $a_1,a_2,\ldots ,a_m$ not all zero such that $a_1\alpha_1+a_2\alpha_2+\cdots +a_m\alpha_m=0$. If no relation of this kind exists then $\alpha_1,\alpha_2,\ldots ,\alpha_m$ are called linearly independent. Thus, if $\beta \neq 0$, the numbers $\alpha$ and $\beta$ are linearly dependent or independent according as $\alpha /\beta$ is rational or irrational. The number $\alpha$ is algebraic if there exists an $n$ (a positive rational integer) such that $1,\alpha ,\alpha^2,\ldots ,\alpha^n$ are linearly dependent.

The numbers (angles) $T,H$ and $O$ are linearly dependent; in fact

(see 197.2; it can be also proved by elementary geometry). Yet $T$ and $O$ are linearly independent.

The numbers $\alpha_1,\alpha_2,\ldots ,\alpha_m$ are called linearly dependent if there exist rational integers $a_1,a_2,\ldots ,a_m$ not all zero such that $a_1\alpha_1+a_2\alpha_2+\cdots +a_m\alpha_m=0$. If no relation of this kind exists then $\alpha_1,\alpha_2,\ldots ,\alpha_m$ are called linearly independent. Thus, if $\beta \neq 0$, the numbers $\alpha$ and $\beta$ are linearly dependent or independent according as $\alpha /\beta$ is rational or irrational. The number $\alpha$ is algebraic if there exists an $n$ (a positive rational integer) such that $1,\alpha ,\alpha^2,\ldots ,\alpha^n$ are linearly dependent.

The numbers (angles) $T,H$ and $O$ are linearly dependent; in fact

(see 197.2; it can be also proved by elementary geometry). Yet $T$ and $O$ are linearly independent.

(continued). Show that $T,H,D$ and $I$ are linearly independent. Also $H,O,D$ and $I$ are linearly independent, and any linear homogeneous relation with rational coefficients between $T,H,O,D$ and $I$ can differ only trivially (that is, only by a numerical factor) from the one given in 197.3.

(continued). Show that $T,H,D$ and $I$ are linearly independent. Also $H,O,D$ and $I$ are linearly independent, and any linear homogeneous relation with rational coefficients between $T,H,O,D$ and $I$ can differ only trivially (that is, only by a numerical factor) from the one given in 197.3.

If both $\vartheta /\pi$ and $\tan^2\vartheta$ are rational then $\tan^2\vartheta$ must have one of the following five values (regarding $\infty =1/0$ as ``rational''):

If both $\vartheta /\pi$ and $\tan^2\vartheta$ are rational then $\tan^2\vartheta$ must have one of the following five values (regarding $\infty =1/0$ as ``rational''):

With the exception of the four cases $n=1,2,4$, and 8, $\tan 2\pi /n$ is irrational for all positive rational integers $n$.

With the exception of the four cases $n=1,2,4$, and 8, $\tan 2\pi /n$ is irrational for all positive rational integers $n$.

There are only finitely many integers of given degree $n$ that together with their conjugates lie in a fixed circle $\left|z\right|<k$ of the complex plane.

The only integer that together with all its conjugates lies in the open unit circle is the number 0.

Let all the zeros of a polynomial with rational integral coefficients and leading coefficient 1 lie in the closed unit circle. Such a zero must lie either at the center or on the boundary of the unit circle and in the latter case it must be a root of unity. [I.e. it must satisfy an equation of the form $x^h=x^k$, where $h,k$ are rational integers, $0<h<k$; 198.]

If the integer $\alpha$ together with all its conjugates is real and in absolute value $<2$, then $\alpha =2\cos 2\pi p/q$, where $p$ and $q$ are rational integers.

The irrational integers $\sqrt{-1}, \sqrt{-3}, \sqrt{-5}$ generate three distinct complex quadratic number fields, in which the integers have the forms

respectively, where $a$ and $b$ are rational integers.

A set $\mathfrak S$ of complex (or real) numbers is called a ``discontinuous integral domain'' if it has the following two properties:

1. If $\zeta '$ and $\zeta ''$ belong to the set $\mathfrak S$, then $\zeta '+\zeta ''$, $\zeta '-\zeta ''$, and $\zeta '\zeta ''$ also belong to the set $\mathfrak S$.

2. The point 0 is not a point of accumulation of the set of numbers $\mathfrak S$. (By (1) the number $0=\zeta '-\zeta '$ belongs to the set $\mathfrak S$.) A discontinuous integral domain consists either of all or of some of the integers of a complex quadratic number field. (E.g. only of the numbers $0,\pm 6,\pm 12,\pm 18,\ldots$ or only of the single number 0.)

If we were to designate as ``integers'' not all integers but only those that belong to the field generated by $\sqrt{-5}$, then the two numbers 3 and $1+2\sqrt{-5}$ would have no g.c. divisor in the sense of the above definition. I.e. it is impossible to find in the field generated by $\sqrt{-5}$, to which the numbers 3 and $1+2\sqrt{-5}$ belong, five integers $\alpha ,\beta ,\gamma ,\delta ,\vartheta$ such that the equations

hold simultaneously. [Investigate the least values of $(a+b\sqrt{-5})(a-b\sqrt{-5})$, for $a,b$ rational integers; 202.]

(continued). The squares $3^2=9$ and $(1+2\sqrt{-5})^2=-19+4\sqrt{-5}$ have a g.c. divisor, even under the restricted definition of ``integer'' given in 204.

(continued). Find the g.c. divisor of 3 and $1+2\sqrt{-5}$ (which is not contained in the field generated by $\sqrt{-5}$!). [194.]

If $\vartheta$ is a g.c. divisor of $\alpha_1,\alpha_2,\ldots ,\alpha_m$, then $\vartheta$ is divisible by every common divisor of $\alpha_1,\alpha_2,\ldots ,\alpha_m$.

Let $\vartheta$ be a g.c. divisor of the numbers $\alpha_1,\alpha_2,\ldots ,\alpha_m$, and $\vartheta '$ another g.c. divisor of the same numbers $\alpha_1,\alpha_2,\ldots ,\alpha_m$. Then $\vartheta '/\vartheta$ is a unit.

If $\vartheta$ is a g.c. divisor of $\alpha_1,\alpha_2,\ldots ,\alpha_m$, then $\gamma \vartheta$ is a g.c. divisor of $\gamma \alpha_1,\gamma \alpha_2,\ldots ,\gamma \alpha_m$.

If $\alpha$ is relatively prime to $\beta \gamma$, then $\alpha$ is relatively prime to $\beta$ as well as to $\gamma$.

If $\alpha$ is relatively prime to $\beta$ as well as to $\gamma$, then $\alpha$ is also relatively prime to $\beta \gamma$.

If $\alpha$ is relatively prime to $\beta$, then a g.c. divisor of $\alpha$ and $\gamma$ is also a g.c. divisor of $\alpha$ and $\beta \gamma$.

If $\delta$ is a g.c. divisor of $\alpha$ and $\beta$, then $\delta^n$ is a g.c. divisor of $\alpha^n$ and $\beta^n$, $n=1,2,3,\ldots$.

If $\delta$ is a g.c. divisor of $\alpha$ and $\beta$, then $\sqrt[n]\delta$ is a g.c. divisor of $\sqrt[n]\alpha$ and $\sqrt[n]\beta$, $n=1,2,3,\ldots$.

If the $m$ numbers $\alpha_1,\alpha_2,\ldots ,\alpha_m$ are coprime (pairwise relatively prime), that is if the g.c. divisor of any two of them is 1, and if $\mu =\alpha_1\alpha_2\cdots \alpha_m$, then 1 is a g.c. divisor of the $m$ numbers

All rational prime numbers, with the exception of finitely many, are relatively prime to a given integer $\alpha$, $\alpha \neq 0$.

Let the integers $a_1,a_2,\ldots ,a_n$ be rational and let $\alpha_1,\alpha_2,\ldots ,\alpha_n$ be the $n$ zeros of the polynomial $x^n+a_1x^{n-1}+a_2x^{n-2}+\cdots +a_{n-1}x+a_n$. Then $\alpha_1,\alpha_2,\ldots ,\alpha_n$ have a g.c. divisor $\delta$; in fact $\delta =\sqrt[N]d$, where $N=n!$ and $d$ is the g.c. divisor of

If $\alpha ,\beta$ belong to $K$ and $\alpha$ is a divisor of $\beta$, then $N(\alpha )$ is a divisor of $N(\beta )$.

Let the field $K$ be such that every pair of integers $\alpha$ and $\beta$ belonging to $K$ have a g.c. divisor belonging to $K$, in the sense that there exist five more integers $\alpha ',\beta ',\gamma ,\delta ,\vartheta$ in $K$ such that

(Not always the case! Cf. 204.) For this to be the case we have the following necessary and sufficient condition: If $\alpha$ and $\beta$ are integers in $K$ and such that $\alpha$ is not a divisor of $\beta$ nor $\beta$ a divisor of $\alpha$, then there exist in $K$ two integers $\xi ,\eta$ such that we have simultaneously

(If $a$ and $b$ are rational integers satisfying $\left|a\right|>\left|b\right|$, and $b$ is not a divisor of $a$, then the remainder on dividing $a$ by $b$, i.e. $r=a\cdot 1-b[a/b]$, has the property required here of $\alpha \xi +\beta \eta$.)

Any two numbers of the field generated by $\sqrt{-1}$ have a g.c. divisor belonging to the same field.

The congruence relation between two algebraic integers with respect to a given modulus is symmetric and transitive. That is we have the following:

From $\alpha \equiv \beta \pmod\mu$ and $\gamma \equiv \delta \pmod\mu$ it follows that

If 1 is the g.c. divisor of $\alpha$ and $\mu$, and $\mu$ is not a unit, then

Let $\alpha ,\beta ,\ldots$ be integers, $f(x,y,\ldots )$ a polynomial with rational integral coefficients and $p$ a rational prime number. Then we have

Let

($\alpha_k\neq 0$, $\omega_k\neq 0$, $\omega_k\neq \omega_l$ for $k\geq l$; $k,l=1,2,3,\ldots ,m$) be integers. Then not all the numbers

can be integers. [Consider suitable prime numbers $p$ and set $n=p,2p,3p,\ldots ,rp$; 216, 224.]

The zeros of the cyclotomic polynomial $K_m(x)$ [36] are also zeros of $x^m-1$, and thus integers. They are given by $\alpha^{r_1},\alpha^{r_2},\ldots ,\alpha^{r_h}$ if we set $e^{\frac{2\pi i}{m}}=\alpha$, $\varphi (m)=h$ (cf. introduction to Chapter 1, § 5), and $r_1,r_2,\ldots ,r_h$ form a reduced remainder system mod $m$, i.e. if the rational integers $r_1,r_2,\ldots ,r_h$ are all relatively prime to $m$ and pair-wise not congruent $\pmod m$. --- Is $K_m(x)$ reducible or irreducible?

By modifying the preceding argument give a proof of the irreducibility of the cyclotomic polynomial $K_m(x)$ without using the Dirichlet theorem concerning arithmetical progressions. It can be avoided by a more laborious but also more elementary construction of the reduced system of remainders mod $m$.

If the rational integers $m$ and $n$ are relatively prime and $n\geq 3$, then $2\cos (2\pi m/n)$ is an algebraic integer of degree $\varphi (n)/2$.

If the rational integers $m$ and $n$ are relatively prime and $n\geq 3$, then $2\cos (2\pi m/n)$ is an algebraic integer of degree $\varphi (n)/2$.

(continued). From this result derive 197.1 [38.1].

(continued). From this result derive 197.1 [38.1].

If the expansion of a rational function both in ascending as well as in descending powers of $z$ is a series with rational integral coefficients, then its poles different from 0 and $\infty$ lie at units.

Let a power series (in ascending powers) be convergent in the unit circle and represent a non-integral rational function. If its coefficients are rational integers, then its poles lie at roots of unity. [156, 200.]

If the power series

represents a rational function and its coefficients $\alpha_0,\alpha_1,\alpha_2,\ldots$ are algebraic integers, then the poles of the function are also algebraic integers.

If

($\alpha_k\neq 0$, $\omega_k\neq 0$, $\omega_k\neq \omega_l$ for $k\geq l$; $k,l=1,2,\ldots ,m$) are algebraic integers, then the coefficients in the power series expansion of

are also algebraic integers. However, the same is not true of the expansion that is obtained from it by term-by-term integration.

Let $f(z)$ be represented by a power series with rational coefficients and let its derivative $f'(z)$ be a rational function. $f(z)$ is rational if its power series satisfies the Eisenstein condition (cf. introduction to 142) and transcendental if it does not.

Let the algebraic function $f(z)$ be defined by the equation

where $P_0(z),P_1(z),\ldots ,P_l(z)$ are polynomials with algebraic coefficients. If $\alpha$ is an algebraic number, then the coefficients in the expansion of $f(z)$ in powers of $(z-\alpha )$ are also algebraic numbers. [Special case: If $z=\alpha$ is a regular point of $f(z)$, then $f(\alpha )$ is an algebraic number.]

Let a rational entire function $F(z,y)$ of the two variables $z,y$ with rational coefficients be termed irreducible if it cannot be factorized into the product of two entire functions with rational coefficients that are both of lower degree with respect to $y$. If $F(z,y)$ is irreducible, then either no solution $y$ of the equation $F(z,y)=0$ is a rational function of $z$, or all solutions are rational functions of $z$.

If the coefficients in the power series expansion of an algebraic function are algebraic numbers, then these coefficients all belong to a finite field (i.e. to a field generated by a single algebraic number, cf. introduction to 202). [151, 152.]

(continued). If the power series in question is

then there exists an integer $\tau$ such that all the numbers $\alpha_1\tau ,\alpha_2\tau^2,\alpha_3\tau^3,\ldots$ are integers. (Generalization of Eisenstein's theorem.)

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.)

The three sides of a triangle are of lengths $l,m$, and $n$ respectively. The numbers $l,m$ and $n$ are positive integers satisfying

Find the number of different triangles of this kind with longest side of given length $n$. [Count lattice points.]

The three sides of a triangle are of lengths $l,m$, and $n$ respectively. The numbers $l,m$ and $n$ are positive integers satisfying

Find the number of different triangles of this kind with longest side of given length $n$. [Count lattice points.]

Draw the figure for some values of the variables, count the solutions in integers of the equation

in all cases. Interesting cases are

Draw the figure for some values of the variables, count the solutions in integers of the equation

in all cases. Interesting cases are