Greatest Common Divisor

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.